A106516 -id:A106516 - cp's OEIS frontend

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

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

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

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

			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

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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.

-- (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)

Flat(List([0..12],n->List([0..n],k->Binomial(n,k)))); # Stefano Spezia, Dec 22 2018

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

/* 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);
```

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 *)

create_list(binomial(n,k),n,0,12,k,0,n); /* Emanuele Munarini, Mar 11 2011 */

C(n,k)=binomial(n,k) \\ Charles R Greathouse IV, Jun 08 2011

# 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

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

def C(n,k): return Subsets(range(n), k).cardinality() # Ralf Stephan, Jan 21 2014

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, nC(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

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

A027649 a(n) = 2*(3^n) - 2^n.

Original entry on oeis.org

1, 4, 14, 46, 146, 454, 1394, 4246, 12866, 38854, 117074, 352246, 1058786, 3180454, 9549554, 28665046, 86027906, 258149254, 774578834, 2323998646, 6972520226, 20918609254, 62757924914, 188277969046, 564842295746, 1694543664454, 5083664547794, 15251060752246

Offset: 0

N. J. A. Sloane

Poly-Bernoulli numbers B_n^(k) with k=-2.
Binomial transform of A007051, if both sequences start at 0. Binomial transform of A000225(n+1). - Paul Barry, Mar 24 2003
Euler expands (1-z)/(1-5z+6z^2) and finds the general term. Section 226 of the Introductio indicates that he could have written down the recursion relation: a(n) = 5 a(n-1)-6 a(n-2). - V. Frederick Rickey (fred-rickey(AT)usma.edu), Feb 10 2006
Let R be a binary relation on the power set P(A) of a set A having n = |A| elements such that for every element x, y of P(A), xRy if x is a subset of y or y is a subset of x. Then a(n) = |R|. - Ross La Haye, Dec 22 2006
With regard to the comment by Ross La Haye: For proper subsets see A056182. - For nonempty subsets see A091344. - For nonempty proper subsets see a(n+1) in A260217. - Manfred Boergens, Aug 02 2023
If x, y are two n-bit binary strings then a(n) gives the number of pairs (x,y) such that XOR(x, y) = ABS(x - y). - Ramasamy Chandramouli, Feb 15 2009
Equals row sums of the triangular version of A038573. - Gary W. Adamson, Jun 04 2009
Inverse binomial transform of A085350. - Paul Curtz, Nov 14 2009
Related to the number of even a's in a nontrivial cycle (should one exist) in the 3x+1 Problem, where a <= floor(log_2(2*(3^n) - 2^n)). The value n correlates to the number of odds in such a nontrivial cycle. See page 1288 of Crandall's paper. Also, this relation gives another proof that the number of odds divided by the number of evens in a nontrivial cycle is bounded by log 2 / log 3 (this observation does not resolve the finite cycles conjecture as the value could be arbitrarily close to this bound). However, the same argument gives that log 2 / log 3 is less than or equal to the number of odds divided by the number of evens in a divergent sequence (should one exist), as log 2 / log 3 is the limit value for a cycle of an arbitrarily large length, where the length is given by the value n. - Jeffrey R. Goodwin, Aug 04 2011
Row sums of Riordan triangle A106516. - Wolfdieter Lang, Jan 09 2015
Number of restricted barred preferential arrangements having 3 bars in which the sections are all restricted sections such that (for fixed sections i and j) section i or section j is empty. - Sithembele Nkonkobe, Oct 12 2015
This is also row 2 of A281891: for n >= 1, when consecutive positive integers are written as a product of primes in nondecreasing order, a factor of 2 or 3 occurs in n-th position a(n) times out of every 6^n. - Peter Munn, May 18 2017
Also row sums of A124929. - Omar E. Pol, Jun 15 2017
This is the sum of A318921(n) for n in the range 2^(k+1) to 2^(k+2)-1. See A318921 for proof. - N. J. A. Sloane, Sep 25 2018
a(n) is also the number of acyclic orientations of the complete bipartite graph K_{2,n}. - Vincent Pilaud, Sep 15 2020
a(n-1) is also the number of n-digit numbers whose largest decimal digit is 2. - Stefano Spezia, Nov 15 2023

Leonhard Euler, Introductio in analysin infinitorum (1748), section 216.

T. D. Noe, Table of n, a(n) for n = 0..200
Taylor Brysiewicz, Holger Eble, and Lukas Kühne, Enumerating chambers of hyperplane arrangements with symmetry, arXiv:2105.14542 [math.CO], 2021.
R. E. Crandall, On the 3x+1 problem, Math. Comp., 32 (1978) 1281-1292.
Shalosh B. Ekhad, N. J. A. Sloane, and Doron Zeilberger, A Meta-Algorithm for Creating Fast Algorithms for Counting ON Cells in Odd-Rule Cellular Automata, arXiv:1503.01796 [math.CO], 2015; see also the Accompanying Maple Package.
Shalosh B. Ekhad, N. J. A. Sloane, and Doron Zeilberger, Odd-Rule Cellular Automata on the Square Grid, arXiv:1503.04249 [math.CO], 2015.
