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 11-20 of 45 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

A001792 a(n) = (n+2)*2^(n-1).

Original entry on oeis.org

1, 3, 8, 20, 48, 112, 256, 576, 1280, 2816, 6144, 13312, 28672, 61440, 131072, 278528, 589824, 1245184, 2621440, 5505024, 11534336, 24117248, 50331648, 104857600, 218103808, 452984832, 939524096, 1946157056, 4026531840, 8321499136, 17179869184, 35433480192
Offset: 0

Views

Author

N. J. A. Sloane

Keywords

Comments

Number of parts in all compositions (ordered partitions) of n + 1. For example, a(2) = 8 because in 3 = 2 + 1 = 1 + 2 = 1 + 1 + 1 we have 8 parts. Also number of compositions (ordered partitions) of 2n + 1 with exactly 1 odd part. For example, a(2) = 8 because the only compositions of 5 with exactly 1 odd part are 5 = 1 + 4 = 2 + 3 = 3 + 2 = 4 + 1 = 1 + 2 + 2 = 2 + 1 + 2 = 2 + 2 + 1. - Emeric Deutsch, May 10 2001
Binomial transform of natural numbers [1, 2, 3, 4, ...].
For n >= 1 a(n) is also the determinant of the n X n matrix with 3's on the diagonal and 1's elsewhere. - Ahmed Fares (ahmedfares(AT)my-deja.com), May 06 2001
The arithmetic mean of first n terms of the sequence is 2^(n-1). - Amarnath Murthy, Dec 25 2001, corrected by M. F. Hasler, Dec 17 2016
Also the number of "winning paths" of length n across an n X n Hex board. Satisfies the recursion a(n) = 2a(n-1) + 2^(n-2). - David Molnar (molnar(AT)stolaf.edu), Apr 10 2002
Diagonal in A053218. - Benoit Cloitre, May 08 2002
Let M_n be the n X n matrix m_(i, j) = 1 + abs(i-j) then det(M_n) = (-1)^(n-1)*a(n-1). - Benoit Cloitre, May 28 2002
Absolute value of determinant of n X n matrix of form: [1 2 3 4 5 / 2 1 2 3 4 / 3 2 1 2 3 / 4 3 2 1 2 / 5 4 3 2 1]. - Benoit Cloitre, Jun 20 2002
Number of ones in all (n+1)-bit integers (cf. A000120). - Ralf Stephan, Aug 02 2003
This sequence also emerges as a floretion force transform of powers of 2 (see program code). Define a(-1) = 0 (as the sequence is returned by FAMP). Then a(n-1) + A098156(n+1) = 2*a(n) (conjecture). - Creighton Dement, Mar 14 2005
This sequence gives the absolute value of the determinant of the Toeplitz matrix with first row containing the first n integers. - Paul Max Payton, May 23 2006
Equals sums of rows right of left edge of A102363 divided by three, + 2^K. - David G. Williams (davidwilliams(AT)paxway.com), Oct 08 2007
If X_1, X_2, ..., X_n are 2-blocks of a (2n+1)-set X then, for n >= 1, a(n) is the number of (n+1)-subsets of X intersecting each X_i, (i = 1, 2, ..., n). - Milan Janjic, Nov 18 2007
Also, a(n-1) is the determinant of the n X n matrix with A[i, j] = n - |i-j|. - M. F. Hasler, Dec 17 2008
1/2 the number of permutations of 1 .. (n+2) arranged in a circle with exactly one local maximum. - R. H. Hardin, Apr 19 2009
The first corrector line for transforming 2^n offset 0 with a leading 1 into the Fibonacci sequence. - Al Hakanson (hawkuu(AT)gmail.com), Jun 01 2009
a(n) is the number of runs of consecutive 1's in all binary sequences of length (n+1). - Geoffrey Critzer, Jul 02 2009
Let X_j (0 < j <= 2^n) all the subsets of N_n; m(i, j) := if {i} in X_j then 1 else 0. Let A = transpose(M).M; Then a(i, j) = (number of elements)|X_i intersect X_j|. Determinant(X*I-A) = (X-(n+1)*2^(n-2))*(X-2^(n-2))^(n-1)*X^(2^n-n).
Eigenvector for (n+1)*2^(n-2) is V_i=|X_i|.
Sum_{k=1..2^n} |X_i intersect X_k|*|X_k| = (n+1)*2^(n-2)*|X_i|.
Eigenvectors for 2^(n-2) are {line(M)[i] - line(M)[j], 1 <= i, j <= n}. - CLARISSE Philippe (clarissephilippe(AT)yahoo.fr), Mar 24 2010
The sequence b(n) = 2*A001792(n), for n >= 1 with b(0) = 1, is an elephant sequence, see A175655. For the central square four A[5] vectors, with decimal values 187, 190, 250 and 442, lead to the b(n) sequence. For the corner squares these vectors lead to the companion sequence A134401. - Johannes W. Meijer, Aug 15 2010
Equals partial sums of A045623: (1, 2, 5, 12, 28, ...); where A045623 = the convolution square of (1, 1, 2, 4, 8, 16, 32, ...). - Gary W. Adamson, Oct 26 2010
Equals (1, 2, 4, 8, 16, ...) convolved with (1, 1, 2, 4, 8, 16, ...); e.g., a(3) = 20 = (1, 1, 2, 4) dot (8, 4, 2, 1) = (8 + 4 + 4 + 4). - Gary W. Adamson, Oct 26 2010
This sequence seems to give the first x+1 nonzero terms in the sequence derived by subtracting the m-th term in the x_binacci sequence (where the first term is one and the y-th term is the sum of x terms immediately preceding it) from 2^(m-2). - Dylan Hamilton, Nov 03 2010
Recursive formulas for a(n) are in many cases derivable from its property wherein delta^k(a(n)) - a(n) = k*2^n where delta^k(a(n)) represents the k-th forward difference of a(n). Provable with a difference table and a little induction. - Ethan Beihl, May 02 2011
Let f(n,k) be the sum of numbers in the subsets of size k of {1, 2, ..., n}. Then a(n-1) is the average of the numbers f(n, 0), ... f(n, n). Example: (f(3, 1), f(3, 2), f(3, 3)) = (6, 12, 6), with average (6+12+6)/3. - Clark Kimberling, Feb 24 2012
a(n) is the number of length-2n binary sequences that contain a subsequence of ones with length n or more. To derive this result, note that there are 2^n sequences where the initial one of the subsequence occurs at entry one. If the initial one of the subsequence occurs at entry 2, 3, ..., or n + 1, there are 2^(n-1) sequences since a zero must precede the initial one. Hence a(n) = 2^n + n*2^(n-1)=(n+2)2^(n-1). An example is given in the example section below. - Dennis P. Walsh, Oct 25 2012
As the total number of parts in all compositions of n+1 (see the first line in Comments) the equivalent sequence for partitions is A006128. On the other hand, as the first differences of A001787 (see the first line in Crossrefs) the equivalent sequence for partitions is A138879. - Omar E. Pol, Aug 28 2013
a(n) is the number of spanning trees of the complete tripartite graph K_{n,1,1}. - James Mahoney, Oct 24 2013
a(n-1) = denominator of the mean (2n/(n+1), after reduction), of the compositions of n; numerator is given by A022998(n). - Clark Kimberling, Mar 11 2014
From Tom Copeland, Nov 09 2014: (Start)
The shifted array belongs to an interpolated family of arrays associated to the Catalan A000108 (t=1), and Riordan, or Motzkin sums A005043 (t=0), with the interpolating o.g.f. (1-sqrt(1-4x/(1+(1-t)x)))/2 and inverse x(1-x)/(1+(t-1)x(1-x)). See A091867 for more info on this family. Here the interpolation is t=-3 (mod signs in the results).
Let C(x) = (1 - sqrt(1-4x))/2, an o.g.f. for the Catalan numbers A000108, with inverse Cinv(x) = x*(1-x) and P(x,t) = x/(1+t*x) with inverse P(x,-t).
Shifted o.g.f: G(x) = x*(1-x)/(1 - 4x*(1-x)) = P[Cinv(x),-4].
Inverse o.g.f: Ginv(x) = [1 - sqrt(1 - 4*x/(1+4x))]/2 = C[P(x, 4)] (signed shifted A001700). Cf. A030528. (End)
For n > 0, element a(n) of the sequence is equal to the gradients of the (n-1)-th row of Pascal triangle multiplied with the square of the integers from n+1,...,1. I.e., row 3 of Pascal's triangle 1,3,3,1 has gradients 1,2,0,-2,-1, so a(4) = 1*(5^2) + 2*(4^2) + 0*(3^2) - 2*(2^2) - 1*(1^2) = 48. - Jens Martin Carlsson, May 18 2017
Number of self-avoiding paths connecting all the vertices of a convex (n+2)-gon. - Ivaylo Kortezov, Jan 19 2020
a(n-1) is the total number of elements of subsets of {1,2,..,n} that contain n. For example, for n = 3, a(2) = 8, and the subsets of {1,2,3} that contain 3 are {3}, {1,3}, {2,3}, {1,2,3}, with a total of 8 elements. - Enrique Navarrete, Aug 01 2020

Examples

			a(0) = 1, a(1) = 2*1 + 1 = 3, a(2) = 2*3 + 2 = 8, a(3) = 2*8 + 4 = 20, a(4) = 2*20 + 8 = 48, a(5) = 2*48 + 16 = 112, a(6) = 2*112 + 32 = 256, ... - _Philippe Deléham_, Apr 19 2009
a(2) = 8 since there are 8 length-4 binary sequences with a subsequence of ones of length 2 or more, namely, 1111, 1110, 1101, 1011, 0111, 1100, 0110, and 0011. - _Dennis P. Walsh_, Oct 25 2012
G.f. = 1 + 3*x + 8*x^2 + 20*x^3 + 48*x^4 + 112*x^5 + 256*x^6 + 576*x^7 + ...

