cp's OEIS Frontend

This is a front-end for the Online Encyclopedia of Integer Sequences, made by Christian Perfect. The idea is to provide OEIS entries in non-ancient HTML, and then to think about how they're presented visually. The source code is on GitHub.

Previous Showing 31-40 of 915 results. Next

A007318 Pascal's triangle read by rows: C(n,k) = binomial(n,k) = n!/(k!*(n-k)!), 0 <= k <= n.

Original entry on oeis.org

1, 1, 1, 1, 2, 1, 1, 3, 3, 1, 1, 4, 6, 4, 1, 1, 5, 10, 10, 5, 1, 1, 6, 15, 20, 15, 6, 1, 1, 7, 21, 35, 35, 21, 7, 1, 1, 8, 28, 56, 70, 56, 28, 8, 1, 1, 9, 36, 84, 126, 126, 84, 36, 9, 1, 1, 10, 45, 120, 210, 252, 210, 120, 45, 10, 1, 1, 11, 55, 165, 330, 462, 462, 330, 165, 55, 11, 1
Offset: 0

Views

Author

N. J. A. Sloane and Mira Bernstein, Apr 28 1994

Keywords

Comments

A. W. F. Edwards writes: "It [the triangle] was first written down long before 1654, the year in which Blaise Pascal wrote his Traité du triangle arithmétique, but it was this work that brought together all the different aspects of the numbers for the first time. In it Pascal developed the properties of the number as a piece of pure mathematics ... and then, in a series of appendices, showed how these properties were relevant to the study of the figurate numbers, to the theory of combinations, to the expansion of binomial expressions, and to the solution of an important problem in the theory of probability." (A. W. F. Edwards, Pascal's Arithmetical Triangle, Johns Hopkins University Press (2002), p. xiii)
Edwards reports that the naming of the triangle after Pascal was done first by Montmort in 1708 as the "Table de M. Pascal pour les combinaisons" and then by De Moivre in 1730 as the "Triangulum Arithmeticum PASCALANIUM". (Edwards, p. xiv)
In China, Yang Hui in 1261 listed the coefficients of (a+b)^n up to n=6, crediting the expansion to Chia Hsein's Shih-so suan-shu circa 1100. Another prominent early use was in Chu Shih-Chieh's Precious Mirror of the Four Elements in 1303. (Edwards, p. 51)
In Persia, Al-Karaji discovered the binomial triangle "some time soon after 1007", and Al-Samawal published it in the Al-bahir some time before 1180. (Edwards, p. 52)
In India, Halayuda's commentary (circa 900) on Pingala's treatise on syllabic combinations (circa 200 B.C.E.) contains a clear description of the additive computation of the triangle. (Amulya Kumar Bag, Binomial Theorem in Ancient India, p. 72)
Also in India, the multiplicative formula for C(n,k) was known to Mahavira in 850 and restated by Bhaskara in 1150. (Edwards, p. 27)
In Italy, Tartaglia published the triangle in his General trattato (1556), and Cardano published it in his Opus novum (1570). (Edwards, p. 39, 44) - Russ Cox, Mar 29 2022
Also sometimes called Omar Khayyam's triangle.
Also sometimes called Yang Hui's triangle.
C(n,k) = number of k-element subsets of an n-element set.
Row n gives coefficients in expansion of (1+x)^n.
Binomial(n+k-1,n-1) is the number of ways of placing k indistinguishable balls into n boxes (the "bars and stars" argument - see Feller).
Binomial(n-1,k-1) is the number of compositions (ordered partitions) of n with k summands.
Binomial(n+k-1,k-1) is the number of weak compositions (ordered weak partitions) of n into exactly k summands. - Juergen Will, Jan 23 2016
Binomial(n,k) is the number of lattice paths from (0,0) to (n,k) using steps (1,0) and (1,1). - Joerg Arndt, Jul 01 2011
If thought of as an infinite lower triangular matrix, inverse begins:
+1
-1 +1
+1 -2 +1
-1 +3 -3 +1
+1 -4 +6 -4 +1
All 2^n palindromic binomial coefficients starting after the A006516(n)-th entry are odd. - Lekraj Beedassy, May 20 2003
Binomial(n+k-1,n-1) is the number of standard tableaux of shape (n,1^k). - Emeric Deutsch, May 13 2004
Can be viewed as an array, read by antidiagonals, where the entries in the first row and column are all 1's and A(i,j) = A(i-1,j) + A(i,j-1) for all other entries. The determinant of each of its n X n subarrays starting at (0,0) is 1. - Gerald McGarvey, Aug 17 2004
Also the lower triangular readout of the exponential of a matrix whose entry {j+1,j} equals j+1 (and all other entries are zero). - Joseph Biberstine (jrbibers(AT)indiana.edu), May 26 2006
Binomial(n-3,k-1) counts the permutations in S_n which have zero occurrences of the pattern 231 and one occurrence of the pattern 132 and k descents. Binomial(n-3,k-1) also counts the permutations in S_n which have zero occurrences of the pattern 231 and one occurrence of the pattern 213 and k descents. - David Hoek (david.hok(AT)telia.com), Feb 28 2007
Inverse of A130595 (as an infinite lower triangular matrix). - Philippe Deléham, Aug 21 2007
Consider integer lists LL of lists L of the form LL = [m#L] = [m#[k#2]] (where '#' means 'times') like LL(m=3,k=3) = [[2,2,2],[2,2,2],[2,2,2]]. The number of the integer list partitions of LL(m,k) is equal to binomial(m+k,k) if multiple partitions like [[1,1],[2],[2]] and [[2],[2],[1,1]] and [[2],[1,1],[2]] are counted only once. For the example, we find 4*5*6/3! = 20 = binomial(6,3). - Thomas Wieder, Oct 03 2007
The infinitesimal generator for Pascal's triangle and its inverse is A132440. - Tom Copeland, Nov 15 2007
Row n>=2 gives the number of k-digit (k>0) base n numbers with strictly decreasing digits; e.g., row 10 for A009995. Similarly, row n-1>=2 gives the number of k-digit (k>1) base n numbers with strictly increasing digits; see A009993 and compare A118629. - Rick L. Shepherd, Nov 25 2007
From Lee Naish (lee(AT)cs.mu.oz.au), Mar 07 2008: (Start)
Binomial(n+k-1, k) is the number of ways a sequence of length k can be partitioned into n subsequences (see the Naish link).
Binomial(n+k-1, k) is also the number of n- (or fewer) digit numbers written in radix at least k whose digits sum to k. For example, in decimal, there are binomial(3+3-1,3)=10 3-digit numbers whose digits sum to 3 (see A052217) and also binomial(4+2-1,2)=10 4-digit numbers whose digits sum to 2 (see A052216). This relationship can be used to generate the numbers of sequences A052216 to A052224 (and further sequences using radix greater than 10). (End)
From Milan Janjic, May 07 2008: (Start)
Denote by sigma_k(x_1,x_2,...,x_n) the elementary symmetric polynomials. Then:
Binomial(2n+1,2k+1) = sigma_{n-k}(x_1,x_2,...,x_n), where x_i = tan^2(i*Pi/(2n+1)), (i=1,2,...,n).
Binomial(2n,2k+1) = 2n*sigma_{n-1-k}(x_1,x_2,...,x_{n-1}), where x_i = tan^2(i*Pi/(2n)), (i=1,2,...,n-1).
Binomial(2n,2k) = sigma_{n-k}(x_1,x_2,...,x_n), where x_i = tan^2((2i-1)Pi/(4n)), (i=1,2,...,n).
Binomial(2n+1,2k) = (2n+1)sigma_{n-k}(x_1,x_2,...,x_n), where x_i = tan^2((2i-1)Pi/(4n+2)), (i=1,2,...,n). (End)
Given matrices R and S with R(n,k) = binomial(n,k)*r(n-k) and S(n,k) = binomial(n,k)*s(n-k), then R*S = T where T(n,k) = binomial(n,k)*[r(.)+s(.)]^(n-k), umbrally. And, the e.g.f.s for the row polynomials of R, S and T are, respectively, exp(x*t)*exp[r(.)*x], exp(x*t)*exp[s(.)*x] and exp(x*t)*exp[r(.)*x]*exp[s(.)*x] = exp{[t+r(.)+s(.)]*x}. The row polynomials are essentially Appell polynomials. See A132382 for an example. - Tom Copeland, Aug 21 2008
As the rectangle R(m,n) = binomial(m+n-2,m-1), the weight array W (defined generally at A144112) of R is essentially R itself, in the sense that if row 1 and column 1 of W=A144225 are deleted, the remaining array is R. - Clark Kimberling, Sep 15 2008
If A007318 = M as an infinite lower triangular matrix, M^n gives A130595, A023531, A007318, A038207, A027465, A038231, A038243, A038255, A027466, A038279, A038291, A038303, A038315, A038327, A133371, A147716, A027467 for n=-1,0,1,2,3,4,5,6,7,8,9,10,11,12,13,14,15 respectively. - Philippe Deléham, Nov 11 2008
The coefficients of the polynomials with e.g.f. exp(x*t)*(cosh(t)+sinh(t)). - Peter Luschny, Jul 09 2009
The triangle or chess sums, see A180662 for their definitions, link Pascal's triangle with twenty different sequences, see the crossrefs. All sums come in pairs due to the symmetrical nature of this triangle. The knight sums Kn14 - Kn110 have been added. It is remarkable that all knight sums are related to the Fibonacci numbers, i.e., A000045, but none of the others. - Johannes W. Meijer, Sep 22 2010
Binomial(n,k) is also the number of ways to distribute n+1 balls into k+1 urns so that each urn gets at least one ball. See example in the example section below. - Dennis P. Walsh, Jan 29 2011
Binomial(n,k) is the number of increasing functions from {1,...,k} to {1,...,n} since there are binomial(n,k) ways to choose the k distinct, ordered elements of the range from the codomain {1,...,n}. See example in the example section below. - Dennis P. Walsh, Apr 07 2011
Central binomial coefficients: T(2*n,n) = A000984(n), T(n, floor(n/2)) = A001405(n). - Reinhard Zumkeller, Nov 09 2011
Binomial(n,k) is the number of subsets of {1,...,n+1} with k+1 as median element. To see this, note that Sum_{j=0..min(k,n-k)}binomial(k,j)*binomial(n-k,j) = binomial(n,k). See example in Example section below. - Dennis P. Walsh, Dec 15 2011
This is the coordinator triangle for the lattice Z^n, see Conway-Sloane, 1997. - N. J. A. Sloane, Jan 17 2012
One of three infinite families of integral factorial ratio sequences of height 1 (see Bober, Theorem 1.2). The other two are A046521 and A068555. For real r >= 0, C_r(n,k) := floor(r*n)!/(floor(r*k)!*floor(r*(n-k))!) is integral. See A211226 for the case r = 1/2. - Peter Bala, Apr 10 2012
Define a finite triangle T(m,k) with n rows such that T(m,0) = 1 is the left column, T(m,m) = binomial(n-1,m) is the right column, and the other entries are T(m,k) = T(m-1,k-1) + T(m-1,k) as in Pascal's triangle. The sum of all entries in T (there are A000217(n) elements) is 3^(n-1). - J. M. Bergot, Oct 01 2012
The lower triangular Pascal matrix serves as a representation of the operator exp(RLR) in a basis composed of a sequence of polynomials p_n(x) characterized by ladder operators defined by R p_n(x) = p_(n+1)(x) and L p_n(x) = n p_(n-1)(x). See A132440, A218272, A218234, A097805, and A038207. The transposed and padded Pascal matrices can be associated to the special linear group SL2. - Tom Copeland, Oct 25 2012
See A193242. - Alexander R. Povolotsky, Feb 05 2013
A permutation p_1...p_n of the set {1,...,n} has a descent at position i if p_i > p_(i+1). Let S(n) denote the subset of permutations p_1...p_n of {1,...,n} such that p_(i+1) - p_i <= 1 for i = 1,...,n-1. Then binomial(n,k) gives the number of permutations in S(n+1) with k descents. Alternatively, binomial(n,k) gives the number of permutations in S(n+1) with k+1 increasing runs. - Peter Bala, Mar 24 2013
Sum_{n=>0} binomial(n,k)/n! = e/k!, where e = exp(1), while allowing n < k where binomial(n,k) = 0. Also Sum_{n>=0} binomial(n+k-1,k)/n! = e * A000262(k)/k!, and for k>=1 equals e * A067764(k)/A067653(k). - Richard R. Forberg, Jan 01 2014
The square n X n submatrix (first n rows and n columns) of the Pascal matrix P(x) defined in the formulas below when multiplying on the left the Vandermonde matrix V(x_1,...,x_n) (with ones in the first row) translates the matrix to V(x_1+x,...,x_n+x) while leaving the determinant invariant. - Tom Copeland, May 19 2014
For k>=2, n>=k, k/((k/(k-1) - Sum_{n=k..m} 1/binomial(n,k))) = m!/((m-k+1)!*(k-2)!). Note: k/(k-1) is the infinite sum. See A000217, A000292, A000332 for examples. - Richard R. Forberg, Aug 12 2014
Let G_(2n) be the subgroup of the symmetric group S_(2n) defined by G_(2n) = {p in S_(2n) | p(i) = i (mod n) for i = 1,2,...,2n}. G_(2n) has order 2^n. Binomial(n,k) gives the number of permutations in G_(2n) having n + k cycles. Cf. A130534 and A246117. - Peter Bala, Aug 15 2014
C(n,k) = the number of Dyck paths of semilength n+1, with k+1 "u"'s in odd numbered positions and k+1 returns to the x axis. Example: {U = u in odd position and = return to x axis} binomial(3,0)=1 (Uudududd); binomial(3,1)=3 [(Uududd_Ud_), (Ud_Uududd_), (Uudd_Uudd_)]; binomial(3,2)=3 [(Ud_Ud_Uudd_), (Uudd_Ud_Ud_), (Ud_Uudd_Ud_)]; binomial(3,3)=1 (Ud_Ud_Ud_Ud_). - Roger Ford, Nov 05 2014
From Daniel Forgues, Mar 12 2015: (Start)
The binomial coefficients binomial(n,k) give the number of individuals of the k-th generation after n population doublings. For each doubling of population, each individual's clone has its generation index incremented by 1, and thus goes to the next row. Just tally up each row from 0 to 2^n - 1 to get the binomial coefficients.
0 1 3 7 15
0: O | . | . . | . . . . | . . . . . . . . |
1: | O | O . | O . . . | O . . . . . . . |
2: | | O | O O . | O O . O . . . |
3: | | | O | O O O . |
4: | | | | O |
This is a fractal process: to get the pattern from 0 to 2^n - 1, append a shifted down (by one row) copy of the pattern from 0 to 2^(n-1) - 1 to the right of the pattern from 0 to 2^(n-1) - 1. (Inspired by the "binomial heap" data structure.)
Sequence of generation indices: 1's-counting sequence: number of 1's in binary expansion of n (or the binary weight of n) (see A000120):
{0, 1, 1, 2, 1, 2, 2, 3, 1, 2, 2, 3, 2, 3, 3, 4, ...}
Binary expansion of 0 to 15:
0 1 10 11 100 101 110 111 1000 1001 1010 1011 1100 1101 1111
(End)
A258993(n,k) = T(n+k,n-k), n > 0. - Reinhard Zumkeller, Jun 22 2015
T(n,k) is the number of set partitions w of [n+1] that avoid 1/2/3 with rb(w)=k. The same holds for ls(w)=k, where avoidance is in the sense of Klazar and ls,rb defined by Wachs and White.
Satisfies Benford's law [Diaconis, 1977] - N. J. A. Sloane, Feb 09 2017
Let {A(n)} be a set with exactly n identical elements, with {A(0)} being the empty set E. Let {A(n,k)} be the k-th iteration of {A(n)}, with {A(n,0)} = {A(n)}. {A(n,1)} = The set of all the subsets of A{(n)}, including {A(n)} and E. {A(n,k)} = The set of all subsets of {A(n,k-1)}, including all of the elements of {A(n,k-1)}. Let A(n,k) be the number of elements in {A(n,k)}. Then A(n,k) = C(n+k,k), with each successive iteration replicating the members of the k-th diagonal of Pascal's Triangle. See examples. - Gregory L. Simay, Aug 06 2018
Binomial(n-1,k) is also the number of permutations avoiding both 213 and 312 with k ascents. - Lara Pudwell, Dec 19 2018
Binomial(n-1,k) is also the number of permutations avoiding both 132 and 213 with k ascents. - Lara Pudwell, Dec 19 2018
Binomial(n,k) is the dimension of the k-th exterior power of a vector space of dimension n. - Stefano Spezia, Dec 22 2018
C(n,k-1) is the number of unoriented colorings of the facets (or vertices) of an n-dimensional simplex using exactly k colors. Each chiral pair is counted as one when enumerating unoriented arrangements. - Robert A. Russell, Oct 20 2020
From Dilcher and Stolarsky: "Two of the most ubiquitous objects in mathematics are the sequence of prime numbers and the binomial coefficients (and thus Pascal's triangle). A connection between the two is given by a well-known characterization of the prime numbers: Consider the entries in the k-th row of Pascal's triangle, without the initial and final entries. They are all divisible by k if and only if k is a prime." - Tom Copeland, May 17 2021
Named "Table de M. Pascal pour les combinaisons" by Pierre Remond de Montmort (1708) after the French mathematician, physicist and philosopher Blaise Pascal (1623-1662). - Amiram Eldar, Jun 11 2021
Consider the n-th diagonal of the triangle as a sequence b(n) with n starting at 0. From it form a new sequence by leaving the 0th term as is, and thereafter considering all compositions of n, taking the product of b(i) over the respective numbers i in each composition, adding terms corresponding to compositions with an even number of parts subtracting terms corresponding to compositions with an odd number of parts. Then the n-th row of the triangle is obtained, with every second term multiplied by -1, followed by infinitely many zeros. For sequences starting with 1, this operation is a special case of a self-inverse operation, and therefore the converse is true. - Thomas Anton, Jul 05 2021
C(n,k) is the number of vertices in an n-dimensional unit hypercube, at an L1 distance of k (or: with a shortest path of k 1d-edges) from a given vertex. - Eitan Y. Levine, May 01 2023
C(n+k-1,k-1) is the number of vertices at an L1 distance from a given vertex in an infinite-dimensional box, which has k sides of length 2^m for each m >= 0. Equivalently, given a set of tokens containing k distinguishable tokens with value 2^m for each m >= 0, C(n+k-1,k-1) is the number of subsets of tokens with a total value of n. - Eitan Y. Levine, Jun 11 2023
Numbers in the k-th column, i.e., numbers of the form C(n,k) for n >= k, are known as k-simplex numbers. - Pontus von Brömssen, Jun 26 2023
Let r(k) be the k-th row and c(k) the k-th column. Denote convolution by * and repeated convolution by ^. Then r(k)*r(m)=r(k+m) and c(k)*c(m)=c(k+m+1). This is because r(k) = r(1) ^ k and c(k) = c(0) ^ k+1. - Eitan Y. Levine, Jul 23 2023
Number of permutations of length n avoiding simultaneously the patterns 231 and 312(resp., 213 and 231; 213 and 312) with k descents (equivalently, with k ascents). An ascent (resp., descent) in a permutation a(1)a(2)...a(n) is position i such that a(i)a(i+1)). - Tian Han, Nov 25 2023
C(n,k) are generalized binomial coefficients of order m=0. Calculated by the formula C(n,k) = Sum_{i=0..n-k} binomial(n+1, n-k-i)*Stirling2(i+ m+ 1, i+1) *(-1)^i, where m = 0 for n>= 0, 0 <= k <= n. - Igor Victorovich Statsenko, Feb 26 2023
The Akiyama-Tanigawa algorithm applied to the diagonals, binomial(n+k,k), yields the powers of n. - Shel Kaphan, May 03 2024

Examples

			Triangle T(n,k) begins:
   n\k 0   1   2   3   4   5   6   7   8   9  10  11 ...
   0   1
   1   1   1
   2   1   2   1
   3   1   3   3   1
   4   1   4   6   4   1
   5   1   5  10  10   5   1
   6   1   6  15  20  15   6   1
   7   1   7  21  35  35  21   7   1
   8   1   8  28  56  70  56  28   8   1
   9   1   9  36  84 126 126  84  36   9   1
  10   1  10  45 120 210 252 210 120  45  10   1
  11   1  11  55 165 330 462 462 330 165  55  11   1
  ...
There are C(4,2)=6 ways to distribute 5 balls BBBBB, among 3 different urns, < > ( ) [ ], so that each urn gets at least one ball, namely, <BBB>(B)[B], <B>(BBB)[B], <B>(B)[BBB], <BB>(BB)[B], <BB>(B)[BB], and <B>(BB)[BB].
There are C(4,2)=6 increasing functions from {1,2} to {1,2,3,4}, namely, {(1,1),(2,2)},{(1,1),(2,3)}, {(1,1),(2,4)}, {(1,2),(2,3)}, {(1,2),(2,4)}, and {(1,3),(2,4)}. - _Dennis P. Walsh_, Apr 07 2011
There are C(4,2)=6 subsets of {1,2,3,4,5} with median element 3, namely, {3}, {1,3,4}, {1,3,5}, {2,3,4}, {2,3,5}, and {1,2,3,4,5}. - _Dennis P. Walsh_, Dec 15 2011
The successive k-iterations of {A(0)} = E are E;E;E;...; the corresponding number of elements are 1,1,1,... The successive k-iterations of {A(1)} = {a} are (omitting brackets) a;a,E; a,E,E;...; the corresponding number of elements are 1,2,3,... The successive k-iterations of {A(2)} = {a,a} are aa; aa,a,E; aa, a, E and a,E and E;...; the corresponding number of elements are 1,3,6,... - _Gregory L. Simay_, Aug 06 2018
Boas-Buck type recurrence for column k = 4: T(8, 4) = (5/4)*(1 + 5 + 15 + 35) = 70. See the Boas-Buck comment above. - _Wolfdieter Lang_, Nov 12 2018

References

  • M. Abramowitz and I. A. Stegun, eds., Handbook of Mathematical Functions, National Bureau of Standards Applied Math. Series 55, 1964 (and various reprintings), p. 828.
  • Amulya Kumar Bag, Binomial theorem in ancient India, Indian Journal of History of Science, vol. 1 (1966), pp. 68-74.
  • Arthur T. Benjamin and Jennifer Quinn, Proofs that really count: the art of combinatorial proof, M.A.A. 2003, p. 63ff.
  • Boris A. Bondarenko, Generalized Pascal Triangles and Pyramids (in Russian), FAN, Tashkent, 1990, ISBN 5-648-00738-8.
  • Louis Comtet, Advanced Combinatorics, Reidel, 1974, p. 306.
  • John H. Conway and Richard K. Guy, The Book of Numbers, New York: Springer-Verlag, 1996. See pp. 68-74.
  • Paul Curtz, Intégration numérique des systèmes différentiels à conditions initiales, Centre de Calcul Scientifique de l'Armement, Arcueil, 1969.
  • A. W. F. Edwards, Pascal's Arithmetical Triangle, 2002.
  • William Feller, An Introduction to Probability Theory and Its Application, Vol. 1, 2nd ed. New York: Wiley, p. 36, 1968.
  • Ronald L. Graham, Donald E. Knuth, and Oren Patashnik, Concrete Mathematics. Addison-Wesley, Reading, MA, 2nd. ed., 1994, p. 155.
  • Jan Gullberg, Mathematics from the Birth of Numbers, W. W. Norton & Co., NY & London, 1997, §4.4 Powers and Roots, pp. 140-141.
  • David Hök, Parvisa mönster i permutationer [Swedish], 2007.
  • Donald E. Knuth, The Art of Computer Programming, Vol. 1, 2nd ed., p. 52.
  • Sergei K. Lando, Lecture on Generating Functions, Amer. Math. Soc., Providence, R.I., 2003, pp. 60-61.
  • Blaise Pascal, Traité du triangle arithmétique, avec quelques autres petits traitez sur la mesme matière, Desprez, Paris, 1665.
  • Clifford A. Pickover, A Passion for Mathematics, Wiley, 2005; see p. 71.
  • Alfred S. Posamentier, Math Charmers, Tantalizing Tidbits for the Mind, Prometheus Books, NY, 2003, pages 271-275.
  • A. P. Prudnikov, Yu. A. Brychkov, and O. I. Marichev, "Integrals and Series", Volume 1: "Elementary Functions", Chapter 4: "Finite Sums", New York, Gordon and Breach Science Publishers, 1986-1992.
  • John Riordan, An Introduction to Combinatorial Analysis, Wiley, 1958, p. 6.
  • John Riordan, Combinatorial Identities, Wiley, 1968, p. 2.
  • Robert Sedgewick and Philippe Flajolet, An Introduction to the Analysis of Algorithms, Addison-Wesley, Reading, MA, 1996, p. 143.
  • N. J. A. Sloane and Simon Plouffe, The Encyclopedia of Integer Sequences, Academic Press, 1995 (includes this sequence).
  • Jerome Spanier and Keith B. Oldham, "Atlas of Functions", Hemisphere Publishing Corp., 1987, chapter 6, pages 43-52.
  • James J. Tattersall, Elementary Number Theory in Nine Chapters, Cambridge University Press, 1999, pages 13, 30-33.
  • David Wells, The Penguin Dictionary of Curious and Interesting Numbers, Penguin Books, 1987, pp. 115-118.
  • Douglas B. West, Combinatorial Mathematics, Cambridge, 2021, p. 25.