John Elias, Illustration of initial terms: Reflected Sierpinski triangle
K. Imatomi, M. Kaneko, and E. Takeda, Multi-Poly-Bernoulli Numbers and Finite Multiple Zeta Values, J. Int. Seq. 17 (2014) # 14.4.5
Ken Kamano, Sums of Products of Poly-Bernoulli Numbers of Negative Index, Journal of Integer Sequences, Vol. 15 (2012), #12.1.3.
K. Kamano, Sums of Products of Bernoulli Numbers, Including Poly-Bernoulli Numbers, J. Int. Seq. 13 (2010), 10.5.2.
Masanobu Kaneko, Poly-Bernoulli numbers, Journal de théorie des nombres de Bordeaux, 9 no. 1 (1997), Pages 221-228.
Takao Komatsu, Some recurrence relations of poly-Cauchy numbers, J. Nonlinear Sci. Appl., (2019) Vol. 12, Issue 12, 829-845.
Ross La Haye, Binary Relations on the Power Set of an n-Element Set, Journal of Integer Sequences, Vol. 12 (2009), Article 09.2.6.
S. Nkonkobe and V. Murali, On some properties and relations between restricted barred preferential arrangements, multi-poly-bernoulli numbers and related numbers, arXiv:1509.07352 [math.CO], 2015.
Eric Weisstein's World of Mathematics, Sierpinski Sieve
Wikipedia, Sierpinski triangle
Index entries for sequences related to Bernoulli numbers.
Index entries for linear recurrences with constant coefficients, signature (5,-6).

Row n = 2 of array A099594.
Also occurs as a row, column, diagonal or as row sums in A038573, A085870, A090888, A106516, A217764, A281891.
Cf. A000079, A000225, A001047, A007051, A053581, A056182, A085350, A091344, A124929, A166060, A260217, A318921.

a027649 n = a027649_list !! n
a027649_list = map fst $ iterate (\(u, v) -> (3 * u + v, 2 * v)) (1, 1)
-- Reinhard Zumkeller, Jun 09 2013

[2*(3^n)-2^n: n in [0..30]]; // Vincenzo Librandi, Jul 17 2011

a(n, k):= (-1)^n*sum( (-1)^'m'*'m'!*Stirling2(n,'m')/('m'+1)^k,'m'=0..n);
seq(a(n, -2), n=0..30);

Table[2(3^n)-2^n,{n,0,30}] (* or *) LinearRecurrence[ {5,-6},{1,4},31]  (* Harvey P. Dale, Apr 22 2011 *)

a(n)=2*(3^n)-2^n \\ Charles R Greathouse IV, Jul 16 2011

Vec((1-x)/((1-2*x)*(1-3*x)) + O(x^50)) \\ Altug Alkan, Oct 12 2015

[2*(3^n - 2^(n-1)) for n in (0..30)] # G. C. Greubel, Aug 01 2022

G.f.: (1-x)/((1-2*x)*(1-3*x)).
a(n) = 3*a(n-1) + 2^(n-1), with a(0) = 1.
a(n) = Sum_{k=0..n} binomial(n, k)*(2^(k+1) - 1). - Paul Barry, Mar 24 2003
Partial sums of A053581. - Paul Barry, Jun 26 2003
Main diagonal of array (A085870) defined by T(i, 1) = 2^i - 1, T(1, j) = 2^j - 1, T(i, j) = T(i-1, j) + T(i-1, j-1). - Benoit Cloitre, Aug 05 2003
a(n) = A090888(n, 3). - Ross La Haye, Sep 21 2004
a(n) = Sum_{k=0..n} binomial(n+2, k+1)*Sum_{j=0..floor(k/2)} A001045(k-2j). - Paul Barry, Apr 17 2005
a(n) = Sum_{k=0..n} Sum_{j=0..n} binomial(n,j)*binomial(j+1,k+1). - Paul Barry, Sep 18 2006
a(n) = A166060(n+1)/6. - Philippe Deléham, Oct 21 2009
a(n) = 5*a(n-1) - 6*a(n-2), a(0)=1, a(1)=4. - Harvey P. Dale, Apr 22 2011
a(n) = A217764(n,2). - Ross La Haye, Mar 27 2013
For n>0, a(n) = 3 * a(n-1) + 2^(n-1) = 2 * (a(n-1) + 3^(n-1)). - J. Conrad, Oct 29 2015
for n>0, a(n) = 2 * (1 + 2^(n-2) + Sum_{x=1..n-2} Sum_{k=0..x-1} (binomial(x-1,k)*(2^(k+1) + 2^(n-x+k)))). - J. Conrad, Dec 10 2015
E.g.f.: exp(2*x)*(2*exp(x) - 1). - Stefano Spezia, May 18 2024

Better formulas from David W. Wilson and Michael Somos
Incorrect formula removed by Charles R Greathouse IV, Mar 18 2010
Duplications (due to corrections to A numbers) removed by Peter Munn, Jun 15 2017

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

Boris Putievskiy, Aug 15 2013

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

			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;

Boris Putievskiy, Rows n = 1..140 of triangle, flattened
Rely Pellicer and David Alvo, Modified Pascal Triangle and Pascal Surfaces p.4
Index entries for triangles and arrays related to Pascal's triangle

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).

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

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

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 *)

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

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

@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

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

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

A055248 Triangle of partial row sums of triangle A007318(n,m) (Pascal's triangle). Triangle A008949 read backwards. Riordan (1/(1-2x), x/(1-x)).