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. 795.
  • 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).
  • A. M. Stepin and A. T. Tagi-Zade, Words with restrictions, pp. 67-74 of Kvant Selecta: Combinatorics I, Amer. Math. Soc., 2001 (G_n on p. 70).

Links

Crossrefs

First differences of A001787.
a(n) = A049600(n, 1), a(n) = A030523(n + 1, 1).
Cf. A053113.
Row sums of triangles A008949 and A055248.
a(n) = -A039991(n+2, 2).
Cf. A130584, A125092, A128064, A000079, A164910, A045623, A000120, A053120.
Cf. A000108, A005043, A091867, A001700, A030528.
If the exponent E in a(n) = Sum_{m=0..n} (Sum_{k=0..m} C(n,k))^E is 1, 2, 3, 4, 5 we get A001792, A003583, A007403, A294435, A294436 respectively.

Programs

  • GAP
    List([0..35],n->(n+2)*2^(n-1)); # Muniru A Asiru, Sep 25 2018
  • Haskell
    a001792 n = a001792_list !! n
a001792_list = scanl1 (+) a045623_list
-- Reinhard Zumkeller, Jul 21 2013
  • Magma
    [(n+2)*2^(n-1): n in [0..40]]; // Vincenzo Librandi, Nov 10 2014
  • Maple
    A001792 := n-> (n+2)*2^(n-1);
spec := [S, {B=Set(Z, 0 <= card), S=Prod(Z, B, B)}, labeled]: seq(combstruct[count](spec, size=n)/4, n=2..30); # Zerinvary Lajos, Oct 09 2006
A001792:=-(-3+4*z)/(2*z-1)^2; # Simon Plouffe in his 1992 dissertation, which gives the sequence without the initial 1
G(x):=1/exp(2*x)*(1-x): f[0]:=G(x): for n from 1 to 54 do f[n]:=diff(f[n-1],x) od: x:=0: seq(abs(f[n]),n=0..28 ); # Zerinvary Lajos, Apr 17 2009
a := n -> hypergeom([-n, 2], [1], -1);
seq(round(evalf(a(n),32)), n=0..31); # Peter Luschny, Aug 02 2014
  • Mathematica
    matrix[n_Integer /; n >= 1] := Table[Abs[p - q] + 1, {q, n}, {p, n}]; a[n_Integer /; n >= 1] := Abs[Det[matrix[n]]] (* Josh Locker (joshlocker(AT)macfora.com), Apr 29 2004 *)
g[n_,m_,r_] := Binomial[n - 1, r - 1] Binomial[m + 1, r] r; Table[1 + Sum[g[n, k - n, r], {r, 1, k}, {n, 1, k - 1}], {k, 1, 29}] (* Geoffrey Critzer, Jul 02 2009 *)
a[n_] := (n + 2)*2^(n - 1); a[Range[0, 40]] (* Vladimir Joseph Stephan Orlovsky, Feb 09 2011 *)
LinearRecurrence[{4, -4}, {1, 3}, 40] (* Harvey P. Dale, Aug 29 2011 *)
CoefficientList[Series[(1 - x) / (1 - 2 x)^2, {x, 0, 40}], x] (* Vincenzo Librandi, Nov 10 2014 *)
b[i_]:=i; a[n_]:=Abs[Det[ToeplitzMatrix[Array[b, n], Array[b, n]]]]; Array[a, 40] (* Stefano Spezia, Sep 25 2018 *)
a[n_]:=Hypergeometric2F1[2,-n+1,1,-1];Array[a,32] (* Giorgos Kalogeropoulos, Jan 04 2022 *)
  • PARI
    A001792(n)=(n+2)<<(n-1) \\ M. F. Hasler, Dec 17 2008
  • Python
    for n in range(0,40): print(int((n+2)*2**(n-1)), end=' ') # Stefano Spezia, Oct 16 2018

Formula

a(n) = (n+2)*2^(n-1).
G.f.: (1 - x)/(1 - 2*x)^2 = 2F1(1,3;2;2x).
a(n) = 4*a(n-1) - 4*a(n-2).
G.f. (-1 + (1-2*x)^(-2))/(x*2^2). - Wolfdieter Lang
a(n) = A018804(2^n). - Matthew Vandermast, Mar 01 2003
a(n) = Sum_{k=0..n+2} binomial(n+2, 2k)*k. - Paul Barry, Mar 06 2003
a(n) = (1/4)*A001787(n+2). - Emeric Deutsch, May 24 2003
With a leading 0, this is ((n+1)2^n - 0^n)/4 = Sum_{m=0..n} binomial(n - 1, m - 1)*m, the binomial transform of A004526(n+1). - Paul Barry, Jun 05 2003
a(n) = Sum_{k=0..n} binomial(n, k)*(k + 1). - Lekraj Beedassy, Jun 24 2004
a(n) = A000244(n) - A066810(n). - Ross La Haye, Apr 29 2006
Row sums of triangle A130585. - Gary W. Adamson, Jun 06 2007
Equals A125092 * [1/1, 1/2, 1/3, ...]. - Gary W. Adamson, Nov 16 2007
a(n) = (n+1)*2^n - n*2^(n-1). Equals A128064 * A000079. - Gary W. Adamson, Dec 28 2007
G.f.: F(3, 1; 2; 2x). - Paul Barry, Sep 03 2008
a(n) = 1 + Sum_{k=1..n} (n - k + 4)2^(n - k - 1). This follows from the result that the number of parts equal to k in all compositions of n is (n - k + 3)2^(n - k - 2) for 0 < k < n. - Geoffrey Critzer, Sep 21 2008
a(n) = 2^(n-1) + 2 a(n-1) ; a(n-1) = det(n - |i - j|){i, j = 1..n}. - _M. F. Hasler, Dec 17 2008
a(n) = 2*a(n-1) + 2^(n-1). - Philippe Deléham, Apr 19 2009
a(n) = A164910(2^n). - Gary W. Adamson, Aug 30 2009
a(n) = Sum_{i=1..2^n} gcd(i, 2^n) = A018804(2^n). So we have: 2^0 * phi(2^n) + ... + 2^n * phi(2^0) = (n + 2)*2^(n-1), where phi is the Euler totient function. - Jeffrey R. Goodwin, Nov 11 2011
a(n) = Sum_{j=0..n} Sum_{i=0..n} binomial(n, i + j). - Yalcin Aktar, Jan 17 2012
Eigensequence of an infinite lower triangular matrix with 2^n as the left border and the rest 1's. - Gary W. Adamson, Jan 30 2012
G.f.: 1 + 2*x*U(0) where U(k) = 1 + (k + 1)/(2 - 8*x/(4*x + (k + 1)/U(k + 1))); (continued fraction, 3 - step). - Sergei N. Gladkovskii, Oct 19 2012
a(n) = Sum_{k=0..n} Sum_{j=0..k} binomial(n,j). - Peter Luschny, Dec 03 2013
a(n) = Hyper2F1([-n, 2], [1], -1). - Peter Luschny, Aug 02 2014
G.f.: 1 / (1 - 3*x / (1 + x / (3 - 4*x))). - Michael Somos, Aug 26 2015
a(n) = -A053120(2+n, n), n >= 0, the negative of the third (sub)diagonal of the triangle of Chebyshev's T polynomials. - Wolfdieter Lang, Nov 26 2019
From Amiram Eldar, Jan 12 2021: (Start)
Sum_{n>=0} 1/a(n) = 8*log(2) - 4.
Sum_{n>=0} (-1)^n/a(n) = 4 - 8*log(3/2). (End)
E.g.f.: exp(2*x)*(1 + x). - Stefano Spezia, Jun 11 2021

A000295 Eulerian numbers (Euler's triangle: column k=2 of A008292, column k=1 of A173018).

Original entry on oeis.org

0, 0, 1, 4, 11, 26, 57, 120, 247, 502, 1013, 2036, 4083, 8178, 16369, 32752, 65519, 131054, 262125, 524268, 1048555, 2097130, 4194281, 8388584, 16777191, 33554406, 67108837, 134217700, 268435427, 536870882, 1073741793, 2147483616, 4294967263, 8589934558
Offset: 0

