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.

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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

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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

A192744 Constant term of the reduction by x^2->x+1 of the polynomial p(n,x) defined below in Comments.

Original entry on oeis.org

1, 1, 3, 8, 29, 133, 762, 5215, 41257, 369032, 3676209, 40333241, 483094250, 6271446691, 87705811341, 1314473334832, 21017294666173, 357096406209005, 6424799978507178, 122024623087820183, 2439706330834135361, 51219771117454755544
Offset: 0

Views

Author

Clark Kimberling, Jul 09 2011

Keywords

Comments

The titular polynomial is defined recursively by p(n,x)=x*p(n-1,x)+n! for n>0, where p(0,x)=1; see the Example. For an introduction to polynomial reduction, see A192232. The discussion at A192232 Comments continues here:
...
Let R(p,q,s) denote the "reduction of polynomial p by q->s" as defined at A192232. Suppose that q(x)=x^k for some k>0 and that s(x)=s(k,0)*x^(k-1)+s(k,1)*x^(k-2)+...+s(k,k-2)x+s(k,k-1).
...
First, we shall take p(x)=x^n, where n>=0; the results will be used to formulate R(p,q,s) for general p. Represent R(x^n,q,s) by
...
R(x^n)=s(n,0)*x^(k-1)+s(n,1)*x^(k-2)+...+s(n,k-2)*x+s(n,k-1).
...
Then each of the sequences u(n)=s(n,h), for h=0,1,...,k-1, satisfies this linear recurrence relation:
...
u(n)=s(k,0)*u(n-1)+s(k,1)*u(n-2)+...+s(k,k-2)*u(n-k-1)+s(k,k-1)*u(n-k), with initial values tabulated here:
...
n: ..s(n,0)...s(n,1)..s(n,2).......s(n,k-2)..s(n,k-1)
0: ....0........0.......0..............0.......1
1: ....0........0.......0..............1.......0
...
k-2: ..0........1.......0..............0.......0
k-1: ..0........0.......0..............0.......0
k: ..s(k,0)...s(k,1)..s(k,2).......s(k,k-2)..s(k,k-1)
...
That completes the formulation for p(x)=x^n. Turning to the general case, suppose that
...
p(n,x)=p(n,0)*x^n+p(n,1)*x^(n-1)+...+p(n,n-1)*x+p(n,n)
...
is a polynomial of degree n>=0. Then the reduction denoted by (R(p(n,x) by x^k -> s(x)) is the polynomial of degree k-1 given by the matrix product P*S*X, where P=(p(n,0)...p(n,1).........p(n-k)...p(n,n-k+1); X has all 0's except for main diagonal (x^(k-1), x^(k-2)...x,1); and S has
...
row 1: ... s(n,0) ... s(n,1) ...... s(n,k-2) . s(n,k-1)
row 2: ... s(n-1,0) . s(n-1,1) .... s(n-1,k-2) s(n-1,k-1)
...
row n-k+1: s(k,0).... s(k,1) ...... s(k,k-2) ..s(k,k-1)
row n-k+2: p(n,n-k+1) p(n,n-k+2) .. p(n,n-1) ..p(n,n)
*****
As a class of examples, suppose that (v(n)), for n>=0, is a sequence, that p(0,x)=1, and p(n,x)=v(n)+p(n-1,x) for n>0. If q(x)=x^2 and s(x)=x+1, and we write the reduction R(p(n,x)) as u1(n)*x+u2(n), then the sequences u1 and u2 are convolutions with the Fibonacci sequence, viz., let F=(0,1,1,2,3,5,8,...)=A000045 and let G=(1,0,1,1,2,3,5,8...); then u1=G**v and u2=F**v, where ** denotes convolution. Examples (with a few exceptions for initial terms):
...
If v(n)=n! then u1=A192744, u2=A192745.
If v(n)=n+1 then u1=A000071, u2=A001924.
If v(n)=2n then u1=A014739, u2=A027181.
If v(n)=2n+1 then u1=A001911, u2=A001891.
If v(n)=3n+1 then u1=A027961, u2=A023537.
If v(n)=3n+2 then u1=A192746, u2=A192747.
If v(n)=3n then u1=A154691, u2=A192748.
If v(n)=4n+1 then u1=A053311, u2=A192749.
If v(n)=4n+2 then u1=A192750, u2=A192751.
If v(n)=4n+3 then u1=A192752, u2=A192753.
If v(n)=4n then u1=A147728, u2=A023654.
If v(n)=5n+1 then u1=A192754, u2=A192755.
If v(n)=5n then u1=A166863, u2=A192756.
If v(n)=floor((n+1)tau) then u1=A192457, u2=A023611.
If v(n)=floor((n+2)/2) then u1=A052952, u2=A129696.
If v(n)=floor((n+3)/3) then u1=A004695, u2=A178982.
If v(n)=floor((n+4)/4) then u1=A080239, u2=A192758.
If v(n)=floor((n+5)/5) then u1=A124502, u2=A192759.
If v(n)=n+2 then u1=A001594, u2=A192760.
If v(n)=n+3 then u1=A022318, u2=A192761.
If v(n)=n+4 then u1=A022319, u2=A192762.
If v(n)=2^n then u1=A027934, u2=A008766.
If v(n)=3^n then u1=A106517, u2=A094688.

Examples

			The first five polynomials and their reductions:
1 -> 1
1+x -> 1+x
2+x+x^2 -> 3+2x
6+2x+x^2+x^3 -> 8+5x
24+6x+2x^2+x^3+x^4 -> 29+13x, so that
A192744=(1,1,3,8,29,...) and A192745=(0,1,2,5,13,...).

Crossrefs

Cf. A192232.

Programs

  • Maple
    A192744p := proc(n,x)
    option remember;
    if n = 0 then
        1;
    else
        x*procname(n-1,x)+n! ;
        expand(%) ;
    end if;
end proc:
A192744 := proc(n)
    local p;
    p := A192744p(n,x) ;
    while degree(p,x) > 1 do
        p := algsubs(x^2=x+1,p) ;
        p := expand(p) ;
    end do:
    coeftayl(p,x=0,0) ;
end proc: # R. J. Mathar, Dec 16 2015
  • Mathematica
    q = x^2; s = x + 1; z = 40;
p[0, n_] := 1; p[n_, x_] := x*p[n - 1, x] + n!;
Table[Expand[p[n, x]], {n, 0, 7}]
reduce[{p1_, q_, s_, x_}] :=
FixedPoint[(s PolynomialQuotient @@ #1 +
       PolynomialRemainder @@ #1 &)[{#1, q, x}] &, p1]
t = Table[reduce[{p[n, x], q, s, x}], {n, 0, z}];
u1 = Table[Coefficient[Part[t, n], x, 0], {n, 1, z}]
  (* A192744 *)
u2 = Table[Coefficient[Part[t, n], x, 1], {n, 1, z}]
  (* A192745 *)

Formula

G.f.: (1-x)/(1-x-x^2)/Q(0), where Q(k)= 1 - x*(k+1)/(1 - x*(k+1)/Q(k+1)); (continued fraction). - Sergei N. Gladkovskii, May 20 2013
Conjecture: a(n) +(-n-2)*a(n-1) +2*(n-1)*a(n-2) +3*a(n-3) +(-n+2)*a(n-4)=0. - R. J. Mathar, May 04 2014
Conjecture: (-n+2)*a(n) +(n^2-n-1)*a(n-1) +(-n^2+3*n-3)*a(n-2) -(n-1)^2*a(n-3)
=0. - R. J. Mathar, Dec 16 2015

A008949 Triangle read by rows of partial sums of binomial coefficients: T(n,k) = Sum_{i=0..k} binomial(n,i) (0 <= k <= n); also dimensions of Reed-Muller codes.

