A014162 -id:A014162 - 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

A001924 Apply partial sum operator twice to Fibonacci numbers.

Original entry on oeis.org

0, 1, 3, 7, 14, 26, 46, 79, 133, 221, 364, 596, 972, 1581, 2567, 4163, 6746, 10926, 17690, 28635, 46345, 75001, 121368, 196392, 317784, 514201, 832011, 1346239, 2178278, 3524546, 5702854, 9227431, 14930317, 24157781, 39088132, 63245948, 102334116, 165580101

Offset: 0

N. J. A. Sloane

Leading coefficients in certain rook polynomials (for n>=2; see p. 18 of the Riordan paper). - Emeric Deutsch, Mar 08 2004
(1, 3, 7, 14, ...) = row sums of triangle A141289. - Gary W. Adamson, Jun 22 2008
a(n) is the number of nonempty subsets of {1,2,...,n} such that the difference of successive elements is at most 2. See example below. Generally, the o.g.f. for the number of nonempty subsets of {1,2,...,n} such that the difference of successive elements is <= k is: x/((1-x)*(1-2*x+x^(k+1))). Cf. A000217 the case for k=1, A001477 the case for k=0 (counts singleton subsets). - Geoffrey Critzer, Feb 17 2012
-Fibonacci(n-2) = p(-1) where p(x) is the unique degree-n polynomial such that p(k) = a(k) for k = 0, 1, ..., n. - Michael Somos, Dec 31 2012
a(n) is the number of bit strings of length n+1 with the pattern 00 and without the pattern 011, see example. - John M. Campbell, Feb 10 2013
From Jianing Song, Apr 28 2025: (Start)
For n >= 2, a(n-2) is the number of subsets of {1,2,...,n} with 2 or more elements that contain no consecutive elements (i.e., such that the difference of successive elements is at least 2). Note that the number of such subsets with k elements is binomial(n+1-k,k), and Sum_{k=2..floor((n+1)/2)} binomial(n+1-k,k) = F(n+2) - binomial(n+1,0) - binomial(n,1) = F(n+2) - (n+1).
If subsets of {1,2,...,n} are required to contain no consecutive elements module n, then the result is A023548(n-3). (End)

			a(5) = 26 because there are 31 nonempty subsets of {1,2,3,4,5} but 5 of these have successive elements that differ by 3 or more: {1,4}, {1,5}, {2,5}, {1,2,5}, {1,4,5}. - _Geoffrey Critzer_, Feb 17 2012
From _John M. Campbell_, Feb 10 2013: (Start)
There are a(5) = 26 bit strings with the pattern 00 and without the pattern 011 of length 5+1:
   000000, 000001, 000010, 000100, 000101, 001000,
   001001, 001010, 010000, 010001, 010010, 010100,
   100000, 100001, 100010, 100100, 100101, 101000, 101001,
   110000, 110001, 110010, 110100, 111000, 111001, 111100.
(End)

J. Riordan, Discordant permutations, Scripta Math., 20 (1954), 14-23.
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).