Original entry on oeis.org

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

Offset: 0

Wolfdieter Lang, May 26 2000

In the language of the Shapiro et al. reference (also given in A053121) such a lower triangular (ordinary) convolution array, considered as matrix, belongs to the Riordan-group. The g.f. for the row polynomials p(n,x) (increasing powers of x) is 1/((1-2*z)*(1-x*z/(1-z))).
Binomial transform of the all 1's triangle: as a Riordan array, it factors to give (1/(1-x),x/(1-x))(1/(1-x),x). Viewed as a number square read by antidiagonals, it has T(n,k) = Sum_{j=0..n} binomial(n+k,n-j) and is then the binomial transform of the Whitney square A004070. - Paul Barry, Feb 03 2005
Riordan array (1/(1-2x), x/(1-x)). Antidiagonal sums are A027934(n+1), n >= 0. - Paul Barry, Jan 30 2005; edited by Wolfdieter Lang, Jan 09 2015
Eigensequence of the triangle = A005493: (1, 3, 10, 37, 151, 674, ...); row sums of triangles A011971 and A159573. - Gary W. Adamson, Apr 16 2009
Read as a square array, this is the generalized Riordan array ( 1/(1 - 2*x), 1/(1 - x) ) as defined in the Bala link (p. 5), which factorizes as ( 1/(1 - x), x/(1 - x) )*( 1/(1 - x), x )*( 1, 1 + x ) = P*U*transpose(P), where P denotes Pascal's triangle, A007318, and U is the lower unit triangular array with 1's on or below the main diagonal. - Peter Bala, Jan 13 2016

			The triangle a(n,m) begins:
n\m    0    1    2   3   4   5   6   7  8  9 10 ...
0:     1
1:     2    1
2:     4    3    1
3:     8    7    4   1
4:    16   15   11   5   1
5:    32   31   26  16   6   1
6:    64   63   57  42  22   7   1
7:   128  127  120  99  64  29   8   1
8:   256  255  247 219 163  93  37   9  1
9:   512  511  502 466 382 256 130  46 10  1
10: 1024 1023 1013 968 848 638 386 176 56 11  1
... Reformatted. - _Wolfdieter Lang_, Jan 09 2015
Fourth row polynomial (n=3): p(3,x)= 8 + 7*x + 4*x^2 + x^3.
The matrix inverse starts
   1;
  -2,   1;
   2,  -3,   1;
  -2,   5,  -4,    1;
   2,  -7,   9,   -5,    1;
  -2,   9, -16,   14,   -6,    1;
   2, -11,  25,-  30,   20,   -7,    1;
  -2,  13, -36,   55,  -50,   27,   -8,    1;
   2, -15,  49,  -91,  105,  -77,   35,   -9,  1;
  -2,  17, -64,  140, -196,  182, -112,   44, -10,   1;
   2, -19,  81, -204,  336, -378,  294, -156,  54, -11, 1;
   ...
which may be related to A029653. - _R. J. Mathar_, Mar 29 2013
From _Peter Bala_, Dec 23 2014: (Start)
With the array M(k) as defined in the Formula section, the infinite product M(0)*M(1)*M(2)*... begins
/1      \ /1        \ /1       \       /1       \
|2 1     ||0 1       ||0 1      |      |2  1     |
|4 3 1   ||0 2 1     ||0 0 1    |... = |4  5 1   |
|8 7 4 1 ||0 4 3 1   ||0 0 2 1  |      |8 19 9 1 |
|...     ||0 8 7 4 1 ||0 0 4 3 1|      |...      |
|...     ||...       ||...      |      |         |
= A143494. (End)
Matrix factorization of square array as P*U*transpose(P):
/1      \ /1        \ /1 1 1 1 ...\    /1  1  1  1 ...\
|1 1     ||1 1       ||0 1 2 3 ... |   |2  3  4  5 ... |
|1 2 1   ||1 1 1     ||0 0 1 3 ... | = |4  7 11 16 ... |
|1 3 3 1 ||1 1 1 1   ||0 0 0 1 ... |   |8 15 26 42 ... |
|...     ||...       ||...         |   |...            |
- _Peter Bala_, Jan 13 2016

Reinhard Zumkeller, Rows n = 0..125 of triangle, flattened
Peter Bala, Notes on generalized Riordan arrays
Peter Bala, A055248: Rapidly converging series for log(2) and Pi
Jean-Luc Baril, Javier F. González, and José L. Ramírez, Last symbol distribution in pattern avoiding Catalan words, Univ. Bourgogne (France, 2022).
Paul Barry, On a Central Transform of Integer Sequences, arXiv:2004.04577 [math.CO], 2020.
Norman Lindquist and Gerard Sierksma, Extensions of set partitions, Journal of Combinatorial Theory, Series A 31.2 (1981): 190-198. See Table I.
L. W. Shapiro, S. Getu, Wen-Jin Woan and L. C. Woodson, The Riordan Group, Discrete Appl. Maths. 34 (1991) 229-239.

Column sequences: A000079 (powers of 2, m=0), A000225 (m=1), A000295 (m=2), A002662 (m=3), A002663 (m=4), A002664 (m=5), A035038 (m=6), A035039 (m=7), A035040 (m=8), A035041 (m=9), A035042 (m=10).
Row sums: A001792(n) = A055249(n, 0).
Alternating row sums: A011782.