Views

Author

N. J. A. Sloane

Keywords

Comments

There are 2 versions of Euler's triangle:
* A008292 Classic version of Euler's triangle used by Comtet (1974).
* A173018 Version of Euler's triangle used by Graham, Knuth and Patashnik in Concrete Math. (1990).
Euler's triangle rows and columns indexing conventions:
* A008292 The rows and columns of the Eulerian triangle are both indexed starting from 1. (Classic version: used in the classic books by Riordan and Comtet.)
* A173018 The rows and columns of the Eulerian triangle are both indexed starting from 0. (Graham et al.)
Number of Dyck paths of semilength n having exactly one long ascent (i.e., ascent of length at least two). Example: a(4)=11 because among the 14 Dyck paths of semilength 4, the paths that do not have exactly one long ascent are UDUDUDUD (no long ascent), UUDDUUDD and UUDUUDDD (two long ascents). Here U=(1,1) and D=(1,-1). Also number of ordered trees with n edges having exactly one branch node (i.e., vertex of outdegree at least two). - Emeric Deutsch, Feb 22 2004
Number of permutations of {1,2,...,n} with exactly one descent (i.e., permutations (p(1),p(2),...,p(n)) such that #{i: p(i)>p(i+1)}=1). E.g., a(3)=4 because the permutations of {1,2,3} with one descent are 132, 213, 231 and 312.
a(n+1) is the convolution of nonnegative integers (A001477) and powers of two (A000079). - Graeme McRae, Jun 07 2006
Partial sum of main diagonal of A125127. - Jonathan Vos Post, Nov 22 2006
Number of partitions of an n-set having exactly one block of size > 1. Example: a(4)=11 because, if the partitioned set is {1,2,3,4}, then we have 1234, 123|4, 124|3, 134|2, 1|234, 12|3|4, 13|2|4, 14|2|3, 1|23|4, 1|24|3 and 1|2|34. - Emeric Deutsch, Oct 28 2006
k divides a(k+1) for k in A014741. - Alexander Adamchuk, Nov 03 2006
(Number of permutations avoiding patterns 321, 2413, 3412, 21534) minus one. - Jean-Luc Baril, Nov 01 2007, Mar 21 2008
The chromatic invariant of the prism graph P_n for n >= 3. - Jonathan Vos Post, Aug 29 2008
Decimal integer corresponding to the result of XORing the binary representation of 2^n - 1 and the binary representation of n with leading zeros. This sequence and a few others are syntactically similar. For n > 0, let D(n) denote the decimal integer corresponding to the binary number having n consecutive 1's. Then D(n).OP.n represents the n-th term of a sequence when .OP. stands for a binary operator such as '+', '-', '*', 'quotentof', 'mod', 'choose'. We then get the various sequences A136556, A082495, A082482, A066524, A000295, A052944. Another syntactically similar sequence results when we take the n-th term as f(D(n)).OP.f(n). For example if f='factorial' and .OP.='/', we get (A136556)(A000295) ; if f='squaring' and .OP.='-', we get (A000295)(A052944). - K.V.Iyer, Mar 30 2009
Chromatic invariant of the prism graph Y_n.
Number of labelings of a full binary tree of height n-1, such that each path from root to any leaf contains each label from {1,2,...,n-1} exactly once. - Michael Vielhaber (vielhaber(AT)gmail.com), Nov 18 2009
Also number of nontrivial equivalence classes generated by the weak associative law X((YZ)T)=(X(YZ))T on words with n open and n closed parentheses. Also the number of join (resp. meet)-irreducible elements in the pruning-grafting lattice of binary trees with n leaves. - Jean Pallo, Jan 08 2010
Nonzero terms of this sequence can be found from the row sums of the third sub-triangle extracted from Pascal's triangle as indicated below by braces:
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;
... - L. Edson Jeffery, Dec 28 2011
For integers a, b, denote by a<+>b the least c >= a, such that the Hamming distance D(a,c) = b (note that, generally speaking, a<+>b differs from b<+>a). Then for n >= 3, a(n) = n<+>n. This has a simple explanation: for n >= 3 in binary we have a(n) = (2^n-1)-n = "anti n". - Vladimir Shevelev, Feb 14 2012
a(n) is the number of binary sequences of length n having at least one pair 01. - Branko Curgus, May 23 2012
Nonzero terms are those integers k for which there exists a perfect (Hamming) error-correcting code. - L. Edson Jeffery, Nov 28 2012
a(n) is the number of length n binary words constructed in the following manner: Select two positions in which to place the first two 0's of the word. Fill in all (possibly none) of the positions before the second 0 with 1's and then complete the word with an arbitrary string of 0's or 1's. So a(n) = Sum_{k=2..n} (k-1)*2^(n-k). - Geoffrey Critzer, Dec 12 2013
Without first 0: a(n)/2^n equals Sum_{k=0..n} k/2^k. For example: a(5)=57, 57/32 = 0/1 + 1/2 + 2/4 + 3/8 + 4/16 + 5/32. - Bob Selcoe, Feb 25 2014
The first barycentric coordinate of the centroid of the first n rows of Pascal's triangle, assuming the numbers are weights, is A000295(n+1)/A000337(n). See attached figure. - César Eliud Lozada, Nov 14 2014
Starting (0, 1, 4, 11, ...), this is the binomial transform of (0, 1, 2, 2, 2, ...). - Gary W. Adamson, Jul 27 2015
Also the number of (non-null) connected induced subgraphs in the n-triangular honeycomb rook graph. - Eric W. Weisstein, Aug 27 2017
a(n) is the number of swaps needed in the worst case to transform a binary tree with n full levels into a heap, using (bottom-up) heapify. - Rudy van Vliet, Sep 19 2017
The utility of large networks, particularly social networks, with n participants is given by the terms a(n) of this sequence. This assertion is known as Reed's Law, see the Wikipedia link. - Johannes W. Meijer, Jun 03 2019
a(n-1) is the number of subsets of {1..n} in which the largest element of the set exceeds by at least 2 the next largest element. For example, for n = 5, a(4) = 11 and the 11 sets are {1,3}, {1,4}, {1,5}, {2,4}, {2,5}, {3,5}, {1,2,4}, {1,2,5}, {1,3,5}, {2,3,5}, {1,2,3,5}. - Enrique Navarrete, Apr 08 2020
a(n-1) is also the number of subsets of {1..n} in which the second smallest element of the set exceeds by at least 2 the smallest element. For example, for n = 5, a(4) = 11 and the 11 sets are {1,3}, {1,4}, {1,5}, {2,4}, {2,5}, {3,5}, {1,3,4}, {1,3,5}, {1,4,5}, {2,4,5}, {1,3,4,5}. - Enrique Navarrete, Apr 09 2020
a(n+1) is the sum of the smallest elements of all subsets of {1..n}. For example, for n=3, a(4)=11; the subsets of {1,2,3} are {1}, {2}, {3}, {1,2}, {1,3}, {2,3}, {1,2,3}, and the sum of smallest elements is 11. - Enrique Navarrete, Aug 20 2020
Number of subsets of an n-set that have more than one element. - Eric M. Schmidt, Mar 13 2021
Number of individual bets in a "full cover" bet on n-1 horses, dogs, etc. in different races. Each horse, etc. can be bet on or not, giving 2^n bets. But, by convention, singles (a bet on only one race) are not included, reducing the total number bets by n. It is also impossible to bet on no horses at all, reducing the number of bets by another 1. A full cover on 4 horses, dogs, etc. is therefore 6 doubles, 4 trebles and 1 four-horse etc. accumulator. In British betting, such a bet on 4 horses etc. is a Yankee; on 5, a super-Yankee. - Paul Duckett, Nov 17 2021
From Enrique Navarrete, May 25 2022: (Start)
Number of binary sequences of length n with at least two 1's.
a(n-1) is the number of ways to choose an odd number of elements greater than or equal to 3 out of n elements.
a(n+1) is the number of ways to split [n] = {1,2,...,n} into two (possibly empty) complementary intervals {1,2,...,i} and {i+1,i+2,...,n} and then select a subset from the first interval (2^i choices, 0 <= i <= n), and one block/cell (i.e., subinterval) from the second interval (n-i choices, 0 <= i <= n).
(End)
Number of possible conjunctions in a system of n planets; for example, there can be 0 conjunctions with one planet, one with two planets, four with three planets (three pairs of planets plus one with all three) and so on. - Wendy Appleby, Jan 02 2023
Largest exponent m such that 2^m divides (2^n-1)!. - Franz Vrabec, Aug 18 2023
It seems that a(n-1) is the number of odd r with 0 < r < 2^n for which there exist u,v,w in the x-independent beginning of the Collatz trajectory of 2^n x + r with u+v = w+1, as detailed in the link "Collatz iteration and Euler numbers?". A better understanding of this might also give a formula for A374527. - Markus Sigg, Aug 02 2024
This sequence has a connection to consecutively halved positional voting (CHPV); see Mendenhall and Switkay. - Hal M. Switkay, Feb 25 2025
a(n) is the number of subsets of size 2 and more of an n-element set. Equivalently, a(n) is the number of (hyper)edges of size 2 and more in a complete hypergraph of n vertices. - Yigit Oktar, Apr 05 2025

Examples

			G.f. = x^2 + 4*x^3 + 11*x^4 + 26*x^5 + 57*x^6 + 120*x^7 + 247*x^8 + 502*x^9 + ...

References

  • O. Bottema, Problem #562, Nieuw Archief voor Wiskunde, 28 (1980) 115.
  • L. Comtet, "Permutations by Number of Rises; Eulerian Numbers." Section 6.5 in Advanced Combinatorics: The Art of Finite and Infinite Expansions, rev. enl. ed. Dordrecht, Netherlands: Reidel, pp. 51 and 240-246, 1974.
  • F. N. David and D. E. Barton, Combinatorial Chance. Hafner, NY, 1962, p. 151.
  • R. L. Graham, D. E. Knuth and O. Patashnik, Concrete Mathematics. Addison-Wesley, Reading, MA, 1990.
  • D. E. Knuth, The Art of Computer Programming. Addison-Wesley, Reading, MA, Vol. 3, p. 34.
  • J. Riordan, An Introduction to Combinatorial Analysis, Wiley, 1958, p. 215.
  • 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. A008292 (classic version of Euler's triangle used by Comtet (1974)).
Cf. A173018 (version of Euler's triangle used by Graham, Knuth and Patashnik in Concrete Math. (1990)).
Cf. A002662 (partial sums).
Partial sums of A000225.
Row sums of A014473 and of A143291.
Second column of triangles A112493 and A112500.
Sequences A125128 and A130103 are essentially the same.
Column k=1 of A124324.
Cf. A000079, A000108, A000217, A000325, A000975, A001511, A002663.
Cf. A002664, A007814, A008292, A008949, A014741, A016031, A028366.
Cf. A035039, A035040, A035041, A035042, A079583, A107907, A130321.
Cf. A130128, A130330, A131768, A131816.