Original entry on oeis.org

1, 1, 2, 1, 3, 4, 1, 4, 7, 8, 1, 5, 11, 15, 16, 1, 6, 16, 26, 31, 32, 1, 7, 22, 42, 57, 63, 64, 1, 8, 29, 64, 99, 120, 127, 128, 1, 9, 37, 93, 163, 219, 247, 255, 256, 1, 10, 46, 130, 256, 382, 466, 502, 511, 512, 1, 11, 56, 176, 386, 638, 848, 968, 1013, 1023, 1024, 1, 12, 67, 232, 562, 1024, 1486, 1816, 1981, 2036, 2047, 2048
Offset: 0

Views

Author

N. J. A. Sloane

Keywords

Comments

The second-left-from-middle column is A000346: T(2n+2, n) = A000346(n). - Ed Catmur (ed(AT)catmur.co.uk), Dec 09 2006
T(n,k) is the maximal number of regions into which n hyperplanes of co-dimension 1 divide R^k (the Cake-Without-Icing numbers). - Rob Johnson, Jul 27 2008
T(n,k) gives the number of vertices within distance k (measured along the edges) of an n-dimensional unit cube, (i.e., the number of vertices on the hypercube graph Q_n whose distance from a reference vertex is <= k). - Robert Munafo, Oct 26 2010
A triangle formed like Pascal's triangle, but with 2^n for n >= 0 on the right border instead of 1. - Boris Putievskiy, Aug 18 2013
For a closed-form formula for generalized Pascal's triangle see A228576. - Boris Putievskiy, Sep 04 2013
Consider each "1" as an apex of two sequences: the first is the set of terms in the same row as the "1", but the rightmost term in the row repeats infinitely. Example: the row (1, 4, 7, 8) becomes (1, 4, 7, 8, 8, 8, ...). The second sequence begins with the same "1" but is the diagonal going down and to the right, thus: (1, 5, 16, 42, 99, 219, 466, ...). It appears that for all such sequence pairs, the binomial transform of the first, (1, 4, 7, 8, 8, 8, ...) in this case; is equal to the second: (1, 5, 16, 42, 99, ...). - Gary W. Adamson, Aug 19 2015
Let T* be the infinite tree with root 0 generated by these rules: if p is in T*, then p+1 is in T* and x*p is in T*. Let q(n) be the sum of polynomials in the n-th generation of T*. For n >= 0, row n of A008949 gives the coefficients of q(n+1); e.g., (row 3) = (1, 4, 7, 8) matches x^3 + 4*x^2 + 7*x + 9, which is the sum of the 8 polynomials in the 4th generation of T*. - Clark Kimberling, Jun 16 2016
T(n,k) is the number of subsets of [n]={1,...,n} of at most size k. Equivalently, T(n,k) is the number of subsets of [n] of at least size n-k. Counting the subsets of at least size (n-k) by conditioning on the largest element m of the smallest (n-k) elements of such a subset provides the formula T(n,k) = Sum_{m=n-k..n} C(m-1,n-k-1)*2^(n-m), and, by letting j=m-n+k, we obtain T(n,k) = Sum_{j=0..k} C(n+j-k-1,j)*2^(k-j). - Dennis P. Walsh, Sep 25 2017
If the interval of integers 1..n is shifted up or down by k, making the new interval 1+k..n+k or 1-k..n-k, then T(n-1,n-1-k) (= 2^(n-1)-T(n-1,k-1)) is the number of subsets of the new interval that contain their own cardinal number as an element. - David Pasino, Nov 01 2018

Examples

			Triangle begins:
  1;
  1,  2;
  1,  3,  4;
  1,  4,  7,   8;
  1,  5, 11,  15,  16;
  1,  6, 16,  26,  31,  32;
  1,  7, 22,  42,  57,  63,  64;
  1,  8, 29,  64,  99, 120, 127, 128;
  1,  9, 37,  93, 163, 219, 247, 255,  256;
  1, 10, 46, 130, 256, 382, 466, 502,  511,  512;
  1, 11, 56, 176, 386, 638, 848, 968, 1013, 1023, 1024;
  ...

References

  • F. J. MacWilliams and N. J. A. Sloane, The Theory of Error-Correcting Codes, Elsevier-North Holland, 1978, p. 376.

Links

Crossrefs

Diagonals are given by A000079, A000225, A000295, A002662, A002663, A002664, A035038-A035042.
Columns are given by A000012, A000027, A000124, A000125, A000127, A006261, A008859, A008860, A008861, A008862, A008863. - Ken Shirriff, Jun 28 2011
Row sums sequence is A001792.
T(n, m)= A055248(n, n-m).
Cf. A110555, A007318, A000346, A171886, A228196, A228576.
Cf. also A027306, A249111, A163866, A261363.
Cf. A000071, A001924

Programs

  • GAP
    T:=Flat(List([0..11],n->List([0..n],k->Sum([0..k],j->Binomial(n+j-k-1,j)*2^(k-j))))); # Muniru A Asiru, Nov 25 2018
  • Haskell
    a008949 n k = a008949_tabl !! n !! k
a008949_row n = a008949_tabl !! n
a008949_tabl = map (scanl1 (+)) a007318_tabl
-- Reinhard Zumkeller, Nov 23 2012
  • Magma
    [[(&+[Binomial(n,j): j in [0..k]]): k in [0..n]]: n in [0..12]]; // G. C. Greubel, Nov 25 2018
  • Maple
    A008949 := proc(n,k) local i; add(binomial(n,i),i=0..k) end; # Typo corrected by R. J. Mathar, Oct 26 2010
  • Mathematica
    Table[Length[Select[Subsets[n], (Length[ # ] <= k) &]], {n, 0, 12}, {k, 0, n}] // Grid (* Geoffrey Critzer, May 13 2009 *)
Flatten[Accumulate/@Table[Binomial[n,i],{n,0,20},{i,0,n}]] (* Harvey P. Dale, Aug 08 2015 *)
T[ n_, k_] := If[ n < 0 || k > n, 0, Binomial[n, k] Hypergeometric2F1[1, -k, n + 1 - k, -1]]; (* Michael Somos, Aug 05 2017 *)
  • PARI
    A008949(n)=T8949(t=sqrtint(2*n-sqrtint(2*n)),n-t*(t+1)/2)
T8949(r,c)={ 2*c > r || return(sum(i=0,c,binomial(r,i))); 1<M. F. Hasler, May 30 2010
  • PARI
    {T(n, k) = if(k>n, 0, sum(i=0, k, binomial(n, i)))}; /* Michael Somos, Aug 05 2017 */
  • PARI
    row(n) = my(v=vector(n+1, k, binomial(n,k-1))); vector(#v, k, sum(i=1, k, v[i])); \\ Michel Marcus, Apr 13 2025
  • Sage
    [[sum(binomial(n,j) for j in range(k+1)) for k in range(n+1)] for n in range(12)] # G. C. Greubel, Nov 25 2018

Formula

From partial sums across rows of Pascal triangle A007318.
T(n, 0) = 1, T(n, n) = 2^n, T(n, k) = T(n-1, k-1) + T(n-1, k), 0 < k < n.
G.f.: (1 - x*y)/((1 - y - x*y)*(1 - 2*x*y)). - Antonio Gonzalez (gonfer00(AT)gmail.com), Sep 08 2009
T(2n,n) = A032443(n). - Philippe Deléham, Sep 16 2009
T(n,k) = 2 T(n-1,k-1) + binomial(n-1,k) = 2 T(n-1,k) - binomial(n-1,k). - M. F. Hasler, May 30 2010
T(n,k) = binomial(n,n-k)* 2F1(1, -k; n+1-k; -1). - Olivier Gérard, Aug 02 2012
For a closed-form formula for arbitrary left and right borders of Pascal like triangle see A228196. - Boris Putievskiy, Aug 18 2013
T(n,floor(n/2)) = A027306(n). - Reinhard Zumkeller, Nov 14 2014
T(n,n) = 2^n, otherwise for 0 <= k <= n-1, T(n,k) = 2^n - T(n,n-k-1). - Bob Selcoe, Mar 30 2017
For fixed j >= 0, lim_{n -> oo} T(n+1,n-j+1)/T(n,n-j) = 2. - Bob Selcoe, Apr 03 2017
T(n,k) = Sum_{j=0..k} C(n+j-k-1,j)*2^(k-j). - Dennis P. Walsh, Sep 25 2017

Extensions

More terms from Larry Reeves (larryr(AT)acm.org), Mar 23 2000

A101220 a(n) = Sum_{k=0..n} Fibonacci(n-k)*n^k.