Cf. A011971, A159573. - Gary W. Adamson, Apr 16 2009
Cf. A007318, A008949, A106516, A143494.

a055248 n k = a055248_tabl !! n !! k
a055248_row n = a055248_tabl !! n
a055248_tabl = map reverse a008949_tabl
-- Reinhard Zumkeller, Jun 20 2015

T := (n,k) -> 2^n - (1/2)*binomial(n, k-1)*hypergeom([1, n + 1], [n-k + 2], 1/2).
seq(seq(simplify(T(n,k)), k=0..n),n=0..10); # Peter Luschny, Oct 10 2019

a[n_, m_] := Sum[ Binomial[n, m + j], {j, 0, n}]; Table[a[n, m], {n, 0, 10}, {m, 0, n}] // Flatten (* Jean-François Alcover, Jul 05 2013, after Paul Barry *)
T[n_, k_] := Binomial[n, k] * Hypergeometric2F1[1, k - n, k + 1, -1];
Flatten[Table[T[n, k], {n, 0, 7}, {k, 0, n}]]  (* Peter Luschny, Oct 06 2023 *)

a(n, m) = A008949(n, n-m), if n > m >= 0.
a(n, m) = Sum_{k=m..n} A007318(n, k) (partial row sums in columns m).
Column m recursion: a(n, m) = Sum_{j=m..n-1} a(j, m) + A007318(n, m) if n >= m >= 0, a(n, m) := 0 if nG.f. for column m: (1/(1-2*x))*(x/(1-x))^m, m >= 0.
a(n, m) = Sum_{j=0..n} binomial(n, m+j). - Paul Barry, Feb 03 2005
Inverse binomial transform (by columns) of A112626. - Ross La Haye, Dec 31 2006
T(2n,n) = A032443(n). - Philippe Deléham, Sep 16 2009
From Peter Bala, Dec 23 2014: (Start)
Exp(x) * e.g.f. for row n = e.g.f. for diagonal n. For example, for n = 3 we have exp(x)*(8 + 7*x + 4*x^2/2! + x^3/3!) = 8 + 15*x + 26*x^2/2! + 42*x^3/3! + 64*x^4/4! + .... The same property holds more generally for Riordan arrays of the form ( f(x), x/(1 - x) ).
Let M denote the present triangle. For k = 0,1,2,... define M(k) to be the lower unit triangular block array
/I_k 0\
\ 0  M/ having the k X k identity matrix I_k as the upper left block; in particular, M(0) = M. The infinite product M(0)*M(1)*M(2)*..., which is clearly well-defined, is equal to A143494 (but with a different offset). See the Example section. Cf. A106516. (End)
a(n,m) = Sum_{p=m..n} 2^(n-p)*binomial(p-1,m-1), n >= m >= 0, else 0. - Wolfdieter Lang, Jan 09 2015
T(n, k) = 2^n - (1/2)*binomial(n, k-1)*hypergeom([1, n+1], [n-k+2], 1/2). - Peter Luschny, Oct 10 2019
T(n, k) = binomial(n, k)*hypergeom([1, k - n], [k + 1], -1). - Peter Luschny, Oct 06 2023
n-th row polynomial R(n, x) = (2^n - x*(1 + x)^n)/(1 - x). These polynomials can be used to find series acceleration formulas for the constants log(2) and Pi. - Peter Bala, Mar 03 2025

A000340 a(0)=1, a(n) = 3*a(n-1) + n + 1.

Original entry on oeis.org

1, 5, 18, 58, 179, 543, 1636, 4916, 14757, 44281, 132854, 398574, 1195735, 3587219, 10761672, 32285032, 96855113, 290565357, 871696090, 2615088290, 7845264891, 23535794695, 70607384108, 211822152348, 635466457069

Offset: 0

N. J. A. Sloane, Simon Plouffe

From Johannes W. Meijer, Feb 20 2009: (Start)
Second right hand column (n-m=1) of the A156920 triangle.
The generating function of this sequence enabled the analysis of the polynomials A156921 and A156925.
(End)
Partial sums of A003462, and thus the second partial sums of A000244 (3^n). Also column k=2 of A106516. - John Keith, Jan 04 2022

			G.f. = 1 + 5*x + 18*x^2 + 58*x^3 + 179*x^4 + 543*x^5 + 1636*x^6 + ...

F. N. David, M. G. Kendall and D. E. Barton, Symmetric Function and Allied Tables, Cambridge, 1966, p. 260.
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).

Vincenzo Librandi, Table of n, a(n) for n = 0..1000
Dillan Agrawal, Selena Ge, Jate Greene, Tanya Khovanova, Dohun Kim, Rajarshi Mandal, Tanish Parida, Anirudh Pulugurtha, Gordon Redwine, Soham Samanta, and Albert Xu, Chip-Firing on Infinite k-ary Trees, arXiv:2501.06675 [math.CO], 2025. See pp. 9, 18.
Shaoshi Chen, Hanqian Fang, Sergey Kitaev, and Candice X.T. Zhang, Patterns in Multi-dimensional Permutations, arXiv:2411.02897 [math.CO], 2024. See p. 7.
INRIA Algorithms Project, Encyclopedia of Combinatorial Structures 389
Simon Plouffe, Approximations de séries génératrices et quelques conjectures, Dissertation, Université du Québec à Montréal, 1992; arXiv:0911.4975 [math.NT], 2009.