T. D. Noe, Table of n, a(n) for n = 0..500
Bader AlBdaiwi, On the Number of Cycles in a Graph, arXiv preprint arXiv:1603.01807 [cs.DM], 2016.
Jean-Luc Baril and Jean-Marcel Pallo, Motzkin subposet and Motzkin geodesics in Tamari lattices, 2013.
Ning-Ning Cao and Feng-Zhen Zhao, Some Properties of Hyperfibonacci and Hyperlucas Numbers, J. Int. Seq. 13 (2010) # 10.8.8.
Hung Viet Chu, Various Sequences from Counting Subsets, arXiv:2005.10081 [math.CO], 2020.
Hung Viet Chu, Partial Sums of the Fibonacci Sequence, arXiv:2106.03659 [math.CO], 2021.
Ligia Loretta Cristea, Ivica Martinjak, and Igor Urbiha, Hyperfibonacci Sequences and Polytopic Numbers, arXiv:1606.06228 [math.CO], 2016.
Emrah Kiliç and Pantelimon Stănică, Generating matrices for weighted sums of second order linear recurrences, JIS 12 (2009) 09.2.7.
Wolfdieter Lang, Problem B-858, Fibonacci Quarterly, 36 (1998), 373-374, Solution, ibid. 37 (1999) 183-184.
Candice A. Marshall, Construction of Pseudo-Involutions in the Riordan Group, Dissertation, Morgan State University, 2017.
Igor Pak, Boris Shapiro, Ilya Smirnov, and Ken-ichi Yoshida, Hilbert-Kunz multiplicity of quadrics via the Ehrhart theory, Stockholm Univ. (Sweden, 2025). See p. 6.
Simon Plouffe, 1031 Generating Functions, Appendix to Thesis, Montreal, 1992.
Simon Plouffe, Approximations de séries génératrices et quelques conjectures, Dissertation, Université du Québec à Montréal, 1992; arXiv:0911.4975 [math.NT], 2009.
J. Riordan, Discordant permutations, Scripta Math., 20 (1954), 14-23. [Annotated scanned copy]
Stacey Wagner, Enumerating Alternating Permutations with One Alternating Descent, DePaul Discoveries: Vol. 2: Iss. 1, Article 2.
Index entries for linear recurrences with constant coefficients, signature (3,-2,-1,1).

Cf. A000045, A023548, A001891, A133640, A141289.
Right-hand column 4 of triangle A011794.
Cf. A014162, A039834, A077880, A122595, A129696.
Cf. A065220.

List([0..40], n-> Fibonacci(n+4) -n-3); # G. C. Greubel, Jul 08 2019

a001924 n = a001924_list !! n
a001924_list = drop 3 $ zipWith (-) (tail a000045_list) [0..]
-- Reinhard Zumkeller, Nov 17 2013

[Fibonacci(n+4)-(n+3): n in [0..40]]; // Vincenzo Librandi, Jun 23 2016

A001924:=-1/(z**2+z-1)/(z-1)**2; # Conjectured by Simon Plouffe in his 1992 dissertation.
##
a:= n-> (<<0|1|0|0>, <0|0|1|0>, <0|0|0|1>, <1|-1|-2|3>>^n.
         <<0, 1, 3, 7>>)[1, 1]:
seq(a(n), n=0..40);  # Alois P. Heinz, Oct 05 2012

a[n_]:= Fibonacci[n+4] -3-n; Array[a, 40, 0]  (* Robert G. Wilson v *)
LinearRecurrence[{3,-2,-1,1},{0,1,3,7},40] (* Harvey P. Dale, Jan 24 2015 *)
Nest[Accumulate,Fibonacci[Range[0,40]],2] (* Harvey P. Dale, Jun 15 2016 *)