Programs

  • Haskell
    a000295 n = 2^n - n - 1  -- Reinhard Zumkeller, Nov 25 2013
  • Magma
    [2^n-n-1: n in [0..40]]; // Vincenzo Librandi, Jul 29 2015
  • Magma
    [EulerianNumber(n, 1): n in [0..40]]; // G. C. Greubel, Oct 02 2024
  • Maple
    [ seq(2^n-n-1, n=1..50) ];
A000295 := -z/(2*z-1)/(z-1)**2; # Simon Plouffe in his 1992 dissertation
# Grammar specification:
spec := [S, { B = Set(Z, 1 <= card), C = Sequence(B, 2 <= card), S = Prod(B, C) }, unlabeled]:
struct := n -> combstruct[count](spec, size = n+1);
seq(struct(n), n = 0..33); # Peter Luschny, Jul 22 2014
  • Mathematica
    a[n_] = If[n==0, 0, n*(HypergeometricPFQ[{1, 1-n}, {2}, -1] - 1)];
Table[a[n], {n,0,40}] (* Olivier Gérard, Mar 29 2011 *)
LinearRecurrence[{4, -5, 2}, {0, 0, 1}, 40] (* Vincenzo Librandi, Jul 29 2015 *)
Table[2^n -n-1, {n,0,40}] (* Eric W. Weisstein, Nov 16 2017 *)
  • PARI
    a(n)=2^n-n-1 \\ Charles R Greathouse IV, Jun 10 2011
  • SageMath
    [2^n -(n+1) for n in range(41)] # G. C. Greubel, Oct 02 2024

Formula

a(n) = 2^n - n - 1.
G.f.: x^2/((1-2*x)*(1-x)^2).
A107907(a(n+2)) = A000079(n+2). - Reinhard Zumkeller, May 28 2005
E.g.f.: exp(x)*(exp(x)-1-x). - Emeric Deutsch, Oct 28 2006
a(0)=0, a(1)=0, a(n) = 3*a(n-1) - 2*a(n-2) + 1. - Miklos Kristof, Mar 09 2005
a(0)=0, a(n) = 2*a(n-1) + n - 1 for all n in Z.
a(n) = Sum_{k=2..n} binomial(n, k). - Paul Barry, Jun 05 2003
a(n+1) = Sum_{i=1..n} Sum_{j=1..i} C(i, j). - Benoit Cloitre, Sep 07 2003
a(n+1) = 2^n*Sum_{k=0..n} k/2^k. - Benoit Cloitre, Oct 26 2003
a(0)=0, a(1)=0, a(n) = Sum_{i=0..n-1} i+a(i) for i > 1. - Gerald McGarvey, Jun 12 2004
a(n+1) = Sum_{k=0..n} (n-k)*2^k. - Paul Barry, Jul 29 2004
a(n) = Sum_{k=0..n} binomial(n, k+2); a(n+2) = Sum_{k=0..n} binomial(n+2, k+2). - Paul Barry, Aug 23 2004
a(n) = Sum_{k=0..floor((n-1)/2)} binomial(n-k-1, k+1)*2^(n-k-2)*(-1/2)^k. - Paul Barry, Oct 25 2004
a(0) = 0; a(n) = Stirling2(n,2) + a(n-1) = A000225(n-1) + a(n-1). - Thomas Wieder, Feb 18 2007
a(n) = A000325(n) - 1. - Jonathan Vos Post, Aug 29 2008
a(0) = 0, a(n) = Sum_{k=0..n-1} 2^k - 1. - Doug Bell, Jan 19 2009
a(n) = A000217(n-1) + A002662(n) for n>0. - Geoffrey Critzer, Feb 11 2009
a(n) = A000225(n) - n. - Zerinvary Lajos, May 29 2009
a(n) = n*(2F1([1,1-n],[2],-1) - 1). - Olivier Gérard, Mar 29 2011
Column k=1 of A173018 starts a'(n) = 0, 1, 4, 11, ... and has the hypergeometric representation n*hypergeom([1, -n+1], [-n], 2). This can be seen as a formal argument to prefer Euler's A173018 over A008292. - Peter Luschny, Sep 19 2014
E.g.f.: exp(x)*(exp(x)-1-x); this is U(0) where U(k) = 1 - x/(2^k - 2^k/(x + 1 - x^2*2^(k+1)/(x*2^(k+1) - (k+1)/U(k+1)))); (continued fraction, 3rd kind, 4-step). - Sergei N. Gladkovskii, Dec 01 2012
a(n) = A079583(n) - A000225(n+1). - Miquel Cerda, Dec 25 2016
a(0) = 0; a(1) = 0; for n > 1: a(n) = Sum_{i=1..2^(n-1)-1} A001511(i). - David Siegers, Feb 26 2019
a(n) = A007814(A028366(n)). - Franz Vrabec, Aug 18 2023
a(n) = Sum_{k=1..floor((n+1)/2)} binomial(n+1, 2*k+1). - Taras Goy, Jan 02 2025

A000346 a(n) = 2^(2*n+1) - binomial(2*n+1, n+1).

Original entry on oeis.org

1, 5, 22, 93, 386, 1586, 6476, 26333, 106762, 431910, 1744436, 7036530, 28354132, 114159428, 459312152, 1846943453, 7423131482, 29822170718, 119766321572, 480832549478, 1929894318332, 7744043540348, 31067656725032, 124613686513778, 499744650202436
Offset: 0

Views

Author

N. J. A. Sloane, Wolfdieter Lang

Keywords

Comments

Also a(n) = 2nd elementary symmetric function of binomial(n,0), binomial(n,1), ..., binomial(n,n).
Also a(n) = one half the sum of the heights, over all Dyck (n+2)-paths, of the vertices that are at even height and terminate an upstep. For example with n=1, these vertices are indicated by asterisks in the 5 Dyck 3-paths: UU*UDDD, UU*DU*DD, UDUU*DD, UDUDUD, UU*DDUD, yielding a(1)=(2+4+2+0+2)/2=5. - David Callan, Jul 14 2006
Hankel transform is (-1)^n*(2n+1); the Hankel transform of sum(k=0..n, C(2*n,k) ) - C(2n,n) is (-1)^n*n. - Paul Barry, Jan 21 2007
Row sums of the Riordan matrix (1/(1-4x),(1-sqrt(1-4x))/2) (A187926). - Emanuele Munarini, Mar 16 2011
From Gus Wiseman, Jul 19 2021: (Start)
For n > 0, a(n-1) is also the number of integer compositions of 2n with nonzero alternating sum, where the alternating sum of a sequence (y_1,...,y_k) is Sum_i (-1)^(i-1) y_i. These compositions are ranked by A053754 /\ A345921. For example, the a(3-1) = 22 compositions of 6 are:
(6) (1,5) (1,1,4) (1,1,1,3) (1,1,1,1,2)
(2,4) (1,2,3) (1,1,3,1) (1,1,2,1,1)
(4,2) (1,4,1) (1,2,1,2) (2,1,1,1,1)
(5,1) (2,1,3) (1,3,1,1)
(2,2,2) (2,1,2,1)
(3,1,2) (3,1,1,1)
(3,2,1)
(4,1,1)
(End)

Examples

			G.f. = 1 + 5*x + 22*x^2 + 93*x^3 + 386*x^4 + 1586*x^5 + 6476*x^6 + ...

References

  • T. Myers and L. Shapiro, Some applications of the sequence 1, 5, 22, 93, 386, ... to Dyck paths and ordered trees, Congressus Numerant., 204 (2010), 93-104.
  • D. Phulara and L. W. Shapiro, Descendants in ordered trees with a marked vertex, Congressus Numerantium, 205 (2011), 121-128.
  • 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. A000108, A014137, A014318. A column of A058893. Subdiagonal of A053979.
Bisection of A058622 and (possibly) A007008.
Cf. A045621, A068551.
Cf. A033504, A038806, A055217, A069731.
Even bisection of A294175 (without the first two terms).
The following relate to compositions of 2n with alternating sum k.
- The k > 0 case is counted by A000302.
- The k <= 0 case is counted by A000302.
- The k != 0 case is counted by A000346 (this sequence).
- The k = 0 case is counted by A001700 or A088218.
- The k < 0 case is counted by A008549.
- The k >= 0 case is counted by A114121.
A011782 counts compositions.
A086543 counts partitions with nonzero alternating sum (bisection: A182616).
A097805 counts compositions by alternating (or reverse-alternating) sum.
A103919 counts partitions by sum and alternating sum (reverse: A344612).
A345197 counts compositions by length and alternating sum.
Cf. A000070, A001791, A007318, A025047, A027306, A032443, A053754, A120452, A163493, A239830, A344611, A345921.