Links

Crossrefs

Equals differences between consecutive terms of A102363. - David G. Williams (davidwilliams(AT)Paxway.com), Jan 23 2006
Row sums give A000079 (powers of 2).
Cf. A083093 (triangle read mod 3), A214292 (first differences of rows).
Partial sums of rows give triangle A008949.
The triangle of the antidiagonals is A011973.
Infinite matrix squared: A038207, cubed: A027465.
Cf. A101164. If rows are sorted we get A061554 or A107430.
Another version: A108044.
Cf. A008277, A132311, A132312, A052216, A052217, A052218, A052219, A052220, A052221, A052222, A052223, A144225, A202750, A211226, A047999, A026729, A052553, A051920, A193242.
Triangle sums (see the comments): A000079 (Row1); A000007 (Row2); A000045 (Kn11 & Kn21); A000071 (Kn12 & Kn22); A001924 (Kn13 & Kn23); A014162 (Kn14 & Kn24); A014166 (Kn15 & Kn25); A053739 (Kn16 & Kn26); A053295 (Kn17 & Kn27); A053296 (Kn18 & Kn28); A053308 (Kn19 & Kn29); A053309 (Kn110 & Kn210); A001519 (Kn3 & Kn4); A011782 (Fi1 & Fi2); A000930 (Ca1 & Ca2); A052544 (Ca3 & Ca4); A003269 (Gi1 & Gi2); A055988 (Gi3 & Gi4); A034943 (Ze1 & Ze2); A005251 (Ze3 & Ze4). - Johannes W. Meijer, Sep 22 2010
Fibonacci-Pascal triangles: A027926, A036355, A037027, A074829, A105809, A109906, A111006, A114197, A162741, A228074, A228196, A228576.
Cf. A137948, A245334.
Cf. A085478, A258993.
Cf. A115940 (pandigital binomial coefficients C(m,k) with k>1).
Cf. (simplex colorings) A325002 (oriented), [k==n+1] (chiral), A325003 (achiral), A325000 (k or fewer colors), A325009 (orthotope facets, orthoplex vertices), A325017 (orthoplex facets, orthotope vertices).
Triangles of generalized binomial coefficients (n,k)_m (or generalized Pascal triangles) for m = 2..12: A001263, A056939, A056940, A056941, A142465, A142467, A142468, A174109, A342889, A342890, A342891.

Programs

  • Axiom
    -- (start)
)set expose add constructor OutputForm
pascal(0,n) == 1
pascal(n,n) == 1
pascal(i,j | 0 < i and i < j) == pascal(i-1,j-1) + pascal(i,j-1)
pascalRow(n) == [pascal(i,n) for i in 0..n]
displayRow(n) == output center blankSeparate pascalRow(n)
for i in 0..20 repeat displayRow i -- (end)
  • GAP
    Flat(List([0..12],n->List([0..n],k->Binomial(n,k)))); # Stefano Spezia, Dec 22 2018
  • Haskell
    a007318 n k = a007318_tabl !! n !! k
a007318_row n = a007318_tabl !! n
a007318_list = concat a007318_tabl
a007318_tabl = iterate (\row -> zipWith (+) ([0] ++ row) (row ++ [0])) [1]
-- Cf. http://www.haskell.org/haskellwiki/Blow_your_mind#Mathematical_sequences
-- Reinhard Zumkeller, Nov 09 2011, Oct 22 2010
  • Magma
    /* As triangle: */ [[Binomial(n, k): k in [0..n]]: n in [0.. 10]]; // Vincenzo Librandi, Jul 29 2015
  • Maple
    A007318 := (n,k)->binomial(n,k);
  • Mathematica
    Flatten[Table[Binomial[n, k], {n, 0, 11}, {k, 0, n}]] (* Robert G. Wilson v, Jan 19 2004 *)
Flatten[CoefficientList[CoefficientList[Series[1/(1 - x - x*y), {x, 0, 12}], x], y]] (* Mats Granvik, Jul 08 2014 *)
  • Maxima
    create_list(binomial(n,k),n,0,12,k,0,n); /* Emanuele Munarini, Mar 11 2011 */
  • PARI
    C(n,k)=binomial(n,k) \\ Charles R Greathouse IV, Jun 08 2011
  • Python
    # See Hobson link. Further programs:
from math import prod,factorial
def C(n,k): return prod(range(n,n-k,-1))//factorial(k) # M. F. Hasler, Dec 13 2019, updated Apr 29 2022, Feb 17 2023
  • Python
    from math import comb, isqrt
def A007318(n): return comb(r:=(m:=isqrt(k:=n+1<<1))-(k<=m*(m+1)),n-comb(r+1,2)) # Chai Wah Wu, Nov 11 2024
  • Sage
    def C(n,k): return Subsets(range(n), k).cardinality() # Ralf Stephan, Jan 21 2014

Formula

a(n, k) = C(n,k) = binomial(n, k).
C(n, k) = C(n-1, k) + C(n-1, k-1).
The triangle is symmetric: C(n,k) = C(n,n-k).
a(n+1, m) = a(n, m) + a(n, m-1), a(n, -1) := 0, a(n, m) := 0, n
C(n, k) = n!/(k!(n-k)!) if 0<=k<=n, otherwise 0.
C(n, k) = ((n-k+1)/k) * C(n, k-1) with C(n, 0) = 1. - Michael B. Porter, Mar 23 2025
G.f.: 1/(1-y-x*y) = Sum_(C(n, k)*x^k*y^n, n, k>=0)
G.f.: 1/(1-x-y) = Sum_(C(n+k, k)*x^k*y^n, n, k>=0).
G.f. for row n: (1+x)^n = Sum_{k=0..n} C(n, k)*x^k.
G.f. for column k: x^k/(1-x)^(k+1); [corrected by Werner Schulte, Jun 15 2022].
E.g.f.: A(x, y) = exp(x+x*y).
E.g.f. for column n: x^n*exp(x)/n!.
In general the m-th power of A007318 is given by: T(0, 0) = 1, T(n, k) = T(n-1, k-1) + m*T(n-1, k), where n is the row-index and k is the column; also T(n, k) = m^(n-k)*C(n, k).
Triangle T(n, k) read by rows; given by A000007 DELTA A000007, where DELTA is Deléham's operator defined in A084938.
Let P(n+1) = the number of integer partitions of (n+1); let p(i) = the number of parts of the i-th partition of (n+1); let d(i) = the number of different parts of the i-th partition of (n+1); let m(i, j) = multiplicity of the j-th part of the i-th partition of (n+1). Define the operator Sum_{i=1..P(n+1), p(i)=k+1} as the sum running from i=1 to i=P(n+1) but taking only partitions with p(i)=(k+1) parts into account. Define the operator Product_{j=1..d(i)} = product running from j=1 to j=d(i). Then C(n, k) = Sum_{p(i)=(k+1), i=1..P(n+1)} p(i)! / [Product_{j=1..d(i)} m(i, j)!]. E.g., C(5, 3) = 10 because n=6 has the following partitions with m=3 parts: (114), (123), (222). For their multiplicities one has: (114): 3!/(2!*1!) = 3; (123): 3!/(1!*1!*1!) = 6; (222): 3!/3! = 1. The sum is 3 + 6 + 1 = 10 = C(5, 3). - Thomas Wieder, Jun 03 2005
C(n, k) = Sum_{j=0..k} (-1)^j*C(n+1+j, k-j)*A000108(j). - Philippe Deléham, Oct 10 2005
G.f.: 1 + x*(1 + x) + x^3*(1 + x)^2 + x^6*(1 + x)^3 + ... . - Michael Somos, Sep 16 2006
Sum_{k=0..floor(n/2)} x^(n-k)*T(n-k,k) = A000007(n), A000045(n+1), A002605(n), A030195(n+1), A057087(n), A057088(n), A057089(n), A057090(n), A057091(n), A057092(n), A057093(n) for x = 0, 1, 2, 3, 4, 5, 6, 7, 8, 9, 10, respectively. Sum_{k=0..floor(n/2)} (-1)^k*x^(n-k)*T(n-k,k) = A000007(n), A010892(n), A009545(n+1), A057083(n), A001787(n+1), A030191(n), A030192(n), A030240(n), A057084(n), A057085(n+1), A057086(n), A084329(n+1) for x = 0, 1, 2, 3, 4, 5, 6, 7, 8, 9, 10, 20, respectively. - Philippe Deléham, Sep 16 2006
C(n,k) <= A062758(n) for n > 1. - Reinhard Zumkeller, Mar 04 2008
C(t+p-1, t) = Sum_{i=0..t} C(i+p-2, i) = Sum_{i=1..p} C(i+t-2, t-1). A binomial number is the sum of its left parent and all its right ancestors, which equals the sum of its right parent and all its left ancestors. - Lee Naish (lee(AT)cs.mu.oz.au), Mar 07 2008
From Paul D. Hanna, Mar 24 2011: (Start)
Let A(x) = Sum_{n>=0} x^(n*(n+1)/2)*(1+x)^n be the g.f. of the flattened triangle:
A(x) = 1 + (x + x^2) + (x^3 + 2*x^4 + x^5) + (x^6 + 3*x^7 + 3*x^8 + x^9) + ...
then A(x) equals the series Sum_{n>=0} (1+x)^n*x^n*Product_{k=1..n} (1-(1+x)*x^(2*k-1))/(1-(1+x)*x^(2*k));
also, A(x) equals the continued fraction 1/(1- x*(1+x)/(1+ x*(1-x)*(1+x)/(1- x^3*(1+x)/(1+ x^2*(1-x^2)*(1+x)/(1- x^5*(1+x)/(1+ x^3*(1-x^3)*(1+x)/(1- x^7*(1+x)/(1+ x^4*(1-x^4)*(1+x)/(1- ...))))))))).
These formulas are due to (1) a q-series identity and (2) a partial elliptic theta function expression. (End)
For n > 0: T(n,k) = A029600(n,k) - A029635(n,k), 0 <= k <= n. - Reinhard Zumkeller, Apr 16 2012
Row n of the triangle is the result of applying the ConvOffs transform to the first n terms of the natural numbers (1, 2, 3, ..., n). See A001263 or A214281 for a definition of this transformation. - Gary W. Adamson, Jul 12 2012
From L. Edson Jeffery, Aug 02 2012: (Start)
Row n (n >= 0) of the triangle is given by the n-th antidiagonal of the infinite matrix P^n, where P = (p_{i,j}), i,j >= 0, is the production matrix
0, 1,
1, 0, 1,
0, 1, 0, 1,
0, 0, 1, 0, 1,
0, 0, 0, 1, 0, 1,
0, 0, 0, 0, 1, 0, 1,
0, 0, 0, 0, 0, 1, 0, 1,
0, 0, 0, 0, 0, 0, 1, 0, 1,
... (End)
Row n of the triangle is also given by the n+1 coefficients of the polynomial P_n(x) defined by the recurrence P_0(x) = 1, P_1(x) = x + 1, P_n(x) = x*P_{n-1}(x) + P_{n-2}(x), n > 1. - L. Edson Jeffery, Aug 12 2013
For a closed-form formula for arbitrary left and right borders of Pascal-like triangles see A228196. - Boris Putievskiy, Aug 18 2013
For a closed-form formula for generalized Pascal's triangle see A228576. - Boris Putievskiy, Sep 04 2013
(1+x)^n = Sum_{k=0..n} (-1)^(n-k)*binomial(n,k)*Sum_{i=0..k} k^(n-i)*binomial(k,i)*x^(n-i)/(n-i)!. - Vladimir Kruchinin, Oct 21 2013
E.g.f.: A(x,y) = exp(x+x*y) = 1 + (x+y*x)/( E(0)-(x+y*x)), where E(k) = 1 + (x+y*x)/(1 + (k+1)/E(k+1) ); (continued fraction). - Sergei N. Gladkovskii, Nov 08 2013
E.g.f.: E(0) -1, where E(k) = 2 + x*(1+y)/(2*k+1 - x*(1+y)/E(k+1) ); (continued fraction). - Sergei N. Gladkovskii, Dec 24 2013
G.f.: 1 + x*(1+x)*(1+x^2*(1+x)/(W(0)-x^2-x^3)), where W(k) = 1 + (1+x)*x^(k+2) - (1+x)*x^(k+3)/W(k+1); (continued fraction). - Sergei N. Gladkovskii, Dec 24 2013
Sum_{n>=0} C(n,k)/n! = e/k!, where e = exp(1), while allowing n < k where C(n,k) = 0. Also Sum_{n>=0} C(n+k-1,k)/n! = e * A000262(k)/k!, and for k>=1 equals e * A067764(k)/A067653(k). - Richard R. Forberg, Jan 01 2014
Sum_{n>=k} 1/C(n,k) = k/(k-1) for k>=1. - Richard R. Forberg, Feb 10 2014
From Tom Copeland, Apr 26 2014: (Start)
Multiply each n-th diagonal of the Pascal lower triangular matrix by x^n and designate the result by A007318(x) = P(x). Then with :xD:^n = x^n*(d/dx)^n and B(n,x), the Bell polynomials (A008277),
A) P(x)= exp(x*dP) = exp[x*(e^M-I)] = exp[M*B(.,x)] = (I+dP)^B(.,x)
with dP = A132440, M = A238385-I, and I = identity matrix, and
B) P(:xD:) = exp(dP:xD:) = exp[(e^M-I):xD:] = exp[M*B(.,:xD:)] = exp[M*xD] = (I+dP)^(xD) with action P(:xD:)g(x) = exp(dP:xD:)g(x) = g[(I+dP)*x] (cf. also A238363).
C) P(x)^y = P(y*x). P(2x) = A038207(x) = exp[M*B(.,2x)], the face vectors of the n-dim hypercubes.
D) P(x) = [St2]*exp(x*M)*[St1] = [St2]*(I+dP)^x*[St1]
E) = [St1]^(-1)*(I+dP)^x*[St1] = [St2]*(I+dP)^x*[St2]^(-1)
where [St1]=padded A008275 just as [St2]=A048993=padded A008277 and exp(x*M) = (I+dP)^x = Sum_{k>=0} C(x,k) dP^k. (End)
T(n,k) = A245334(n,k) / A137948(n,k), 0 <= k <= n. - Reinhard Zumkeller, Aug 31 2014
From Peter Bala, Dec 21 2014: (Start)
Recurrence equation: T(n,k) = T(n-1,k)*(n + k)/(n - k) - T(n-1,k-1) for n >= 2 and 1 <= k < n, with boundary conditions T(n,0) = T(n,n) = 1. Note, changing the minus sign in the recurrence to a plus sign gives a recurrence for the square of the binomial coefficients - see A008459.
There is a relation between the e.g.f.'s of the rows and the diagonals of the triangle, namely, exp(x) * e.g.f. for row n = e.g.f. for diagonal n. For example, for n = 3 we have exp(x)*(1 + 3*x + 3*x^2/2! + x^3/3!) = 1 + 4*x + 10*x^2/2! + 20*x^3/3! + 35*x^4/4! + .... This property holds more generally for the Riordan arrays of the form ( f(x), x/(1 - x) ), where f(x) is an o.g.f. of the form 1 + f_1*x + f_2*x^2 + .... See, for example, A055248 and A106516.
Let P denote the present triangle. For k = 0,1,2,... define P(k) to be the lower unit triangular block array
/I_k 0\
\ 0 P/ having the k X k identity matrix I_k as the upper left block; in particular, P(0) = P. The infinite product P(0)*P(1)*P(2)*..., which is clearly well-defined, is equal to the triangle of Stirling numbers of the second kind A008277. The infinite product in the reverse order, that is, ...*P(2)*P(1)*P(0), is equal to the triangle of Stirling cycle numbers A130534. (End)
C(a+b,c) = Sum_{k=0..a} C(a,k)*C(b,b-c+k). This is a generalization of equation 1 from section 4.2.5 of the Prudnikov et al. reference, for a=b=c=n: C(2*n,n) = Sum_{k=0..n} C(n,k)^2. See Links section for animation of new formula. - Hermann Stamm-Wilbrandt, Aug 26 2015
The row polynomials of the Pascal matrix P(n,x) = (1+x)^n are related to the Bernoulli polynomials Br(n,x) and their umbral compositional inverses Bv(n,x) by the umbral relation P(n,x) = (-Br(.,-Bv(.,x)))^n = (-1)^n Br(n,-Bv(.,x)), which translates into the matrix relation P = M * Br * M * Bv, where P is the Pascal matrix, M is the diagonal matrix diag(1,-1,1,-1,...), Br is the matrix for the coefficients of the Bernoulli polynomials, and Bv that for the umbral inverse polynomials defined umbrally by Br(n,Bv(.,x)) = x^n = Bv(n,Br(.,x)). Note M = M^(-1). - Tom Copeland, Sep 05 2015
1/(1-x)^k = (r(x) * r(x^2) * r(x^4) * ...) where r(x) = (1+x)^k. - Gary W. Adamson, Oct 17 2016
Boas-Buck type recurrence for column k for Riordan arrays (see the Aug 10 2017 remark in A046521, also for the reference) with the Boas-Buck sequence b(n) = {repeat(1)}. T(n, k) = ((k+1)/(n-k))*Sum_{j=k..n-1} T(j, k), for n >= 1, with T(n, n) = 1. This reduces, with T(n, k) = binomial(n, k), to a known binomial identity (e.g, Graham et al. p. 161). - Wolfdieter Lang, Nov 12 2018
C((p-1)/a, b) == (-1)^b * fact_a(a*b-a+1)/fact_a(a*b) (mod p), where fact_n denotes the n-th multifactorial, a divides p-1, and the denominator of the fraction on the right side of the equation represents the modular inverse. - Isaac Saffold, Jan 07 2019
C(n,k-1) = A325002(n,k) - [k==n+1] = (A325002(n,k) + A325003(n,k)) / 2 = [k==n+1] + A325003(n,k). - Robert A. Russell, Oct 20 2020
From Hermann Stamm-Wilbrandt, May 13 2021: (Start)
Binomial sums are Fibonacci numbers A000045:
Sum_{k=0..n} C(n + k, 2*k + 1) = F(2*n).
Sum_{k=0..n} C(n + k, 2*k) = F(2*n + 1). (End)
C(n,k) = Sum_{i=0..k} A000108(i) * C(n-2i-1, k-i), for 0 <= k <= floor(n/2)-1. - Tushar Bansal, May 17 2025

Extensions

Checked all links, deleted 8 that seemed lost forever and were probably not of great importance. - N. J. A. Sloane, May 08 2018

A000244 Powers of 3: a(n) = 3^n.