a(n)=fibonacci(n+4)-n-3 \\ Charles R Greathouse IV, Feb 24 2011

[fibonacci(n+4) -n-3 for n in (0..40)] # G. C. Greubel, Jul 08 2019

From Wolfdieter Lang: (Start)
G.f.: x/((1-x-x^2)*(1-x)^2).
Convolution of natural numbers n >= 1 with Fibonacci numbers F(k).
a(n) = Fibonacci(n+4) - (3+n). (End)
From Henry Bottomley, Jan 03 2003: (Start)
a(n) = a(n-1) + a(n-2) + n = a(n-1) + A000071(n+2).
a(n) = A001891(n) - a(n-1) = n + A001891(n-1).
a(n) = A065220(n+4) + 1 = A000126(n+1) - 1. (End)
a(n) = Sum_{k=0..n} Sum_{i=0..k} Fibonacci(i). - Benoit Cloitre, Jan 26 2003
a(n) = (sqrt(5)/2 + 1/2)^n*(7*sqrt(5)/10 + 3/2) + (3/2 - 7*sqrt(5)/10)*(sqrt(5)/2 - 1/2)^n*(-1)^n - n - 3. - Paul Barry, Mar 26 2003
a(n) = Sum_{k=0..n} Fibonacci(k)*(n-k). - Benoit Cloitre, Jun 07 2004
A107909(a(n)) = A000225(n) = 2^n - 1. - Reinhard Zumkeller, May 28 2005
a(n) - a(n-1) = A101220(1,1,n). - Ross La Haye, May 31 2006
F(n) + a(n-3) = A133640(n). - Gary W. Adamson, Sep 19 2007
a(n) = A077880(-3-n) = 2*a(n-1) - a(n-3) + 1. - Michael Somos, Dec 31 2012
INVERT transform is A122595. PSUM transform is A014162. PSUMSIGN transform is A129696. BINOMIAL transform of A039834 with 0,1 prepended is this sequence. - Michael Somos, Dec 31 2012
a(n) = A228074(n+1,3) for n > 1. - Reinhard Zumkeller, Aug 15 2013
a(n) = Sum_{k=0..n} Sum_{i=0..n} i * C(n-k,k-i). - Wesley Ivan Hurt, Sep 21 2017
E.g.f.: exp(x/2)*(15*cosh(sqrt(5)*x/2) + 7*sqrt(5)*sinh(sqrt(5)*x/2))/5 - exp(x)*(3 + x). - Stefano Spezia, Jun 25 2022

Description improved by N. J. A. Sloane, Jan 01 1997

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

Clark Kimberling

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

			.   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

Reinhard Zumkeller, Rows n = 0..100 of table, flattened
Index entries for triangles and arrays related to Pascal's triangle

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.

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

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

[&+[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

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

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 *)

{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 */

{T(n, k) = if( k<0 || k>2*n, 0, sum( j=max(0, k-n), k\2, binomial(k-j, j)))}; /* Michael Somos */

[[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

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

Incorporates comments from Michael Somos.

Example extended by Reinhard Zumkeller, Aug 15 2013

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

Reinhard Zumkeller, Aug 15 2013

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

			.    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 .

Reinhard Zumkeller, Rows n = 0..120 of table, flattened
Index entries for triangles and arrays related to Pascal's triangle

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.

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

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])

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

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 *)

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

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

A105809 Riordan array (1/(1 - x - x^2), x/(1 - x)).

Original entry on oeis.org

1, 1, 1, 2, 2, 1, 3, 4, 3, 1, 5, 7, 7, 4, 1, 8, 12, 14, 11, 5, 1, 13, 20, 26, 25, 16, 6, 1, 21, 33, 46, 51, 41, 22, 7, 1, 34, 54, 79, 97, 92, 63, 29, 8, 1, 55, 88, 133, 176, 189, 155, 92, 37, 9, 1, 89, 143, 221, 309, 365, 344, 247, 129, 46, 10, 1, 144, 232, 364, 530, 674, 709, 591

Offset: 0

Paul Barry, May 04 2005

Previous name was: A Fibonacci-Pascal matrix.
From Wolfdieter Lang, Oct 04 2014: (Start)
In the column k of this triangle (without leading zeros) is the k-fold iterated partial sums of the Fibonacci numbers, starting with 1. A000045(n+1), A000071(n+3), A001924(n+1), A014162(n+1), A014166(n+1), ..., n >= 0. See the Riordan property.
For a combinatorial interpretation of these iterated partial sums see the H. Belbachir and A. Belkhir link. There table 1 shows in the rows these columns. In their notation (with r = k) f^(k)(n) = T(k, n+k).
The A-sequence of this Riordan triangle is [1, 1] (see the recurrence for T(n, k), k >= 1, given in the formula section). The Z-sequence is A165326 = [1, repeat(1, -1)]. See the W. Lang link under A006232 for Riordan A- and Z-sequences. (End)