Simon Plouffe, 1031 Generating Functions, Appendix to Thesis, Montreal, 1992
László Tóth, On Schizophrenic Patterns in b-ary Expansions of Some Irrational Numbers, arXiv:2002.06584 [math.NT], 2020. Mentions this sequence. See also Proc. Amer. Math. Soc. 148 (2020), 461-469.
Index entries for linear recurrences with constant coefficients, signature (5,-7,3).

From Johannes W. Meijer, Feb 20 2009: (Start)
Cf. A156921, A156925, A156927, A156933. Other columns A156922, A156923, A156924.
Equals A156920 second right hand column.
Equals A142963 second right hand column divided by 2^n.
Equals A156919 second right hand column divided by 2.
(End)
Cf. A014915.
Equals column k=1 of A008971 (shifted). - Jeremy Dover, Jul 11 2021
Cf. A000340, A003462 (first differences), A106516.

[(3^(n+2)-2*n-5)/4: n in [0..30]]; // Vincenzo Librandi, Aug 15 2011

a[ -1]:=0:a[0]:=1:for n from 1 to 50 do a[n]:=4*a[n-1]-3*a[n-2]+1 od: seq(a[n],n=0..50); # Miklos Kristof, Mar 09 2005
A000340:=-1/(3*z-1)/(z-1)**2; # conjectured by Simon Plouffe in his 1992 dissertation

a[ n_] := MatrixPower[ {{1, 0, 0}, {1, 1, 0}, {1, 1, 3}}, n + 1][[3, 1]]; (* Michael Somos, May 28 2014 *)
RecurrenceTable[{a[0]==1,a[n]==3a[n-1]+n+1},a,{n,30}] (* or *) LinearRecurrence[{5,-7,3},{1,5,18},30] (* Harvey P. Dale, Jan 31 2017 *)

G.f.: 1/((1-3*x)*(1-x)^2).
a(n) = (3^(n+2) - 2*n - 5)/4.
a(n) = Sum_{k=0..n+1} (n-k+1)*3^k = Sum_{k=0..n+1} k*3^(n-k+1). - Paul Barry, Jul 30 2004
a(n) = Sum_{k=0..n} binomial(n+2, k+2)*2^k. - Paul Barry, Jul 30 2004
a(-1)=0, a(0)=1, a(n) = 4*a(n-1) - 3*a(n-2) + 1. - Miklos Kristof, Mar 09 2005
a(n) = 5*a(n-1) - 7*a(n-2) + 3*a(n-3). - Johannes W. Meijer, Feb 20 2009
a(-2 - n) = 3^-n * A014915(n). - Michael Somos, May 28 2014
E.g.f.: exp(x)*(9*exp(2*x) - 2*x - 5)/4. - Stefano Spezia, Nov 09 2024

A143495 Triangle read by rows: 3-Stirling numbers of the second kind.

Original entry on oeis.org

1, 3, 1, 9, 7, 1, 27, 37, 12, 1, 81, 175, 97, 18, 1, 243, 781, 660, 205, 25, 1, 729, 3367, 4081, 1890, 380, 33, 1, 2187, 14197, 23772, 15421, 4550, 644, 42, 1, 6561, 58975, 133057, 116298, 47271, 9702, 1022, 52, 1, 19683, 242461, 724260, 830845, 447195

Offset: 3

Peter Bala, Aug 20 2008

This is the case r = 3 of the r-Stirling numbers of the second kind. The 3-Stirling numbers of the second kind give the number of ways of partitioning the set {1,2,...,n} into k nonempty disjoint subsets with the restriction that the elements 1, 2 and 3 belong to distinct subsets. For remarks on the general case see A143494 (r = 2). The corresponding array of 3-Stirling numbers of the first kind is A143492. The theory of r-Stirling numbers of both kinds is developed in [Broder]. For 3-Lah numbers refer to A143498.
From Peter Bala, Sep 19 2008: (Start)
Let D be the derivative operator d/dx and E the Euler operator x*d/dx. Then x^(-3)*E^n*x^3 = Sum_{k = 0..n} T(n+3,k+3)*x^k*D^k.
The row generating polynomials R_n(x) := Sum_{k= 3..n} T(n,k)*x^k satisfy the recurrence R_(n+1)(x) = x*R_n(x) + x*d/dx(R_n(x)) with R_3(x) = x^3. It follows that the polynomials R_n(x) have only real zeros (apply Corollary 1.2. of [Liu and Wang]).
Relation with the 3-Eulerian numbers E_3(n,j) := A144697(n,j): T(n,k) = (3!/k!)*Sum_{j = n-k..n-3} E_3(n,j)*binomial(j,n-k) for n >= k >= 3.
(End)
T(n,k) = S(n,k,3), n>=k>=3, in Mikhailov's first paper, eq.(28) or (A3). E.g.f. column k from (A20) with k->3, r->k. Therefore, with offset [0,0], this triangle is the Sheffer triangle (exp(3*x),exp(x)-1) with e.g.f. of column no. m>=0: exp(3*x)*((exp(x)-1)^m)/m!. See one of the formulas given below. For Sheffer matrices see the W. Lang link under A006232 with the S. Roman reference, also found in A132393. - Wolfdieter Lang, Sep 29 2011

			Triangle begins
  n\k|....3....4....5....6....7....8
  ==================================
  3..|....1
  4..|....3....1
  5..|....9....7....1
  6..|...27...37...12....1
  7..|...81..175...97...18....1
  8..|..243..781..660..205...25....1
  ...