Original entry on oeis.org

1, 3, 9, 27, 81, 243, 729, 2187, 6561, 19683, 59049, 177147, 531441, 1594323, 4782969, 14348907, 43046721, 129140163, 387420489, 1162261467, 3486784401, 10460353203, 31381059609, 94143178827, 282429536481, 847288609443, 2541865828329, 7625597484987
Offset: 0

Views

Author

N. J. A. Sloane

Keywords

Comments

Same as Pisot sequences E(1, 3), L(1, 3), P(1, 3), T(1, 3). Essentially same as Pisot sequences E(3, 9), L(3, 9), P(3, 9), T(3, 9). See A008776 for definitions of Pisot sequences.
Number of (s(0), s(1), ..., s(2n+2)) such that 0 < s(i) < 6 and |s(i) - s(i-1)| = 1 for i = 1, 2, ..., 2n + 2, s(0) = 1, s(2n+2) = 3. - Herbert Kociemba, Jun 10 2004
a(1) = 1, a(n+1) is the least number such that there are a(n) even numbers between a(n) and a(n+1). Generalization for the sequence of powers of k: 1, k, k^2, k^3, k^4, ... There are a(n) multiples of k-1 between a(n) and a(n+1). - Amarnath Murthy, Nov 28 2004
a(n) = sum of (n+1)-th row in Triangle A105728. - Reinhard Zumkeller, Apr 18 2005
With p(n) being the number of integer partitions of n, p(i) being the number of parts of the i-th partition of n, d(i) being the number of different parts of the i-th partition of n, m(i, j) being the multiplicity of the j-th part of the i-th partition of n, Sum_{i = 1..p(n)} being the sum over i and Product_{j = 1..d(i)} being the product over j, one has: a(n) = Sum_{i = 1..p(n)} (p(i)!/(Product_{j = 1..d(i)} m(i, j)!))*2^(p(i) - 1). - Thomas Wieder, May 18 2005
For any k > 1 in the sequence, k is the first prime power appearing in the prime decomposition of repunit R_k, i.e., of A002275(k). - Lekraj Beedassy, Apr 24 2006
a(n-1) is the number of compositions of compositions. In general, (k+1)^(n-1) is the number of k-levels nested compositions (e.g., 4^(n-1) is the number of compositions of compositions of compositions, etc.). Each of the n - 1 spaces between elements can be a break for one of the k levels, or not a break at all. - Franklin T. Adams-Watters, Dec 06 2006
Let S be a binary relation on the power set P(A) of a set A having n = |A| elements such that for every element x, y of P(A), xSy if x is a subset of y. Then a(n) = |S|. - Ross La Haye, Dec 22 2006
From Manfred Boergens, Mar 28 2023: (Start)
With regard to the comment by Ross La Haye:
Cf. A001047 if either nonempty subsets are considered or x is a proper subset of y.
Cf. a(n+1) in A028243 if nonempty subsets are considered and x is a proper subset of y. (End)
If X_1, X_2, ..., X_n is a partition of the set {1, 2, ..., 2*n} into blocks of size 2 then, for n >= 1, a(n) is equal to the number of functions f : {1, 2, ..., 2*n} -> {1, 2} such that for fixed y_1, y_2, ..., y_n in {1, 2} we have f(X_i) <> {y_i}, (i = 1, 2, ..., n). - Milan Janjic, May 24 2007
This is a general comment on all sequences of the form a(n) = [(2^k)-1]^n for all positive integers k. Example 1.1.16 of Stanley's "Enumerative Combinatorics" offers a slightly different version. a(n) in the number of functions f:[n] into P([k]) - {}. a(n) is also the number of functions f:[k] into P([n]) such that the generalized intersection of f(i) for all i in [k] is the empty set. Where [n] = {1, 2, ..., n}, P([n]) is the power set of [n] and {} is the empty set. - Geoffrey Critzer, Feb 28 2009
a(n) = A064614(A000079(n)) and A064614(m)A000079(n). - Reinhard Zumkeller, Feb 08 2010
3^(n+1) = (1, 2, 2, 2, ...) dot (1, 1, 3, 9, ..., 3^n); e.g., 3^3 = 27 = (1, 2, 2, 2) dot (1, 1, 3, 9) = (1 + 2 + 6 + 18). - Gary W. Adamson, May 17 2010
a(n) is the number of generalized compositions of n when there are 3*2^i different types of i, (i = 1, 2, ...). - Milan Janjic, Sep 24 2010
For n >= 1, a(n-1) is the number of generalized compositions of n when there are 2^(i-1) different types of i, (i = 1, 2, ...). - Milan Janjic, Sep 24 2010
The sequence in question ("Powers of 3") also describes the number of moves of the k-th disk solving the [RED ; BLUE ; BLUE] or [RED ; RED ; BLUE] pre-colored Magnetic Tower of Hanoi puzzle (cf. A183111 - A183125).
a(n) is the number of Stern polynomials of degree n. See A057526. - T. D. Noe, Mar 01 2011
Positions of records in the number of odd prime factors, A087436. - Juri-Stepan Gerasimov, Mar 17 2011
Sum of coefficients of the expansion of (1+x+x^2)^n. - Adi Dani, Jun 21 2011
a(n) is the number of compositions of n elements among {0, 1, 2}; e.g., a(2) = 9 since there are the 9 compositions 0 + 0, 0 + 1, 1 + 0, 0 + 2, 1 + 1, 2 + 0, 1 + 2, 2 + 1, and 2 + 2. [From Adi Dani, Jun 21 2011; modified by editors.]
Except the first two terms, these are odd numbers n such that no x with 2 <= x <= n - 2 satisfy x^(n-1) == 1 (mod n). - Arkadiusz Wesolowski, Jul 03 2011
The compositions of n in which each natural number is colored by one of p different colors are called p-colored compositions of n. For n >= 1, a(n) equals the number of 3-colored compositions of n such that no adjacent parts have the same color. - Milan Janjic, Nov 17 2011
Explanation from David Applegate, Feb 20 2017: (Start)
Since the preceding comment appears in a large number of sequences, it might be worth adding a proof.
The number of compositions of n into exactly k parts is binomial(n-1,k-1).
For a p-colored composition of n such that no adjacent parts have the same color, there are exactly p choices for the color of the first part, and p-1 choices for the color of each additional part (any color other than the color of the previous one). So, for a partition into k parts, there are p (p-1)^(k-1) valid colorings.
Thus the number of p-colored compositions of n into exactly k parts such that no adjacent parts have the same color is binomial(n-1,k-1) p (p-1)^(k-1).
The total number of p-colored compositions of n such that no adjacent parts have the same color is then
Sum_{k=1..n} binomial(n-1,k-1) * p * (p-1)^(k-1) = p^n.
To see this, note that the binomial expansion of ((p - 1) + 1)^(n - 1) = Sum_{k = 0..n - 1} binomial(n - 1, k) (p - 1)^k 1^(n - 1 - k) = Sum_{k = 1..n} binomial(n - 1, k - 1) (p - 1)^(k - 1).
(End)
Also, first and least element of the matrix [1, sqrt(2); sqrt(2), 2]^(n+1). - M. F. Hasler, Nov 25 2011
One-half of the row sums of the triangular version of A035002. - J. M. Bergot, Jun 10 2013
Form an array with m(0,n) = m(n,0) = 2^n; m(i,j) equals the sum of the terms to the left of m(i,j) and the sum of the terms above m(i,j), which is m(i,j) = Sum_{k=0..j-1} m(i,k) + Sum_{k=0..i-1} m(k,j). The sum of the terms in antidiagonal(n+1) = 4*a(n). - J. M. Bergot, Jul 10 2013
a(n) = A007051(n+1) - A007051(n), and A007051 are the antidiagonal sums of an array defined by m(0,k) = 1 and m(n,k) = Sum_{c = 0..k - 1} m(n, c) + Sum_{r = 0..n - 1} m(r, k), which is the sum of the terms to left of m(n, k) plus those above m(n, k). m(1, k) = A000079(k); m(2, k) = A045623(k + 1); m(k + 1, k) = A084771(k). - J. M. Bergot, Jul 16 2013
Define an array to have m(0,k) = 2^k and m(n,k) = Sum_{c = 0..k - 1} m(n, c) + Sum_{r = 0..n - 1} m(r, k), which is the sum of the terms to the left of m(n, k) plus those above m(n, k). Row n = 0 of the array comprises A000079, column k = 0 comprises A011782, row n = 1 comprises A001792. Antidiagonal sums of the array are a(n): 1 = 3^0, 1 + 2 = 3^1, 2 + 3 + 4 = 3^2, 4 + 7 + 8 + 8 = 3^3. - J. M. Bergot, Aug 02 2013
The sequence with interspersed zeros and o.g.f. x/(1 - 3*x^2), A(2*k) = 0, A(2*k + 1) = 3^k = a(k), k >= 0, can be called hexagon numbers. This is because the algebraic number rho(6) = 2*cos(Pi/6) = sqrt(3) of degree 2, with minimal polynomial C(6, x) = x^2 - 3 (see A187360, n = 6), is the length ratio of the smaller diagonal and the side in the hexagon. Hence rho(6)^n = A(n-1)*1 + A(n)*rho(6), in the power basis of the quadratic number field Q(rho(6)). One needs also A(-1) = 1. See also a Dec 02 2010 comment and the P. Steinbach reference given in A049310. - Wolfdieter Lang, Oct 02 2013
Numbers k such that sigma(3k) = 3k + sigma(k). - Jahangeer Kholdi, Nov 23 2013
All powers of 3 are perfect totient numbers (A082897), since phi(3^n) = 2 * 3^(n - 1) for n > 0, and thus Sum_{i = 0..n} phi(3^i) = 3^n. - Alonso del Arte, Apr 20 2014
The least number k > 0 such that 3^k ends in n consecutive decreasing digits is a 3-term sequence given by {1, 13, 93}. The consecutive increasing digits are {3, 23, 123}. There are 100 different 3-digit endings for 3^k. There are no k-values such that 3^k ends in '012', '234', '345', '456', '567', '678', or '789'. The k-values for which 3^k ends in '123' are given by 93 mod 100. For k = 93 + 100*x, the digit immediately before the run of '123' is {9, 5, 1, 7, 3, 9, 5, 1, 3, 7, ...} for x = {0, 1, 2, 3, 4, 5, 6, 7, 8, 9, ...}, respectively. Thus we see the digit before '123' will never be a 0. So there are no further terms. - Derek Orr, Jul 03 2014
All elements of A^n where A = (1, 1, 1; 1, 1, 1; 1, 1, 1). - David Neil McGrath, Jul 23 2014
Counts all walks of length n (open or closed) on the vertices of a triangle containing a loop at each vertex starting from any given vertex. - David Neil McGrath, Oct 03 2014
a(n) counts walks (closed) on the graph G(1-vertex;1-loop,1-loop,1-loop). - David Neil McGrath, Dec 11 2014
2*a(n-2) counts all permutations of a solitary closed walk of length (n) from the vertex of a triangle that contains 2 loops on each of the remaining vertices. In addition, C(m,k)=2*(2^m)*B(m+k-2,m) counts permutations of walks that contain (m) loops and (k) arcs. - David Neil McGrath, Dec 11 2014
a(n) is the sum of the coefficients of the n-th layer of Pascal's pyramid (a.k.a., Pascal's tetrahedron - see A046816). - Bob Selcoe, Apr 02 2016
Numbers n such that the trinomial x^(2*n) + x^n + 1 is irreducible over GF(2). Of these only the trinomial for n=1 is primitive. - Joerg Arndt, May 16 2016
Satisfies Benford's law [Berger-Hill, 2011]. - N. J. A. Sloane, Feb 08 2017
a(n-1) is also the number of compositions of n if the parts can be runs of any length from 1 to n, and can contain any integers from 1 to n. - Gregory L. Simay, May 26 2017
Also the number of independent vertex sets and vertex covers in the n-ladder rung graph n P_2. - Eric W. Weisstein, Sep 21 2017
Also the number of (not necessarily maximal) cliques in the n-cocktail party graph. - Eric W. Weisstein, Nov 29 2017
a(n-1) is the number of 2-compositions of n; see Hopkins & Ouvry reference. - Brian Hopkins, Aug 15 2020
a(n) is the number of faces of any dimension (vertices, edges, square faces, etc.) of the n-dimensional hypercube. For example, the 0-dimensional hypercube is a point, and its only face is itself. The 1-dimensional hypercube is a line, which has two vertices and an edge. The 2-dimensional hypercube is a square, which has four vertices, four edges, and a square face. - Kevin Long, Mar 14 2023
Number of pairs (A,B) of subsets of M={1,2,...,n} with union(A,B)=M. For nonempty subsets cf. A058481. - Manfred Boergens, Mar 28 2023
From Jianing Song, Sep 27 2023: (Start)
a(n) is the number of disjunctive clauses of n variables up to equivalence. A disjunctive clause is a propositional formula of the form l_1 OR ... OR l_m, where l_1, ..., l_m are distinct elements in {x_1, ..., x_n, NOT x_1, ..., NOT x_n} for n variables x_1, ... x_n, and no x_i and NOT x_i appear at the same time. For each 1 <= i <= n, we can have neither of x_i or NOT x_i, only x_i or only NOT x_i appearing in a disjunctive clause, so the number of such clauses is 3^n. Viewing the propositional formulas of n variables as functions {0,1}^n -> {0,1}, a disjunctive clause corresponds to a function f such that the inverse image of 0 is of the form A_1 X ... X A_n, where A_i is nonempty for all 1 <= i <= n. Since each A_i has 3 choices ({0}, {1} or {0,1}), we also find that the number of disjunctive clauses of n variables is 3^n.
Equivalently, a(n) is the number of conjunctive clauses of n variables. (End)
The finite subsequence a(2), a(3), a(4), a(5) = 9, 27, 81, 243 is one of only two geometric sequences that can be formed with all interior angles (all integer, in degrees) of a simple polygon. The other sequence is a subsequence of A007283 (see comment there). - Felix Huber, Feb 15 2024

Examples

			G.f. = 1 + 3*x + 9*x^2 + 27*x^3 + 81*x^4 + 243*x^5 + 729*x^6 + 2187*x^7 + ...

References

  • N. J. A. Sloane, A Handbook of Integer Sequences, Academic Press, 1973 (includes this sequence).
  • N. J. A. Sloane and Simon Plouffe, The Encyclopedia of Integer Sequences, Academic Press, 1995 (includes this sequence).

Links

Crossrefs

Cf. A008776 (2*a(n), and first differences).
a(n) = A092477(n, 2) for n > 0.
a(n) = A159991(n) / A009964(n).
Cf. A100772, A035002. Row sums of A125076 and A153279.
a(n) = A217764(0, n).
Cf. A046816, A006521, A014945, A275414 (multisets).
The following are parallel families: A000079 (2^n), A004094 (2^n reversed), A028909 (2^n sorted up), A028910 (2^n sorted down), A036447 (double and reverse), A057615 (double and sort up), A263451 (double and sort down); A000244 (3^n), A004167 (3^n reversed), A321540 (3^n sorted up), A321539 (3^n sorted down), A163632 (triple and reverse), A321542 (triple and sort up), A321541 (triple and sort down).
Cf. A001047, A028243, A058481.

Programs

Formula

a(n) = 3^n.
a(0) = 1; a(n) = 3*a(n-1).
G.f.: 1/(1-3*x).
E.g.f.: exp(3*x).
a(n) = n!*Sum_{i + j + k = n, i, j, k >= 0} 1/(i!*j!*k!). - Benoit Cloitre, Nov 01 2002
a(n) = Sum_{k = 0..n} 2^k*binomial(n, k), binomial transform of A000079.
a(n) = A090888(n, 2). - Ross La Haye, Sep 21 2004
a(n) = 2^(2n) - A005061(n). - Ross La Haye, Sep 10 2005
a(n) = A112626(n, 0). - Ross La Haye, Jan 11 2006
Hankel transform of A007854. - Philippe Deléham, Nov 26 2006
a(n) = 2*StirlingS2(n+1,3) + StirlingS2(n+2,2) = 2*(StirlingS2(n+1,3) + StirlingS2(n+1,2)) + 1. - Ross La Haye, Jun 26 2008
a(n) = 2*StirlingS2(n+1, 3) + StirlingS2(n+2, 2) = 2*(StirlingS2(n+1, 3) + StirlingS2(n+1, 2)) + 1. - Ross La Haye, Jun 09 2008
Sum_{n >= 0} 1/a(n) = 3/2. - Gary W. Adamson, Aug 29 2008
If p(i) = Fibonacci(2i-2) and if A is the Hessenberg matrix of order n defined by A(i, j) = p(j-i+1), (i <= j), A(i, j) = -1, (i = j+1), and A(i, j) = 0 otherwise, then, for n >= 1, a(n-1) = det A. - Milan Janjic, May 08 2010
G.f. A(x) = M(x)/(1-M(x))^2, M(x) - o.g.f for Motzkin numbers (A001006). - Vladimir Kruchinin, Aug 18 2010
a(n) = A133494(n+1). - Arkadiusz Wesolowski, Jul 27 2011
2/3 + 3/3^2 + 2/3^3 + 3/3^4 + 2/3^5 + ... = 9/8. [Jolley, Summation of Series, Dover, 1961]
a(n) = Sum_{k=0..n} A207543(n,k)*4^(n-k). - Philippe Deléham, Feb 25 2012
a(n) = Sum_{k=0..n} A125185(n,k). - Philippe Deléham, Feb 26 2012
Sum_{n > 0} Mobius(n)/a(n) = 0.181995386702633887827... (see A238271). - Alonso del Arte, Aug 09 2012. See also the sodium 3s orbital energy in table V of J. Chem. Phys. 53 (1970) 348.
a(n) = (tan(Pi/3))^(2*n). - Bernard Schott, May 06 2022
a(n-1) = binomial(2*n-1, n) + Sum_{k >= 1} binomial(2*n, n+3*k)*(-1)^k. - Greg Dresden, Oct 14 2022
G.f.: Sum_{k >= 0} x^k/(1-2*x)^(k+1). - Kevin Long, Mar 14 2023

A000670 Fubini numbers: number of preferential arrangements of n labeled elements; or number of weak orders on n labeled elements; or number of ordered partitions of [n].

Original entry on oeis.org

1, 1, 3, 13, 75, 541, 4683, 47293, 545835, 7087261, 102247563, 1622632573, 28091567595, 526858348381, 10641342970443, 230283190977853, 5315654681981355, 130370767029135901, 3385534663256845323, 92801587319328411133, 2677687796244384203115, 81124824998504073881821
Offset: 0

Views

Author

N. J. A. Sloane

Keywords

Comments

Number of ways n competitors can rank in a competition, allowing for the possibility of ties.
Also number of asymmetric generalized weak orders on n points.
Also called the ordered Bell numbers.
A weak order is a relation that is transitive and complete.
Called Fubini numbers by Comtet: counts formulas in Fubini theorem when switching the order of summation in multiple sums. - Olivier Gérard, Sep 30 2002 [Named after the Italian mathematician Guido Fubini (1879-1943). - Amiram Eldar, Jun 17 2021]
If the points are unlabeled then the answer is a(0) = 1, a(n) = 2^(n-1) (cf. A011782).
For n>0, a(n) is the number of elements in the Coxeter complex of type A_{n-1}. The corresponding sequence for type B is A080253 and there one can find a worked example as well as a geometric interpretation. - Tim Honeywill and Paul Boddington, Feb 10 2003
Also number of labeled (1+2)-free posets. - Detlef Pauly, May 25 2003
Also the number of chains of subsets starting with the empty set and ending with a set of n distinct objects. - Andrew Niedermaier, Feb 20 2004
From Michael Somos, Mar 04 2004: (Start)
Stirling transform of A007680(n) = [3,10,42,216,...] gives [3,13,75,541,...].
Stirling transform of a(n) = [1,3,13,75,...] is A083355(n) = [1,4,23,175,...].
Stirling transform of A000142(n) = [1,2,6,24,120,...] is a(n) = [1,3,13,75,...].
Stirling transform of A005359(n-1) = [1,0,2,0,24,0,...] is a(n-1) = [1,1,3,13,75,...].
Stirling transform of A005212(n-1) = [0,1,0,6,0,120,0,...] is a(n-1) = [0,1,3,13,75,...].
(End)
Unreduced denominators in convergent to log(2) = lim_{n->infinity} n*a(n-1)/a(n).
a(n) is congruent to a(n+(p-1)p^(h-1)) (mod p^h) for n >= h (see Barsky).
Stirling-Bernoulli transform of 1/(1-x^2). - Paul Barry, Apr 20 2005
This is the sequence of moments of the probability distribution of the number of tails before the first head in a sequence of fair coin tosses. The sequence of cumulants of the same probability distribution is A000629. That sequence is twice the result of deletion of the first term of this sequence. - Michael Hardy (hardy(AT)math.umn.edu), May 01 2005
With p(n) = the number of integer partitions of n, p(i) = the number of parts of the i-th partition of n, d(i) = the number of different parts of the i-th partition of n, p(j,i) = the j-th part of the i-th partition of n, m(i,j) = multiplicity of the j-th part of the i-th partition of n, one has: a(n) = Sum_{i=1..p(n)} (n!/(Product_{j=1..p(i)} p(i,j)!)) * (p(i)!/(Product_{j=1..d(i)} m(i,j)!)). - Thomas Wieder, May 18 2005
The number of chains among subsets of [n]. The summed term in the new formula is the number of such chains of length k. - Micha Hofri (hofri(AT)wpi.edu), Jul 01 2006
Occurs also as first column of a matrix-inversion occurring in a sum-of-like-powers problem. Consider the problem for any fixed natural number m>2 of finding solutions to the equation Sum_{k=1..n} k^m = (k+1)^m. Erdős conjectured that there are no solutions for n, m > 2. Let D be the matrix of differences of D[m,n] := Sum_{k=1..n} k^m - (k+1)^m. Then the generating functions for the rows of this matrix D constitute a set of polynomials in n (for varying n along columns) and the m-th polynomial defining the m-th row. Let GF_D be the matrix of the coefficients of this set of polynomials. Then the present sequence is the (unsigned) first column of GF_D^-1. - Gottfried Helms, Apr 01 2007
Assuming A = log(2), D is d/dx and f(x) = x/(exp(x)-1), we have a(n) = (n!/2*A^(n+1)) Sum_{k=0..n} (A^k/k!) D^n f(-A) which gives Wilf's asymptotic value when n tends to infinity. Equivalently, D^n f(-a) = 2*( A*a(n) - 2*a(n-1) ). - Martin Kochanski (mjk(AT)cardbox.com), May 10 2007
List partition transform (see A133314) of (1,-1,-1,-1,...). - Tom Copeland, Oct 24 2007
First column of A154921. - Mats Granvik, Jan 17 2009
A slightly more transparent interpretation of a(n) is as the number of 'factor sequences' of N for the case in which N is a product of n distinct primes. A factor sequence of N of length k is of the form 1 = x(1), x(2), ..., x(k) = N, where {x(i)} is an increasing sequence such that x(i) divides x(i+1), i=1,2,...,k-1. For example, N=70 has the 13 factor sequences {1,70}, {1,2,70}, {1,5,70}, {1,7,70}, {1,10,70}, {1,14,70}, {1,35,70}, {1,2,10,70}, {1,2,14,70}, {1,5,10,70}, {1,5,35,70}, {1,7,14,70}, {1,7,35,70}. - Martin Griffiths, Mar 25 2009
Starting (1, 3, 13, 75, ...) = row sums of triangle A163204. - Gary W. Adamson, Jul 23 2009
Equals double inverse binomial transform of A007047: (1, 3, 11, 51, ...). - Gary W. Adamson, Aug 04 2009
If f(x) = Sum_{n>=0} c(n)*x^n converges for every x, then Sum_{n>=0} f(n*x)/2^(n+1) = Sum_{n>=0} c(n)*a(n)*x^n. Example: Sum_{n>=0} exp(n*x)/2^(n+1) = Sum_{n>=0} a(n)*x^n/n! = 1/(2-exp(x)) = e.g.f. - Miklos Kristof, Nov 02 2009
Hankel transform is A091804. - Paul Barry, Mar 30 2010
It appears that the prime numbers greater than 3 in this sequence (13, 541, 47293, ...) are of the form 4n+1. - Paul Muljadi, Jan 28 2011
The Fi1 and Fi2 triangle sums of A028246 are given by the terms of this sequence. For the definitions of these triangle sums, see A180662. - Johannes W. Meijer, Apr 20 2011
The modified generating function A(x) = 1/(2-exp(x))-1 = x + 3*x^2/2! + 13*x^3/3! + ... satisfies the autonomous differential equation A' = 1 + 3*A + 2*A^2 with initial condition A(0) = 0. Applying [Bergeron et al., Theorem 1] leads to two combinatorial interpretations for this sequence: (A) a(n) gives the number of plane-increasing 0-1-2 trees on n vertices, where vertices of outdegree 1 come in 3 colors and vertices of outdegree 2 come in 2 colors. (B) a(n) gives the number of non-plane-increasing 0-1-2 trees on n vertices, where vertices of outdegree 1 come in 3 colors and vertices of outdegree 2 come in 4 colors. Examples are given below. - Peter Bala, Aug 31 2011
Starting with offset 1 = the eigensequence of A074909 (the beheaded Pascal's triangle), and row sums of triangle A208744. - Gary W. Adamson, Mar 05 2012
a(n) = number of words of length n on the alphabet of positive integers for which the letters appearing in the word form an initial segment of the positive integers. Example: a(2) = 3 counts 11, 12, 21. The map "record position of block containing i, 1<=i<=n" is a bijection from lists of sets on [n] to these words. (The lists of sets on [2] are 12, 1/2, 2/1.) - David Callan, Jun 24 2013
This sequence was the subject of one of the earliest uses of the database. Don Knuth, who had a computer printout of the database prior to the publication of the 1973 Handbook, wrote to N. J. A. Sloane on May 18, 1970, saying: "I have just had my first real 'success' using your index of sequences, finding a sequence treated by Cayley that turns out to be identical to another (a priori quite different) sequence that came up in connection with computer sorting." A000670 is discussed in Exercise 3 of Section 5.3.1 of The Art of Computer Programming, Vol. 3, 1973. - N. J. A. Sloane, Aug 21 2014
Ramanujan gives a method of finding a continued fraction of the solution x of an equation 1 = x + a2*x^2 + ... and uses log(2) as the solution of 1 = x + x^2/2 + x^3/6 + ... as an example giving the sequence of simplified convergents as 0/1, 1/1, 2/3, 9/13, 52/75, 375/541, ... of which the sequence of denominators is this sequence, while A052882 is the numerators. - Michael Somos, Jun 19 2015
For n>=1, a(n) is the number of Dyck paths (A000108) with (i) n+1 peaks (UD's), (ii) no UUDD's, and (iii) at least one valley vertex at every nonnegative height less than the height of the path. For example, a(2)=3 counts UDUDUD (of height 1 with 2 valley vertices at height 0), UDUUDUDD, UUDUDDUD. These paths correspond, under the "glove" or "accordion" bijection, to the ordered trees counted by Cayley in the 1859 reference, after a harmless pruning of the "long branches to a leaf" in Cayley's trees. (Cayley left the reader to infer the trees he was talking about from examples for small n and perhaps from his proof.) - David Callan, Jun 23 2015
From David L. Harden, Apr 09 2017: (Start)
Fix a set X and define two distance functions d,D on X to be metrically equivalent when d(x_1,y_1) <= d(x_2,y_2) iff D(x_1,y_1) <= D(x_2,y_2) for all x_1, y_1, x_2, y_2 in X.
Now suppose that we fix a function f from unordered pairs of distinct elements of X to {1,...,n}. Then choose positive real numbers d_1 <= ... <= d_n such that d(x,y) = d_{f(x,y)}; the set of all possible choices of the d_i's makes this an n-parameter family of distance functions on X. (The simplest example of such a family occurs when n is a triangular number: When that happens, write n = (k 2). Then the set of all distance functions on X, when |X| = k, is such a family.) The number of such distance functions, up to metric equivalence, is a(n).
It is easy to see that an equivalence class of distance functions gives rise to a well-defined weak order on {d_1, ..., d_n}. To see that any weak order is realizable, choose distances from the set of integers {n-1, ..., 2n-2} so that the triangle inequality is automatically satisfied. (End)
a(n) is the number of rooted labeled forests on n nodes that avoid the patterns 213, 312, and 321. - Kassie Archer, Aug 30 2018
From A.H.M. Smeets, Nov 17 2018: (Start)
Also the number of semantic different assignments to n variables (x_1, ..., x_n) including simultaneous assignments. From the example given by Joerg Arndt (Mar 18 2014), this is easily seen by replacing
"{i}" by "x_i := expression_i(x_1, ..., x_n)",
"{i, j}" by "x_i, x_j := expression_i(x_1, .., x_n), expression_j(x_1, ..., x_n)", i.e., simultaneous assignment to two different variables (i <> j),
similar for simultaneous assignments to more variables, and
"<" by ";", i.e., the sequential constructor. These examples are directly related to "Number of ways n competitors can rank in a competition, allowing for the possibility of ties." in the first comment.
From this also the number of different mean definitions as obtained by iteration of n different mean functions on n initial values. Examples:
the AGM(x1,x2) = AGM(x2,x1) is represented by {arithmetic mean, geometric mean}, i.e., simultaneous assignment in any iteration step;
Archimedes's scheme (for Pi) is represented by {geometric mean} < {harmonic mean}, i.e., sequential assignment in any iteration step;
the geometric mean of two values can also be observed by {arithmetic mean, harmonic mean};
the AGHM (as defined in A319215) is represented by {arithmetic mean, geometric mean, harmonic mean}, i.e., simultaneous assignment, but there are 12 other semantic different ways to assign the values in an AGHM scheme.
By applying power means (also called Holder means) this can be extended to any value of n. (End)
Total number of faces of all dimensions in the permutohedron of order n. For example, the permutohedron of order 3 (a hexagon) has 6 vertices + 6 edges + 1 2-face = 13 faces, and the permutohedron of order 4 (a truncated octahedron) has 24 vertices + 36 edges + 14 2-faces + 1 3-face = 75 faces. A001003 is the analogous sequence for the associahedron. - Noam Zeilberger, Dec 08 2019
Number of odd multinomial coefficients N!/(a_1!*a_2!*...*a_k!). Here each a_i is positive, and Sum_{i} a_i = N (so 2^{N-1} multinomial coefficients in all), where N is any positive integer whose binary expansion has n 1's. - Richard Stanley, Apr 05 2022 (edited Oct 19 2022)
From Peter Bala, Jul 08 2022: (Start)
Conjecture: Let k be a positive integer. The sequence obtained by reducing a(n) modulo k is eventually periodic with the period dividing phi(k) = A000010(k). For example, modulo 16 we obtain the sequence [1, 1, 3, 13, 11, 13, 11, 13, 11, 13, ...], with an apparent period of 2 beginning at a(4). Cf. A354242.
More generally, we conjecture that the same property holds for integer sequences having an e.g.f. of the form G(exp(x) - 1), where G(x) is an integral power series. (End)
a(n) is the number of ways to form a permutation of [n] and then choose a subset of its descent set. - Geoffrey Critzer, Apr 29 2023
This is the Akiyama-Tanigawa transform of A000079, the powers of two. - Shel Kaphan, May 02 2024