			The triangle T(n,k) begins:
n\k   0   1   2    3    4    5    6    7    8   9  10 11 12 13 ...
0:    1
1:    1   1
2:    2   2   1
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
11: 144 232 364  530  674  709  591  376  175  56  11  1
12: 233 376 596  894 1204 1383 1300  967  551 231  67 12  1
13: 377 609 972 1490 2098 2587 2683 2267 1518 782 298 79 13  1
... reformatted and extended - _Wolfdieter Lang_, Oct 03 2014
------------------------------------------------------------------
Recurrence from Z-sequence (see a comment above): 8 = T(0,5) = (+1)*5 + (+1)*7 + (-1)*7 + (+1)*4 + (-1)*1 = 8. - _Wolfdieter Lang_, Oct 04 2014

Reinhard Zumkeller, Rows n = 0..120 of table, flattened
H. Belbachir and A. Belkhir, Combinatorial Expressions Involving Fibonacci, Hyperfibonacci, and Incomplete Fibonacci Numbers, Journal of Integer Sequences, Vol. 17 (2014), Article 14.4.3.
Hung Viet Chu, Partial Sums of the Fibonacci Sequence, arXiv:2106.03659 [math.CO], 2021.
Anton Vladimirovich Shutov, On some analogue of the Gelfond problem for Zeckendorf representations, Chebyshevskii Sbornik (2024) Vol. 25, No. 5, 195-215. See p. 215. (In Russian)
Index entries for triangles and arrays related to Pascal's triangle

Cf. A165326 (Z-sequence), A027934 (row sums), A010049(n+1) (antidiagonal sums), A212804 (alternating row sums), inverse is A105810.

Some other Fibonacci-Pascal triangles: A027926, A036355, A037027, A074829, A109906, A111006, A114197, A162741, A228074.

a105809 n k = a105809_tabl !! n !! k
a105809_row n = a105809_tabl !! n
a105809_tabl = map fst $ iterate
   (\(u:_, vs) -> (vs, zipWith (+) ([u] ++ vs) (vs ++ [0]))) ([1], [1,1])
-- Reinhard Zumkeller, Aug 15 2013

T := (n,k) -> `if`(n=0,1,binomial(n,k)*hypergeom([1,k/2-n/2,k/2-n/2+1/2], [k+1,-n], -4)); for n from 0 to 13 do seq(simplify(T(n,k)),k=0..n) od; # Peter Luschny, Oct 10 2014

T[n_, k_] := Sum[Binomial[n-j, k+j], {j, 0, n}]; Table[T[n, k], {n, 0, 11}, {k, 0, n}] (* Jean-François Alcover, Jun 11 2019 *)

Riordan array (1/(1-x-x^2), x/(1-x)).
Triangle T(n, k) = Sum_{j=0..n} binomial(n-j, k+j); T(n, 0) = A000045(n+1);
T(n, m) = T(n-1, m-1) + T(n-1, m).
T(n, k) = Sum_{j=0..n} binomial(j, n+k-j). - Paul Barry, Oct 23 2006
G.f. of row polynomials Sum_{k=0..n} T(n, k)*x^k is (1 - z)/((1 - z - z^2)*(1 - (1 + x)*z)) (Riordan property). - Wolfdieter Lang, Oct 04 2014
T(n, k) = binomial(n, k)*hypergeom([1, k/2 - n/2, k/2 - n/2 + 1/2],[k + 1, -n], -4) for n > 0. - Peter Luschny, Oct 10 2014
From Wolfdieter Lang, Feb 13 2025: (Start)
Array A(k, n) = Sum_{j=0..n} F(j+1)*binomial(k-1+n-j, k-1), k >= 0, n >= 0, with F = A000045, (from Riordan triangle k-th convolution in columns without leading 0s).
A(k, n) = F(n+1+2*k) - Sum_{j=0..k-1} F(2*(k-j)-1) * binomial(n+1+j, j), (from iteration of partial sums).
Triangle T(n, k) = A(k, n-k) = Sum_{j=k..n} F(n-j+1) * binomial(j-1, k-1), 0 <= k <= n.
T(n, k) = F(n+1+k) - Sum_{j=0..k-1} F(2*(k-j)-1) * binomial(n - (k-1-j), j). (End)
T(n, k) =  A027926(n, n+k), for 0 <= k <= n. - Wolfdieter Lang, Mar 08 2025

Use first formula as a more descriptive name, Joerg Arndt, Jun 08 2021

A011794 Triangle defined by T(n+1, k) = T(n, k-1) + T(n-1, k), T(n,1) = 1, T(1,k) = 1, T(2,k) = min(2,k).