Original entry on oeis.org

0, 1, 3, 14, 91, 820, 9650, 140601, 2440317, 49109632, 1123595495, 28792920872, 816742025772, 25402428294801, 859492240650847, 31427791175659690, 1234928473553777403, 51893300561135516404, 2322083099525697299278
Offset: 0

Views

Author

Ross La Haye, Dec 14 2004

Keywords

Comments

In what follows a(i,j,k) denotes a three-dimensional array, the terms a(n) are defined as a(n,n,n) in that array. - Joerg Arndt, Jan 03 2021
Previous name was: Three-dimensional array: a(i,j,k) = expansion of x*(1 + (i-j)*x)/((1-j*x)*(1-x-x^2)), read by a(n,n,n).
a(i,j,k) = the k-th value of the convolution of the Fibonacci numbers (A000045) with the powers of i = Sum_{m=0..k} a(i-1,j,m), both for i = j and i > 0; a(i,j,k) = a(i-1,j,k) + a(j,j,k-1), for i,k > 0; a(i,1,k) = Sum_{m=0..k} a(i-1,0,m), for i > 0. With F = Fibonacci and L = Lucas, then a(1,1,k) = F(k+2) - 1; a(2,1,k) = F(k+3) - 2; a(3,1,k) = L(k+2) - 3; a(4,1,k) = 4*F(k+1) + F(k) - 4; a(1,2,k) = 2^k - F(k+1); a(2,2,k) = 2^(k+1) - F(k+3); a(3,2,k) = 3(2^k - F(k+2)) + F(k); a(4,2,k) = 2^(k+2) - F(k+4) - F(k+2); a(1,3,k) = (3^k + L(k-1))/5, for k > 0; a(2,3,k) = (2 * 3^k - L(k)) /5, for k > 0; a(3,3,k) = (3^(k+1) - L(k+2))/5; a(4,3,k) = (4 * 3^k - L(k+2) - L(k+1))/5, etc..

Examples

			a(1,3,3) = 6 because a(1,3,0) = 0, a(1,3,1) = 1, a(1,3,2) = 2 and 4*2 - 2*1 - 3*0 = 6.

Links

Crossrefs

a(0, j, k) = A000045(k).
a(1, 2, k+1) - a(1, 2, k) = A099036(k).
a(3, 2, k+1) - a(3, 2, k) = A104004(k).
a(4, 2, k+1) - a(4, 2, k) = A027973(k).
a(1, 3, k+1) - a(1, 3, k) = A099159(k).
a(i, 0, k) = A109754(i, k).
a(i, i+1, 3) = A002522(i+1).
a(i, i+1, 4) = A071568(i+1).
a(2^i-2, 0, k+1) = A118654(i, k), for i > 0.
Sequences of the form a(n, 0, k): A000045(k+1) (n=1), A000032(k) (n=2), A000285(k-1) (n=3), A022095(k-1) (n=4), A022096(k-1) (n=5), A022097(k-1) (n=6), A022098(k-1) (n=7), A022099(k-1) (n=8), A022100(k-1) (n=9), A022101(k-1) (n=10), A022102(k-1) (n=11), A022103(k-1) (n=12), A022104(k-1) (n=13), A022105(k-1) (n=14), A022106(k-1) (n=15), A022107(k-1) (n=16), A022108(k-1) (n=17), A022109(k-1) (n=18), A022110(k-1) (n=19), A088209(k-2) (n=k-2), A007502(k) (n=k-1), A094588(k) (n=k).
Sequences of the form a(1, n, k): A000071(k+2) (n=1), A027934(k-1) (n=2), A098703(k) (n=3).
Sequences of the form a(2, n, k): A001911(k) (n=1), A008466(k+1) (n=2), A106517(k-1) (n=3).
Sequences of the form a(3, n, k): A027961(k) (n=1), A094688(k) (n=3).
Sequences of the form a(4, n, k): A053311(k-1) (n=1), A027974(k-1) (n=2).
Cf. A001622, A001891, A001924, A023537, A104161, A106517.

Programs

  • Magma
    A101220:= func< n | (&+[n^k*Fibonacci(n-k): k in [0..n]]) >;
[A101220(n): n in [0..30]]; // G. C. Greubel, Jun 01 2025
  • Mathematica
    Join[{0}, Table[Sum[Fibonacci[n-k]*n^k, {k, 0, n}], {n, 1, 20}]] (* Vaclav Kotesovec, Jan 03 2021 *)
  • PARI
    a(n)=sum(k=0,n,fibonacci(n-k)*n^k) \\ Joerg Arndt, Jan 03 2021
  • SageMath
    def A101220(n): return sum(n^k*fibonacci(n-k) for k in range(n+1))
print([A101220(n) for n in range(31)]) # G. C. Greubel, Jun 01 2025

Formula

a(i, j, 0) = 0, a(i, j, 1) = 1, a(i, j, 2) = i+1; a(i, j, k) = ((j+1)*a(i, j, k-1)) - ((j-1)*a(i, j, k-2)) - (j*a(i, j, k-3)), for k > 2.
a(i, j, k) = Fibonacci(k) + i*a(j, j, k-1), for i, k > 0.
a(i, j, k) = (Phi^k - (-Phi)^-k + i(((j^k - Phi^k) / (j - Phi)) - ((j^k - (-Phi)^-k) / (j - (-Phi)^-1)))) / sqrt(5), where Phi denotes the golden mean/ratio (A001622).
i^k = a(i-1, i, k) + a(i-2, i, k+1).
A104161(k) = Sum_{m=0..k} a(k-m, 0, m).
a(i, j, 0) = 0, a(i, j, 1) = 1, a(i, j, 2) = i+1, a(i, j, 3) = i*(j+1) + 2; a(i, j, k) = (j+2)*a(i, j, k-1) - 2*j*a(i, j, k-2) - a(i, j, k-3) + j*a(i, j, k-4), for k > 3. a(i, j, 0) = 0, a(i, j, 1) = 1; a(i, j, k) = a(i, j, k-1) + a(i, j, k-2) + i * j^(k-2), for k > 1.
G.f.: x*(1 + (i-j)*x)/((1-j*x)*(1-x-x^2)).
a(n, n, n) = Sum_{k=0..n} Fibonacci(n-k) * n^k. - Ross La Haye, Jan 14 2006
Sum_{m=0..k} binomial(k,m)*(i-1)^m = a(i-1,i,k) + a(i-2,i,k+1), for i > 1. - Ross La Haye, May 29 2006
From Ross La Haye, Jun 03 2006: (Start)
a(3, 3, k+1) - a(3, 3, k) = A106517(k).
a(1, 1, k) = A001924(k) - A001924(k-1), for k > 0.
a(2, 1, k) = A001891(k) - A001891(k-1), for k > 0.
a(3, 1, k) = A023537(k) - A023537(k-1), for k > 0.
Sum_{j=0..i+1} a(i-j+1, 0, j) - Sum_{j=0..i} a(i-j, 0, j) = A001595(i). (End)
a(i,j,k) = a(j,j,k) + (i-j)*a(j,j,k-1), for k > 0.
a(n) ~ n^(n-1). - Vaclav Kotesovec, Jan 03 2021

Extensions

New name from Joerg Arndt, Jan 03 2021

A027926 Triangular array T read by rows: T(n,0) = T(n,2n) = 1 for n >= 0; T(n,1) = 1 for n >= 1; T(n,k) = T(n-1,k-2) + T(n-1,k-1) for k = 2..2n-1, n >= 2.