Programs

  • Magma
    [2^(2*n+1) - Binomial(2*n+1,n+1): n in [0..30]]; // Vincenzo Librandi, Jun 07 2011
  • Maple
    seq(2^(2*n+1)-binomial(2*n,n)*(2*n+1)/(n+1), n=0..12); # Emanuele Munarini, Mar 16 2011
  • Mathematica
    Table[2^(2n+1)-Binomial[2n,n](2n+1)/(n+1),{n,0,20}] (* Emanuele Munarini, Mar 16 2011 *)
a[ n_] := If[ n<-4, 0, (4^(n + 1) - Binomial[2 n + 2, n + 1]) / 2]; (* Michael Somos, Jan 25 2014 *)
  • Maxima
    makelist(2^(2*n+1)-binomial(2*n,n)*(2*n+1)/(n+1),n,0,12); /* Emanuele Munarini, Mar 16 2011 */
  • PARI
    {a(n) = if( n<-4, 0, n++; (2^(2*n) - binomial(2*n, n)) / 2)}; /* Michael Somos, Jan 25 2014 */

Formula

G.f.: c(x)/(1-4x), c(x) = g.f. of Catalan numbers.
Convolution of Catalan numbers and powers of 4.
Also one half of convolution of central binomial coeffs. A000984(n), n=0, 1, 2, ... with shifted central binomial coeffs. A000984(n), n=1, 2, 3, ...
a(n) = A045621(2n+1) = (1/2)*A068551(n+1).
a(n) = Sum_{k=0..n} A000984(k)*A001700(n-k). - Philippe Deléham, Jan 22 2004
a(n) = Sum_{k=0..n+1} binomial(n+k, k-1)2^(n-k+1). - Paul Barry, Nov 13 2004
a(n) = Sum_{i=0..n} binomial(2n+2, i). See A008949. - Ed Catmur (ed(AT)catmur.co.uk), Dec 09 2006
a(n) = Sum_{k=0..n+1, C(2n+2,k)} - C(2n+2,n+1). - Paul Barry, Jan 21 2007
Logarithm g.f. log(1/(2-C(x)))=sum(n>0, a(n)/n*x^n), C(x)=(1-sqrt(1-4*x))/2x (A000108). - Vladimir Kruchinin, Aug 10 2010
D-finite with recurrence: (n+3) a(n+2) - 2(4n+9) a(n+1) + 8(2n+3) a(n) = 0. - Emanuele Munarini, Mar 16 2011
E.g.f.:exp(2*x)*(2*exp(2*x) - BesselI(0,2*x) - BesselI(1,2*x)).
This is the first derivative of exp(2*x)*(exp(2*x) - BesselI(0,2*x))/2. See the e.g.f. of A032443 (which has a plus sign) and the remarks given there. - Wolfdieter Lang, Jan 16 2012
a(n) = Sum_{0<=iMircea Merca, Apr 05 2012
A000346 = A004171 - A001700. See A032443 for the sum. - M. F. Hasler, Jan 02 2014
0 = a(n) * (256*a(n+1) - 224*a(n+2) + 40*a(n+3)) + a(n+1) * (-32*a(n+1) + 56*a(n+2) - 14*a(n+3)) + a(n+2) * (-2*a(n+2) + a(n+3)) if n>-5. - Michael Somos, Jan 25 2014
REVERT transform is [1,-5,28,-168,1056,...] = alternating signed version of A069731. - Michael Somos, Jan 25 2014
Convolution square is A038806. - Michael Somos, Jan 25 2014
BINOMIAL transform of A055217(n-1) is a(n-1). - Michael Somos, Jan 25 2014
(n+1) * a(n) = A033504(n). - Michael Somos, Jan 25 2014
Recurrence: (n+1)*a(n) = 512*(2*n-7)*a(n-5) + 256*(13-5*n)*a(n-4) + 64*(10*n-17)*a(n-3) + 32*(4-5*n)*a(n-2) + 2*(10*n+1)*a(n-1), n>=5. - Fung Lam, Mar 21 2014
Asymptotic approximation: a(n) ~ 2^(2n+1)*(1-1/sqrt(n*Pi)). - Fung Lam, Mar 21 2014
a(n) = Sum_{m = n+2..2*(n+1)} binomial(2*(n+1), m), n >= 0. - Wolfdieter Lang, May 22 2015
a(n) = A000302(n) + A008549(n). - Gus Wiseman, Jul 19 2021
a(n) = Sum_{j=1..n+1} Sum_{k=1..j} 2^(j-k)*binomial(n+k-1, n). - Fabio Visonà, May 04 2022
a(n) = (1/2)*(-1)^n*binomial(-(n+1), n+2)*hypergeom([1, 2*n + 3], [n + 3], 1/2). - Peter Luschny, Nov 29 2023

Extensions

Corrected by Christian G. Bower

A027306 a(n) = 2^(n-1) + ((1 + (-1)^n)/4)*binomial(n, n/2).

Original entry on oeis.org

1, 1, 3, 4, 11, 16, 42, 64, 163, 256, 638, 1024, 2510, 4096, 9908, 16384, 39203, 65536, 155382, 262144, 616666, 1048576, 2449868, 4194304, 9740686, 16777216, 38754732, 67108864, 154276028, 268435456, 614429672, 1073741824, 2448023843
Offset: 0

Views

Author

Clark Kimberling

Keywords

Comments

Inverse binomial transform of A027914. Hankel transform (see A001906 for definition) is {1, 2, 3, 4, ..., n, ...}. - Philippe Deléham, Jul 21 2005
Number of walks of length n on a line that starts at the origin and ends at or above 0. - Benjamin Phillabaum, Mar 05 2011
Number of binary integers (i.e., with a leading 1 bit) of length n+1 which have a majority of 1-bits. E.g., for n+1=4: (1011, 1101, 1110, 1111) a(3)=4. - Toby Gottfried, Dec 11 2011
Number of distinct symmetric staircase walks connecting opposite corners of a square grid of side n > 1. - Christian Barrientos, Nov 25 2018
From Gus Wiseman, Aug 20 2021: (Start)
Also the number of integer compositions of n + 1 with alternating sum > 0, where the alternating sum of a sequence (y_1,...,y_k) is Sum_i (-1)^(i-1) y_i. These compositions are ranked by A345917. For example, the a(0) = 1 through a(4) = 11 compositions are:
(1) (2) (3) (4) (5)
(21) (31) (32)
(111) (112) (41)
(211) (113)
(122)
(212)
(221)
(311)
(1121)
(2111)
(11111)
The following relate to these compositions:
- The unordered version is A027193.
- The complement is counted by A058622.
- The reverse unordered version is A086543.
- The version for alternating sum >= 0 is A116406.
- The version for alternating sum < 0 is A294175.
- Ranked by A345917. (End)
The Gauss congruences a(n*p^k) == a(n^p^(k-1)) (mod p^k) hold for prime p and positive integers n and k. - Peter Bala, Jan 07 2022

Examples

			From _Gus Wiseman_, Aug 20 2021: (Start)
The a(0) = 1 through a(4) = 11 binary numbers with a majority of 1-bits (Gottfried's comment) are:
  1   11   101   1011   10011
           110   1101   10101
           111   1110   10110
                 1111   10111
                        11001
                        11010
                        11011
                        11100
                        11101
                        11110
                        11111
The version allowing an initial zero is A058622.
(End)

References

  • 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, Eq. (4.2.1.6)

Links

Crossrefs

a(n) = Sum{(k+1)T(n, m-k)}, 0<=k<=[ (n+1)/2 ], T given by A008315.
Column k=2 of A226873. - Alois P. Heinz, Jun 21 2013
Cf. A008949, A248574, A001405, A246437.
The even bisection is A000302.
The odd bisection appears to be A032443.
Cf. A000984, A001700, A001791, A008549, A011782, A088218, A097805, A163493, A182616, A345197.

Programs

  • GAP
    List([0..35],n->Sum([0..Int(n/2)],k->Binomial(n,k))); # Muniru A Asiru, Nov 27 2018
  • Haskell
    a027306 n = a008949 n (n `div` 2)  -- Reinhard Zumkeller, Nov 14 2014
  • Magma
    [2^(n-1)+(1+(-1)^n)/4*Binomial(n, n div 2): n in [0..40]]; // Vincenzo Librandi, Jun 19 2016
  • Maple
    a:= proc(n) add(binomial(n, j), j=0..n/2) end:
seq(a(n), n=0..32); # Zerinvary Lajos, Mar 29 2009
  • Mathematica
    Table[Sum[Binomial[n, k], {k, 0, Floor[n/2]}], {n, 1, 35}]
(* Second program: *)
a[0] = a[1] = 1; a[2] = 3; a[n_] := a[n] = (2(n-1)(2a[n-2] + a[n-1]) - 8(n-2) a[n-3])/n; Array[a, 33, 0] (* Jean-François Alcover, Sep 04 2016 *)
  • PARI
    a(n)=if(n<0,0,(2^n+if(n%2,0,binomial(n, n/2)))/2)

Formula

a(n) = Sum_{k=0..floor(n/2)} binomial(n,k).
Odd terms are 2^(n-1). Also a(2n) - 2^(2n-1) is given by A001700. a(n) = 2^n + (n mod 2)*binomial(n, (n-1)/2).
E.g.f.: (exp(2x) + I_0(2x))/2.
O.g.f.: 2*x/(1-2*x)/(1+2*x-((1+2*x)*(1-2*x))^(1/2)). - Vladeta Jovovic, Apr 27 2003
a(n) = A008949(n, floor(n/2)); a(n) + a(n-1) = A248574(n), n > 0. - Reinhard Zumkeller, Nov 14 2014
From Peter Bala, Jul 21 2015: (Start)
a(n) = [x^n]( 2*x - 1/(1 - x) )^n.
O.g.f.: (1/2)*( 1/sqrt(1 - 4*x^2) + 1/(1 - 2*x) ).
Inverse binomial transform is (-1)^n*A246437(n).
exp( Sum_{n >= 1} a(n)*x^n/n ) = 1 + x + 2*x^2 + 3*x^3 + 6*x^4 + 10*x^5 + ... is the o.g.f. for A001405. (End)
a(n) = Sum_{k=1..floor((n+1)/2)} binomial(n-1,(2n+1-(-1)^n)/4 -k). - Anthony Browne, Jun 18 2016
D-finite with recurrence: n*a(n) + 2*(-n+1)*a(n-1) + 4*(-n+1)*a(n-2) + 8*(n-2)*a(n-3) = 0. - R. J. Mathar, Aug 09 2017

Extensions

Better description from Robert G. Wilson v, Aug 30 2000 and from Yong Kong (ykong(AT)curagen.com), Dec 28 2000

A228196 A triangle formed like Pascal's triangle, but with n^2 on the left border and 2^n on the right border instead of 1.