T(5,4) = 7. The set {1,2,3,4,5} can be partitioned into four subsets such that 1, 2 and 3 belong to different subsets in 7 ways: {{1}{2}{3}{4,5}}, {{1}{2}{5}{3,4}}, {{1}{2}{4}{3,5}}, {{1}{3}{4}{2,5}}, {{1}{3}{5}{2,4}}, {{2}{3}{4}{1,5}} and {{2}{3}{5}{1,4}}.
From _Peter Bala_, Feb 23 2025: (Start)
The array factorizes as
/ 1               \       /1             \ /1             \ /1            \
| 3    1           |     | 3   1          ||0  1           ||0  1          |
| 9    7   1       |  =  | 9   4   1      ||0  3   1       ||0  0  1       | ...
|27   37  12   1   |     |27  13   5  1   ||0  9   4  1    ||0  0  3  1    |
|81  175  97  18  1|     |81  40  18  6  1||0 27  13  5  1 ||0  0  9  4  1 |
|...               |     |...             ||...            ||...           |
where, in the infinite product on the right-hand side, the first array is the Riordan array (1/(1 - 3*x), x/(1 - x)). See A106516. (End)

Peter Bala, Factorising (r,b)-Stirling arrays
Andrei Z. Broder, The r-Stirling numbers, Report Number: CS-TR-82-949, Stanford University, Department of Computer Science; see also, Discrete Math. 49, 241-259 (1984).
A. Dzhumadildaev and D. Yeliussizov, Path decompositions of digraphs and their applications to Weyl algebra, arXiv preprint arXiv:1408.6764v1 [math.CO], 2014. [Version 1 contained many references to the OEIS, which were removed in Version 2. - _N. J. A. Sloane_, Mar 28 2015]
Askar Dzhumadil’daev and Damir Yeliussizov, Walks, partitions, and normal ordering, Electronic Journal of Combinatorics, 22(4) (2015), #P4.10.
V. V. Mikhailov, Ordering of some boson operator functions, J. Phys A: Math. Gen. 16 (1983) 3817-3827.
V. V. Mikhailov, Normal ordering and generalised Stirling numbers, J. Phys A: Math. Gen. 18 (1985) 231-235.
Erich Neuwirth, Recursively defined combinatorial functions: Extending Galton's board, Discrete Math. 239 No. 1-3, 33-51 (2001).
L. Liu and Y. Wang, A unified approach to polynomial sequences with only real zeros, arXiv:math/0509207 [math.CO], 2005-2006.
Michael J. Schlosser and Meesue Yoo, Elliptic Rook and File Numbers, Electronic Journal of Combinatorics, 24(1) (2017), #P1.31.

Cf. A005061 (column 4), A005494 (row sums), A008277, A016753 (column 5), A028025 (column 6), A049458 (matrix inverse), A106516, A143492, A143494, A143496, A143498.

A143495 := (n, k) -> (1/(k-3)!)*add((-1)^(k-i-1)*binomial(k-3,i)*(i+3)^(n-3), i = 0..k-3): for n from 3 to 12 do seq(A143495(n, k), k = 3..n) end do;

nmax = 12; t[n_, k_] := 1/(k-3)!* Sum[ (-1)^(k-j-1)*Binomial[k-3, j]*(j+3)^(n-3), {j, 0, k-3}]; Flatten[ Table[ t[n, k], {n, 3, nmax}, {k, 3, n}]] (* Jean-François Alcover, Dec 07 2011, after Maple *)

@CachedFunction
def stirling2r(n, k, r) :
    if n < r: return 0
    if n == r: return 1 if k == r else 0
    return stirling2r(n-1, k-1, r) + k*stirling2r(n-1, k, r)
A143495 = lambda n, k: stirling2r(n, k, 3)
for n in (3..8): [A143495(n, k) for k in (3..n)] # Peter Luschny, Nov 19 2012

T(n+3,k+3) = (1/k!)*Sum_{i = 0..k} (-1)^(k-i)*binomial(k,i)*(i+3)^n, n,k >= 0.
T(n,k) = Stirling2(n,k) - 3*Stirling2(n-1,k) + 2*Stirling2(n-2,k), n,k >= 3.
Recurrence relation: T(n,k) = T(n-1,k-1) + k*T(n-1,k) for n > 3, with boundary conditions: T(n,2) = T(2,n) = 0 for all n; T(3,3) = 1; T(3,k) = 0 for k > 3.
Special cases: T(n,3) = 3^(n-3); T(n,4) = 4^(n-3) - 3^(n-3).
E.g.f. (k+3) column (with offset 3): (1/k!)*exp(3x)*(exp(x)-1)^k.
O.g.f. k-th column: Sum_{n >= k} T(n,k)*x^n = x^k/((1-3*x)*(1-4*x)*...*(1-k*x)).
E.g.f.: exp(3*t + x*(exp(t)-1)) = Sum_{n >= 0} Sum_{k = 0..n} T(n+3,k+3)*x^k*t^n/n! = Sum_{n >= 0} B_n(3;x)*t^n/n! = 1 + (3+x)*t/1! + (9+7*x+x^2)*t^2/2! + ..., where the row polynomials, B_n(3;x) := Sum_{k = 0..n} T(n+3,k+3)*x^k, may be called the 3-Bell polynomials.