Original entry on oeis.org

1, 1, 1, 1, 1, 1, 2, 2, 1, 1, 1, 2, 3, 4, 3, 1, 1, 1, 2, 3, 5, 7, 7, 4, 1, 1, 1, 2, 3, 5, 8, 12, 14, 11, 5, 1, 1, 1, 2, 3, 5, 8, 13, 20, 26, 25, 16, 6, 1, 1, 1, 2, 3, 5, 8, 13, 21, 33, 46, 51, 41, 22, 7, 1, 1, 1, 2, 3, 5, 8, 13, 21, 34, 54, 79, 97, 92, 63, 29, 8, 1
Offset: 0

Views

Author

Clark Kimberling

Keywords

Comments

T(n,k) = number of strings s(0),...,s(n) such that s(0)=0, s(n)=n-k and for 1<=i<=n, s(i)=s(i-1)+d, with d in {0,1,2} if i=0, in {0,2} if s(i)=2i, in {0,1,2} if s(i)=2i-1, in {0,1} if 0<=s(i)<=2i-2.
Can be seen as concatenation of triangles A104763 and A105809, with identifying column of Fibonacci numbers, see example. - Reinhard Zumkeller, Aug 15 2013

Examples

			.   0:                           1
.   1:                        1  1   1
.   2:                     1  1  2   2   1
.   3:                  1  1  2  3   4   3   1
.   4:               1  1  2  3  5   7   7   4   1
.   5:            1  1  2  3  5  8  12  14  11   5   1
.   6:          1 1  2  3  5  8 13  20  26  25  16   6   1
.   7:        1 1 2  3  5  8 13 21  33  46  51  41  22   7   1
.   8:      1 1 2 3  5  8 13 21 34  54  79  97  92  63  29   8  1
.   9:    1 1 2 3 5  8 13 21 34 55  88 133 176 189 155  92  37  9  1
.  10:  1 1 2 3 5 8 13 21 34 55 89 143 221 309 365 344 247 129 46 10  1
.
.   1:                           1
.   2:                        1  1
.   3:                     1  1  2
.   4:                  1  1  2  3
.   5:               1  1  2  3  5      columns = A000045, > 0
.   6:            1  1  2  3  5  8     +---------+
.   7:          1 1  2  3  5  8 13     | A104763 |
.   8:        1 1 2  3  5  8 13 21     +---------+
.   9:      1 1 2 3  5  8 13 21 34
.  10:    1 1 2 3 5  8 13 21 34 55
.  11:  1 1 2 3 5 8 13 21 34 55 89
.
.   0:                           1
.   1:                           1   1                +---------+
.   2:                           2   2   1            | A105809 |
.   3:                           3   4   3   1        +---------+
.   4:                           5   7   7   4   1
.   5:                           8  12  14  11   5   1
.   6:                          13  20  26  25  16   6   1
.   7:                          21  33  46  51  41  22   7   1
.   8:                          34  54  79  97  92  63  29   8  1
.   9:                          55  88 133 176 189 155  92  37  9  1
.  10:                          89 143 221 309 365 344 247 129 46 10  1

Links

Crossrefs

Many columns of T are A000045 (Fibonacci sequence), also in T: A001924, A004006, A000071, A000124, A014162, A014166, A027927-A027933.
Some other Fibonacci-Pascal triangles: A036355, A037027, A074829, A105809, A109906, A111006, A114197, A162741, A228074.

Programs

  • GAP
    Flat(List([0..10], n-> List([0..2*n], k-> Sum([0..Int((2*n-k+1)/2) ], j-> Binomial(n-j, 2*n-k-2*j) )))); # G. C. Greubel, Sep 05 2019
  • Haskell
    a027926 n k = a027926_tabf !! n !! k
a027926_row n = a027926_tabf !! n
a027926_tabf = iterate (\xs -> zipWith (+)
                               ([0] ++ xs ++ [0]) ([1,0] ++ xs)) [1]
-- Variant, cf. example:
a027926_tabf' = zipWith (++) a104763_tabl (map tail a105809_tabl)
-- Reinhard Zumkeller, Aug 15 2013
  • Magma
    [&+[Binomial(n-j, 2*n-k-2*j): j in [0..Floor((2*n-k+1)/2)]]: k in [0..2*n], n in [0..10]]; // G. C. Greubel, Sep 05 2019
  • Maple
    A027926 := proc(n,k)
    add(binomial(n-j,2*n-k-2*j),j=0..(2*n-k+1)/2) ;
end proc: # R. J. Mathar, Apr 11 2016
  • Mathematica
    z = 15; t[n_, 0] := 1; t[n_, k_] := 1 /; k == 2 n; t[n_, 1] := 1;
t[n_, k_] := t[n, k] = t[n - 1, k - 2] + t[n - 1, k - 1];
u = Table[t[n, k], {n, 0, z}, {k, 0, 2 n}];
TableForm[u] (* A027926 array *)
v = Flatten[u] (* A027926 sequence *)
(* Clark Kimberling, Aug 31 2014 *)
Table[Sum[Binomial[n-j, 2*n-k-2*j], {j, 0, Floor[(2*n-k+1)/2]}], {n, 0, 10}, {k, 0, 2*n}]//Flatten (* G. C. Greubel, Sep 05 2019 *)
  • PARI
    {T(n, k) = if( k<0 || k>2*n, 0, if( k<=1 || k==2*n, 1, T(n-1, k-2) + T(n-1, k-1)))}; /* _Michael Somos, Feb 26 1999 */
  • PARI
    {T(n, k) = if( k<0 || k>2*n, 0, sum( j=max(0, k-n), k\2, binomial(k-j, j)))}; /* Michael Somos */
  • Sage
    [[sum(binomial(n-j, 2*n-k-2*j) for j in (0..floor((2*n-k+1)/2))) for k in (0..2*n)] for n in (0..10)] # G. C. Greubel, Sep 05 2019

Formula

T(n, k) = Sum_{j=0..floor((2*n-k+1)/2)} binomial(n-j, 2*n-k-2*j). - Len Smiley, Oct 21 2001

Extensions

Incorporates comments from Michael Somos.
Example extended by Reinhard Zumkeller, Aug 15 2013

A000126 A nonlinear binomial sum.