Original entry on oeis.org

1, 1, 2, 1, 2, 3, 1, 3, 4, 5, 1, 3, 6, 7, 8, 1, 4, 7, 11, 12, 13, 1, 4, 10, 14, 19, 20, 21, 1, 5, 11, 21, 26, 32, 33, 34, 1, 5, 15, 25, 40, 46, 53, 54, 55, 1, 6, 16, 36, 51, 72, 79, 87, 88, 89, 1, 6, 21, 41, 76, 97, 125, 133, 142, 143, 144, 1, 7, 22, 57, 92, 148, 176, 212, 221, 231, 232, 233

Offset: 1

N. J. A. Sloane and David Broadhurst

			matrix(10,10,n,k,a(n-1,k-1))
  [ 0 0 0 0 0 0 0 0 0 0 ]
  [ 0 1 1 1 1 1 1 1 1 1 ]
  [ 0 1 2 2 2 2 2 2 2 2 ]
  [ 0 1 2 3 3 3 3 3 3 3 ]
  [ 0 1 3 4 5 5 5 5 5 5 ]
  [ 0 1 3 6 7 8 8 8 8 8 ]
Triangle begins as:
  1;
  1, 2;
  1, 2,  3;
  1, 3,  4,  5;
  1, 3,  6,  7,  8;
  1, 4,  7, 11, 12, 13;
  1, 4, 10, 14, 19, 20, 21;
  1, 5, 11, 21, 26, 32, 33, 34;
  1, 5, 15, 25, 40, 46, 53, 54, 55;
  1, 6, 16, 36, 51, 72, 79, 87, 88, 89;

G. C. Greubel, Rows n = 1..100 of the triangle, flattened
D. J. Broadhurst, On the enumeration of irreducible k-fold Euler sums and their roles in knot theory and field theory, arXiv:hep-th/9604128, 1996.

Columns include A008619 and (essentially) A055802, A055803, A055804, A055805, A055806.
Right-hand columns 1-14 are A000045, A000071, A001911, A001924, A001891, A014162, A053808, A014166, A053809, A053739, A054469, A053295, A054470, A053296.
Essentially a reflected version of A055801.
Sums include: A039834 (signed row), A131913 (row).
Cf. A008466, A074878, A103609.

function T(n,k) // T = A011794(n,k)
  if k eq 1 or n eq 1 then return 1;
  elif n eq 2 then return Min(2, k);
  else return T(n-1,k-1) + T(n-2,k);
  end if;
end function;
[T(n,k): k in [1..n], n in [1..15]]; // G. C. Greubel, Oct 21 2024

T[n_, k_]:= T[n, k]= T[n-1, k-1] + T[n-2, k]; T[n_, 1] = 1; T[1, k_] = 1; T[2, k_] := Min[2, k]; Table[T[n, k], {n,15}, {k,n}]//Flatten (* Jean-François Alcover, Feb 26 2013 *)

T(n,k)=if(n<=0 || k<=0,0, if(n<=2 || k==1, min(n,k), T(n-1,k-1)+T(n-2,k)))

def T(n, k): # T = A011794
    if (k==1 or n==1): return 1
    elif (n==2): return min(2,k)
    else: return T(n-1, k-1) + T(n-2, k)
flatten([[T(n, k) for k in range(1,n+1)] for n in range(1,16)]) # G. C. Greubel, Oct 21 2024