Dobinski-type identities: Row polynomial B_n(3;x) = exp(-x)*Sum_{i >= 0} (i+3)^n*x^i/i!; Sum_{k = 0..n} k!*T(n+3,k+3)*x^k = Sum_{i >= 0} (i+3)^n*x^i/(1+x)^(i+1).
The T(n,k) are the connection coefficients between the falling factorials and the shifted monomials (x+3)^(n-3). For example, 9 + 7*x + x*(x-1) = (x+3)^2 and 27 + 37*x + 12x*(x-1) + x*(x-1)*(x-2) = (x+3)^3.
This array is the matrix product P^2 * S, where P denotes Pascal's triangle, A007318 and S denotes the lower triangular array of Stirling numbers of the second kind, A008277 (apply Theorem 10 of [Neuwirth]). The inverse array is A049458, the signed 3-Stirling numbers of the first kind.

A093560 (3,1) Pascal triangle.

Original entry on oeis.org

1, 3, 1, 3, 4, 1, 3, 7, 5, 1, 3, 10, 12, 6, 1, 3, 13, 22, 18, 7, 1, 3, 16, 35, 40, 25, 8, 1, 3, 19, 51, 75, 65, 33, 9, 1, 3, 22, 70, 126, 140, 98, 42, 10, 1, 3, 25, 92, 196, 266, 238, 140, 52, 11, 1, 3, 28, 117, 288, 462, 504, 378, 192, 63, 12, 1, 3, 31, 145, 405, 750, 966, 882, 570, 255, 75, 13, 1

Offset: 0

Wolfdieter Lang, Apr 22 2004

The array F(3;n,m) gives in the columns m >= 1 the figurate numbers based on A016777, including the pentagonal numbers A000326 (see the W. Lang link).
This is the third member, d=3, in the family of triangles of figurate numbers, called (d,1) Pascal triangles: A007318 (Pascal (d=1), A029653 (d=2).
This is an example of a Riordan triangle (see A053121 for a comment and the 1991 Shapiro et al. reference on the Riordan group) with o.g.f. of column no. m of the type g(x)*(x*f(x))^m with f(0)=1. Therefore the o.g.f. for the row polynomials p(n,x):=Sum_{m=0..n} a(n,m)*x^m is G(z,x)=g(z)/(1-x*z*f(z)). Here: g(x)=(1+2*x)/(1-x), f(x)=1/(1-x), hence G(z,x)=(1+2*z)/(1-(1+x)*z).
The SW-NE diagonals give the Lucas numbers A000032: L(n) = Sum_{k=0..ceiling((n-1)/2)} a(n-1-k,k), n >= 1, with L(0)=2. Observation by Paul Barry, Apr 29 2004. Proof via recursion relations and comparison of inputs.
Triangle T(n,k), read by rows, given by [3,-2,0,0,0,0,0,0,...] DELTA [1,0,0,0,0,0,0,0,...] where DELTA is the operator defined in A084938. - Philippe Deléham, Sep 17 2009
For a closed-form formula for generalized Pascal's triangle see A228576. - Boris Putievskiy, Sep 09 2013
From Wolfdieter Lang, Jan 09 2015: (Start)
The signed lower triangular matrix (-1)^(n-1)*a(n,m) is the inverse of the Riordan matrix A106516; that is Riordan ((1-2*x)/(1+x),x/(1+x)).
See the Peter Bala comment from Dec 23 2014 in A106516 for general Riordan triangles of the type (g(x), x/(1-x)): exp(x)*r(n,x) = d(n,x) with the e.g.f. r(n,x) of row n and the e.g.f. of diagonal n.
Similarly, for general Riordan triangles of the type (g(x), x/(1+x)): exp(x)*r(n,-x) = d(n,x). (End)
The n-th row polynomial is (3 + x)*(1 + x)^(n-1) for n >= 1. More generally, the n-th row polynomial of the Riordan array ( (1-a*x)/(1-b*x), x/(1-b*x) ) is (b - a + x)*(b + x)^(n-1) for n >= 1. - Peter Bala, Mar 02 2018
Binomial(n-2,k)+2*Binomial(n-3,k) is also the number of permutations avoiding both 123 and 132 with k double descents, i.e., positions with w[i]>w[i+1]>w[i+2]. - Lara Pudwell, Dec 19 2018

			Triangle begins
  1,
  3,  1,
  3,  4,  1,
  3,  7,  5,   1,
  3, 10, 12,   6,   1,
  3, 13, 22,  18,   7,   1,
  3, 16, 35,  40,  25,   8,   1,
  3, 19, 51,  75,  65,  33,   9,  1,
  3, 22, 70, 126, 140,  98,  42, 10,  1,
  3, 25, 92, 196, 266, 238, 140, 52, 11, 1,

Kurt Hawlitschek, Johann Faulhaber 1580-1635, Veroeffentlichung der Stadtbibliothek Ulm, Band 18, Ulm, Germany, 1995, Ch. 2.1.4. Figurierte Zahlen.
Ivo Schneider, Johannes Faulhaber 1580-1635, Birkhäuser, Basel, Boston, Berlin, 1993, ch.5, pp. 109-122.