Examples

			Let the points be labeled 1,2,3,...
a(2) = 3: 1<2, 2<1, 1=2.
a(3) = 13 from the 13 arrangements: 1<2<3, 1<3<2, 2<1<3, 2<3<1, 3<1<2, 3<2<1, 1=2<3 1=3<2, 2=3<1, 1<2=3, 2<1=3, 3<1=2, 1=2=3.
Three competitors can finish in 13 ways: 1,2,3; 1,3,2; 2,1,3; 2,3,1; 3,1,2; 3,2,1; 1,1,3; 2,2,1; 1,3,1; 2,1,2; 3,1,1; 1,2,2; 1,1,1.
a(3) = 13. The 13 plane increasing 0-1-2 trees on 3 vertices, where vertices of outdegree 1 come in 3 colors and vertices of outdegree 2 come in 2 colors, are:
........................................................
........1 (x3 colors).....1(x2 colors)....1(x2 colors)..
........|................/.\............./.\............
........2 (x3 colors)...2...3...........3...2...........
........|...............................................
........3...............................................
......====..............====............====............
.Totals 9......+..........2....+..........2....=..13....
........................................................
a(4) = 75. The 75 non-plane increasing 0-1-2 trees on 4 vertices, where vertices of outdegree 1 come in 3 colors and vertices of outdegree 2 come in 4 colors, are:
...............................................................
.....1 (x3).....1(x4).......1(x4).....1(x4)........1(x3).......
.....|........./.\........./.\......./.\...........|...........
.....2 (x3)...2...3.(x3)..3...2(x3).4...2(x3)......2(x4).......
.....|.............\...........\.........\......../.\..........
.....3.(x3).........4...........4.........3......3...4.........
.....|.........................................................
.....4.........................................................
....====......=====........====......====.........====.........
Tots 27....+....12......+...12....+...12.......+...12...=...75.
From _Joerg Arndt_, Mar 18 2014: (Start)
The a(3) = 13 strings on the alphabet {1,2,3} containing all letters up to the maximal value appearing and the corresponding ordered set partitions are:
01:  [ 1 1 1 ]     { 1, 2, 3 }
02:  [ 1 1 2 ]     { 1, 2 } < { 3 }
03:  [ 1 2 1 ]     { 1, 3 } < { 2 }
04:  [ 2 1 1 ]     { 2, 3 } < { 1 }
05:  [ 1 2 2 ]     { 1 } < { 2, 3 }
06:  [ 2 1 2 ]     { 2 } < { 1, 3 }
07:  [ 2 2 1 ]     { 3 } < { 1, 2 }
08:  [ 1 2 3 ]     { 1 } < { 2 } < { 3 }
09:  [ 1 3 2 ]     { 1 } < { 3 } < { 2 }
00:  [ 2 1 3 ]     { 2 } < { 1 } < { 3 }
11:  [ 2 3 1 ]     { 3 } < { 1 } < { 2 }
12:  [ 3 1 2 ]     { 2 } < { 3 } < { 1 }
13:  [ 3 2 1 ]     { 3 } < { 2 } < { 1 }
(End)

References

  • Mohammad K. Azarian, Geometric Series, Problem 329, Mathematics and Computer Education, Vol. 30, No. 1, Winter 1996, p. 101. Solution published in Vol. 31, No. 2, Spring 1997, pp. 196-197.
  • Norman Biggs, E. Keith Lloyd and Robin J. Wilson, Graph Theory 1736-1936, Oxford, 1976, p. 44 (P(x)).
  • Miklos Bona, editor, Handbook of Enumerative Combinatorics, CRC Press, 2015, page 183 (see R_n).
  • Kenneth S. Brown, Buildings, Springer-Verlag, 1988.
  • Louis Comtet, Advanced Combinatorics, Reidel, 1974, p. 228.
  • Jean-Marie De Koninck, Ces nombres qui nous fascinent, Entry 13, pp 4, Ellipses, Paris 2008.
  • P. J. Freyd, On the size of Heyting semi-lattices, preprint, 2002.
  • Ian P. Goulden and David M. Jackson, Combinatorial Enumeration, John Wiley and Sons, N.Y., 1983.
  • Ronald L. Graham, Donald E. Knuth, and Oren Patashnik, Concrete Mathematics. Addison-Wesley, Reading, MA, 2nd Ed., 1994, exercise 7.44 (pp. 378, 571).
  • Silvia Heubach and Toufik Mansour, Combinatorics of Compositions and Words, CRC Press, 2010.
  • Donald E. Knuth, The Art of Computer Programming. Addison-Wesley, Reading, MA, Vol. 3, 1973, Section 5.3.1, Problem 3.
  • M. Muresan, Generalized Fubini numbers, Stud. Cerc. Mat., Vol. 37, No. 1 (1985), pp. 70-76.
  • Paul Peart, Hankel determinants via Stieltjes matrices. Proceedings of the Thirty-first Southeastern International Conference on Combinatorics, Graph Theory and Computing (Boca Raton, FL, 2000). Congr. Numer. 144 (2000), 153-159.
  • S. Ramanujan, Notebooks, Tata Institute of Fundamental Research, Bombay 1957 Vol. 1, see page 19.
  • Ulrike Sattler, Decidable classes of formal power series with nice closure properties, Diplomarbeit im Fach Informatik, Univ. Erlangen - Nuernberg, Jul 27 1994.
  • N. J. A. Sloane, A Handbook of Integer Sequences, Academic Press, 1973 (includes this sequence).
  • N. J. A. Sloane and Simon Plouffe, The Encyclopedia of Integer Sequences, Academic Press, 1995 (includes this sequence).
  • Richard P. Stanley, Enumerative Combinatorics, Wadsworth, Vol. 1, 1986; see Example 3.15.10, p. 146.
  • Jack van der Elsen, Black and White Transformations, Shaker Publishing, Maastricht, 2005, p. 18.

Links

Crossrefs

See A240763 for a list of the actual preferential arrangements themselves.
A000629, this sequence, A002050, A032109, A052856, A076726 are all more-or-less the same sequence. - N. J. A. Sloane, Jul 04 2012
Binomial transform of A052841. Inverse binomial transform of A000629.
Asymptotic to A034172.
Cf. A002144, A002869, A004121, A004122, A007047, A007318, A048144, A053525, A080253, A080254, A011782, A154921, A162312, A163204, A242280, A261959, A290376, A074206.
Row r=1 of A094416. Row 0 of array in A226513. Row n=1 of A262809.
Main diagonal of: A135313, A261781, A276890, A327245, A327583, A327584.
Row sums of triangles A019538, A131689, A208744 and A276891.
A217389 and A239914 give partial sums.
Column k=1 of A326322.

Programs

  • Haskell
    a000670 n = a000670_list !! n
a000670_list = 1 : f [1] (map tail $ tail a007318_tabl) where
   f xs (bs:bss) = y : f (y : xs) bss where y = sum $ zipWith (*) xs bs
-- Reinhard Zumkeller, Jul 26 2014
  • Magma
    R:=PowerSeriesRing(Rationals(), 40);
Coefficients(R!(Laplace( 1/(2-Exp(x)) ))); // G. C. Greubel, Jun 11 2024
  • Maple
    A000670 := proc(n) option remember; local k; if n <=1 then 1 else add(binomial(n,k)*A000670(n-k),k=1..n); fi; end;
with(combstruct); SeqSetL := [S, {S=Sequence(U), U=Set(Z,card >= 1)},labeled]; seq(count(SeqSetL,size=j),j=1..12);
with(combinat): a:=n->add(add((-1)^(k-i)*binomial(k, i)*i^n, i=0..n), k=0..n): seq(a(n), n=0..18); # Zerinvary Lajos, Jun 03 2007
a := n -> add(combinat:-eulerian1(n,k)*2^k,k=0..n): # Peter Luschny, Jan 02 2015
a := n -> (polylog(-n, 1/2)+`if`(n=0,1,0))/2: seq(round(evalf(a(n),32)), n=0..20); # Peter Luschny, Nov 03 2015
# next Maple program:
b:= proc(n, k) option remember;
     `if`(n=0, k!, k*b(n-1, k)+b(n-1, k+1))
    end:
a:= n-> b(n, 0):
seq(a(n), n=0..20);  # Alois P. Heinz, Aug 04 2021
  • Mathematica
    Table[(PolyLog[-z, 1/2] + KroneckerDelta[z])/2, {z, 0, 20}] (* Wouter Meeussen *)
a[0] = 1; a[n_]:= a[n]= Sum[Binomial[n, k]*a[n-k], {k, 1, n}]; Table[a[n], {n, 0, 30}] (* Roger L. Bagula and Gary W. Adamson, Sep 13 2008 *)
t = 30; Range[0, t]! CoefficientList[Series[1/(2 - Exp[x]), {x, 0, t}], x] (* Vincenzo Librandi, Mar 16 2014 *)
a[ n_] := If[ n < 0, 0, n! SeriesCoefficient[ 1 / (2 - Exp@x), {x, 0, n}]]; (* Michael Somos, Jun 19 2015 *)
Table[Sum[k^n/2^(k+1),{k,0,Infinity}],{n,0,20}] (* Vaclav Kotesovec, Jun 26 2015 *)
Table[HurwitzLerchPhi[1/2, -n, 0]/2, {n, 0, 20}] (* Jean-François Alcover, Jan 31 2016 *)
Fubini[n_, r_] := Sum[k!*Sum[(-1)^(i+k+r)*((i+r)^(n-r)/(i!*(k-i-r)!)), {i, 0, k-r}], {k, r, n}]; Fubini[0, 1] = 1; Table[Fubini[n, 1], {n, 0, 20}] (* Jean-François Alcover, Mar 31 2016 *)
Eulerian1[0, 0] = 1; Eulerian1[n_, k_] := Sum[(-1)^j (k-j+1)^n Binomial[n+1, j], {j, 0, k+1}]; Table[Sum[Eulerian1[n, k] 2^k, {k, 0, n}], {n, 0, 20}] (* Jean-François Alcover, Jul 13 2019, after Peter Luschny *)
Prepend[Table[-(-1)^k HurwitzLerchPhi[2, -k, 0]/2, {k, 1, 50}], 1] (* Federico Provvedi,Sep 05 2020 *)
Table[Sum[k!*StirlingS2[n,k], {k, 0, n}], {n, 0, 20}] (* Vaclav Kotesovec, Nov 22 2020 *)
  • Maxima
    makelist(sum(stirling2(n,k)*k!,k,0,n),n,0,12); /* Emanuele Munarini, Jul 07 2011 */
  • Maxima
    a[0]:1$ a[n]:=sum(binomial(n,k)*a[n-k],k,1,n)$ A000670(n):=a[n]$ makelist(A000670(n),n,0,30); /* Martin Ettl, Nov 05 2012 */
  • PARI
    {a(n) = if( n<0, 0, n! * polcoeff( subst( 1 / (1 - y), y, exp(x + x*O(x^n)) - 1), n))}; /* Michael Somos, Mar 04 2004 */
  • PARI
    Vec(serlaplace(1/(2-exp('x+O('x^66))))) /* Joerg Arndt, Jul 10 2011 */
  • PARI
    {a(n)=polcoeff(sum(m=0,n,m!*x^m/prod(k=1,m,1-k*x+x*O(x^n))),n)} /* Paul D. Hanna, Jul 20 2011 */
  • PARI
    {a(n) = if( n<1, n==0, sum(k=1, n, binomial(n, k) * a(n-k)))}; /* Michael Somos, Jul 16 2017 */
  • Python
    from math import factorial
from sympy.functions.combinatorial.numbers import stirling
def A000670(n): return sum(factorial(k)*stirling(n,k) for k in range(n+1)) # Chai Wah Wu, Nov 08 2022
  • Sage
    @CachedFunction
def A000670(n) : return 1 if n == 0 else add(A000670(k)*binomial(n,k) for k in range(n))
[A000670(n) for n in (0..20)] # Peter Luschny, Jul 14 2012