T(n,n) = Fibonacci(n+1). - Jean-François Alcover, Feb 26 2013
From G. C. Greubel, Oct 21 2024: (Start)
Sum_{k=1..n} T(n, k) = A131913(n-1).
Sum_{k=1..n} (-1)^(k-1)*T(n, k) = A039834(n).
Sum_{k=1..floor((n+1)/2)} T(n-k+1,k) = (1/2)*((1-(-1)^n)*A074878((n+3)/2) + (1+(-1)^n)*A008466((n+6)/2)) (diagonal row sums).
Sum_{k=1..floor((n+1)/2)} (-1)^(k-1)*T(n-k+1,k) = (-1)^floor((n-1)/2)*A103609(n) + [n=1] (signed diagonal row sums). (End)

Entry improved by comments from Michael Somos

More terms added by G. C. Greubel, Oct 21 2024

A136431 Hyperfibonacci square number array a(k,n) = F(n)^(k), read by ascending antidiagonals (k, n >= 0).

Original entry on oeis.org

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

Offset: 0

Jonathan Vos Post, Apr 01 2008

Main diagonal is A108081. Antidiagonal sums form A027934. - Gerald McGarvey, Oct 01 2008

Seen as triangle read by rows: T(n,0) = 1, T(n,n) = A000045(n) and for 0 < k < n: T(n,k) = T(n-1,k-1) + T(n-1,k). - Reinhard Zumkeller, Jul 16 2013

			The array F(n)^(k) begins:
.....|n=0|n=1|.n=2|.n=3|.n=4.|.n=5.|..n=6.|.n=7..|..n=8..|..n=9..|.n=10..|.in.OEIS
k=0..|.0.|.1.|..1.|..2.|...3.|...5.|....8.|...13.|....21.|....34.|....55.|.A000045
k=1..|.0.|.1.|..2.|..4.|...7.|..12.|...20.|...33.|....54.|....88.|...143.|.A000071
k=2..|.0.|.1.|..3.|..7.|..14.|..26.|...46.|...79.|...133.|...221.|...364.|.A001924
k=3..|.0.|.1.|..4.|.11.|..25.|..51.|...97.|..176.|...309.|...530.|...894.|.A014162
k=4..|.0.|.1.|..5.|.16.|..41.|..92.|..189.|..365.|...674.|..1204.|..2098.|.A014166
k=5..|.0.|.1.|..6.|.22.|..63.|.155.|..344.|..709.|..1383.|..2587.|..4685.|.A053739
k=6..|.0.|.1.|..7.|.29.|..92.|.247.|..591.|.1300.|..2683.|..5270.|..9955.|.A053295
k=7..|.0.|.1.|..8.|.37.|.129.|.376.|..967.|.2267.|..4950.|.10220.|.20175.|.A053296
k=8..|.0.|.1.|..9.|.46.|.175.|.551.|.1518.|.3785.|..8735.|.18955.|.39130.|.A053308
k=9..|.0.|.1.|.10.|.56.|.231.|.782.|.2300.|.6085.|.14820.|.33775.|.72905.|.A053309

Reinhard Zumkeller, Rows n = 0..125 of triangle, flattened
H. Belbachir, A. Belkhir, Combinatorial Expressions Involving Fibonacci, Hyperfibonacci, and Incomplete Fibonacci Numbers, J. Int. Seq. 17 (2014), Article #14.4.3.
Ayhan Dil and Istvan Mezo, A Symmetric Algorithm for Hyperharmonic and Fibonacci Numbers, arXiv:0803.4388 [math.NT], 2008.
Ayhan Dil and Istvan Mezo, A Symmetric Algorithm for Hyperharmonic and Fibonacci Numbers, Applied Mathematics and Computation 206(2) (2008), 942-951.
Eric Weisstein's World of Mathematics, Steenrod Algebra

Cf. A000045, A000071, A001924, A014162, A014166, A053739, A053295, A053296, A053308, A053309, A123736, A137176.