Original entry on oeis.org

1, 2, 4, 8, 15, 27, 47, 80, 134, 222, 365, 597, 973, 1582, 2568, 4164, 6747, 10927, 17691, 28636, 46346, 75002, 121369, 196393, 317785, 514202, 832012, 1346240, 2178279, 3524547, 5702855, 9227432, 14930318, 24157782, 39088133, 63245949
Offset: 1

Views

Author

N. J. A. Sloane

Keywords

Comments

a(n)-1 counts ternary numbers with no 0 digit (A007931) and at least one 2 digit, where the total of ternary digits is <= n. E.g., a(4)-1 = 7: 2 12 21 22 112 121 211. - Frank Ellermann, Dec 02 2001
A107909(a(n-1)) = A000079(n-1) = 2^(n-1). - Reinhard Zumkeller, May 28 2005
a(n) is the permanent of the n X n 0-1 matrix whose (i,j) entry is 1 iff i=1 or j=n or |i-j|<=1. For example, a(5)=15 is per([[1, 1, 1, 1, 1], [1, 1, 1, 0, 1], [0, 1, 1, 1, 1], [0, 0, 1, 1, 1], [0, 0, 0, 1, 1]]). - David Callan, Jun 07 2006
Conjecture. Let S(1)={1} and, for n>1, let S(n) be the smallest set containing x+1 and 2x+1 for each element x in S(n-1). Then a(n) is the sum of the elements in S(n). (See A122554 for a sequence defined in this way.) - John W. Layman, Nov 21 2007
a(n+1) indexes the corner blocks on the Fibonacci spiral built from blocks of unit area (using F(1) and F(2) as the sides of the first block). - Paul Barry, Mar 06 2008
The number of length n binary words with fewer than 2 0-digits between any pair of consecutive 1-digits. - Jeffrey Liese, Dec 23 2010
If b(n) = a(n+1) then b(0) = 1 and 2*b(n) >= b(n+1) for all n > 1 which is sufficient for b(n) to be a complete sequence. - Frank M Jackson, Mar 17 2013
From Gus Wiseman, Feb 10 2019: (Start)
Also the number of non-singleton subsets of {1, ..., n + 1} with no successive elements. For example, the a(5) = 15 subsets are:
{},
{1,3}, {1,4}, {1,5}, {1,6}, {2,4}, {2,5}, {2,6}, {3,5}, {3,6}, {4,6},
{1,3,5}, {1,3,6}, {1,4,6}, {2,4,6}.
Also the number of binary sequences with all zeros or at least 2 ones and no adjacent ones. For example, the a(1) = 1 through a(4) = 8 sequences are:
(00) (000) (0000) (00000)
(101) (0101) (00101)
(1001) (01001)
(1010) (01010)
(10001)
(10010)
(10100)
(10101)
(End)

References

  • Ralph P. Grimaldi, A generalization of the Fibonacci sequence. Proceedings of the seventeenth Southeastern international conference on combinatorics, graph theory, and computing (Boca Raton, Fla., 1986). Congr. Numer. 54 (1986), 123--128. MR0885268 (89f:11030). - N. J. A. Sloane, Apr 08 2012
  • 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

Heap-transform of A000071. - John W. Layman
Cf. A066067, A001924, A065220.
Cf. A007931: binary strings with leading 0's, or ternary strings without 0's.
Differences are A000071.
Cf. A122554.
Cf. A005251, A066982, A169985.
Cf. A000045.

Programs

  • GAP
    List([1..40], n-> Fibonacci(n+3)-(n+1)); # G. C. Greubel, Jul 09 2019
  • Magma
    [Fibonacci(n+3)-(n+1): n in [1..40]]; // G. C. Greubel, Jul 09 2019
  • Maple
    a:= n-> (Matrix([[1,1,1,2]]). Matrix(4, (i,j)-> if (i=j-1) then 1 elif j=1 then [3,-2,-1,1][i] else 0 fi)^n)[1,2]; seq(a(n), n=1..36); # Alois P. Heinz, Aug 26 2008
# alternative
A000126 := proc(n)
    combinat[fibonacci](n+3)-n-1 ;
end proc:
seq(A000126(n),n=1..40) ; # R. J. Mathar, Aug 05 2022
  • Mathematica
    LinearRecurrence[{3,-2,-1,1},{1,2,4,8},40] (* or *) CoefficientList[ Series[-(1-x+x^3)/((x^2+x-1)(x-1)^2),{x,0,40}],x]  (* Harvey P. Dale, Apr 24 2011 *)
Table[Length[Select[Subsets[Range[n]],Min@@Abs[Subtract@@@Partition[#,2,1,1]]>1&]],{n,15}] (* Gus Wiseman, Feb 10 2019 *)
  • PARI
    Vec((1-x+x^3)/(1-x-x^2)/(1-x)^2+O(x^40)) \\ Charles R Greathouse IV, Oct 06 2011
  • PARI
    vector(40, n, fibonacci(n+3) -(n+1)) \\ G. C. Greubel, Jul 09 2019
  • Python
    def seq(n):
    if n < 0:
        return 1
    a, b = 1, 1
    for i in range(n + 1):
        a, b = b, a + b + i
    return a
[seq(i) for i in range(n)] # Reza K Ghazi, Mar 03 2019
  • Sage
    [fibonacci(n+3)-(n+1) for n in (1..40)] # G. C. Greubel, Jul 09 2019

Formula

G.f.: (1 - x + x^3 ) / (( 1 - x - x^2 )*( 1 - x )^2). - Simon Plouffe in his 1992 dissertation.
From Henry Bottomley, Oct 22 2001: (Start)
a(n) = Fibonacci(n+3) - (n+1) = a(n-1) + a(n-2) + n - 2
a(n) = A001924(n-1) + 1 = A065220(n+3) + 2. (End)
a(n) = 2*a(n-1) - a(n-3) + 1. - Franklin T. Adams-Watters, Jan 13 2006
a(n+1) = 1 + Sum_{k=0..n} (Fibonacci(k+2) - 1) = Sum_{k=0..n} Fibonacci(k+2) - n. - Paul Barry, Mar 06 2008
a(n) = 3*a(n-1)-2*a(n-2)-a(n-3)+a(n-4). - Harvey P. Dale, May 05 2011
Closed-form without extra leading 1: ((15+7*sqrt(5))*((1+sqrt(5))/2)^n+(15-7*sqrt(5))*((1-sqrt(5))/2)^n-10*n-20)/10; closed-form with extra leading 1: ((20+8*sqrt(5))*((1+sqrt(5))/2)^n+(20-8*sqrt(5))*((1-sqrt(5))/2)^n-20*n-20)/20. - Tim Monahan, Jul 16 2011
G.f. for closed-form with extra leading 1: (1-2*x+x^2+x^3)/((1-x-x^2)*(x-1)^2). - Tim Monahan, Jul 17 2011

A228074 A Fibonacci-Pascal triangle read by rows: T(n,0) = Fibonacci(n), T(n,n) = n and for n > 0: T(n,k) = T(n-1,k-1) + T(n-1,k), 0 < k < n.

Original entry on oeis.org

0, 1, 1, 1, 2, 2, 2, 3, 4, 3, 3, 5, 7, 7, 4, 5, 8, 12, 14, 11, 5, 8, 13, 20, 26, 25, 16, 6, 13, 21, 33, 46, 51, 41, 22, 7, 21, 34, 54, 79, 97, 92, 63, 29, 8, 34, 55, 88, 133, 176, 189, 155, 92, 37, 9, 55, 89, 143, 221, 309, 365, 344, 247, 129, 46, 10
Offset: 0

Views

Author

Reinhard Zumkeller, Aug 15 2013

Keywords

Comments

Sum of n-th row is 2^(n+1) - F(n+1) - 1 = A228078(n+1). - Greg Dresden and Sadek Mohammed, Aug 30 2022

Examples

			.    0:                                 0
.    1:                               1   1
.    2:                             1   2   2
.    3:                          2    3    4   3
.    4:                       3    5    7    7   4
.    5:                     5    8   12   14   11   5
.    6:                  8   13   20   26   25   16   6
.    7:               13   21   33   46   51   41   22   7
.    8:            21   34   54   79   97   92   63   29   8
.    9:          34   55   88  133  176  189  155   92   37   9
.   10:       55   89  143  221  309  365  344  247  129   46  10
.   11:     89  144  232  364  530  674  709  591  376  175  56   11
.   12:  144 233  376  596  894 1204 1383 1300  967  551  231  67   12 .