Formula

a(n) = Sum_{k=0..n} k! * StirlingS2(n,k) (whereas the Bell numbers A000110(n) = Sum_{k=0..n} StirlingS2(n,k)).
E.g.f.: 1/(2-exp(x)).
a(n) = Sum_{k=1..n} binomial(n, k)*a(n-k), a(0) = 1.
The e.g.f. y(x) satisfies y' = 2*y^2 - y.
a(n) = A052856(n) - 1, if n>0.
a(n) = A052882(n)/n, if n>0.
a(n) = A076726(n)/2.
a(n) is asymptotic to (1/2)*n!*log_2(e)^(n+1), where log_2(e) = 1.442695... [Barthelemy80, Wilf90].
For n >= 1, a(n) = (n!/2) * Sum_{k=-infinity..infinity} of (log(2) + 2 Pi i k)^(-n-1). - Dean Hickerson
a(n) = ((x*d/dx)^n)(1/(2-x)) evaluated at x=1. - Karol A. Penson, Sep 24 2001
For n>=1, a(n) = Sum_{k>=1} (k-1)^n/2^k = A000629(n)/2. - Benoit Cloitre, Sep 08 2002
Value of the n-th Eulerian polynomial (cf. A008292) at x=2. - Vladeta Jovovic, Sep 26 2003
First Eulerian transform of the powers of 2 [A000079]. See A000142 for definition of FET. - Ross La Haye, Feb 14 2005
a(n) = Sum_{k=0..n} (-1)^k*k!*Stirling2(n+1, k+1)*(1+(-1)^k)/2. - Paul Barry, Apr 20 2005
a(n) + a(n+1) = 2*A005649(n). - Philippe Deléham, May 16 2005 - Thomas Wieder, May 18 2005
Equals inverse binomial transform of A000629. - Gary W. Adamson, May 30 2005
a(n) = Sum_{k=0..n} k!*( Stirling2(n+2, k+2) - Stirling2(n+1, k+2) ). - Micha Hofri (hofri(AT)wpi.edu), Jul 01 2006
Recurrence: 2*a(n) = (a+1)^n where superscripts are converted to subscripts after binomial expansion - reminiscent of Bernoulli numbers' B_n = (B+1)^n. - Martin Kochanski (mjk(AT)cardbox.com), May 10 2007
a(n) = (-1)^n * n! * Laguerre(n,P((.),2)), umbrally, where P(j,t) are the polynomials in A131758. - Tom Copeland, Sep 27 2007
Formula in terms of the hypergeometric function, in Maple notation: a(n) = hypergeom([2,2...2],[1,1...1],1/2)/4, n=1,2..., where in the hypergeometric function there are n upper parameters all equal to 2 and n-1 lower parameters all equal to 1 and the argument is equal to 1/2. Example: a(4) = evalf(hypergeom([2,2,2,2],[1,1,1],1/2)/4) = 75. - Karol A. Penson, Oct 04 2007
a(n) = Sum_{k=0..n} A131689(n,k). - Philippe Deléham, Nov 03 2008
From Peter Bala, Jul 01 2009: (Start)
Analogy with the Bernoulli numbers.
We enlarge upon the above comment of M. Kochanski.
The Bernoulli polynomials B_n(x), n = 0,1,..., are given by the formula
(1)... B_n(x) := Sum_{k=0..n} binomial(n,k)*B(k)*x^(n-k),
where B(n) denotes the sequence of Bernoulli numbers B(0) = 1,
B(1) = -1/2, B(2) = 1/6, B(3) = 0, ....
By analogy, we associate with the present sequence an Appell sequence of polynomials {P_n(x)} n >= 0 defined by
(2)... P_n(x) := Sum_{k=0..n} binomial(n,k)*a(k)*x^(n-k).
These polynomials have similar properties to the Bernoulli polynomials.
The first few values are P_0(x) = 1, P_1(x) = x + 1,
P_2(x) = x^2 + 2*x + 3, P_3(x) = x^3 + 3*x^2 + 9*x + 13 and
P_4(x) = x^4 + 4*x^3 + 18*x^2 + 52*x + 75. See A154921 for the triangle of coefficients of these polynomials.
The e.g.f. for this polynomial sequence is
(3)... exp(x*t)/(2 - exp(t)) = 1 + (x + 1)*t + (x^2 + 2*x + 3)*t^2/2! + ....
The polynomials satisfy the difference equation
(4)... 2*P_n(x - 1) - P_n(x) = (x - 1)^n,
and so may be used to evaluate the weighted sums of powers of integers
(1/2)*1^m + (1/2)^2*2^m + (1/2)^3*3^m + ... + (1/2)^(n-1)*(n-1)^m
via the formula
(5)... Sum_{k=1..n-1} (1/2)^k*k^m = 2*P_m(0) - (1/2)^(n-1)*P_m(n),
analogous to the evaluation of the sums 1^m + 2^m + ... + (n-1)^m in terms of Bernoulli polynomials.
This last result can be generalized to
(6)... Sum_{k=1..n-1} (1/2)^k*(k+x)^m = 2*P_m(x)-(1/2)^(n-1)*P_m(x+n).
For more properties of the polynomials P_n(x), refer to A154921.
For further information on weighted sums of powers of integers and the associated polynomial sequences, see A162312.
The present sequence also occurs in the evaluation of another sum of powers of integers. Define
(7)... S_m(n) := Sum_{k=1..n-1} (1/2)^k*((n-k)*k)^m, m = 1,2,....
Then
(8)... S_m(n) = (-1)^m *[2*Q_m(-n) - (1/2)^(n-1)*Q_m(n)],
where Q_m(x) are polynomials in x given by
(9)... Q_m(x) = Sum_{k=0..m} a(m+k)*binomial(m,k)*x^(m-k).
The first few values are Q_1(x) = x + 3, Q_2(x) = 3*x^2 + 26*x + 75
and Q_3(x) = 13*x^3 + 225*x^2 + 1623*x + 4683.
For example, m = 2 gives
(10)... S_2(n) := Sum_{k=1..n-1} (1/2)^k*((n-k)*k)^2
= 2*(3*n^2 - 26*n + 75) - (1/2)^(n-1)*(3*n^2 + 26*n + 75).
(End)
G.f.: 1/(1-x/(1-2*x/(1-2*x/(1-4*x/(1-3*x/(1-6*x/(1-4*x/(1-8*x/(1-5*x/(1-10*x/(1-6*x/(1-... (continued fraction); coefficients of continued fraction are given by floor((n+2)/2)*(3-(-1)^n)/2 (A029578(n+2)). - Paul Barry, Mar 30 2010
G.f.: 1/(1-x-2*x^2/(1-4*x-8*x^2/(1-7*x-18*x^2/(1-10*x-32*x^2/(1../(1-(3*n+1)*x-2*(n+1)^2*x^2/(1-... (continued fraction). - Paul Barry, Jun 17 2010
G.f.: A(x) = Sum_{n>=0} n!*x^n / Product_{k=1..n} (1-k*x). - Paul D. Hanna, Jul 20 2011
a(n) = A074206(q_1*q_2*...*q_n), where {q_i} are distinct primes. - Vladimir Shevelev, Aug 05 2011
The adjusted e.g.f. A(x) := 1/(2-exp(x))-1, has inverse function A(x)^-1 = Integral_{t=0..x} 1/((1+t)*(1+2*t)). Applying [Dominici, Theorem 4.1] to invert the integral yields a formula for a(n): Let f(x) = (1+x)*(1+2*x). Let D be the operator f(x)*d/dx. Then a(n) = D^(n-1)(f(x)) evaluated at x = 0. Compare with A050351. - Peter Bala, Aug 31 2011
a(n) = D^n*(1/(1-x)) evaluated at x = 0, where D is the operator (1+x)*d/dx. Cf. A052801. - Peter Bala, Nov 25 2011
From Sergei N. Gladkovskii, from Oct 2011 to Oct 2013: (Start)
Continued fractions:
G.f.: 1+x/(1-x+2*x*(x-1)/(1+3*x*(2*x-1)/(1+4*x*(3*x-1)/(1+5*x*(4*x-1)/(1+... or 1+x/(U(0)-x), U(k) = 1+(k+2)*(k*x+x-1)/U(k+1).
E.g.f.: 1 + x/(G(0)-2*x) where G(k) = x + k + 1 - x*(k+1)/G(k+1).
E.g.f. (2 - 2*x)*(1 - 2*x^3/(8*x^2 - 4*x + (x^2 - 4*x + 2)*G(0)))/(x^2 - 4*x + 2) where G(k) = k^2 + k*(x+4) + 2*x + 3 - x*(k+1)*(k+3)^2 /G(k+1).
G.f.: 1 + x/G(0) where G(k) = 1 - 3*x*(k+1) - 2*x^2*(k+1)*(k+2)/G(k+1).
G.f.: 1/G(0) where G(k) = 1 - x*(k+1)/( 1 - 2*x*(k+1)/G(k+1) ).
G.f.: 1 + x/Q(0), where Q(k) = 1 - 3*x*(2*k+1) - 2*x^2*(2*k+1)*(2*k+2)/( 1 - 3*x*(2*k+2) - 2*x^2*(2*k+2)*(2*k+3)/Q(k+1) ).
G.f.: T(0)/(1-x), where T(k) = 1 - 2*x^2*(k+1)^2/( 2*x^2*(k+1)^2 - (1-x-3*x*k)*(1-4*x-3*x*k)/T(k+1) ). (End)
a(n) is always odd. For odd prime p and n >= 1, a((p-1)*n) = 0 (mod p). - Peter Bala, Sep 18 2013
a(n) = log(2)* Integral_{x>=0} floor(x)^n * 2^(-x) dx. - Peter Bala, Feb 06 2015
For n > 0, a(n) = Re(polygamma(n, i*log(2)/(2*Pi))/(2*Pi*i)^(n+1)) - n!/(2*log(2)^(n+1)). - Vladimir Reshetnikov, Oct 15 2015
a(n) = Sum_{k=1..n} (k*b2(k-1)*(k)!*Stirling2(n, k)), n>0, a(0)=1, where b2(n) is the n-th Bernoulli number of the second kind. - Vladimir Kruchinin, Nov 21 2016
Conjecture: a(n) = Sum_{k=0..2^(n-1)-1} A284005(k) for n > 0 with a(0) = 1. - Mikhail Kurkov, Jul 08 2018
a(n) = A074206(k) for squarefree k with n prime factors. In particular a(n) = A074206(A002110(n)). - Amiram Eldar, May 13 2019
For n > 0, a(n) = -(-1)^n / 2 * PHI(2, -n, 0), where PHI(z, s, a) is the Lerch zeta function. - Federico Provvedi, Sep 05 2020
a(n) = Sum_{s in S_n} Product_{i=1..n} binomial(i,s(i)-1), where s ranges over the set S_n of permutations of [n]. - Jose A. Rodriguez, Feb 02 2021
Sum_{n>=0} 1/a(n) = 2.425674839121428857970063350500499393706641093287018840857857170864211946122664... - Vaclav Kotesovec, Jun 17 2021
From Jacob Sprittulla, Oct 05 2021: (Start)
The following identities hold for sums over Stirling numbers of the second kind with even or odd second argument:
a(n) = 2 * Sum_{k=0..floor(n/2)} ((2k)! * Stirling2(n,2*k) ) - (-1)^n = 2*A052841-(-1)^n
a(n) = 2 * Sum_{k=0..floor(n/2)} ((2k+1)!* Stirling2(n,2*k+1))+ (-1)^n = 2*A089677+(-1)^n
a(n) = Sum_{k=1..floor((n+1)/2)} ((2k-1)!* Stirling2(n+1,2*k))
a(n) = Sum_{k=0..floor((n+1)/2)} ((2k)! * Stirling2(n+1,2*k+1)). (End)

A010060 Thue-Morse sequence: let A_k denote the first 2^k terms; then A_0 = 0 and for k >= 0, A_{k+1} = A_k B_k, where B_k is obtained from A_k by interchanging 0's and 1's.

Original entry on oeis.org

0, 1, 1, 0, 1, 0, 0, 1, 1, 0, 0, 1, 0, 1, 1, 0, 1, 0, 0, 1, 0, 1, 1, 0, 0, 1, 1, 0, 1, 0, 0, 1, 1, 0, 0, 1, 0, 1, 1, 0, 0, 1, 1, 0, 1, 0, 0, 1, 0, 1, 1, 0, 1, 0, 0, 1, 1, 0, 0, 1, 0, 1, 1, 0, 1, 0, 0, 1, 0, 1, 1, 0, 0, 1, 1, 0, 1, 0, 0, 1, 0, 1, 1, 0, 1, 0, 0, 1, 1, 0, 0, 1, 0, 1, 1, 0, 0, 1, 1, 0, 1, 0, 0, 1, 1
Offset: 0

Views

Author

N. J. A. Sloane

Keywords

Comments

Named after Axel Thue, whose name is pronounced as if it were spelled "Tü" where the ü sound is roughly as in the German word üben. (It is incorrect to say "Too-ee" or "Too-eh".) - N. J. A. Sloane, Jun 12 2018
Also called the Thue-Morse infinite word, or the Morse-Hedlund sequence, or the parity sequence.
Fixed point of the morphism 0 --> 01, 1 --> 10, see example. - Joerg Arndt, Mar 12 2013
The sequence is cubefree (does not contain three consecutive identical blocks) [see Offner for a direct proof] and is overlap-free (does not contain XYXYX where X is 0 or 1 and Y is any string of 0's and 1's).
a(n) = "parity sequence" = parity of number of 1's in binary representation of n.
To construct the sequence: alternate blocks of 0's and 1's of successive lengths A003159(k) - A003159(k-1), k = 1, 2, 3, ... (A003159(0) = 0). Example: since the first seven differences of A003159 are 1, 2, 1, 1, 2, 2, 2, the sequence starts with 0, 1, 1, 0, 1, 0, 0, 1, 1, 0, 0. - Emeric Deutsch, Jan 10 2003
Characteristic function of A000069 (odious numbers). - Ralf Stephan, Jun 20 2003
a(n) = S2(n) mod 2, where S2(n) = sum of digits of n, n in base-2 notation. There is a class of generalized Thue-Morse sequences: Let Sk(n) = sum of digits of n; n in base-k notation. Let F(t) be some arithmetic function. Then a(n)= F(Sk(n)) mod m is a generalized Thue-Morse sequence. The classical Thue-Morse sequence is the case k=2, m=2, F(t)= 1*t. - Ctibor O. Zizka, Feb 12 2008 (with correction from Daniel Hug, May 19 2017)
More generally, the partial sums of the generalized Thue-Morse sequences a(n) = F(Sk(n)) mod m are fractal, where Sk(n) is sum of digits of n, n in base k; F(t) is an arithmetic function; m integer. - Ctibor O. Zizka, Feb 25 2008
Starting with offset 1, = running sums mod 2 of the kneading sequence (A035263, 1, 0, 1, 1, 1, 0, 1, 0, 1, 0, 1, 1, 1, ...); also parity of A005187: (1, 3, 4, 7, 8, 10, 11, 15, 16, 18, 19, ...). - Gary W. Adamson, Jun 15 2008
Generalized Thue-Morse sequences mod n (n>1) = the array shown in A141803. As n -> infinity the sequences -> (1, 2, 3, ...). - Gary W. Adamson, Jul 10 2008
The Thue-Morse sequence for N = 3 = A053838, (sum of digits of n in base 3, mod 3): (0, 1, 2, 1, 2, 0, 2, 0, 1, 1, 2, ...) = A004128 mod 3. - Gary W. Adamson, Aug 24 2008
For all positive integers k, the subsequence a(0) to a(2^k-1) is identical to the subsequence a(2^k+2^(k-1)) to a(2^(k+1)+2^(k-1)-1). That is to say, the first half of A_k is identical to the second half of B_k, and the second half of A_k is identical to the first quarter of B_{k+1}, which consists of the k/2 terms immediately following B_k.
Proof: The subsequence a(2^k+2^(k-1)) to a(2^(k+1)-1), the second half of B_k, is by definition formed from the subsequence a(2^(k-1)) to a(2^k-1), the second half of A_k, by interchanging its 0's and 1's. In turn, the subsequence a(2^(k-1)) to a(2^k-1), the second half of A_k, which is by definition also B_{k-1}, is by definition formed from the subsequence a(0) to a(2^(k-1)-1), the first half of A_k, which is by definition also A_{k-1}, by interchanging its 0's and 1's. Interchanging the 0's and 1's of a subsequence twice leaves it unchanged, so the subsequence a(2^k+2^(k-1)) to a(2^(k+1)-1), the second half of B_k, must be identical to the subsequence a(0) to a(2^(k-1)-1), the first half of A_k.
Also, the subsequence a(2^(k+1)) to a(2^(k+1)+2^(k-1)-1), the first quarter of B_{k+1}, is by definition formed from the subsequence a(0) to a(2^(k-1)-1), the first quarter of A_{k+1}, by interchanging its 0's and 1's. As noted above, the subsequence a(2^(k-1)) to a(2^k-1), the second half of A_k, which is by definition also B_{k-1}, is by definition formed from the subsequence a(0) to a(2^(k-1)-1), which is by definition A_{k-1}, by interchanging its 0's and 1's, as well. If two subsequences are formed from the same subsequence by interchanging its 0's and 1's then they must be identical, so the subsequence a(2^(k+1)) to a(2^(k+1)+2^(k-1)-1), the first quarter of B_{k+1}, must be identical to the subsequence a(2^(k-1)) to a(2^k-1), the second half of A_k.
Therefore the subsequence a(0), ..., a(2^(k-1)-1), a(2^(k-1)), ..., a(2^k-1) is identical to the subsequence a(2^k+2^(k-1)), ..., a(2^(k+1)-1), a(2^(k+1)), ..., a(2^(k+1)+2^(k-1)-1), QED.
According to the German chess rules of 1929 a game of chess was drawn if the same sequence of moves was repeated three times consecutively. Euwe, see the references, proved that this rule could lead to infinite games. For his proof he reinvented the Thue-Morse sequence. - Johannes W. Meijer, Feb 04 2010
"Thue-Morse 0->01 & 1->10, at each stage append the previous with its complement. Start with 0, 1, 2, 3 and write them in binary. Next calculate the sum of the digits (mod 2) - that is divide the sum by 2 and use the remainder." Pickover, The Math Book.
Let s_2(n) be the sum of the base-2 digits of n and epsilon(n) = (-1)^s_2(n), the Thue-Morse sequence, then prod(n >= 0, ((2*n+1)/(2*n+2))^epsilon(n) ) = 1/sqrt(2). - Jonathan Vos Post, Jun 06 2012
Dekking shows that the constant obtained by interpreting this sequence as a binary expansion is transcendental; see also "The Ubiquitous Prouhet-Thue-Morse Sequence". - Charles R Greathouse IV, Jul 23 2013
Drmota, Mauduit, and Rivat proved that the subsequence a(n^2) is normal--see A228039. - Jonathan Sondow, Sep 03 2013
Although the probability of a 0 or 1 is equal, guesses predicated on the latest bit seen produce a correct match 2 out of 3 times. - Bill McEachen, Mar 13 2015
From a(0) to a(2n+1), there are n+1 terms equal to 0 and n+1 terms equal to 1 (see Hassan Tarfaoui link, Concours Général 1990). - Bernard Schott, Jan 21 2022

Examples

			The evolution starting at 0 is:
  0
  0, 1
  0, 1, 1, 0
  0, 1, 1, 0, 1, 0, 0, 1
  0, 1, 1, 0, 1, 0, 0, 1, 1, 0, 0, 1, 0, 1, 1, 0
  0, 1, 1, 0, 1, 0, 0, 1, 1, 0, 0, 1, 0, 1, 1, 0, 1, 0, 0, 1, 0, 1, 1, 0, 0, 1, 1, 0, 1, 0, 0, 1
  .......
A_2 = 0 1 1 0, so B_2 = 1 0 0 1 and A_3 = A_2 B_2 = 0 1 1 0 1 0 0 1.
From _Joerg Arndt_, Mar 12 2013: (Start)
The first steps of the iterated substitution are
Start: 0
Rules:
  0 --> 01
  1 --> 10
-------------
0:   (#=1)
  0
1:   (#=2)
  01
2:   (#=4)
  0110
3:   (#=8)
  01101001
4:   (#=16)
  0110100110010110
5:   (#=32)
  01101001100101101001011001101001
6:   (#=64)
  0110100110010110100101100110100110010110011010010110100110010110
(End)
From _Omar E. Pol_, Oct 28 2013: (Start)
Written as an irregular triangle in which row lengths is A011782, the sequence begins:
  0;
  1;
  1,0;
  1,0,0,1;
  1,0,0,1,0,1,1,0;
  1,0,0,1,0,1,1,0,0,1,1,0,1,0,0,1;
  1,0,0,1,0,1,1,0,0,1,1,0,1,0,0,1,0,1,1,0,1,0,0,1,1,0,0,1,0,1,1,0;
It appears that: row j lists the first A011782(j) terms of A010059, with j >= 0; row sums give A166444 which is also 0 together with A011782; right border gives A000035.
(End)

References

  • J.-P. Allouche and J. Shallit, Automatic Sequences, Cambridge Univ. Press, 2003, p. 15.
  • Jason Bell, Michael Coons, and Eric Rowland, "The Rational-Transcendental Dichotomy of Mahler Functions", Journal of Integer Sequences, Vol. 16 (2013), #13.2.10.
  • J. Berstel and J. Karhumaki, Combinatorics on words - a tutorial, Bull. EATCS, #79 (2003), pp. 178-228.
  • B. Bollobas, The Art of Mathematics: Coffee Time in Memphis, Cambridge, 2006, p. 224.
  • S. Brlek, Enumeration of factors in the Thue-Morse word, Discrete Applied Math., 24 (1989), 83-96. doi:10.1016/0166-218X(92)90274-E.
  • Yann Bugeaud and Guo-Niu Han, A combinatorial proof of the non-vanishing of Hankel determinants of the Thue-Morse sequence, Electronic Journal of Combinatorics 21(3) (2014), #P3.26.
  • Y. Bugeaud and M. Queffélec, On Rational Approximation of the Binary Thue-Morse-Mahler Number, Journal of Integer Sequences, 16 (2013), #13.2.3.
  • Currie, James D. "Non-repetitive words: Ages and essences." Combinatorica 16.1 (1996): 19-40
  • Colin Defant, Anti-Power Prefixes of the Thue-Morse Word, Journal of Combinatorics, 24(1) (2017), #P1.32
  • F. M. Dekking, Transcendance du nombre de Thue-Morse, Comptes Rendus de l'Academie des Sciences de Paris 285 (1977), pp. 157-160.
  • F. M. Dekking, On repetitions of blocks in binary sequences. J. Combinatorial Theory Ser. A 20 (1976), no. 3, pp. 292-299. MR0429728(55 #2739)
  • Dekking, Michel, Michel Mendès France, and Alf van der Poorten. "Folds." The Mathematical Intelligencer, 4.3 (1982): 130-138 & front cover, and 4:4 (1982): 173-181 (printed in two parts).
  • Dubickas, Artūras. On a sequence related to that of Thue-Morse and its applications. Discrete Math. 307 (2007), no. 9-10, 1082--1093. MR2292537 (2008b:11086).
  • Fabien Durand, Julien Leroy, and Gwenaël Richomme, "Do the Properties of an S-adic Representation Determine Factor Complexity?", Journal of Integer Sequences, Vol. 16 (2013), #13.2.6.
  • M. Euwe, Mengentheoretische Betrachtungen Über das Schachspiel, Proceedings Koninklijke Nederlandse Akademie van Wetenschappen, Amsterdam, Vol. 32 (5): 633-642, 1929.
  • S. Ferenczi, Complexity of sequences and dynamical systems, Discrete Math., 206 (1999), 145-154.
  • S. R. Finch, Mathematical Constants, Cambridge, 2003, Section 6.8.
  • W. H. Gottschalk and G. A. Hedlund, Topological Dynamics. American Mathematical Society, Colloquium Publications, Vol. 36, Providence, RI, 1955, p. 105.
  • J. Grytczuk, Thue type problems for graphs, points and numbers, Discrete Math., 308 (2008), 4419-4429.
  • A. Hof, O. Knill and B. Simon, Singular continuous spectrum for palindromic Schroedinger operators, Commun. Math. Phys. 174 (1995), 149-159.
  • Mari Huova and Juhani Karhumäki, "On Unavoidability of k-abelian Squares in Pure Morphic Words", Journal of Integer Sequences, Vol. 16 (2013), #13.2.9.
  • B. Kitchens, Review of "Computational Ergodic Theory" by G. H. Choe, Bull. Amer. Math. Soc., 44 (2007), 147-155.
  • Le Breton, Xavier, Linear independence of automatic formal power series. Discrete Math. 306 (2006), no. 15, 1776-1780.
  • M. Lothaire, Combinatorics on Words. Addison-Wesley, Reading, MA, 1983, p. 23.
  • Donald MacMurray, A mathematician gives an hour to chess, Chess Review 6 (No. 10, 1938), 238. [Discusses Marston's 1938 article]
  • Mauduit, Christian. Multiplicative properties of the Thue-Morse sequence. Period. Math. Hungar. 43 (2001), no. 1-2, 137--153. MR1830572 (2002i:11081)
  • C. A. Pickover, Wonders of Numbers, Adventures in Mathematics, Mind and Meaning, Chapter 17, 'The Pipes of Papua,' Oxford University Press, Oxford, England, 2000, pages 34-38.
  • C. A. Pickover, A Passion for Mathematics, Wiley, 2005; see p. 60.
  • Clifford A. Pickover, The Math Book, From Pythagoras to the 57th Dimension, 250 Milestones in the History of Mathematics, Sterling Publ., NY, 2009, page 316.
  • Narad Rampersad and Elise Vaslet, "On Highly Repetitive and Power Free Words", Journal of Integer Sequences, Vol. 16 (2013), #13.2.7.
  • G. Richomme, K. Saari, L. Q. Zamboni, Abelian complexity in minimal subshifts, J. London Math. Soc. 83(1) (2011) 79-95.
  • Michel Rigo, Formal Languages, Automata and Numeration Systems, 2 vols., Wiley, 2014. Mentions this sequence - see "List of Sequences" in Vol. 2.
  • M. Rigo, P. Salimov, and E. Vandomme, "Some Properties of Abelian Return Words", Journal of Integer Sequences, Vol. 16 (2013), #13.2.5.
  • Benoit Rittaud, Elise Janvresse, Emmanuel Lesigne and Jean-Christophe Novelli, Quand les maths se font discrètes, Le Pommier, 2008 (ISBN 978-2-7465-0370-0).
  • A. Salomaa, Jewels of Formal Language Theory. Computer Science Press, Rockville, MD, 1981, p. 6.
  • Shallit, J. O. "On Infinite Products Associated with Sums of Digits." J. Number Th. 21, 128-134, 1985.
  • Ian Stewart, "Feedback", Mathematical Recreations Column, Scientific American, 274 (No. 3, 1996), page 109 [Historical notes on this sequence]
  • Thomas Stoll, On digital blocks of polynomial values and extractions in the Rudin-Shapiro sequence, RAIRO - Theoretical Informatics and Applications (RAIRO: ITA), EDP Sciences, 2016, 50, pp. 93-99. .
  • A. Thue. Über unendliche Zeichenreihen, Norske Vid. Selsk. Skr. I. Mat. Nat. Kl. Christiania, No. 7 (1906), 1-22.
  • A. Thue, Über die gegenseitige Lage gleicher Teile gewisser Zeichenreihen, Norske Vid. Selsk. Skr. I. Mat. Nat. Kl. Christiania, 1 (1912), 1-67.
  • S. Wolfram, A New Kind of Science, Wolfram Media, 2002; p. 890.

Links

Crossrefs

Cf. A001285 (for 1, 2 version), A010059 (for 1, 0 version), A106400 (for +1, -1 version), A048707. A010060(n)=A000120(n) mod 2.
Cf. A007413, A080813, A080814, A036581, A108694. See also the Thue (or Roth) constant A014578, also A014571.
Cf. also A001969, A035263, A005187, A115384, A132680, A141803, A104248, A193231.
Run lengths give A026465. Backward first differences give A029883.
Cf. A004128, A053838, A059448, A171900, A161916, A214212, A005942 (subword complexity), A010693 (Abelian complexity), A225186 (squares), A228039 (a(n^2)), A282317.
Sequences mentioned in the Allouche et al. "Taxonomy" paper, listed by example number: 1: A003849, 2: A010060, 3: A010056, 4: A020985 and A020987, 5: A191818, 6: A316340 and A273129, 18: A316341, 19: A030302, 20: A063438, 21: A316342, 22: A316343, 23: A003849 minus its first term, 24: A316344, 25: A316345 and A316824, 26: A020985 and A020987, 27: A316825, 28: A159689, 29: A049320, 30: A003849, 31: A316826, 32: A316827, 33: A316828, 34: A316344, 35: A043529, 36: A316829, 37: A010060.