Original entry on oeis.org

0, 1, 2, 4, 3, 4, 9, 7, 7, 8, 16, 16, 14, 15, 16, 25, 32, 30, 29, 31, 32, 36, 57, 62, 59, 60, 63, 64, 49, 93, 119, 121, 119, 123, 127, 128, 64, 142, 212, 240, 240, 242, 250, 255, 256, 81, 206, 354, 452, 480, 482, 492, 505, 511, 512, 100, 287, 560, 806, 932, 962, 974, 997, 1016, 1023, 1024
Offset: 1

Views

Author

Boris Putievskiy, Aug 15 2013

Keywords

Comments

The third row is (n^4 - n^2 + 24*n + 24)/12.
For a closed-form formula for generalized Pascal's triangle see A228576. - Boris Putievskiy, Sep 04 2013

Examples

			The start of the sequence as a triangular array read by rows:
   0;
   1,  2;
   4,  3,  4;
   9,  7,  7,  8;
  16, 16, 14, 15, 16;
  25, 32, 30, 29, 31, 32;
  36, 57, 62, 59, 60, 63, 64;

Links

Crossrefs

Cf. We denote Pascal-like triangle with L(n) on the left border and R(n) on the right border by (L(n),R(n)). A007318 (1,1), A008949 (1,2^n), A029600 (2,3), A029618 (3,2), A029635 (1,2), A029653 (2,1), A037027 (Fibonacci(n),1), A051601 (n,n) n>=0, A051597 (n,n) n>0, A051666 (n^2,n^2), A071919 (1,0), A074829 (Fibonacci(n), Fibonacci(n)), A074909 (1,n), A093560 (3,1), A093561 (4,1), A093562 (5,1), A093563 (6,1), A093564 (7,1), A093565 (8,1), A093644 (9,1), A093645 (10,1), A095660 (1,3), A095666 (1,4), A096940 (1,5), A096956 (1,6), A106516 (3^n,1), A108561(1,(-1)^n), A132200 (4,4), A134636 (2n+1,2n+1), A137688 (2^n,2^n), A160760 (3^(n-1),1), A164844(1,10^n), A164847 (100^n,1), A164855 (101*100^n,1), A164866 (101^n,1), A172171 (1,9), A172185 (9,11), A172283 (-9,11), A177954 (int(n/2),1), A193820 (1,2^n), A214292 (n,-n), A227074 (4^n,4^n), A227075 (3^n,3^n), A227076 (5^n,5^n), A227550 (n!,n!), A228053 ((-1)^n,(-1)^n), A228074 (Fibonacci(n), n).
Cf. A000290 (row 1), A153056 (row 2), A000079 (column 1), A000225 (column 2), A132753 (column 3), A118885 (row sums of triangle array + 1), A228576 (generalized Pascal's triangle).

Programs

  • GAP
    T:= function(n,k)
    if k=0 then return n^2;
    elif k=n then return 2^n;
    else return T(n-1,k-1) + T(n-1,k);
    fi;
  end;
Flat(List([0..12], n-> List([0..n], k-> T(n,k) ))); # G. C. Greubel, Nov 12 2019
  • Maple
    T:= proc(n, k) option remember;
      if k=0 then n^2
    elif k=n then 2^k
    else T(n-1, k-1) + T(n-1, k)
      fi
    end:
seq(seq(T(n, k), k=0..n), n=0..10); # G. C. Greubel, Nov 12 2019
  • Mathematica
    T[n_, k_]:= T[n, k] = If[k==0, n^2, If[k==n, 2^k, T[n-1, k-1] + T[n-1, k]]]; Table[T[n, k], {n,0,10}, {k,0,n}]//Flatten (* G. C. Greubel, Nov 12 2019 *)
Flatten[Table[Sum[i^2 Binomial[n-1-i, n-k-i], {i,1,n-k}] + Sum[2^i Binomial[n-1-i, k-i], {i,1,k}], {n,0,10}, {k,0,n}]] (* Greg Dresden, Aug 06 2022 *)
  • PARI
    T(n,k) = if(k==0, n^2, if(k==n, 2^k, T(n-1, k-1) + T(n-1, k) )); \\ G. C. Greubel, Nov 12 2019
  • Python
    def funcL(n):
   q = n**2
   return q
def funcR(n):
   q = 2**n
   return q
for n in range (1,9871):
   t=int((math.sqrt(8*n-7) - 1)/ 2)
   i=n-t*(t+1)/2-1
   j=(t*t+3*t+4)/2-n-1
   sum1=0
   sum2=0
   for m1 in range (1,i+1):
      sum1=sum1+funcR(m1)*binomial(i+j-m1-1,i-m1)
   for m2 in range (1,j+1):
      sum2=sum2+funcL(m2)*binomial(i+j-m2-1,j-m2)
   sum=sum1+sum2
  • Sage
    @CachedFunction
def T(n, k):
    if (k==0): return n^2
    elif (k==n): return 2^n
    else: return T(n-1, k-1) + T(n-1, k)
[[T(n, k) for k in (0..n)] for n in (0..12)] # G. C. Greubel, Nov 12 2019

Formula

T(n,0) = n^2, n>0; T(0,k) = 2^k; T(n, k) = T(n-1, k-1) + T(n-1, k) for n,k > 0. [corrected by G. C. Greubel, Nov 12 2019]
Closed-form formula for general case. Let L(m) and R(m) be the left border and the right border of Pascal like triangle, respectively. We denote binomial(n,k) by C(n,k).
As table read by antidiagonals T(n,k) = Sum_{m1=1..n} R(m1)*C(n+k-m1-1, n-m1) + Sum_{m2=1..k} L(m2)*C(n+k-m2-1, k-m2); n,k >=0.
As linear sequence a(n) = Sum_{m1=1..i} R(m1)*C(i+j-m1-1, i-m1) + Sum_{m2=1..j} L(m2)*C(i+j-m2-1, j-m2), where i=n-t*(t+1)/2-1, j=(t*t+3*t+4)/2-n-1, t=floor((-1+sqrt(8*n-7))/2); n>0.
Some special cases. If L(m)={b,b,b...} b*A000012, then the second sum takes form b*C(n+k-1,j). If L(m) is {0,b,2b,...} b*A001477, then the second sum takes form b*C(n+k,n-1). Similarly for R(m) and the first sum.
For this sequence L(m)=m^2 and R(m)=2^m.
As table read by antidiagonals T(n,k) = Sum_{m1=1..n} (2^m1)*C(n+k-m1-1, n-m1) + Sum_{m2=1..k} (m2^2)*C(n+k-m2-1, k-m2); n,k >=0.
As linear sequence a(n) = Sum_{m1=1..i} (2^m1)*C(i+j-m1-1, i-m1) + Sum_{m2=1..j} (m2^2)*C(i+j-m2-1, j-m2), where i=n-t*(t+1)/2-1, j=(t*t+3*t+4)/2-n-1, t=floor((-1+sqrt(8*n-7))/2).
As a triangular array read by rows, T(n,k) = Sum_{i=1..n-k} i^2*C(n-1-i, n-k-i) + Sum_{i=1..k} 2^i*C(n-1-i, k-i); n,k >=0. - Greg Dresden, Aug 06 2022

Extensions

Cross-references corrected and extended by Philippe Deléham, Dec 27 2013

A004070 Table of Whitney numbers W(n,k) read by antidiagonals, where W(n,k) is maximal number of pieces into which n-space is sliced by k hyperplanes, n >= 0, k >= 0.

Original entry on oeis.org

1, 1, 1, 1, 2, 1, 1, 2, 3, 1, 1, 2, 4, 4, 1, 1, 2, 4, 7, 5, 1, 1, 2, 4, 8, 11, 6, 1, 1, 2, 4, 8, 15, 16, 7, 1, 1, 2, 4, 8, 16, 26, 22, 8, 1, 1, 2, 4, 8, 16, 31, 42, 29, 9, 1, 1, 2, 4, 8, 16, 32, 57, 64, 37, 10, 1, 1, 2, 4, 8, 16, 32, 63, 99, 93, 46, 11, 1, 1, 2, 4, 8, 16, 32, 64, 120, 163
Offset: 0

Views

Author

N. J. A. Sloane

Keywords

Comments

As a number triangle, this is given by T(n,k)=sum{j=0..n, C(n,j)(-1)^(n-j)sum{i=0..j, C(j+k,i-k)}}. - Paul Barry, Aug 23 2004
As a number triangle, this is the Riordan array (1/(1-x), x(1+x)) with T(n,k)=sum{i=0..n, binomial(k,i-k)}. Diagonal sums are then A023434(n+1). - Paul Barry, Feb 16 2005
Form partial sums across rows of square array of binomial coefficients A026729; see also A008949. - Philippe Deléham, Aug 28 2005
Square array A026729 -> Partial sums across rows
1 0 0 0 0 0 0 . . . . 1 1 1 1 1 1 1 . . . . . .
1 1 0 0 0 0 0 . . . . 1 2 2 2 2 2 2 . . . . . .
1 2 1 0 0 0 0 . . . . 1 3 4 4 4 4 4 . . . . . .
1 3 3 1 0 0 0 . . . . 1 4 7 8 8 8 8 . . . . . .
For other Whitney numbers see A007799.
W(n,k) is the number of length k binary sequences containing no more than n 1's. - Geoffrey Critzer, Mar 15 2010
From Emeric Deutsch, Jun 15 2010: (Start)
Viewed as a number triangle, T(n,k) is the number of internal nodes of the Fibonacci tree of order n+2 at level k. A Fibonacci tree of order n (n>=2) is a complete binary tree whose left subtree is the Fibonacci tree of order n-1 and whose right subtree is the Fibonacci tree of order n-2; each of the Fibonacci trees of order 0 and 1 is defined as a single node.
(End)
Named after the American mathematician Hassler Whitney (1907-1989). - Amiram Eldar, Jun 13 2021

Examples

			Table W(n,k) begins:
  1 1 1 1  1  1  1 ...
  1 2 3 4  5  6  7 ...
  1 2 4 7 11 16 22 ...
  1 2 4 8 15 26 42 ...
W(2,4) = 11 because there are 11 length 4 binary sequences containing no more than 2 1's: {0, 0, 0, 0}, {0, 0, 0, 1}, {0, 0, 1, 0}, {0, 0, 1, 1}, {0, 1, 0, 0}, {0, 1, 0, 1}, {0, 1, 1, 0}, {1, 0, 0, 0}, {1, 0, 0, 1}, {1, 0, 1, 0}, {1, 1, 0, 0}. - _Geoffrey Critzer_, Mar 15 2010
Table T(n, k) begins:
  1
  1  1
  1  2  1
  1  2  3  1
  1  2  4  4  1
  1  2  4  7  5  1
  1  2  4  8 11  6  1
...