a136431 n k = a136431_tabl !! n !! k
a136431_row n = a136431_tabl !! n
a136431_tabl = map fst $ iterate h ([0], 1) where
   h (row, fib) = (zipWith (+) ([0] ++ row) (row ++ [fib]), last row)
-- Reinhard Zumkeller, Jul 16 2013

A136431 := proc(k,n) local x ; coeftayl(x/(1-x-x^2)/(1-x)^k,x=0,n) ; end: for d from 0 to 20 do for n from 0 to d do printf("%d,",A136431(d-n,n)) ; od: od: # R. J. Mathar, Apr 25 2008

t[n_, k_] := CoefficientList[Series[x/(1 - x - x^2)/(1 - x)^k, {x, 0, n + 1}], x][[n + 1]]; Table[ t[n, k - n], {k, 0, 11}, {n, 0, k}] // Flatten
(* To view the table above *) Table[ t[n, k], {k, 0, 9}, {n, 0, 10}] // TableForm

a(k,n) = Apply partial sum operator k times to Fibonacci numbers.

For k > 0 and n > 1, a(k,n) = a(k-1,n) + a(k,n-1). - Gerald McGarvey, Oct 01 2008

A255874 Triangular array T: T(n,k) = number of subset S of {1,2,...,n+1} such that |S| > 1 and max(S) = k, where S is the set {x(2)-x(1), x(3)-x(2), ..., x(h+1)-x(h)} when the elements of S are written as x(1) < x(2) < ... < x(h+1).

Original entry on oeis.org

1, 3, 1, 6, 4, 1, 10, 11, 4, 1, 15, 25, 12, 4, 1, 21, 51, 31, 12, 4, 1, 28, 97, 73, 32, 12, 4, 1, 36, 176, 162, 79, 32, 12, 4, 1, 45, 309, 345, 185, 80, 32, 12, 4, 1, 55, 530, 713, 418, 191, 80, 32, 12, 4, 1, 66, 894, 1441, 920, 441, 192, 80, 32, 12, 4, 1

Offset: 1

Clark Kimberling, Mar 08 2015

Column 1: A000217. Conjectures: Column 2 = A014162, and the rows have a limiting tail (1,4,12,32,...) = A001787.

			First nine rows:
1
3   1
6   4   1
10  11  4   1
15  25  12  4   1
21  51  31  12  4   1
28  97  73  32  12  4   1
36  172 162 79  32  12  4  1
45  309 345 185 80  32  12  4  1
T(3,1) counts these 6 subsets:  {1,2}, {2,3}, {3,4}, {1,2,3}, {2,3,4}, {1,2,3,4};
T(3,2) counts these 4 subsets:  {1,3}, {2,4}, {1,2,4}, {1,3,4};
T(3,3) = counts this subset: {1,4}.

Cf. A000217, A014162.

s[n_] := Subsets[Range[1, n]]; v[n_] := Map[Max, Map[Differences, s[n]]]
t = Table[Count[v[n], k], {n, 1, 15}, {k, 1, n - 1}]
Flatten[t]   (* A255874 sequence *)
TableForm[t] (* A255874 array *)

A257838 Main diagonal of iterated partial sums array of Fibonacci numbers (starting with the first partial sums).

Original entry on oeis.org

0, 1, 4, 16, 63, 247, 967, 3785, 14820, 58060, 227612, 892926, 3505386, 13770404, 54129602, 212904952, 837885495, 3299264407, 12997784803, 51230474669, 202014314769, 796928589755, 3145066003589, 12416625685891, 49037912997003, 193734379979677, 765632076098287, 3026670770970925, 11968378998073935

Offset: 0

Luciano Ancora, May 10 2015

The array used here starts in row n=0 with the first partial sums of A000045. The array which starts with the Fibonacci numbers in row k=0 is shown in A136431. The diagonal of that array is given in A176085. - Wolfdieter Lang, Jun 03 2015

			This sequence is the main diagonal of the following array (see the comment and Example field of A136431):
0, 1, 2,  4,  7,  12, ...  A000071
0, 1, 3,  7, 14,  26, ...  A001924
0, 1, 4, 11, 25,  51, ...  A014162
0, 1, 5, 16, 41,  92, ...  A014166
0, 1, 6, 22, 63, 155, ...  A053739
0, 1, 7, 29, 92, 247, ...  A053295

G. C. Greubel, Table of n, a(n) for n = 0..1000

Cf. A000045, A000071, A001924, A014162, A014166, A136431, A176085.