Programs

  • Haskell
    a010060 n = a010060_list !! n
a010060_list =
   0 : interleave (complement a010060_list) (tail a010060_list)
   where complement = map (1 - )
         interleave (x:xs) ys = x : interleave ys xs
-- Doug McIlroy (doug(AT)cs.dartmouth.edu), Jun 29 2003
-- Edited by Reinhard Zumkeller, Oct 03 2012
  • Maple
    s := proc(k) local i, ans; ans := [ 0,1 ]; for i from 0 to k do ans := [ op(ans),op(map(n->(n+1) mod 2, ans)) ] od; return ans; end; t1 := s(6); A010060 := n->t1[n]; # s(k) gives first 2^(k+2) terms.
a := proc(k) b := [0]: for n from 1 to k do b := subs({0=[0,1], 1=[1,0]},b) od: b; end; # a(k), after the removal of the brackets, gives the first 2^k terms. # Example: a(3); gives [[[[0, 1], [1, 0]], [[1, 0], [0, 1]]]]
A010060:=proc(n)
    add(i,i=convert(n, base, 2)) mod 2 ;
end proc:
seq(A010060(n),n=0..104); # Emeric Deutsch, Mar 19 2005
map(`-`,convert(StringTools[ThueMorse](1000),bytes),48); # Robert Israel, Sep 22 2014
  • Mathematica
    Table[ If[ OddQ[ Count[ IntegerDigits[n, 2], 1]], 1, 0], {n, 0, 100}];
mt = 0; Do[ mt = ToString[mt] <> ToString[(10^(2^n) - 1)/9 - ToExpression[mt] ], {n, 0, 6} ]; Prepend[ RealDigits[ N[ ToExpression[mt], 2^7] ] [ [1] ], 0]
Mod[ Count[ #, 1 ]& /@Table[ IntegerDigits[ i, 2 ], {i, 0, 2^7 - 1} ], 2 ] (* Harlan J. Brothers, Feb 05 2005 *)
Nest[ Flatten[ # /. {0 -> {0, 1}, 1 -> {1, 0}}] &, {0}, 7] (* Robert G. Wilson v Sep 26 2006 *)
a[n_] := If[n == 0, 0, If[Mod[n, 2] == 0, a[n/2], 1 - a[(n - 1)/2]]] (* Ben Branman, Oct 22 2010 *)
a[n_] := Mod[Length[FixedPointList[BitAnd[#, # - 1] &, n]], 2] (* Jan Mangaldan, Jul 23 2015 *)
Table[2/3 (1 - Cos[Pi/3 (n - Sum[(-1)^Binomial[n, k], {k, 1, n}])]), {n, 0, 100}] (* or, for version 10.2 or higher *) Table[ThueMorse[n], {n, 0, 100}] (* Vladimir Reshetnikov, May 06 2016 *)
ThueMorse[Range[0, 100]] (* The program uses the ThueMorse function from Mathematica version 11 *) (* Harvey P. Dale, Aug 11 2016 *)
Nest[Join[#, 1 - #] &, {0}, 7] (* Paolo Xausa, Oct 25 2024 *)
  • PARI
    a(n)=if(n<1,0,sum(k=0,length(binary(n))-1,bittest(n,k))%2)
  • PARI
    a(n)=if(n<1,0,subst(Pol(binary(n)), x,1)%2)
  • PARI
    default(realprecision, 6100); x=0.0; m=20080; for (n=1, m-1, x=x+x; x=x+sum(k=0, length(binary(n))-1, bittest(n, k))%2); x=2*x/2^m; for (n=0, 20000, d=floor(x); x=(x-d)*2; write("b010060.txt", n, " ", d)); \\ Harry J. Smith, Apr 28 2009
  • PARI
    a(n)=hammingweight(n)%2 \\ Charles R Greathouse IV, Mar 22 2013
  • Python
    A010060_list = [0]
for _ in range(14):
    A010060_list += [1-d for d in A010060_list] # Chai Wah Wu, Mar 04 2016
  • Python
    def A010060(n): return n.bit_count()&1 # Chai Wah Wu, Mar 01 2023
  • R
    maxrow <- 8 # by choice
b01 <- 1
for(m in 0:maxrow) for(k in 0:(2^m-1)){
b01[2^(m+1)+    k] <-   b01[2^m+k]
b01[2^(m+1)+2^m+k] <- 1-b01[2^m+k]
}
(b01 <- c(0,b01))
# Yosu Yurramendi, Apr 10 2017

Formula

a(2n) = a(n), a(2n+1) = 1 - a(n), a(0) = 0. Also, a(k+2^m) = 1 - a(k) if 0 <= k < 2^m.
If n = Sum b_i*2^i is the binary expansion of n then a(n) = Sum b_i (mod 2).
Let S(0) = 0 and for k >= 1, construct S(k) from S(k-1) by mapping 0 -> 01 and 1 -> 10; sequence is S(infinity).
G.f.: (1/(1 - x) - Product_{k >= 0} (1 - x^(2^k)))/2. - Benoit Cloitre, Apr 23 2003
a(0) = 0, a(n) = (n + a(floor(n/2))) mod 2; also a(0) = 0, a(n) = (n - a(floor(n/2))) mod 2. - Benoit Cloitre, Dec 10 2003
a(n) = -1 + (Sum_{k=0..n} binomial(n,k) mod 2) mod 3 = -1 + A001316(n) mod 3. - Benoit Cloitre, May 09 2004
Let b(1) = 1 and b(n) = b(ceiling(n/2)) - b(floor(n/2)) then a(n-1) = (1/2)*(1 - b(2n-1)). - Benoit Cloitre, Apr 26 2005
a(n) = 1 - A010059(n) = A001285(n) - 1. - Ralf Stephan, Jun 20 2003
a(n) = A001969(n) - 2n. - Franklin T. Adams-Watters, Aug 28 2006
a(n) = A115384(n) - A115384(n-1) for n > 0. - Reinhard Zumkeller, Aug 26 2007
For n >= 0, a(A004760(n+1)) = 1 - a(n). - Vladimir Shevelev, Apr 25 2009
a(A160217(n)) = 1 - a(n). - Vladimir Shevelev, May 05 2009
a(n) == A000069(n) (mod 2). - Robert G. Wilson v, Jan 18 2012
a(n) = A000035(A000120(n)). - Omar E. Pol, Oct 26 2013
a(n) = A000035(A193231(n)). - Antti Karttunen, Dec 27 2013
a(n) + A181155(n-1) = 2n for n >= 1. - Clark Kimberling, Oct 06 2014
G.f. A(x) satisfies: A(x) = x / (1 - x^2) + (1 - x) * A(x^2). - Ilya Gutkovskiy, Jul 29 2021
From Bernard Schott, Jan 21 2022: (Start)
a(n) = a(n*2^k) for k >= 0.
a((2^m-1)^2) = (1-(-1)^m)/2 (see Hassan Tarfaoui link, Concours Général 1990). (End)

A003242 Number of compositions of n such that no two adjacent parts are equal (these are sometimes called Carlitz compositions).

Original entry on oeis.org

1, 1, 1, 3, 4, 7, 14, 23, 39, 71, 124, 214, 378, 661, 1152, 2024, 3542, 6189, 10843, 18978, 33202, 58130, 101742, 178045, 311648, 545470, 954658, 1670919, 2924536, 5118559, 8958772, 15680073, 27443763, 48033284, 84069952, 147142465, 257534928, 450748483, 788918212
Offset: 0

Views

Author

E. Rodney Canfield

Keywords

Examples

			From _Joerg Arndt_, Oct 27 2012:  (Start)
The 23 such compositions of n=7 are
[ 1]  1 2 1 2 1
[ 2]  1 2 1 3
[ 3]  1 2 3 1
[ 4]  1 2 4
[ 5]  1 3 1 2
[ 6]  1 3 2 1
[ 7]  1 4 2
[ 8]  1 5 1
[ 9]  1 6
[10]  2 1 3 1
[11]  2 1 4
[12]  2 3 2
[13]  2 4 1
[14]  2 5
[15]  3 1 2 1
[16]  3 1 3
[17]  3 4
[18]  4 1 2
[19]  4 2 1
[20]  4 3
[21]  5 2
[22]  6 1
[23]  7
(End)

References

  • Miklos Bona, editor, Handbook of Enumerative Combinatorics, CRC Press, 2015, page 191.

Links

Crossrefs

Cf. A106351, A114900, A114902.
Cf. A096568, A011782, A106356. - Franklin T. Adams-Watters, May 27 2010
Row sums of A232396, A241701.
Cf. A241902.
Column k=1 of A261960.
Cf. A048272.
Compositions with adjacent parts coprime are A167606.
The complement is counted by A261983.
Cf. A000740, A005251, A032020, A114901, A178470, A261041, A274174, A329738, A329863.

Programs

Formula

a(n) = Sum_{k=1..n} A048272(k)*a(n-k), n>1, a(0)=1. - Vladeta Jovovic, Feb 05 2002
G.f.: 1/(1 - Sum_{k>0} x^k/(1+x^k)).
a(n) ~ c r^n where c is approximately 0.456387 and r is approximately 1.750243. (Formula from Knopfmacher and Prodinger reference.) - Franklin T. Adams-Watters, May 27 2010. With better precision: r = 1.7502412917183090312497386246398158787782058181381590561316586... (see A241902), c = 0.4563634740588133495321001859298593318027266156100046548066205... - Vaclav Kotesovec, Apr 30 2014
G.f. is the special case p=2 of 1/(1 - Sum_{k>0} (z^k/(1-z^k) - p*z^(k*p)/(1-z^(k*p)))), see A129922. - Joerg Arndt, Apr 28 2013
G.f.: 1/(1 - x * (d/dx) log(Product_{k>=1} (1 + x^k)^(1/k))). - Ilya Gutkovskiy, Oct 18 2018
Moebius transform of A329738. - Gus Wiseman, Nov 27 2019
For n>=2, a(n) = A128695(n) - A091616(n). - Vaclav Kotesovec, Jul 07 2020

Extensions

More terms from David W. Wilson

A001511 The ruler function: exponent of the highest power of 2 dividing 2n. Equivalently, the 2-adic valuation of 2n.

Original entry on oeis.org

1, 2, 1, 3, 1, 2, 1, 4, 1, 2, 1, 3, 1, 2, 1, 5, 1, 2, 1, 3, 1, 2, 1, 4, 1, 2, 1, 3, 1, 2, 1, 6, 1, 2, 1, 3, 1, 2, 1, 4, 1, 2, 1, 3, 1, 2, 1, 5, 1, 2, 1, 3, 1, 2, 1, 4, 1, 2, 1, 3, 1, 2, 1, 7, 1, 2, 1, 3, 1, 2, 1, 4, 1, 2, 1, 3, 1, 2, 1, 5, 1, 2, 1, 3, 1, 2, 1, 4, 1, 2, 1, 3, 1, 2, 1, 6, 1, 2, 1, 3, 1, 2, 1, 4, 1
Offset: 1

Views

Author

N. J. A. Sloane

Keywords

Comments

Number of 2's dividing 2*n.
a(n) is equivalently the exponent of the smallest power of 2 which does not divide n. - David James Sycamore, Oct 02 2023
a(n) - 1 is the number of trailing zeros in the binary expansion of n.
If you are counting in binary and the least significant bit is numbered 1, the next bit is 2, etc., a(n) is the bit that is incremented when increasing from n-1 to n. - Jud McCranie, Apr 26 2004
Number of steps to reach an integer starting with (n+1)/2 and using the map x -> x*ceiling(x) (cf. A073524).
a(n) is the number of the disk to be moved at the n-th step of the optimal solution to Towers of Hanoi problem (comment from Andreas M. Hinz).
Shows which bit to flip when creating the binary reflected Gray code (bits are numbered from the right, offset is 1). This is essentially equivalent to Hinz's comment. - Adam Kertesz, Jul 28 2001
a(n) is the Hamming distance between n and n-1 (in binary). This is equivalent to Kertesz's comments above. - Tak-Shing Chan (chan12(AT)alumni.usc.edu), Feb 25 2003
Let S(0) = {1}, S(n) = {S(n-1), S(n-1)-{x}, x+1} where x = last term of S(n-1); sequence gives S(infinity). - Benoit Cloitre, Jun 14 2003
The sum of all terms up to and including the first occurrence of m is 2^m-1. - Donald Sampson (marsquo(AT)hotmail.com), Dec 01 2003
m appears every 2^m terms starting with the 2^(m-1)th term. - Donald Sampson (marsquo(AT)hotmail.com), Dec 08 2003
Sequence read mod 4 gives A092412. - Philippe Deléham, Mar 28 2004
If q = 2n/2^A001511(n) and if b(m) is defined by b(0)=q-1 and b(m)=2*b(m-1)+1, then 2n = b(A001511(n)) + 1. - Gerald McGarvey, Dec 18 2004
Repeating pattern ABACABADABACABAE ... - Jeremy Gardiner, Jan 16 2005
Relation to C(n) = Collatz function iteration using only odd steps: a(n) is the number of right bits set in binary representation of A004767(n) (numbers of the form 4*m+3). So for m=A004767(n) it follows that there are exactly a(n) recursive steps where m
Between every two instances of any positive integer m there are exactly m distinct values (1 through m-1 and one value greater than m). - Franklin T. Adams-Watters, Sep 18 2006
Number of divisors of n of the form 2^k. - Giovanni Teofilatto, Jul 25 2007
Every prefix up to (but not including) the first occurrence of some k >= 2 is a palindrome. - Gary W. Adamson, Sep 24 2008
1 interleaved with (2 interleaved with (3 interleaved with ( ... ))). - Eric D. Burgess (ericdb(AT)gmail.com), Oct 17 2009
A054525 (Möbius transform) * A001511 = A036987 = A047999^(-1) * A001511. - Gary W. Adamson, Oct 26 2009
Equals A051731 * A036987, (inverse Möbius transform of the Fredholm-Rueppel sequence) = A047999 * A036987. - Gary W. Adamson, Oct 26 2009
Cf. A173238, showing links between generalized ruler functions and A000041. - Gary W. Adamson, Feb 14 2010
Given A000041, P(x) = A(x)/A(x^2) with P(x) = (1 + x + 2x^2 + 3x^3 + 5x^4 + 7x^5 + ...), A(x) = (1 + x + 3x^2 + 4x^3 + 10x^4 + 13x^5 + ...), A(x^2) = (1 + x^2 + 3x^4 + 4x^6 + 10x^8 + ...), where A092119 = (1, 1, 3, 4, 10, ...) = Euler transform of the ruler sequence, A001511. - Gary W. Adamson, Feb 11 2010
Subtracting 1 from every term and deleting any 0's yields the same sequence, A001511. - Ben Branman, Dec 28 2011
In the listing of the compositions of n as lists in lexicographic order, a(k) is the last part of composition(k) for all k <= 2^(n-1) and all n, see example. - Joerg Arndt, Nov 12 2012
According to Hinz, et al. (see links), this sequence was studied by Louis Gros in his 1872 pamphlet "Théorie du Baguenodier" and has therefore been called the Gros sequence.
First n terms comprise least squarefree word of length n using positive integers, where "squarefree" means that the word contains no consecutive identical subwords; e.g., 1 contains no square; 11 contains a square but 12 does not; 121 contains no square; both 1211 and 1212 have squares but 1213 does not; etc. - Clark Kimberling, Sep 05 2013
Length of 0-run starting from 2 (10, 100, 110, 1000, 1010, ...), or length of 1-run starting from 1 (1, 11, 101, 111, 1001, 1011, ...) of every second number, from right to left in binary representation. - Armands Strazds, Apr 13 2017
a(n) is also the frequency of the largest part in the integer partition having viabin number n. The viabin number of an integer partition is defined in the following way. Consider the southeast border of the Ferrers board of the integer partition and consider the binary number obtained by replacing each east step with 1 and each north step, except the last one, with 0. The corresponding decimal form is, by definition, the viabin number of the given integer partition. "Viabin" is coined from "via binary". For example, consider the integer partition [2,2,2,1]. The southeast border of its Ferrers board yields 10100, leading to the viabin number 20. - Emeric Deutsch, Jul 24 2017
As A000005(n) equals the number of even divisors of 2n and A001227(n) = A001227(2n), the formula A001511(n) = A000005(n)/A001227(n) might be read as "The number of even divisors of 2n is always divisible by the number of odd divisors of 2n" (where number of divisors means sum of zeroth powers of divisors). Conjecture: For any nonnegative integer k, the sum of the k-th powers of even divisors of n is always divisible by the sum of the k-th powers of odd divisors of n. - Ivan N. Ianakiev, Jul 06 2019
From Benoit Cloitre, Jul 14 2022: (Start)
To construct the sequence, start from 1's separated by a place 1,,1,,1,,1,,1,,1,,1,,1,,1,,1,,1,,1,,1,,1,...
Then put the 2's in every other remaining place
1,2,1,,1,2,1,,1,2,1,,1,2,1,,1,2,1,,1,2,1,,1,2,1,...
Then the 3's in every other remaining place
1,2,1,3,1,2,1,,1,2,1,3,1,2,1,,1,2,1,3,1,2,1,,1,2,1,...
Then the 4's in every other remaining place
1,2,1,3,1,2,1,4,1,2,1,3,1,2,1,,1,2,1,3,1,2,1,4,1,2,1,...
By iterating this process, we get the ruler function 1,2,1,3,1,2,1,4,1,2,1,3,1,2,1,5,1,2,1,3,1,2,1,4,1,2,1,... (End)
a(n) is the least positive integer k for which there does not exist i+j=n and a(i)=a(j)=k (cf. A322523). - Rémy Sigrist and Jianing Song, Aug 23 2022
a(n) is the smallest positive integer that does not occur in the coincidences of the sequence so far a(1..n-1) and its reverse. - Neal Gersh Tolunsky, Jan 18 2023
The geometric mean of this sequence approaches the Somos constant (A112302). - Jwalin Bhatt, Jan 31 2025

Examples

			For example, 2^1|2, 2^2|4, 2^1|6, 2^3|8, 2^1|10, 2^2|12, ... giving the initial terms 1, 2, 1, 3, 1, 2, ...
From _Omar E. Pol_, Jun 12 2009: (Start)
Triangle begins:
1;
2,1;
3,1,2,1;
4,1,2,1,3,1,2,1;
5,1,2,1,3,1,2,1,4,1,2,1,3,1,2,1;
6,1,2,1,3,1,2,1,4,1,2,1,3,1,2,1,5,1,2,1,3,1,2,1,4,1,2,1,3,1,2,1;
7,1,2,1,3,1,2,1,4,1,2,1,3,1,2,1,5,1,2,1,3,1,2,1,4,1,2,1,3,1,2,1,6,1,2,1,3,...
(End)
S(0) = {} S(1) = 1 S(2) = 1, 2, 1 S(3) = 1, 2, 1, 3, 1, 2, 1 S(4) = 1, 2, 1, 3, 1, 2, 1, 4, 1, 2, 1, 3, 1, 2, 1. - Yann David (yann_david(AT)hotmail.com), Mar 21 2010
From _Joerg Arndt_, Nov 12 2012: (Start)
The 16 compositions of 5 as lists in lexicographic order:
[ n]  a(n)  composition
[ 1]  [ 1]  [ 1 1 1 1 1 ]
[ 2]  [ 2]  [ 1 1 1 2 ]
[ 3]  [ 1]  [ 1 1 2 1 ]
[ 4]  [ 3]  [ 1 1 3 ]
[ 5]  [ 1]  [ 1 2 1 1 ]
[ 6]  [ 2]  [ 1 2 2 ]
[ 7]  [ 1]  [ 1 3 1 ]
[ 8]  [ 4]  [ 1 4 ]
[ 9]  [ 1]  [ 2 1 1 1 ]
[10]  [ 2]  [ 2 1 2 ]
[11]  [ 1]  [ 2 2 1 ]
[12]  [ 3]  [ 2 3 ]
[13]  [ 1]  [ 3 1 1 ]
[14]  [ 2]  [ 3 2 ]
[15]  [ 1]  [ 4 1 ]
[16]  [ 5]  [ 5 ]
a(n) is the last part in each list.
(End)
From _Omar E. Pol_, Aug 20 2013: (Start)
Also written as a triangle in which the right border gives A000027 and row lengths give A011782 and row sums give A000079 the sequence begins:
1;
2;
1,3;
1,2,1,4;
1,2,1,3,1,2,1,5;
1,2,1,3,1,2,1,4,1,2,1,3,1,2,1,6;
1,2,1,3,1,2,1,4,1,2,1,3,1,2,1,5,1,2,1,3,1,2,1,4,1,2,1,3,1,2,1,7;
(End)
G.f. = x + 2*x^2 + x^3 + 3*x^4 + x^5 + 2*x^6 + x^7 + 4*x^8 + x^9 + 2*x^10 + ...

References

  • J.-P. Allouche and J. Shallit, Automatic Sequences, Cambridge Univ. Press, 2003.
  • E. R. Berlekamp, J. H. Conway and R. K. Guy, Winning Ways, Academic Press, NY, 2 vols., 2nd ed., 2001-2003; see Dim- and Dim+ on p. 98; Dividing Rulers, on pp. 436-437; The Ruler Game, pp. 469-470; Ruler Fours, Fives, ... Fifteens on p. 470.
  • L. Gros, Théorie du Baguenodier, Aimé Vingtrinier, Lyon, 1872.
  • R. K. Guy, Unsolved Problems in Number Theory, Springer, 1st edition, 1981. See section E22.
  • A. M. Hinz, The Tower of Hanoi, in Algebras and combinatorics (Hong Kong, 1997), 277-289, Springer, Singapore, 1999.
  • D. E. Knuth, The Art of Computer Programming, vol. 4A, Combinatorial Algorithms, Section 7.1.3, Problem 41, p. 589.
  • Andrew Schloss, "Towers of Hanoi" composition, in The Digital Domain. Elektra/Asylum Records 9 60303-2, 1983. Works by Jaffe (Finale to "Silicon Valley Breakdown"), McNabb ("Love in the Asylum"), Schloss ("Towers of Hanoi"), Mattox ("Shaman"), Rush, Moorer ("Lions are Growing") and others.
  • N. J. A. Sloane, A Handbook of Integer Sequences, Academic Press, 1973 (includes this sequence).
  • N. J. A. Sloane and Simon Plouffe, The Encyclopedia of Integer Sequences, Academic Press, 1995 (includes this sequence).

Links

Crossrefs

Column 1 of table A050600.
Sequence read mod 2 gives A035263.
Sequence is bisection of A007814, A050603, A050604, A067029, A089309.
This is Guy Steele's sequence GS(4, 2) (see A135416).
Cf. A005187 (partial sums), A085058 (bisection), A112302 (geometric mean), A171977 (2^a(n)).
Cf. A000005, A000041, A000079, A003188, A003278, A003602, A007949, A018238, A047999, A051731, A054525, A054852, A065176, A082850, A089080, A092119, A117303, A129360, A130093, A173238, A181988, A220466.
Cf. A287896, A002487, A209229 (Mobius trans.), A092673 (Dirichlet inv.).
Cf. generalized ruler functions for k=3,4,5: A051064, A115362, A055457.