References

  • Donald E. Knuth, The Art of Computer Programming, Vol. 3, 2nd edition, Addison-Wesley, Reading, MA, 1998, p. 417.

Links

Crossrefs

Cf. A007799. As a triangle, mirror A052509.
Rows converge to powers of two (A000079). Subdiagonals include A000225, A000295, A002662, A002663, A002664, A035038, A035039, A035040, A035041, A035042. Antidiagonal sums are A000071.
Rows are: A000012, A000027, A000124, A000125, A000127, A006261, A008859, A008860, A008861, A008862, A008863.
Cf. A178522, A178524.

Programs

  • Mathematica
    Transpose[ Table[Table[Sum[Binomial[n, k], {k, 0, m}], {m, 0, 15}], {n, 0, 15}]] // Grid (* Geoffrey Critzer, Mar 15 2010 *)
T[ n_, k_] := Sum[ Binomial[n, j] (-1)^(n - j) Sum[ Binomial[j + k, i - k], {i, 0, j}], {j, 0, n}]; (* Michael Somos, May 31 2016 *)
  • PARI
    /* array read by antidiagonals up coordinate index functions */
t1(n) = binomial(floor(3/2 + sqrt(2+2*n)), 2) - (n+1); /* A025581 */
t2(n) = n - binomial(floor(1/2 + sqrt(2+2*n)), 2); /* A002262 */
/* define the sequence array function for A004070 */
W(n, k) = sum(i=0, n, binomial(k, i));
/* visual check ( origin 0,0 ) */
printp(matrix(7, 7, n, k, W(n-1, k-1)));
/* print the sequence entries by antidiagonals going up ( origin 0,0 ) */
print1("S A004070 "); for(n=0, 32, print1(W(t1(n), t2(n))","));
print1("T A004070 "); for(n=33, 61, print1(W(t1(n), t2(n))","));
print1("U A004070 "); for(n=62, 86, print1(W(t1(n), t2(n))",")); /* Michael Somos, Apr 28 2000 */
  • PARI
    T(n, k)=sum(m=0, n-k, binomial(k, m)) \\ Jianing Song, May 30 2022

Formula

W(n, k) = Sum_{i=0..n} binomial(k, i). - Bill Gosper
W(n, k) = if k=0 or n=0 then 1 else W(n, k-1)+W(n-1, k-1). - David Broadhurst, Jan 05 2000
The table W(n,k) = A000012 * A007318(transform), where A000012 = (1; 1,1; 1,1,1; ...). - Gary W. Adamson, Nov 15 2007
E.g.f. for row n: (1 + x + x^2/2! + ... + x^n/n!)* exp(x). - Geoffrey Critzer, Mar 15 2010
G.f.: 1 / (1 - x - x*y*(1 - x^2)) = Sum_{0 <= k <= n} x^n * y^k * T(n, k). - Michael Somos, May 31 2016
W(n, n) = 2^n. - Michael Somos, May 31 2016
From Jianing Song, May 30 2022: (Start)
T(n, 0) = T(n, n) = 1 for n >= 0; T(n, k) = T(n-1, k-1) + T(n-2, k-1) for k=1, 2, ..., n-1, n >= 2.
T(n, k) = Sum_{m=0..n-k} binomial(k, m).
T(n,k) = 2^k for 0 <= k <= floor(n/2). (End)

Extensions

More terms from Larry Reeves (larryr(AT)acm.org), Mar 20 2000

A052509 Knights-move Pascal triangle: T(n,k), n >= 0, 0 <= k <= n; T(n,0) = T(n,n) = 1, T(n,k) = T(n-1,k) + T(n-2,k-1) for k = 1,2,...,n-1, n >= 2.

Original entry on oeis.org

1, 1, 1, 1, 2, 1, 1, 3, 2, 1, 1, 4, 4, 2, 1, 1, 5, 7, 4, 2, 1, 1, 6, 11, 8, 4, 2, 1, 1, 7, 16, 15, 8, 4, 2, 1, 1, 8, 22, 26, 16, 8, 4, 2, 1, 1, 9, 29, 42, 31, 16, 8, 4, 2, 1, 1, 10, 37, 64, 57, 32, 16, 8, 4, 2, 1, 1, 11, 46, 93, 99, 63, 32, 16, 8, 4, 2, 1
Offset: 0

Views

Author

N. J. A. Sloane, Mar 17 2000

Keywords

Comments

Also square array T(n,k) (n >= 0, k >= 0) read by antidiagonals: T(n,k) = Sum_{i=0..k} binomial(n,i).
As a number triangle read by rows, this is T(n,k) = Sum_{i=n-2*k..n-k} binomial(n-k,i), with T(n,k) = T(n-1,k) + T(n-2,k-1). Row sums are A000071(n+2). Diagonal sums are A023435(n+1). It is the reverse of the Whitney triangle A004070. - Paul Barry, Sep 04 2005
Also, twice number of orthants intersected by a generic k-dimensional subspace of R^n [Naiman and Scheinerman, 2017]. - N. J. A. Sloane, Mar 03 2018

Examples

			Triangle begins:
[0] 1;
[1] 1, 1;
[2] 1, 2,  1;
[3] 1, 3,  2,  1;
[4] 1, 4,  4,  2,  1;
[5] 1, 5,  7,  4,  2,  1;
[6] 1, 6, 11,  8,  4,  2, 1;
[7] 1, 7, 16, 15,  8,  4, 2, 1;
[8] 1, 8, 22, 26, 16,  8, 4, 2, 1;
[9] 1, 9, 29, 42, 31, 16, 8, 4, 2, 1;
As a square array, this begins:
  1  1  1  1  1  1 ...
  1  2  2  2  2  2 ...
  1  3  4  4  4  4 ...
  1  4  7  8  8  8 ...
  1  5 11 15 16 ...
  1  6 16 26 31 32 ...

Links

Crossrefs

Cf. A054123, A054124, A007318, A008949.
Row sums A000071; Diagonal sums A023435; Mirror A004070.
Columns give A000027, A000124, A000125, A000127, A006261, ...
Cf. A052509, A054123, A054124, A007318, A008949, A052553.
Partial sums across rows of (extended) Pascal's triangle A052553.

Programs

  • GAP
    A052509:=Flat(List([0..100],n->List([0..n],k->Sum([0..n],m->Binomial(n-k,k-m))))); # Muniru A Asiru, Sat Feb 17 2018
  • Haskell
    a052509 n k = a052509_tabl !! n !! k
a052509_row n = a052509_tabl !! n
a052509_tabl = [1] : [1,1] : f [1] [1,1] where
   f row' row = rs : f row rs where
     rs = zipWith (+) ([0] ++ row' ++ [1]) (row ++ [0])
-- Reinhard Zumkeller, Nov 22 2012
  • Magma
    [[(&+[Binomial(n-k, k-j): j in [0..n]]): k in [0..n]]: n in [0..10]]; // G. C. Greubel, May 13 2019
  • Maple
    a := proc(n::nonnegint, k::nonnegint) option remember: if k=0 then RETURN(1) fi: if k=n then RETURN(1) fi: a(n-1,k)+a(n-2,k-1) end: for n from 0 to 11 do for k from 0 to n do printf(`%d,`,a(n,k)) od: od: # James Sellers, Mar 17 2000
with(combinat): for s from 0 to 11 do for n from s to 0 by -1 do if n=0 or s-n=0 then printf(`%d,`,1) else printf(`%d,`,sum(binomial(n, i), i=0..s-n)) fi; od: od: # James Sellers, Mar 17 2000
  • Mathematica
    Table[Sum[Binomial[n-k, k-m], {m, 0, n}], {n, 0, 10}, {k, 0, n}]
T[n_, k_] := Hypergeometric2F1[-k, -n + k, -k, -1];
Table[T[n, k], {n, 0, 10}, {k, 0, n}] // Flatten (* Peter Luschny, Nov 28 2021 *)
  • PARI
    T(n,k)=sum(m=0,n,binomial(n-k,k-m));
for(n=0,10,for(k=0,n,print1(T(n,k),", "););print();); /* show triangle */
  • Sage
    [[sum(binomial(n-k, k-j) for j in (0..n)) for k in (0..n)] for n in (0..10)] # G. C. Greubel, May 13 2019

Formula

T(n, k) = Sum_{m=0..n} binomial(n-k, k-m). - Wouter Meeussen, Oct 03 2002
From Werner Schulte, Feb 15 2018: (Start)
Referring to the square array T(i,j):
G.f. of row n: Sum_{k>=0} T(n,k) * x^k = (1+x)^n / (1-x).
G.f. of T(i,j): Sum_{k>=0, n>=0} T(n,k) * x^k * y^n = 1 / ((1-x)*(1-y-x*y)).
Let a_i(n) be multiplicative with a_i(p^e) = T(i, e), p prime and e >= 0, then Sum_{n>0} a_i(n)/n^s = (zeta(s))^(i+1) / (zeta(2*s))^i for i >= 0.
(End)
T(n, k) = hypergeom([-k, -n + k], [-k], -1). - Peter Luschny, Nov 28 2021
From Jianing Song, May 30 2022: (Start)
Referring to the triangle, G.f.: Sum_{n>=0, 0<=k<=n} T(n,k) * x^n * y^k = 1 / ((1-x*y)*(1-x-x^2*y)).
T(n,k) = 2^(n-k) for ceiling(n/2) <= k <= n. (End)

Extensions

More terms from James Sellers, Mar 17 2000
Entry formed by merging two earlier entries. - N. J. A. Sloane, Jun 17 2007
Edited by Johannes W. Meijer, Jul 24 2011

A108561 Triangle read by rows: T(n,0)=1, T(n,n)=(-1)^n, T(n+1,k)=T(n,k-1)+T(n,k) for 0 < k < n.