Table[DifferenceRoot[Function[{a, n},{(2*n + 4*n^2)*a[n] + (2 + 7*n + 15*n^2)*a[1 + n] + (8 - 6*n - 8*n^2)*a[2 + n] + (-2 + n + n^2)*a[3 + n] == 0, a[1] == 0, a[2] == 1, a[3] == 4, a[4] == 16}]][n], {n, 30}]

a(n):=sum(binomial(2*n-k,n-k)*fib(k),k,0,n); /* Vladimir Kruchinin, Oct 09 2016 */

x='x+O('x^50); concat([0], Vec(-(4*x+sqrt(1-4*x)-1)/(8*x^2+sqrt(1-4*x)*(8*x-2)-2*x))) \\ G. C. Greubel, Apr 08 2017

a(n) = F^{n+1}(n), n >= 0, with the k-th iterated partial sum F^{k} of the Fibonacci number A000045. - Wolfdieter Lang, Jun 03 2015
Conjecture: n*(n-3)*a(n) +2*(-4*n^2+13*n-6)*a(n-1) +(15*n^2-53*n+48)*a(n-2) +2*(2*n-3)*(n-2)*a(n-3)=0. - R. J. Mathar, Dec 10 2015
G.f.: -(4*x+sqrt(1-4*x)-1)/(8*x^2+sqrt(1-4*x)*(8*x-2)-2*x). - Vladimir Kruchinin, Oct 09 2016
a(n) = Sum_{k=0..n} binomial(2*n-k,n-k)*F(k), where F(k) = A000045(k). - Vladimir Kruchinin, Oct 09 2016
a(n) ~ 2^(2*n+1)/sqrt(Pi*n). - Vaclav Kotesovec, Oct 09 2016

Name edited by Wolfdieter Lang, Jun 03 2015

A104797 Triangle T(n,k) = Fib(n-k+4)-n-k-3, n>=1, 0<=k

Original entry on oeis.org

1, 3, 1, 7, 3, 1, 14, 7, 3, 1, 26, 14, 7, 3, 1, 46, 26, 14, 7, 3, 1, 79, 46, 26, 14, 7, 3, 1, 133, 79, 46, 26, 14, 7, 3, 1, 221, 133, 79, 46, 26, 14, 7, 3, 1, 364, 221, 133, 79, 46, 26, 14, 7, 3, 1, 596, 364, 221, 133, 79, 46, 26, 14, 7, 3, 1, 972, 596, 364, 221, 133, 79, 46, 26

Offset: 1

Gary W. Adamson, Mar 26 2005

Repeatedly writing the sequence A001924 backwards.

			First few rows of the triangle are:
1;
3, 1;
7, 3, 1;
14, 7, 3, 1;
26, 14, 7, 3, 1;
46, 26, 14, 7, 3, 1;
...

Row sums are in A014162.

Cf. A104732.

Edited by Ralf Stephan, Apr 05 2009

Showing 1-10 of 12 results. Next

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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A001924 Apply partial sum operator twice to Fibonacci numbers.

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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.

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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.

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A105809 Riordan array (1/(1 - x - x^2), x/(1 - x)).

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A255874 Triangular array T: T(n,k) = number of subset S of {1,2,...,n+1} such that |S| > 1 and max(S) = k, where S is the set {x(2)-x(1), x(3)-x(2), ..., x(h+1)-x(h)} when the elements of S are written as x(1) < x(2) < ... < x(h+1).