Links

Crossrefs

Cf. A000045 (left edge), A001477 (right edge), A228078 (row sums), A027988 (maxima per row);
diagonals T(*,k): A000045, A000071, A001924, A014162, A014166, A053739, A053295, A053296, A053308, A053309;
diagonals T(k,*): A001477, A001245, A004006, A027927, A027928, A027929, A027930, A027931, A027932, A027933;
some other Fibonacci-Pascal triangles: A027926, A036355, A037027, A074829, A105809, A109906, A111006, A114197, A162741.

Programs

  • GAP
    T:= function(n,k)
    if k=0 then return Fibonacci(n);
    elif k=n then return 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, Sep 05 2019
  • Haskell
    a228074 n k = a228074_tabl !! n !! k
a228074_row n = a228074_tabl !! n
a228074_tabl = map fst $ iterate
   (\(u:_, vs) -> (vs, zipWith (+) ([u] ++ vs) (vs ++ [1]))) ([0], [1,1])
  • Maple
    with(combinat);
T:= proc (n, k) option remember;
if k = 0 then fibonacci(n)
elif k = n then n
else T(n-1, k-1) + T(n-1, k)
end if
end proc;
seq(seq(T(n, k), k = 0..n), n = 0..12); # G. C. Greubel, Sep 05 2019
  • Mathematica
    T[n_, k_]:= T[n, k]= If[k==0, Fibonacci[n], If[k==n, n, T[n-1, k-1] + T[n -1, k]]]; Table[T[n, k], {n,0,12}, {k,0,n}] (* G. C. Greubel, Sep 05 2019 *)
  • PARI
    T(n,k) = if(k==0, fibonacci(n), if(k==n, n, T(n-1, k-1) + T(n-1, k)));
for(n=0, 12, for(k=0, n, print1(T(n,k), ", "))) \\ G. C. Greubel, Sep 05 2019
  • Sage
    def T(n, k):
    if (k==0): return fibonacci(n)
    elif (k==n): return n
    else: return T(n-1, k) + T(n-1, k-1)
[[T(n, k) for k in (0..n)] for n in (0..12)] # G. C. Greubel, Sep 05 2019

A050447 Table T(n,m) giving total degree of n-th-order elementary symmetric polynomials in m variables, -1 <= n, 1 <= m, transposed and read by upward antidiagonals.

Original entry on oeis.org

1, 1, 1, 1, 2, 1, 1, 3, 3, 1, 1, 4, 6, 5, 1, 1, 5, 10, 14, 8, 1, 1, 6, 15, 30, 31, 13, 1, 1, 7, 21, 55, 85, 70, 21, 1, 1, 8, 28, 91, 190, 246, 157, 34, 1, 1, 9, 36, 140, 371, 671, 707, 353, 55, 1, 1, 10, 45, 204, 658, 1547, 2353, 2037, 793, 89, 1, 1, 11, 55, 285, 1086, 3164, 6405, 8272, 5864, 1782, 144, 1
Offset: 0

Views

Author

N. J. A. Sloane, Dec 23 1999

Keywords

Examples

			Table begins
.    1   1   1    1     1      1       1       1        1         1
.    1   2   3    5     8     13      21      34       55        89
.    1   3   6   14    31     70     157     353      793      1782
.    1   4  10   30    85    246     707    2037     5864     16886
.    1   5  15   55   190    671    2353    8272    29056    102091
.    1   6  21   91   371   1547    6405   26585   110254    457379
.    1   7  28  140   658   3164   15106   72302   345775   1654092
.    1   8  36  204  1086   5916   31998  173502   940005   5094220
.    1   9  45  285  1695  10317   62349  377739  2286648  13846117
.    1  10  55  385  2530  17017  113641  760804  5089282  34053437

References

  • J. Berman and P. Koehler, Cardinalities of finite distributive lattices, Mitteilungen aus dem Mathematischen Seminar Giessen, 121 (1976), 103-124.
  • S. J. Cyvin and I. Gutman, Kekulé structures in benzenoid hydrocarbons, Lecture Notes in Chemistry, No. 46, Springer, New York, 1988 (see p. 120).
  • Manfred Goebel, Rewriting Techniques and Degree Bounds for Higher Order Symmetric Polynomials, Applicable Algebra in Engineering, Communication and Computing (AAECC), Volume 9, Issue 6 (1999), 559-573.

Links

Crossrefs

Rows give A000012, A000045, A006356, A006357, A006358, ...
Columns give A000012, A000027, A000217, A000330, A006322, ...
Cf. A001924, A050446.

Programs

  • Mathematica
    nmax = 12; t[n_, m_?Positive] := t[n, m] = t[n, m-1] + Sum[t[2k, m-1]*t[n-1-2k, m], {k, 0, (n-1)/2}]; t[n_, 0]=1; Flatten[ Table[ t[k-1, n-k], {n, 1, nmax}, {k, 1, n}]] (* Jean-François Alcover, Nov 14 2011 *)
nmax = 10; f[0, x_] := 1; f[1, x_] := 1/(1 - x); f[n_, x_] := (x + f[n - 2, x])/(1 - x^2 - x*f[n - 2, x]); t[n_, m_] := Coefficient[Series[f[n, x], {x, 0, m}], x, m]; Grid[Table[t[n, m], {n, nmax}, {m, 0, nmax - 1}]] (* L. Edson Jeffery, Oct 19 2017 *)
  • PARI
    M(n)=matrix(n,n,i,j,if(sign(i+j-n)-1,0,1)); V(n)=vector(n,i,1); P(r,n)=vecmax(V(r)*M(r)^n) \\ P(r,n) is T(n,k); Benoit Cloitre, Jan 27 2003

Formula

See PARI code. See A050446 for recurrence.
G.f. for row n >= 0: f(n, x) = (x + f(n-2, x))/(1 - x^2 - x*f(n-2, x)), where f(0, x) = 1 and f(1, x) = 1/(1 - x) [R. P. Stanley]. - L. Edson Jeffery, Oct 19 2017

Extensions

More terms from Naohiro Nomoto, Jul 03 2001

A104712 Pascal's triangle, with the first two columns removed.

Original entry on oeis.org

1, 3, 1, 6, 4, 1, 10, 10, 5, 1, 15, 20, 15, 6, 1, 21, 35, 35, 21, 7, 1, 28, 56, 70, 56, 28, 8, 1, 36, 84, 126, 126, 84, 36, 9, 1, 45, 120, 210, 252, 210, 120, 45, 10, 1, 55, 165, 330, 462, 462, 330, 165, 55, 11, 1, 66, 220, 495, 792, 924, 792, 495, 220, 66, 12, 1, 78, 286, 715
Offset: 2