Original entry on oeis.org

1, 1, -1, 1, 0, 1, 1, 1, 1, -1, 1, 2, 2, 0, 1, 1, 3, 4, 2, 1, -1, 1, 4, 7, 6, 3, 0, 1, 1, 5, 11, 13, 9, 3, 1, -1, 1, 6, 16, 24, 22, 12, 4, 0, 1, 1, 7, 22, 40, 46, 34, 16, 4, 1, -1, 1, 8, 29, 62, 86, 80, 50, 20, 5, 0, 1, 1, 9, 37, 91, 148, 166, 130, 70, 25, 5, 1, -1, 1, 10, 46, 128, 239, 314, 296, 200, 95, 30, 6, 0, 1, 1, 11, 56, 174, 367
Offset: 0

Views

Author

Reinhard Zumkeller, Jun 10 2005

Keywords

Comments

Sum_{k=0..n} T(n,k) = A078008(n);
Sum_{k=0..n} abs(T(n,k)) = A052953(n-1) for n > 0;
T(n,1) = n - 2 for n > 1;
T(n,2) = A000124(n-3) for n > 2;
T(n,3) = A003600(n-4) for n > 4;
T(n,n-6) = A001753(n-6) for n > 6;
T(n,n-5) = A001752(n-5) for n > 5;
T(n,n-4) = A002623(n-4) for n > 4;
T(n,n-3) = A002620(n-1) for n > 3;
T(n,n-2) = A008619(n-2) for n > 2;
T(n,n-1) = n mod 2 for n > 0;
T(2*n,n) = A072547(n+1).
Sum_{k=0..n} T(n,k)*x^k = A232015(n), A078008(n), A000012(n), A040000(n), A001045(n+2), A140725(n+1) for x = 2, 1, 0, -1, -2, -3 respectively. - Philippe Deléham, Nov 17 2013, Nov 19 2013
(1,a^n) Pascal triangle with a = -1. - Philippe Deléham, Dec 27 2013
T(n,k) = A112465(n,n-k). - Reinhard Zumkeller, Jan 03 2014

Examples

			From _Philippe Deléham_, Nov 17 2013: (Start)
Triangle begins:
  1;
  1, -1;
  1,  0,  1;
  1,  1,  1, -1;
  1,  2,  2,  0,  1;
  1,  3,  4,  2,  1, -1;
  1,  4,  7,  6,  3,  0,  1; (End)

Links

Crossrefs

Cf. A007318 (a=1), A008949(a=2), A164844(a=10).
Similar to the triangles A035317, A059259, A080242, A112555.
Cf. A228196
Cf. A072547 (central terms).

Programs

  • GAP
    Flat(List([0..13],n->List([0..n],k->Sum([0..k],i->Binomial(n,i)*(-2)^(k-i))))); # Muniru A Asiru, Feb 19 2018
  • Haskell
    a108561 n k = a108561_tabl !! n !! k
a108561_row n = a108561_tabl !! n
a108561_tabl = map reverse a112465_tabl
-- Reinhard Zumkeller, Jan 03 2014
  • Maple
    A108561 := (n, k) -> add(binomial(n, i)*(-2)^(k-i), i = 0..k):
seq(seq(A108561(n,k), k = 0..n), n = 0..12); # Peter Bala, Feb 18 2018
  • Mathematica
    Clear[t]; t[n_, 0] = 1; t[n_, n_] := t[n, n] = (-1)^Mod[n, 2]; t[n_, k_] := t[n, k] = t[n-1, k] + t[n-1, k-1]; Table[t[n, k], {n, 0, 13}, {k, 0, n}] // Flatten (* Jean-François Alcover, Mar 06 2013 *)
  • Sage
    def A108561_row(n):
    @cached_function
    def prec(n, k):
        if k==n: return 1
        if k==0: return 0
        return -prec(n-1,k-1)-sum(prec(n,k+i-1) for i in (2..n-k+1))
    return [(-1)^k*prec(n, k) for k in (1..n-1)]+[(-1)^(n+1)]
for n in (1..12): print(A108561_row(n)) # Peter Luschny, Mar 16 2016

Formula

G.f.: (1-y*x)/(1-x-(y+y^2)*x). - Philippe Deléham, Nov 17 2013
T(n,k) = T(n-1,k) + T(n-2,k-1) + T(n-2,k-2), T(0,0)=T(1,0)=1, T(1,1)=-1, T(n,k)=0 if k < 0 or if k > n. - Philippe Deléham, Nov 17 2013
From Peter Bala, Feb 18 2018: (Start)
T(n,k) = Sum_{i = 0..k} binomial(n,i)*(-2)^(k-i), 0 <= k <= n.
The n-th row polynomial is the n-th degree Taylor polynomial of the rational function (1 + x)^n/(1 + 2*x) about 0. For example, for n = 4, (1 + x)^4/(1 + 2*x) = 1 + 2*x + 2*x^2 + x^4 + O(x^5). (End)

Extensions

Definition corrected by Philippe Deléham, Dec 26 2013

A110555 Triangle of partial sums of alternating binomial coefficients: T(n, k) = Sum_{j=0..k} binomial(n, j)*(-1)^j, for n >= 0, 0 <= k <= n.

Original entry on oeis.org

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

Views

Author

Reinhard Zumkeller, Jul 27 2005

Keywords

Examples

			Triangle T(n, k) starts:
  [0] 1;
  [1] 1,  0;
  [2] 1, -1,  0;
  [3] 1, -2,  1,   0;
  [4] 1, -3,  3,  -1,  0;
  [5] 1, -4,  6,  -4,  1,   0;
  [6] 1, -5, 10, -10,  5,  -1,  0;
  [7] 1, -6, 15, -20, 15,  -6,  1,  0;
  [8] 1, -7, 21, -35, 35, -21,  7, -1,  0.

Links

Crossrefs

T(n,1) = -n + 1 for n>0;
T(n,2) = A000217(n-2) for n > 1;
T(n,3) = -A000292(n-4) for n > 2;
T(n,4) = A000332(n-1) for n > 3;
T(n,5) = -A000389(n-1) for n > 5;
T(n,6) = A000579(n-1) for n > 6;
T(n,7) = -A000580(n-1) for n > 7;
T(n,8) = A000581(n-1) for n > 8;
T(n,9) = -A000582(n-1) for n > 9;
T(n,10) = A001287(n-1) for n > 10;
T(n,11) = -A001288(n-1) for n > 11;
T(n,12) = A010965(n-1) for n > 12;
T(n,13) = -A010966(n-1) for n > 13;
T(n,14) = A010967(n-1) for n > 14;
T(n,15) = -A010968(n-1) for n > 15;
T(n,16) = A010969(n-1) for n > 16.
Cf. A071919 (variant), A000007 (row sums), A110556 (central terms).
Cf. A008949, A007318.

Programs

  • Maple
    T := (n, k) -> (-1)^k * binomial(n-1, k):
seq(print(seq(T(n, k), k = 0..n)), n = 0..7); # Peter Luschny, Apr 13 2023
  • Mathematica
    T[0, 0] := 1;  T[n_, n_] := 0; T[n_, k_] := (-1)^k*Binomial[n - 1, k]; Table[T[n, k], {n, 0, 20}, {k, 0, n}] // Flatten (* G. C. Greubel, Aug 31 2017 *)
  • PARI
    concat(1, for(n=1,10, for(k=0,n, print1(if(k != n, (-1)^k*binomial(n-1,k), 0), ", ")))) \\ G. C. Greubel, Aug 31 2017

Formula

T(n, 0) = 1, T(n, n) = 0^n, T(n, k) = -T(n-1, k-1) + T(n-1, k), for 0 < k < n.
T(n, k) = binomial(n-1, k)*(-1)^k, 0 <= k < n, T(n, n) = 0^n.
T(n, n-k-1) = -T(n, k), for 0 < k < n.
T(n, k) = A071919(n, k)*(-1)^k and A071919(n, k) = abs(T(n, k)).
Triangle T(n,k), 0 <= k <= n, read by rows, given by [1, 0, 0, 0, 0, 0, 0, 0, ...] DELTA [0, -1, 0, 0, 0, 0, 0, 0, ...] where DELTA is the operator defined in A084938. - Philippe Deléham, Sep 05 2005
G.f.: (1 + x*y) / (1 + x*y - x). - R. J. Mathar, Aug 11 2015

Extensions

Typo in name corrected by Andrey Zabolotskiy, Feb 22 2022
Offset corrected by Peter Luschny, Apr 13 2023
Previous Showing 11-20 of 45 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.