Reinhard Zumkeller, Rows n = 0..125 of triangle, flattened
Peter Bala, A note on the diagonals of a proper Riordan Array
M. Bukata, R. Kulwicki, N. Lewandowski, L. Pudwell, J. Roth, and T. Wheeland, Distributions of Statistics over Pattern-Avoiding Permutations, arXiv preprint arXiv:1812.07112 [math.CO], 2018.
W. Lang, First 10 rows and array of figurate numbers .

Cf. Column sequences for m=1..9: A016777, A000326 (pentagonal), A002411, A001296, A051836, A051923, A050494, A053367, A053310;

A007318 (Pascal's triangle), A029653 ((2,1) Pascal triangle), A093561 ((4,1) Pascal triangle), A228196, A228576.

Concatenation([1],Flat(List([1..11],n->List([0..n],k->Binomial(n,k)+2*Binomial(n-1,k))))); # Muniru A Asiru, Dec 20 2018

a093560 n k = a093560_tabl !! n !! k
a093560_row n = a093560_tabl !! n
a093560_tabl = [1] : iterate
               (\row -> zipWith (+) ([0] ++ row) (row ++ [0])) [3, 1]
-- Reinhard Zumkeller, Aug 31 2014

from math import comb, isqrt
def A093560(n): return comb(r:=(m:=isqrt(k:=n+1<<1))-(k<=m*(m+1)),a:=n-comb(r+1,2))*(r+(r-a<<1))//r if n else 1 # Chai Wah Wu, Nov 12 2024

a(n, m)=F(3;n-m, m) for 0<= m <= n, otherwise 0, with F(3;0, 0)=1, F(3;n, 0)=3 if n>=1 and F(3;n, m):=(3*n+m)*binomial(n+m-1, m-1)/m if m>=1.
G.f. column m (without leading zeros): (1+2*x)/(1-x)^(m+1), m>=0.
Recursion: a(n, m)=0 if m>n, a(0, 0)= 1; a(n, 0)=3 if n>=1; a(n, m)= a(n-1, m) + a(n-1, m-1).
T(n, k) = C(n, k) + 2*C(n-1, k). - Philippe Deléham, Aug 28 2005
Equals M * A007318, where M = an infinite triangular matrix with all 1's in the main diagonal and all 2's in the subdiagonal. - Gary W. Adamson, Dec 01 2007
Sum_{k=0..n} T(n,k) = A151821(n+1). - Philippe Deléham, Sep 17 2009
exp(x) * e.g.f. for row n = e.g.f. for diagonal n. For example, for n = 3 we have exp(x)*(3 + 7*x + 5*x^2/2! + x^3/3!) = 3 + 10*x + 22*x^2/2! + 40*x^3/3! + 65*x^4/4! + .... The same property holds more generally for Riordan arrays of the form ( f(x), x/(1 - x) ). - Peter Bala, Dec 22 2014
G.f.: (-1-2*x)/(-1+x+x*y). - R. J. Mathar, Aug 11 2015

Incorrect connection with A046055 deleted by N. J. A. Sloane, Jul 08 2009

A106517 Convolution of Fibonacci(n-1) and 3^n.

Original entry on oeis.org

1, 3, 10, 31, 95, 288, 869, 2615, 7858, 23595, 70819, 212512, 637625, 1913019, 5739290, 17218247, 51655351, 154967040, 464902717, 1394710735, 4184136386, 12552415923, 37657258715, 112971793856, 338915410225, 1016746277043

Offset: 0

Paul Barry, May 05 2005

G. C. Greubel, Table of n, a(n) for n = 0..1000
Index entries for linear recurrences with constant coefficients, signature (4,-2,-3).

Cf. A000032, A000045, A000244, A094688, A101220.

Diagonal sums of number triangle A106516.

I:=[1,3,10]; [n le 3 select I[n] else 4*Self(n-1) -2*Self(n-2) -3*Self(n-3): n in [1..41]]; // G. C. Greubel, Aug 05 2021

LinearRecurrence[{4,-2,-3},{1,3,10},30] (* Harvey P. Dale, Oct 08 2014 *)

a(n) = sum(k=0, n, fibonacci(n-k-1) * 3^k); \\ Michel Marcus, Aug 06 2021

[(2*3^(n+1) - lucas_number2(n+1, 1, -1))/5 for n in (0..40)] # G. C. Greubel, Aug 05 2021

G.f.: (1-x)/((1-x-x^2)*(1-3*x)).
a(n) = Sum_{k=0..n} Fibonacci(n-k-1) * 3^k.
a(n) = A101220(2, 3, n+1). - Ross La Haye, Jul 25 2005
a(n) = A101220(3, 3, n+1) - A101220(3, 3, n). - Ross La Haye, May 31 2006
a(n) = (1/5)*(6*3^n - Lucas(n+1)). - Ralf Stephan, Nov 16 2010
Sum_{k=0..n} a(k) = A094688(n+1). - G. C. Greubel, Aug 05 2021

Showing 1-8 of 8 results.

cp's OEIS Frontend

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

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A027649 a(n) = 2*(3^n) - 2^n.

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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.

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A055248 Triangle of partial row sums of triangle A007318(n,m) (Pascal's triangle). Triangle A008949 read backwards. Riordan (1/(1-2x), x/(1-x)).

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A000340 a(0)=1, a(n) = 3*a(n-1) + n + 1.

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