Programs

  • Haskell
    a001511 n = length $ takeWhile ((== 0) . (mod n)) a000079_list
-- Reinhard Zumkeller, Sep 27 2011
  • Haskell
    a001511 n | odd n = 1 | otherwise = 1 + a001511 (n `div` 2)
-- Walt Rorie-Baety, Mar 22 2013
  • MATLAB
    nmax=5;r=1;for n=2:nmax;r=[r n r];end % Adriano Caroli, Feb 26 2016
  • Magma
    [Valuation(2*n,2): n in [1..105]]; // Bruno Berselli, Nov 23 2015
  • Maple
    A001511 := n->2-wt(n)+wt(n-1); # where wt is defined in A000120
# This is the binary logarithm of the denominator of (256^n-1)B_{8n}/n, in Maple parlance a := n -> log[2](denom((256^n-1)*bernoulli(8*n)/n)). - Peter Luschny, May 31 2009
A001511 := n -> padic[ordp](2*n,2): seq(A001511(n), n=1..105);  # Peter Luschny, Nov 26 2010
a:= n-> ilog2((Bits[Xor](2*n, 2*n-1)+1)/2): seq(a(n), n=1..50);  # Gary Detlefs, Dec 13 2018
  • Mathematica
    Array[ If[ Mod[ #, 2] == 0, FactorInteger[ # ][[1, 2]], 0] &, 105] + 1 (* or *)
Nest[ Flatten[ # /. a_Integer -> {1, a + 1}] &, {1}, 7] (* Robert G. Wilson v, Mar 04 2005 *)
IntegerExponent[2*n, 2] (* Alexander R. Povolotsky, Aug 19 2011 *)
myHammingDistance[n_, m_] := Module[{g = Max[m, n], h = Min[m, n]}, b1 = IntegerDigits[g, 2]; b2 = IntegerDigits[h, 2, Length[b1]]; HammingDistance[b1, b2]] (* Vladimir Shevelev A206853 *) Table[ myHammingDistance[n, n - 1], {n, 111}] (* Robert G. Wilson v, Apr 05 2012 *)
Table[Position[Reverse[IntegerDigits[n,2]],1,1,1],{n,110}]//Flatten (* Harvey P. Dale, Aug 18 2017 *)
  • PARI
    a(n) = sum(k=0,floor(log(n)/log(2)),floor(n/2^k)-floor((n-1)/2^k)) /* Ralf Stephan */
  • PARI
    a(n)=if(n%2,1,factor(n)[1,2]+1) /* Jon Perry, Jun 06 2004 */
  • PARI
    {a(n) = if( n, valuation(n, 2) + 1, 0)}; /* Michael Somos, Sep 30 2006 */
  • PARI
    {a(n)=if(n==1,1,polcoeff(x-sum(k=1, n-1, a(k)*x^k*(1-x^k)*(1-x+x*O(x^n))), n))} /* Paul D. Hanna, Jun 22 2007 */
  • Python
    def a(n): return bin(n)[2:][::-1].index("1") + 1 # Indranil Ghosh, May 11 2017
  • Python
    A001511 = lambda n: (n&-n).bit_length() # M. F. Hasler, Apr 09 2020
  • Python
    def A001511(n): return (~n & n-1).bit_length()+1 # Chai Wah Wu, Jul 01 2022
  • Sage
    [valuation(2*n,2) for n in (1..105)]  # Bruno Berselli, Nov 23 2015
  • Scheme
    (define (A001511 n) (let loop ((n n) (e 1)) (if (odd? n) e (loop (/ n 2) (+ 1 e))))) ;; Antti Karttunen, Oct 06 2017

Formula

a(n) = A007814(n) + 1 = A007814(2*n).
a(2*n+1) = 1; a(2*n) = 1 + a(n). - Philippe Deléham, Dec 08 2003
a(n) = 2 - A000120(n) + A000120(n-1), n >= 1. - Daniele Parisse
a(n) = 1 + log_2(abs(A003188(n) - A003188(n-1))).
Multiplicative with a(p^e) = e+1 if p = 2; 1 if p > 2. - David W. Wilson, Aug 01 2001
For any real x > 1/2: lim_{N->infinity} (1/N)*Sum_{n=1..N} x^(-a(n)) = 1/(2*x-1); also lim_{N->infinity} (1/N)*Sum_{n=1..N} 1/a(n) = log(2). - Benoit Cloitre, Nov 16 2001
s(n) = Sum_{k=1..n} a(k) is asymptotic to 2*n since s(n) = 2*n - A000120(n). - Benoit Cloitre, Aug 31 2002
For any n >= 0, for any m >= 1, a(2^m*n + 2^(m-1)) = m. - Benoit Cloitre, Nov 24 2002
a(n) = Sum_{d divides n and d is odd} mu(d)*tau(n/d). - Vladeta Jovovic, Dec 04 2002
G.f.: A(x) = Sum_{k>=0} x^(2^k)/(1-x^(2^k)). - Ralf Stephan, Dec 24 2002
a(1) = 1; for n > 1, a(n) = a(n-1) + (-1)^n*a(floor(n/2)). - Vladeta Jovovic, Apr 25 2003
A fixed point of the mapping 1->12; 2->13; 3->14; 4->15; 5->16; ... . - Philippe Deléham, Dec 13 2003
Product_{k>0} (1+x^k)^a(k) is g.f. for A000041(). - Vladeta Jovovic, Mar 26 2004
G.f. A(x) satisfies A(x) = A(x^2) + x/(1-x). - Franklin T. Adams-Watters, Feb 09 2006
a(A118413(n,k)) = A002260(n,k); = a(A118416(n,k)) = A002024(n,k); a(A014480(n)) = A003602(A014480(n)). - Reinhard Zumkeller, Apr 27 2006
Ordinal transform of A003602. - Franklin T. Adams-Watters, Aug 28 2006 (The ordinal transform of a sequence b_0, b_1, b_2, ... is the sequence a_0, a_1, a_2, ... where a_n is the number of times b_n has occurred in {b_0 ... b_n}.)
Could be extended to n <= 0 using a(-n) = a(n), a(0) = 0, a(2*n) = a(n)+1 unless n=0. - Michael Somos, Sep 30 2006
A094267(2*n) = A050603(2*n) = A050603(2*n + 1) = a(n). - Michael Somos, Sep 30 2006
Sequence = A129360 * A000005 = M*V, where M = an infinite lower triangular matrix and V = d(n) as a vector: [1, 2, 2, 3, 2, 4, ...]. - Gary W. Adamson, Apr 15 2007
Row sums of triangle A130093. - Gary W. Adamson, May 13 2007
Dirichlet g.f.: zeta(s)*2^s/(2^s-1). - Ralf Stephan, Jun 17 2007
a(n) = -Sum_{d divides n} mu(2*d)*tau(n/d). - Benoit Cloitre, Jun 21 2007
G.f.: x/(1-x) = Sum_{n>=1} a(n)*x^n*( 1 - x^n ). - Paul D. Hanna, Jun 22 2007
2*n = 2^a(n)* A000265(n). - Eric Desbiaux, May 14 2009 [corrected by Alejandro Erickson, Apr 17 2012]
Multiplicative with a(2^k) = k + 1, a(p^k) = 1 for any odd prime p. - Franklin T. Adams-Watters, Jun 09 2009
With S(n): 2^n - 1 first elements of the sequence then S(0) = {} (empty list) and if n > 0, S(n) = S(n-1), n, S(n-1). - Yann David (yann_david(AT)hotmail.com), Mar 21 2010
a(n) = log_2(A046161(n)/A046161(n-1)). - Johannes W. Meijer, Nov 04 2012
a((2*n-1)*2^p) = p+1, p >= 0 and n >= 1. - Johannes W. Meijer, Feb 05 2013
a(n+1) = 1 + Sum_{j=0..ceiling(log_2(n+1))} (j * (1 - abs(sign((n mod 2^(j + 1)) - 2^j + 1)))). - Enrico Borba, Oct 01 2015
Conjecture: a(n) = A181988(n)/A003602(n). - L. Edson Jeffery, Nov 21 2015
a(n) = log_2(A006519(n)) + 1. - Doug Bell, Jun 02 2017
Inverse Moebius transform of A209229. - Andrew Howroyd, Aug 04 2018
a(n) = 1 + (A183063(n)/A001227(n)). - Omar E. Pol, Nov 06 2018 (after Franklin T. Adams-Watters)
a(n) = log_2((Xor(2*n,2*n-1)+1)/2). - Gary Detlefs, Dec 13 2018
(2^(a(n)-1)-1)*(n mod 4) = 2*floor(((n+1) mod 4)/3). - Gary Detlefs, Dec 14 2018
a(n) = A000005(n)/A001227(n). - Ivan N. Ianakiev, Jul 05 2019
a(n) = Sum_{j=1..r} (j/2^j)*(Product_{k=1..j} (1 - (-1)^floor( (n+2^(j-1))/2^(k-1) ))), for n < a predefined 2^r. - Adriano Caroli, Sep 30 2019

Extensions

Name edited following suggestion by David James Sycamore, Oct 05 2023

A001787 a(n) = n*2^(n-1).

Original entry on oeis.org

0, 1, 4, 12, 32, 80, 192, 448, 1024, 2304, 5120, 11264, 24576, 53248, 114688, 245760, 524288, 1114112, 2359296, 4980736, 10485760, 22020096, 46137344, 96468992, 201326592, 419430400, 872415232, 1811939328, 3758096384, 7784628224, 16106127360, 33285996544
Offset: 0

Views

Author

N. J. A. Sloane

Keywords

Comments

Number of edges in an n-dimensional hypercube.
Number of 132-avoiding permutations of [n+2] containing exactly one 123 pattern. - Emeric Deutsch, Jul 13 2001
Number of ways to place n-1 nonattacking kings on a 2 X 2(n-1) chessboard for n >= 2. - Antonio G. Astudillo (afg_astudillo(AT)hotmail.com), May 22 2001
Arithmetic derivative of 2^n: a(n) = A003415(A000079(n)). - Reinhard Zumkeller, Feb 26 2002
(-1) times the determinant of matrix A_{i,j} = -|i-j|, 0 <= i,j <= n.
a(n) is the number of ones in binary numbers 1 to 111...1 (n bits). a(n) = A000337(n) - A000337(n-1) for n = 2,3,... . - Emeric Deutsch, May 24 2003
The number of 2 X n 0-1 matrices containing n+1 1's and having no zero row or column. The number of spanning trees of the complete bipartite graph K(2,n). This is the case m = 2 of K(m,n). See A072590. - W. Edwin Clark, May 27 2003
Binomial transform of 0,1,2,3,4,5,... (A001477). Without the initial 0, binomial transform of odd numbers.
With an additional leading zero, [0,0,1,4,...] this is the binomial transform of the integers repeated A004526. Its formula is then (2^n*(n-1) + 0^n)/4. - Paul Barry, May 20 2003
Number of zeros in all different (n+1)-bit integers. - Ralf Stephan, Aug 02 2003
From Lekraj Beedassy, Jun 03 2004: (Start)
Final element of a summation table (as opposed to a difference table) whose first row consists of integers 0 through n (or first n+1 nonnegative integers A001477); illustrating the case n=5:
0 1 2 3 4 5
1 3 5 7 9
4 8 12 16
12 20 28
32 48
80
and the final element is a(5)=80. (End)
This sequence and A001871 arise in counting ordered trees of height at most k where only the rightmost branch at the root actually achieves this height and the count is by the number of edges, with k = 3 for this sequence and k = 4 for A001871.
Let R be a binary relation on the power set P(A) of a set A having n = |A| elements such that for all elements x,y of P(A), xRy if x is a proper subset of y and there are no z in P(A) such that x is a proper subset of z and z is a proper subset of y. Then a(n) = |R|. - Ross La Haye, Sep 21 2004
Number of 2 X n binary matrices avoiding simultaneously the right-angled numbered polyomino patterns (ranpp) (00;1) and (10;1). An occurrence of a ranpp (xy;z) in a matrix A=(a(i,j)) is a triple (a(i1,j1), a(i1,j2), a(i2,j1)) where i1 < i2, j1 < j2 and these elements are in same relative order as those in the triple (x,y,z). - Sergey Kitaev, Nov 11 2004
Number of subsequences 00 in all binary words of length n+1. Example: a(2)=4 because in 000,001,010,011,100,101,110,111 the sequence 00 occurs 4 times. - Emeric Deutsch, Apr 04 2005
If you expand the n-factor expression (a+1)*(b+1)*(c+1)*...*(z+1), there are a(n) variables in the result. For example, the 3-factor expression (a+1)*(b+1)*(c+1) expands to abc+ab+ac+bc+a+b+c+1 with a(3) = 12 variables. - David W. Wilson, May 08 2005
An inverse Chebyshev transform of n^2, where g(x)->(1/sqrt(1-4*x^2))*g(x*c(x^2)), c(x) the g.f. of A000108. - Paul Barry, May 13 2005
Sequences A018215 and A058962 interleaved. - Graeme McRae, Jul 12 2006
The number of never-decreasing positive integer sequences of length n with a maximum value of 2*n. - Ben Paul Thurston, Nov 13 2006
Total size of all the subsets of an n-element set. For example, a 2-element set has 1 subset of size 0, 2 subsets of size 1 and 1 of size 2. - Ross La Haye, Dec 30 2006
Convolution of the natural numbers [A000027] and A045623 beginning [0,1,2,5,...]. - Ross La Haye, Feb 03 2007
If M is the matrix (given by rows) [2,1;0,2] then the sequence gives the (1,2) entry in M^n. - Antonio M. Oller-Marcén, May 21 2007
If X_1,X_2,...,X_n is a partition of a 2n-set X into 2-blocks then, for n > 0, a(n) is equal to the number of (n+1)-subsets of X intersecting each X_i (i=1,2,...,n). - Milan Janjic, Jul 21 2007
Number of n-permutations of 3 objects u,v,w, with repetition allowed, containing exactly one u. Example: a(2)=4 because we have uv, vu, uw and wu. - Zerinvary Lajos, Dec 27 2007
A member of the family of sequences defined by a(n) = n*[c(1)*...*c(r)]^(n-1); c(i) integer. This sequence has c(1)=2, A027471 has c(1)=3. - Ctibor O. Zizka, Feb 23 2008
a(n) is the number of ways to split {1,2,...,n-1} into two (possibly empty) complementary intervals {1,2,...,i} and {i+1,i+2,...,n-1} and then select a subset from each interval. - Geoffrey Critzer, Jan 31 2009
Equals the Jacobsthal sequence A001045 convolved with A003945: (1, 3, 6, 12, ...). - Gary W. Adamson, May 23 2009
Starting with offset 1 = A059570: (1, 2, 6, 14, 34, ...) convolved with (1, 2, 2, 2, ...). - Gary W. Adamson, May 23 2009
Equals the first left hand column of A167591. - Johannes W. Meijer, Nov 12 2009
The number of tatami tilings of an n X n square with n monomers is n*2^(n-1). - Frank Ruskey, Sep 25 2010
Under T. D. Noe's variant of the hypersigma function, this sequence gives hypersigma(2^n): a(n) = A191161(A000079(n)). - Alonso del Arte, Nov 04 2011
Number of Dyck (n+2)-paths with exactly one valley at height 1 and no higher valley. - David Scambler, Nov 07 2011
Equals triangle A059260 * A016777 as a vector, where A016777 = (3n + 1): [1, 4, 7, 10, 13, ...]. - Gary W. Adamson, Mar 06 2012
Main transitions in systems of n particles with spin 1/2 (see A212697 with b=2). - Stanislav Sykora, May 25 2012
Let T(n,k) be the triangle with (first column) T(n,1) = 2*n-1 for n >= 1, otherwise T(n,k) = T(n,k-1) + T(n-1,k-1), then a(n) = T(n,n). - J. M. Bergot, Jan 17 2013
Sum of all parts of all compositions (ordered partitions) of n. The equivalent sequence for partitions is A066186. - Omar E. Pol, Aug 28 2013
Starting with a(1)=1: powers of 2 (A000079) self-convolved. - Bob Selcoe, Aug 05 2015
Coefficients of the series expansion of the normalized Schwarzian derivative -S{p(x)}/6 of the polynomial p(x) = -(x-x1)*(x-x2) with x1 + x2 = 1 (cf. A263646). - Tom Copeland, Nov 02 2015
a(n) is the number of North-East lattice paths from (0,0) to (n+1,n+1) that have exactly one east step below y = x-1 and no east steps above y = x+1. Details can be found in Pan and Remmel's link. - Ran Pan, Feb 03 2016
Also the number of maximal and maximum cliques in the n-hypercube graph for n > 0. - Eric W. Weisstein, Dec 01 2017
Let [n]={1,2,...,n}; then a(n-1) is the total number of elements missing in proper subsets of [n] that contain n to form [n]. For example, for n = 3, a(2) = 4 since the proper subsets of [3] that contain 3 are {3}, {1,3}, {2,3} and the total number of elements missing in these subsets to form [3] is 4: 2 in the first subset, 1 in the second, and 1 in the third. - Enrique Navarrete, Aug 08 2020
Number of 3-permutations of n elements avoiding the patterns 132, 231. See Bonichon and Sun. - Michel Marcus, Aug 19 2022

Examples

			a(2)=4 since 2314, 2341,3124 and 4123 are the only 132-avoiding permutations of 1234 containing exactly one increasing subsequence of length 3.
x + 4*x^2 + 12*x^3 + 32*x^4 + 80*x^5 + 192*x^6 + 448*x^7 + ...
a(5) = 1*0 + 5*1 + 10*2 + 10*3 + 5*4 + 1*5 = 80, with 1,5,10,10,5,1 the 5th row of Pascal's triangle. - _J. M. Bergot_, Apr 29 2014

References

  • M. Abramowitz and I. A. Stegun, eds., Handbook of Mathematical Functions, National Bureau of Standards Applied Math. Series 55, 1964 (and various reprintings), p. 796.
  • A. T. Benjamin and J. J. Quinn, Proofs that really count: the art of combinatorial proof, M.A.A. 2003, id. 131.
  • Clifford A. Pickover, The Math Book, From Pythagoras to the 57th Dimension, 250 Milestones in the History of Mathematics, Sterling Publ., NY, 2009, page 282.
  • N. J. A. Sloane, A Handbook of Integer Sequences, Academic Press, 1973 (includes this sequence).
  • N. J. A. Sloane and Simon Plouffe, The Encyclopedia of Integer Sequences, Academic Press, 1995 (includes this sequence).

Links

Crossrefs

Three other versions, essentially identical, are A085750, A097067, A118442.
Partial sums of A001792.
A058922(n+1) = 4*A001787(n).
Equals A090802(n, 1).
Column k=1 of A038207.
Row sums of A003506, A322427, A322428.
Cf. A053109, A001788, A001789, A000337, A130300, A134083, A002064, A027471, A003945, A059670, A167591, A059260, A016777, A212697, A000079, A263646.

Programs

  • Haskell
    a001787 n = n * 2 ^ (n - 1)
a001787_list = zipWith (*) [0..] $ 0 : a000079_list
-- Reinhard Zumkeller, Jul 11 2014
  • Magma
    [n*2^(n-1): n in [0..40]]; // Vincenzo Librandi, Feb 04 2016
  • Maple
    spec := [S, {B=Set(Z, 0 <= card), S=Prod(Z, B, B)}, labeled]: seq(combstruct[count](spec, size=n), n=0..29); # Zerinvary Lajos, Oct 09 2006
A001787:=1/(2*z-1)^2; # Simon Plouffe in his 1992 dissertation, dropping the initial zero
  • Mathematica
    Table[Sum[Binomial[n, i] i, {i, 0, n}], {n, 0, 30}] (* Geoffrey Critzer, Mar 18 2009 *)
f[n_] := n 2^(n - 1); f[Range[0, 40]] (* Vladimir Joseph Stephan Orlovsky, Feb 09 2011 *)
Array[# 2^(# - 1) &, 40, 0] (* Harvey P. Dale, Jul 26 2011 *)
Join[{0}, Table[n 2^(n - 1), {n, 20}]] (* Eric W. Weisstein, Dec 01 2017 *)
Join[{0}, LinearRecurrence[{4, -4}, {1, 4}, 20]] (* Eric W. Weisstein, Dec 01 2017 *)
CoefficientList[Series[x/(-1 + 2 x)^2, {x, 0, 20}], x] (* Eric W. Weisstein, Dec 01 2017 *)
  • PARI
    {a(n) = if( n<0, 0, n * 2^(n-1))}
  • PARI
    concat(0, Vec(x/(1-2*x)^2 + O(x^50))) \\ Altug Alkan, Nov 03 2015
  • Python
    def A001787(n): return n*(1<Chai Wah Wu, Nov 14 2022

Formula

a(n) = Sum_{k=1..n} k*binomial(n, k). - Benoit Cloitre, Dec 06 2002
E.g.f.: x*exp(2x). - Paul Barry, Apr 10 2003
G.f.: x/(1-2*x)^2.
G.f.: x / (1 - 4*x / (1 + x / (1 - x))). - Michael Somos, Apr 07 2012
A108666(n) = Sum_{k=0..n} binomial(n, k)^2 * a(n). - Michael Somos, Apr 07 2012
PSumSIGN transform of A053220. PSumSIGN transform is A045883. Binomial transform is A027471(n+1). - Michael Somos, Jul 10 2003
Starting at a(1)=1, INVERT transform is A002450, INVERT transform of A049072, MOBIUS transform of A083413, PSUM transform is A000337, BINOMIAL transform is A081038, BINOMIAL transform of A005408. - Michael Somos, Apr 07 2012
a(n) = 2*a(n-1)+2^(n-1).
a(2*n) = n*4^n, a(2*n+1) = (2*n+1)4^n.
G.f.: x/det(I-x*M) where M=[1,i;i,1], i=sqrt(-1). - Paul Barry, Apr 27 2005
Starting 1, 1, 4, 12, ... this is 0^n + n2^(n-1), the binomial transform of the 'pair-reversed' natural numbers A004442. - Paul Barry, Jul 24 2003
Convolution of [1, 2, 4, 8, ...] with itself. - Jon Perry, Aug 07 2003
The signed version of this sequence, n(-2)^(n-1), is the inverse binomial transform of n(-1)^(n-1) (alternating sign natural numbers). - Paul Barry, Aug 20 2003
a(n-1) = (Sum_{k=0..n} 2^(n-k-1)*C(n-k, k)*C(1,(k+1)/2)*(1-(-1)^k)/2) - 0^n/4. - Paul Barry, Oct 15 2004
a(n) = Sum_{k=0..floor(n/2)} binomial(n, k)(n-2k)^2. - Paul Barry, May 13 2005
a(n+2) = A049611(n+2) - A001788(n).
a(n) = n! * Sum_{k=0..n} 1/((k - 1)!(n - k)!). - Paul Barry, Mar 26 2003
a(n+1) = Sum_{k=0..n} 4^k * A109466(n,k). - Philippe Deléham, Nov 13 2006
Row sums of A130300 starting (1, 4, 12, 32, ...). - Gary W. Adamson, May 20 2007
Equals row sums of triangle A134083. Equals A002064(n) + (2^n - 1). - Gary W. Adamson, Oct 07 2007
a(n) = 4*a(n-1) - 4*a(n-2), a(0)=0, a(1)=1. - Philippe Deléham, Nov 16 2008
Sum_{n>0} 1/a(n) = 2*log(2). - Jaume Oliver Lafont, Feb 10 2009
a(n) = A000788(A000225(n)) = A173921(A000225(n)). - Reinhard Zumkeller, Mar 04 2010
a(n) = n * A011782(n). - Omar E. Pol, Aug 28 2013
a(n-1) = Sum_{t_1+2*t_2+...+n*t_n=n} (t_1+t_2+...+t_n-1)*multinomial(t_1+t_2 +...+t_n,t_1,t_2,...,t_n). - Mircea Merca, Dec 06 2013
a(n+1) = Sum_{r=0..n} (2*r+1)*C(n,r). - J. M. Bergot, Apr 07 2014
a(n) = A007283(n)*n/6. - Enxhell Luzhnica, Apr 16 2016
a(n) = (A000225(n) + A000337(n))/2. - Anton Zakharov, Sep 17 2016
Sum_{n>0} (-1)^(n+1)/a(n) = 2*log(3/2) = 2*A016578. - Ilya Gutkovskiy, Sep 17 2016
a(n) = Sum_{k=0..n-1} Sum_{i=0..n-1} (i+1) * C(k,i). - Wesley Ivan Hurt, Sep 21 2017
a(n) = Sum_{i=1..n} Sum_{j=1..n} phi(i)*binomial(n, i*j). - Ridouane Oudra, Feb 17 2024

A006519 Highest power of 2 dividing n.