Views

Author

Gary W. Adamson, Mar 19 2005

Keywords

Comments

A000295 (Eulerian numbers) gives the row sums.
Write A004736 and Pascal's triangle as infinite lower triangular matrices A and B; then A*B is this triangle.
From Peter Luschny, Apr 10 2011: (Start)
A slight variation has a combinatorial interpretation: remove the last column and the second one from Pascal's triangle. Let P(m, k) denote the set partitions of {1,2,..,n} with the following properties:
(a) Each partition has at least one singleton block;
(c) k is the size of the largest block of the partition;
(b) m = n - k + 1 is the number of parts of the partition.
Then A000295(n) = Sum_{k=1..n} card(P(n-k+1,k)).
For instance, A000295(4) = P(4,1) + P(3,2) + P(2,3) + P(1,4) = card({1|2|3|4}) + card({1|2|34, 1|3|24,1|4|23, 2|3|14, 2|4|13, 3|4|12}) + card({1|234, 2|134, 3|124, 4|123}) = 1 + 6 + 4 = 11.
This interpretation can be superimposed on the sequence by changing the offset to 1 and adding the value 1 in front. The triangle then starts
1;
1, 3;
1, 6, 4;
1, 10, 10, 5;
1, 15, 20, 15, 6;
...
(End)
Diagonal sums are A001924(n+1). - Philippe Deléham, Jan 11 2014
Relation to K-theory: T acting on the column vector (d,-d^2,d^3,...) generates the Euler classes for a hypersurface of degree d in CP^n. Cf. Dugger p. 168, A111492, A238363, and A135278. - Tom Copeland, Apr 11 2014

Examples

			The triangle a(n, k) begins:
  n\k  2   3   4    5    6    7    8   9  10 11 12 13
  2:   1
  3:   3   1
  4:   6   4   1
  5:  10  10   5    1
  6:  15  20  15    6    1
  7:  21  35  35   21    7    1
  8:  28  56  70   56   28    8    1
  9:  36  84 126  126   84   36    9   1
  10: 45 120 210  252  210  120   45  10   1
  11: 55 165 330  462  462  330  165  55  11  1
  12: 66 220 495  792  924  792  495 220  66 12  1
  13: 78 286 715 1287 1716 1716 1287 715 286 78 13  1
... reformatted. - _Wolfdieter Lang_, Mar 20 2015

Links

Crossrefs

Cf. A000295, A007318, A008292, A104713, A027641/A027642 (first Bernoulli numbers B-), A164555/A027642 (second Bernoulli numbers B+), A176327/A176289.

Programs

  • Magma
    /* As triangle */ [[Binomial(n, k): k in [2..n]]: n in [2..10]]; // G. C. Greubel, May 15 2018
  • Mathematica
    t[n_, k_] := Binomial[n, k]; Table[ t[n, k], {n, 2, 13}, {k, 2, n}] // Flatten (* Robert G. Wilson v, Apr 16 2011 *)
  • PARI
    for(n=2, 10, for(k=2,n, print1(binomial(n,k), ", "))) \\ G. C. Greubel, May 15 2018

Formula

T(n,k) = binomial(n,k), for 2 <= k <= n.
From Peter Bala, Jul 16 2013: (Start)
The following remarks assume an offset of 0.
Riordan array (1/(1 - x)^3, x/(1 - x)).
O.g.f.: 1/(1 - t)^2*1/(1 - (1 + x)*t) = 1 + (3 + x)*t + (6 + 4*x + x^2)*t^2 + ....
E.g.f.: (1/x*d/dt)^2 (exp(t)*(exp(x*t) - 1 - x*t)) = 1 + (3 + x)*t + (6 + 4*x + x^2)*t^2/2! + ....
The infinitesimal generator for this triangle has the sequence [3,4,5,...] on the main subdiagonal and 0's elsewhere. (End)
As triangle T(n,k), 0<=k<=n: T(n,k) = 3*T(n-1,k) + T(n-1,k-1) - 3*T(n-2,k) - 2*T(n-2,k-1) + T(n-3,k) + T(n-3,k-1), T(0,0)=1, T(n,k)=0 if k<0 or if k>n. - Philippe Deléham, Jan 11 2014
From Tom Copeland, Apr 11 2014: (Start)
A) The infinitesimal generator for this matrix is given in A132681 with m=2. See that entry for numerous relations to differential operators and the Laguerre polynomials of order m=2, i.e., Lag(n,t,2) = Sum_{j=0..n} binomial(n+2,n-j)*(-t)^j/j!.
B) O.g.f.: 1 / { [ 1 - t * x/(1-x) ] * (1-x)^3 }
C) O.g.f. of row e.g.f.s: exp[t*x/(1-x)]/(1-x)^3 = [Sum_{n>=0} x^n * Lag(n,-t,2)] = 1 + (3 + t)*x + (6 + 4t + t^2/2!)*x^2 + (10 + 10t + 5t^2/2! + t^3/3!)*x^3 + ....
D) E.g.f. of row o.g.f.s: [(1+t)*exp((1+t)*x) - (1+t+t*x)exp(x)]/t^2. (End)
O.g.f. for m-th row (m=n-2): [(1+x)^(m+2)-(1+(m+2)*x)]/x^2. - Tom Copeland, Apr 16 2014
Reverse T = [St2]*dP*[St1]- dP = [St2]*(exp(x*M)-I)*[St1]-(exp(x*M)-I) with top two rows of zeros removed, [St1]=padded A008275 just as [St2]=A048993=padded A008277, dP= A132440, M=A238385-I, and I=identity matrix. Cf. A238363. - Tom Copeland, Apr 26 2014
O.g.f. of column k (with k leading zeros): (x^k)/(1-x)^(k+1), k >= 2. - Wolfdieter Lang, Mar 20 2015

Extensions

Edited and extended by David Wasserman, Jul 03 2007

A205497 Triangle read by rows: Zig-zag Eulerian number triangle T(n, k).

Original entry on oeis.org

1, 1, 1, 1, 1, 1, 3, 1, 1, 7, 7, 1, 1, 14, 31, 14, 1, 1, 26, 109, 109, 26, 1, 1, 46, 334, 623, 334, 46, 1, 1, 79, 937, 2951, 2951, 937, 79, 1, 1, 133, 2475, 12331, 20641, 12331, 2475, 133, 1, 1, 221, 6267, 47191, 123216, 123216, 47191, 6267, 221, 1
Offset: 0

Views

Author

L. Edson Jeffery, Jan 27 2012

Keywords

Comments

From Kyle Petersen, Jun 02 2024: (Start)
Coefficients of the "P-Eulerian" polynomial of a naturally labeled zig-zag poset, which counts linear extensions according to number of descents. T(n, k) is the number of linear extensions of the n-element zig-zag poset with k descents.
Also T(n, k) is the number of up-down permutations of length n with k "big returns". A big return is a pair (i, i+1) for which i appears more than one place to the right of i+1 in the permutation. This interpretation implies row sums are given by A000111. (End)
From L. Edson Jeffery, Jan 27 2012: (Start)
(Previous name:) Omitting the first two ones, a rectangular array M read by antidiagonals in which entry M_{n-k, k} in row n-k and column k, 0 <= k <= n, gives the coefficient of x^k in the numerator of the conjectured generating function for row n + 3 of the tabular form of A050446.
In the following, let M_{n, k} denote the entry in row n and column k of M, n, k in {0, 1, ...}.
Conjecture: 1. M_{n, k} = M_{k, n}, for all n and k; that is, M is symmetric about the central terms {1, 3, 31, 623,...}. (This has been verified for the first 100 antidiagonals of M.)
Conjecture: 2. For m in {3, 4,...}, row m of array A050446 has generating function of the form H_m(x)/(1 - x)^m, in which the numerator H_m(x) is a polynomial of degree m - 3 in x with coefficients given by the entries of the (m - 3)-th antidiagonal of M containing the sequence of entries {M_{m-3-j,j}}, j=0..m-3 (see the example below). It is known that H_1(x) = H_2(x) = 1.
Conjecture: 3. Define the Chebyshev polynomials of the second kind by U_0(t) = 1, U_1(t) = 2*t and U_r(t) = 2*t*U_(r-1)(t) - U_(r-2)(t) (r > 1). Assuming Conjecture 1, lim_{n -> infinity} M_{n+1, k}/M_{n, k} = U_k(cos(Pi/(2*k+3))) = spectral radius of the (k+1) X (k+1) unit-primitive matrix (see [Jeffery]) A_{2*k+3, k} = [0,...,0,1; 0,...,0,1,1; ...; 0,1,...,1; 1,...,1], with identical limits for the columns of the transpose M^T of M.
Conjecture: 4. Let S(u, v) denote the entry in row u and column v of triangle S = A187660, 0 <= v <= u. Define the polynomials P_u(x) = Sum[S(u, v)*x^v]. Assuming Conjecture 1, then (i) the generating function for row (or column) n of M is of the form
G_n(x)/((P_1(x))^(n+1) * (P_2(x))^n * ... * (P_n(x))^2 * P_(n+1)(x)),
in which (ii) the numerator G_n(x) is a polynomial of degree A005586(n), and (iii) the denominator is a polynomial of degree A000292(n+1).
Remarks: The coefficients in the numerators G_n(x) appear to have no pattern. The polynomial P_j(x), j in {1,...,n+1}, of Conjecture 4 is also obtained from the characteristic polynomial of the unit-primitive matrix A_{2*j+3,j} of Conjecture 3 by taking the exponents of the latter in reverse order; and P_j(x) is otherwise identical to the characteristic polynomial of the unit-primitive matrix A_{2*j+3,1}.
(End)
Conjecture: The Eulerian zig-zag polynomials have only negative and simple real roots and form a Sturm sequence, that is, p(n+1, x) has n real roots separated by the roots of p(n, x). This property was proved by Frobenius for the classical Eulerian polynomials. - Peter Luschny, Jun 04 2024