Original entry on oeis.org

1, 2, 1, 4, 1, 2, 1, 8, 1, 2, 1, 4, 1, 2, 1, 16, 1, 2, 1, 4, 1, 2, 1, 8, 1, 2, 1, 4, 1, 2, 1, 32, 1, 2, 1, 4, 1, 2, 1, 8, 1, 2, 1, 4, 1, 2, 1, 16, 1, 2, 1, 4, 1, 2, 1, 8, 1, 2, 1, 4, 1, 2, 1, 64, 1, 2, 1, 4, 1, 2, 1, 8, 1, 2, 1, 4, 1, 2, 1, 16, 1, 2, 1, 4, 1, 2, 1, 8, 1, 2, 1, 4, 1, 2, 1, 32, 1, 2, 1, 4, 1, 2
Offset: 1

Views

Author

N. J. A. Sloane, Simon Plouffe

Keywords

Comments

Least positive k such that m^k + 1 divides m^n + 1 (with fixed base m). - Vladimir Baltic, Mar 25 2002
To construct the sequence: start with 1, concatenate 1, 1 and double last term gives 1, 2. Concatenate those 2 terms, 1, 2, 1, 2 and double last term 1, 2, 1, 2 -> 1, 2, 1, 4. Concatenate those 4 terms: 1, 2, 1, 4, 1, 2, 1, 4 and double last term -> 1, 2, 1, 4, 1, 2, 1, 8, etc. - Benoit Cloitre, Dec 17 2002
a(n) = gcd(seq(binomial(2*n, 2*m+1)/2, m = 0 .. n - 1)) (odd numbered entries of even numbered rows of Pascal's triangle A007318 divided by 2), where gcd() denotes the greatest common divisor of a set of numbers. Due to the symmetry of the rows it suffices to consider m = 0 .. floor((n-1)/2). - Wolfdieter Lang, Jan 23 2004
Equals the continued fraction expansion of a constant x (cf. A100338) such that the continued fraction expansion of 2*x interleaves this sequence with 2's: contfrac(2*x) = [2; 1, 2, 2, 2, 1, 2, 4, 2, 1, 2, 2, 2, 1, 2, 8, 2, ...].
Simon Plouffe observes that this sequence and A003484 (Radon function) are very similar, the difference being all zeros except for every 16th term (see A101119 for nonzero differences). Dec 02 2004
This sequence arises when calculating the next odd number in a Collatz sequence: Next(x) = (3*x + 1) / A006519, or simply (3*x + 1) / BitAnd(3*x + 1, -3*x - 1). - Jim Caprioli, Feb 04 2005
a(n) = n if and only if n = 2^k. This sequence can be obtained by taking a(2^n) = 2^n in place of a(2^n) = n and using the same sequence building approach as in A001511. - Amarnath Murthy, Jul 08 2005
Also smallest m such that m + n - 1 = m XOR (n - 1); A086799(n) = a(n) + n - 1. - Reinhard Zumkeller, Feb 02 2007
Number of 1's between successive 0's in A159689. - Philippe Deléham, Apr 22 2009
Least number k such that all coefficients of k*E(n, x), the n-th Euler polynomial, are integers (cf. A144845). - Peter Luschny, Nov 13 2009
In the binary expansion of n, delete everything left of the rightmost 1 bit. - Ralf Stephan, Aug 22 2013
The equivalent sequence for partitions is A194446. - Omar E. Pol, Aug 22 2013
Also the 2-adic value of 1/n, n >= 1. See the Mahler reference, definition on p. 7. This is a non-archimedean valuation. See Mahler, p. 10. Sometimes called 2-adic absolute value of 1/n. - Wolfdieter Lang, Jun 28 2014
First 2^(k-1) - 1 terms are also the heights of the successive rectangles and squares of width 2 that are adjacent to any of the four sides of the toothpick structure of A139250 after 2^k stages, with k >= 2. For example: if k = 5 the heights after 32 stages are [1, 2, 1, 4, 1, 2, 1, 8, 1, 2, 1, 4, 1, 2, 1] respectively, the same as the first 15 terms of this sequence. - Omar E. Pol, Dec 29 2020

Examples

			2^3 divides 24, but 2^4 does not divide 24, so a(24) = 8.
2^0 divides 25, but 2^1 does not divide 25, so a(25) = 1.
2^1 divides 26, but 2^2 does not divide 26, so a(26) = 2.
Per _Marc LeBrun_'s 2000 comment, a(n) can also be determined with bitwise operations in two's complement. For example, given n = 48, we see that n in binary in an 8-bit byte is 00110000 while -n is 11010000. Then 00110000 AND 11010000 = 00010000, which is 16 in decimal, and therefore a(48) = 16.
G.f. = x + 2*x^2 + x^3 + 4*x^4 + x^5 + 2*x^6 + x^7 + 8*x^8 + x^9 + ...

References

  • Kurt Mahler, p-adic numbers and their functions, second ed., Cambridge University Press, 1981.
  • N. J. A. Sloane and Simon Plouffe, The Encyclopedia of Integer Sequences, Academic Press, 1995 (includes this sequence).

Links

Crossrefs

Partial sums are in A006520, second partial sums in A022560.
Sequences used in definitions of this sequence: A000079, A001511, A004198, A007814.
Cf. A000265, A100338, A003484, A001620, A101119, A002487, A139250.
Sequences with related definitions: A038712, A171977, A135481 (GS(1, 6)).
This is Guy Steele's sequence GS(5, 2) (see A135416).
Related to A007913 via A225546.
A059897 is used to express relationship between sequence terms.
Cf. A091476 (Dgf at s=2).

Programs

  • Haskell
    import Data.Bits ((.&.))
a006519 n = n .&. (-n) :: Integer
-- Reinhard Zumkeller, Mar 11 2012, Dec 29 2011
  • Julia
    using IntegerSequences
[EvenPart(n) for n in 1:102] |> println  # Peter Luschny, Sep 25 2021
  • Magma
    [2^Valuation(n, 2): n in [1..100]]; // Vincenzo Librandi, Mar 27 2015
  • Maple
    with(numtheory): for n from 1 to 200 do if n mod 2 = 1 then printf(`%d,`,1) else printf(`%d,`,2^ifactors(n)[2][1][2]) fi; od:
A006519 := proc(n) if type(n,'odd') then 1 ; else for f in ifactors(n)[2] do if op(1,f) = 2 then return 2^op(2,f) ; end if; end do: end if; end proc: # R. J. Mathar, Oct 25 2010
A006519 := n -> 2^padic[ordp](n,2): # Peter Luschny, Nov 26 2010
  • Mathematica
    lowestOneBit[n_] := Block[{k = 0}, While[Mod[n, 2^k] == 0, k++]; 2^(k - 1)]; Table[lowestOneBit[n], {n, 102}] (* Robert G. Wilson v Nov 17 2004 *)
Table[2^IntegerExponent[n, 2], {n, 128}] (* Jean-François Alcover, Feb 10 2012 *)
Table[BitAnd[BitNot[i - 1], i], {i, 1, 102}] (* Peter Luschny, Oct 10 2019 *)
  • PARI
    {a(n) = 2^valuation(n, 2)};
  • PARI
    a(n)=1<Joerg Arndt, Jun 10 2011
  • PARI
    a(n)=bitand(n,-n); \\ Joerg Arndt, Jun 10 2011
  • PARI
    a(n)=direuler(p=2,n,if(p==2,1/(1-2*X),1/(1-X)))[n] \\ Ralf Stephan, Mar 27 2015
  • Python
    def A006519(n): return n&-n # Chai Wah Wu, Jul 06 2022
  • Scala
    (1 to 128).map(Integer.lowestOneBit()) // _Alonso del Arte, Mar 04 2020

Formula

a(n) = n AND -n (where "AND" is bitwise, and negative numbers are represented in two's complement in a suitable bit width). - Marc LeBrun, Sep 25 2000, clarified by Alonso del Arte, Mar 16 2020
Also: a(n) = gcd(2^n, n). - Labos Elemer, Apr 22 2003
Multiplicative with a(p^e) = p^e if p = 2; 1 if p > 2. - David W. Wilson, Aug 01 2001
G.f.: Sum_{k>=0} 2^k*x^2^k/(1 - x^2^(k+1)). - Ralf Stephan, May 06 2003
Dirichlet g.f.: zeta(s)*(2^s - 1)/(2^s - 2) = zeta(s)*(1 - 2^(-s))/(1 - 2*2^(-s)). - Ralf Stephan, Jun 17 2007
a(n) = 2^floor(A002487(n - 1) / A002487(n)). - Reikku Kulon, Oct 05 2008
a(n) = 2^A007814(n). - R. J. Mathar, Oct 25 2010
a((2*k - 1)*2^e) = 2^e, k >= 1, e >= 0. - Johannes W. Meijer, Jun 07 2011
a(n) = denominator of Euler(n-1, 1). - Arkadiusz Wesolowski, Jul 12 2012
a(n) = A011782(A001511(n)). - Omar E. Pol, Sep 13 2013
a(n) = (n XOR floor(n/2)) XOR (n-1 XOR floor((n-1)/2)) = n - (n AND n-1) (where "AND" is bitwise). - Gary Detlefs, Jun 12 2014
a(n) = ((n XOR n-1)+1)/2. - Gary Detlefs, Jul 02 2014
a(n) = A171977(n)/2. - Peter Kern, Jan 04 2017
a(n) = 2^(A001511(n)-1). - Doug Bell, Jun 02 2017
a(n) = abs(A003188(n-1) - A003188(n)). - Doug Bell, Jun 02 2017
Conjecture: a(n) = (1/(A000203(2*n)/A000203(n)-2)+1)/2. - Velin Yanev, Jun 30 2017
a(n) = (n-1) o n where 'o' is the bitwise converse nonimplication. 'o' is not commutative. n o (n+1) = A135481(n). - Peter Luschny, Oct 10 2019
From Peter Munn, Dec 13 2019: (Start)
a(A225546(n)) = A225546(A007913(n)).
a(A059897(n,k)) = A059897(a(n), a(k)). (End)
Sum_{k=1..n} a(k) ~ (1/(2*log(2)))*n*log(n) + (3/4 + (gamma-1)/(2*log(2)))*n, where gamma is Euler's constant (A001620). - Amiram Eldar, Nov 15 2022
a(n) = n / A000265(n). - Amiram Eldar, May 22 2025

Extensions

More terms from James Sellers, Jun 20 2000

A029931 If 2n = Sum 2^e_i, a(n) = Sum e_i.

Original entry on oeis.org

0, 1, 2, 3, 3, 4, 5, 6, 4, 5, 6, 7, 7, 8, 9, 10, 5, 6, 7, 8, 8, 9, 10, 11, 9, 10, 11, 12, 12, 13, 14, 15, 6, 7, 8, 9, 9, 10, 11, 12, 10, 11, 12, 13, 13, 14, 15, 16, 11, 12, 13, 14, 14, 15, 16, 17, 15, 16, 17, 18, 18, 19, 20, 21, 7, 8, 9, 10, 10, 11, 12, 13, 11, 12, 13, 14, 14, 15, 16
Offset: 0

Views

Author

N. J. A. Sloane

Keywords

Comments

Write n in base 2, n = sum b(i)*2^(i-1), then a(n) = sum b(i)*i. - Benoit Cloitre, Jun 09 2002
May be regarded as a triangular array read by rows, giving weighted sum of compositions in standard order. The standard order of compositions is given by A066099. - Franklin T. Adams-Watters, Nov 06 2006
Sum of all positive integer roots m_i of polynomial {m,k} - see link [Shevelev]; see also A264613. - Vladimir Shevelev, Dec 13 2015
Also the sum of binary indices of n, where a binary index of n (A048793) is any position of a 1 in its reversed binary expansion. For example, the binary indices of 11 are {1,2,4}, so a(11) = 7. - Gus Wiseman, May 22 2024

Examples

			14 = 8+4+2 so a(7) = 3+2+1 = 6.
Composition number 11 is 2,1,1; 1*2+2*1+3*1 = 7, so a(11) = 7.
The triangle starts:
  0
  1
  2 3
  3 4 5 6
The reversed binary expansion of 18 is (0,1,0,0,1) with 1's at positions {2,5}, so a(18) = 2 + 5 = 7. - _Gus Wiseman_, Jul 22 2019

Links

Crossrefs

Other sequences that are built by replacing 2^k in the binary representation with other numbers: A022290 (Fibonacci), A059590 (factorials), A073642, A089625 (primes), A116549, A326031.
Cf. A001793 (row sums), A011782 (row lengths), A059867, A066099, A124757.
Row sums of A048793 and A272020.
Cf. A035327, A056239, A291166, A295235, A326669, A326674, A326675, A326699/A326700.
Contains exactly A000009(n) copies of n.
For length instead of sum we have A000120, complement A023416.
For minimum instead of sum we have A001511, opposite A000012.
For maximum instead of sum we have A029837 or A070939, opposite A070940.
For product instead of sum we have A096111.
The reverse version is A230877, row sums of A371572.
The reverse complement is A359359, row sums of A371571.
The complement is A359400, row sums of A368494.
Numbers k such that a(k) is prime are A372689.
A014499 lists binary indices of prime numbers.
A019565 gives Heinz number of binary indices, inverse A048675.
A372471 lists binary indices of primes, row-sums A372429.
Cf. A000009, A030101, A359401, A359402.

Programs

  • Haskell
    a029931 = sum . zipWith (*) [1..] . a030308_row
-- Reinhard Zumkeller, Feb 28 2014
  • Maple
    HammingWeight := n -> add(i, i = convert(n, base, 2)):
a := proc(n) option remember; `if`(n = 0, 0,
ifelse(n::even, a(n/2) + HammingWeight(n/2), a(n-1) + 1)) end:
seq(a(n), n = 0..78); # Peter Luschny, Oct 30 2021
  • Mathematica
    a[n_] := (b = IntegerDigits[n, 2]).Reverse @ Range[Length @ b]; Array[a,78,0] (* Jean-François Alcover, Apr 28 2011, after B. Cloitre *)
  • PARI
    for(n=0,100,l=length(binary(n)); print1(sum(i=1,l, component(binary(n),i)*(l-i+1)),","))
  • PARI
    a(n) = my(b=binary(n)); b*-[-#b..-1]~; \\ Ruud H.G. van Tol, Oct 17 2023
  • Python
    def A029931(n): return sum(i if j == '1' else 0 for i, j in enumerate(bin(n)[:1:-1],1)) # Chai Wah Wu, Dec 20 2022
(C#)
ulong A029931(ulong n) {
    ulong result = 0, counter = 1;
    while(n > 0) {
        if (n % 2 == 1)
          result += counter;
        counter++;
        n /= 2;
    }
    return result;
} // Frank Hollstein, Jan 07 2023

Formula

a(n) = a(n - 2^L(n)) + L(n) + 1 [where L(n) = floor(log_2(n)) = A000523(n)] = sum of digits of A048794 [at least for n < 512]. - Henry Bottomley, Mar 09 2001
a(0) = 0, a(2n) = a(n) + e1(n), a(2n+1) = a(2n) + 1, where e1(n) = A000120(n). a(n) = log_2(A029930(n)). - Ralf Stephan, Jun 19 2003
G.f.: (1/(1-x)) * Sum_{k>=0} (k+1)*x^2^k/(1+x^2^k). - Ralf Stephan, Jun 23 2003
a(n) = Sum_{k>=0} A030308(n,k)*A000027(k+1). - Philippe Deléham, Oct 15 2011
a(n) = sum of n-th row of the triangle in A213629. - Reinhard Zumkeller, Jun 17 2012
From Reinhard Zumkeller, Feb 28 2014: (Start)
a(A089633(n)) = n and a(m) != n for m < A089633(n).
a(n) = Sum_{k=1..A070939(n)} k*A030308(n,k-1). (End)
a(n) = A073642(n) + A000120(n). - Peter Kagey, Apr 04 2016

Extensions

More terms from Erich Friedman

A238279 Triangle read by rows: T(n,k) is the number of compositions of n into nonzero parts with k parts directly followed by a different part, n>=0, 0<=k<=A004523(n-1).

Original entry on oeis.org

1, 1, 2, 2, 2, 3, 4, 1, 2, 10, 4, 4, 12, 14, 2, 2, 22, 29, 10, 1, 4, 26, 56, 36, 6, 3, 34, 100, 86, 31, 2, 4, 44, 148, 200, 99, 16, 1, 2, 54, 230, 374, 278, 78, 8, 6, 58, 322, 680, 654, 274, 52, 2, 2, 74, 446, 1122, 1390, 814, 225, 22, 1, 4, 88, 573, 1796, 2714, 2058, 813, 136, 10, 4, 88, 778, 2694, 4927
Offset: 0

Views

Author

Joerg Arndt and Alois P. Heinz, Feb 22 2014

Keywords

Comments

Same as A238130, with zeros omitted.
Last elements in rows are 1, 1, 2, 2, 1, 4, 2, 1, 6, 2, 1, 8, ... with g.f. -(x^6+x^4-2*x^2-x-1)/(x^6-2*x^3+1).
For n > 0, also the number of compositions of n with k + 1 runs. - Gus Wiseman, Apr 10 2020

Examples

			Triangle starts:
  00:  1;
  01:  1;
  02:  2;
  03:  2,   2;
  04:  3,   4,   1;
  05:  2,  10,   4;
  06:  4,  12,  14,    2;
  07:  2,  22,  29,   10,    1;
  08:  4,  26,  56,   36,    6;
  09:  3,  34, 100,   86,   31,    2;
  10:  4,  44, 148,  200,   99,   16,    1;
  11:  2,  54, 230,  374,  278,   78,    8;
  12:  6,  58, 322,  680,  654,  274,   52,    2;
  13:  2,  74, 446, 1122, 1390,  814,  225,   22,   1;
  14:  4,  88, 573, 1796, 2714, 2058,  813,  136,  10;
  15:  4,  88, 778, 2694, 4927, 4752, 2444,  618,  77,  2;
  16:  5, 110, 953, 3954, 8531, 9930, 6563, 2278, 415, 28, 1;
  ...
Row n=5 is 2, 10, 4 because in the 16 compositions of 5
  ##:  [composition]  no. of changes
  01:  [ 1 1 1 1 1 ]   0
  02:  [ 1 1 1 2 ]   1
  03:  [ 1 1 2 1 ]   2
  04:  [ 1 1 3 ]   1
  05:  [ 1 2 1 1 ]   2
  06:  [ 1 2 2 ]   1
  07:  [ 1 3 1 ]   2
  08:  [ 1 4 ]   1
  09:  [ 2 1 1 1 ]   1
  10:  [ 2 1 2 ]   2
  11:  [ 2 2 1 ]   1
  12:  [ 2 3 ]   1
  13:  [ 3 1 1 ]   1
  14:  [ 3 2 ]   1
  15:  [ 4 1 ]   1
  16:  [ 5 ]   0
there are 2 with no changes, 10 with one change, and 4 with two changes.

Links

Crossrefs

Columns k=0-10 give: A000005 (for n>0), 2*A002133, A244714, A244715, A244716, A244717, A244718, A244719, A244720, A244721, A244722.
Row lengths are A004523.
Row sums are A011782.
The version counting adjacent equal parts is A106356.
The version for ascents/descents is A238343.
The version for weak ascents/descents is A333213.
The k-th composition in standard-order has A124762(k) adjacent equal parts, A124767(k) maximal runs, A333382(k) adjacent unequal parts, and A333381(k) maximal anti-runs.
Cf. A064113, A333214, A333216.

Programs

  • Maple
    b:= proc(n, v) option remember; `if`(n=0, 1, expand(
      add(b(n-i, i)*`if`(v=0 or v=i, 1, x), i=1..n)))
    end:
T:= n-> (p-> seq(coeff(p, x, i), i=0..degree(p)))(b(n, 0)):
seq(T(n), n=0..14);
  • Mathematica
    b[n_, v_] := b[n, v] = If[n == 0, 1, Expand[Sum[b[n-i, i]*If[v == 0 || v == i, 1, x], {i, 1, n}]]]; T[n_] := Function[{p}, Table[Coefficient[p, x, i], {i, 0, Exponent[p, x]}]][b[n, 0]]; Table[T[n], {n, 0, 14}] // Flatten (* Jean-François Alcover, Feb 11 2015, after Maple *)
Table[If[n==0,1,Length[Select[Join@@Permutations/@IntegerPartitions[n],Length[Split[#]]==k+1&]]],{n,0,12},{k,0,If[n==0,0,Floor[2*(n-1)/3]]}] (* Gus Wiseman, Apr 10 2020 *)
  • PARI
    T_xy(max_row) = {my(N=max_row+1, x='x+O('x^N),h=(1+ sum(i=1,N,(x^i-y*x^i)/(1+y*x^i-x^i)))/(1-sum(i=1,N, y*x^i/(1+y*x^i-x^i)))); for(n=0,N-1, print(Vecrev(polcoeff(h,n))))}
T_xy(16) \\ John Tyler Rascoe, Jul 10 2024

Formula

G.f.: A(x,y) = ( 1 + Sum_{i>0} ((x^i)*(1 - y)/(1 + y*x^i - x^i)) )/( 1 - Sum_{i>0} ((y*x^i)/(1 + y*x^i - x^i)) ). - John Tyler Rascoe, Jul 10 2024
Previous Showing 31-40 of 915 results. Next

Started in 1964 by Neil J. A. Sloane | Maintained by The OEIS Foundation Inc.

Content is available under The OEIS End-User License Agreement.