Examples

			From _Kyle Petersen_, Jun 02 2024: (Start)
Triangle T(n, k) begins:
  1;
  1;
  1;
  1,  1;
  1,  3,   1;
  1,  7,   7,    1;
  1, 14,  31,   14,    1;
  1, 26, 109,  109,   26,   1;
  1, 46, 334,  623,  334,  46,  1;
  1, 79, 937, 2951, 2951, 937, 79, 1;
  ...
For n=4, the naturally labeled zig-zag poset 1<3>2<4 has five linear extensions: 1234, 1243, 2134, 2143, 2413, and their descent numbers are (respectively) 0, 1, 1, 2, 1. Thus T(4,0) = 1, T(4,1) = 3, and T(4,2) = 1. Also with n=4, there are five up-down permutations: 1324, 1423, 2314, 2413, 3412, and their big return numbers are (respectively) 0, 1, 1, 2, 1. (End)
Without the first two ones the data can be seen as an array M read by antidiagonals. Christopher H. Gribble kindly calculated the first 100 antidiagonals which starts as:
  1,  1,   1,     1,      1,       1, ...
  1,  3,   7,    14,     26,      46, ...
  1,  7,  31,   109,    334,     937, ...
  1, 14, 109,   623,   2951,   12331, ...
  1, 26, 334,  2951,  20641,  123216, ...
  1, 46, 937, 12331, 123216, 1019051, ...
  ...
The antidiagonals of M written as the rows of a triangle, yielding then, by the conjectures and the definition of H_m(x), row m = 7 of table A050446 has generating function H_7(x)/(1-x)^7 = (Sum_{j=0..4} M_{4-j,j}*x^j)/(1-x)^7 = (1 + 14*x + 31*x^2 + 14*x^3 + x^4)/(1-x)^7.

Links

Crossrefs

Cf. A000012, A000292, A001924, A050446, A050447, A187660, row sums are A000111.
Cf. A350354, A373388, A373389.

Programs

  • Maple
    Gn := proc(n) local F;
    if n = 0 then p*q*x/(1 - q*x);
    elif n > 0 then
        F := Gn(n - 1);
        simplify(p/(p - q)*(subs({p = q, q = p}, F) - subs(p = q, F)));
    fi;
end:
Zn := proc(n) expand(simplify(subs({p = 1, q = 1}, Gn(n))*(1 - x)^(n + 1))) end:
seq( coeffs(Zn(n)), n=0..15);  # Kyle Petersen, Jun 02 2024
# Alternative:
A205497row := proc(n) local k, j; ifelse(n < 2, 1,
seq(add((-1)^j * binomial(n + 1, j) * A050446(n, k - j), j = 0..k), k = 0..n-2)) end:  # Peter Luschny, Jun 17 2024
  • Mathematica
    Gn[n_] := Module[{F}, If[n == 0, p*q*x/(1-q*x), If[n > 0, F = Gn[n-1]; Simplify[p/(p-q)*(ReplaceAll[F, {p -> q, q -> p}] - ReplaceAll[F, p -> q])]]]];
Zn[n_] := Expand[Simplify[ReplaceAll[Gn[n], {p -> 1, q -> 1}]*(1-x)^(n+1)]];
Table[Rest@CoefficientList[Zn[n], x], {n, 0, 15}] // Flatten (* Jean-François Alcover, Jun 04 2024, after Kyle Petersen *)
  • Python
    from functools import cache
from math import comb as binomial
@cache
def S(n, k):
    return (S(n, k - 1) + sum(S(2 * j, k - 1) * S(n - 1 - 2 * j, k)
            for j in range(1 + (n - 1) // 2)) if k > 0 else 1)
def A205497(dim):  # returns [row(0), ..., row(dim-1)]
    if dim < 4: return [[1]] * dim
    Y = [[0 for _ in range(n - 2)] for n in range(dim + 1)]
    for n in range(dim + 1):
        for k in range(n - 2):
            for j in range(k + 1):
                Y[n][k] += (-1)**j * binomial(n, j) * S(n - 1, k - j)
    Y[1] = Y[2] = [1]
    return Y[1::]
print(A205497(9))  # Peter Luschny, Jun 14 2024

Formula

Conjecture: 5.1. G.f. for column 0 of M is 1/(1-x) (A000012).
Conjecture: 5.2. G.f. for column 1 of M is 1/((1-x)^2*(1-x-x^2)) (A001924).
Conjecture: 5.3. G.f. for column 2 of M is (1 - x^2 - x^3 - x^4 + x^5)/((1-x)^3*(1-x-x^2)^2*(1 - 2*x - x^2 + x^3)) (A205492).
Conjecture: 5.4. G.f. for column 3 of M is (1 + x - 6*x^2 - 15*x^3 + 21*x^4 + 35*x^5 - 13*x^6 - 51*x^7 + 3*x^8 + 21*x^9 + 5*x^10 + x^11 - 5*x^12 - x^13 - x^14)/((1-x)^4*(1-x-x^2)^3*(1 - 2*x - x^2 + x^3)^2*(1 - 2*x - 3*x^2 + x^3 + x^4)) (A205493).
Conjecture: 5.5. G.f. for column 4 of M is (1 + 4*x - 31*x^2 - 67*x^3 + 348*x^4 + 418*x^5 - 1893*x^6 - 1084*x^7 + 4326*x^8 + 4295*x^9 - 7680*x^10 - 9172*x^11 + 9104*x^12 + 11627*x^13 - 5483*x^14 - 10773*x^15 + 1108*x^16 + 7255*x^17 + 315*x^18 - 3085*x^19 - 228*x^20 + 669*x^21 + 102*x^22 - 23*x^23 - 45*x^24 - 16*x^25 + 11*x^26 + 2*x^27 - x^28)/((1-x)^5*(1-x-x^2)^4*(1 - 2*x - x^2 + x^3)^3*(1 - 2*x - 3*x^2 + x^3 + x^4)^2*(1 - 3*x - 3*x^2 + 4*x^3 + x^4 - x^5)) (A205494).

Extensions

Two 1's prepended and new name by Kyle Petersen Jun 02 2024
Edited by Peter Luschny, Jun 02 2024
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