cp's OEIS Frontend

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

Showing 1-10 of 35 results. Next

A104726 Triangle generated as the matrix product of A026729 and A000012 (triangular views), read by rows.

Original entry on oeis.org

1, 1, 1, 2, 2, 1, 3, 3, 3, 1, 5, 5, 5, 4, 1, 8, 8, 8, 8, 5, 1, 13, 13, 13, 13, 12, 6, 1, 21, 21, 21, 21, 21, 17, 7, 1, 34, 34, 34, 34, 34, 33, 23, 8, 1, 55, 55, 55, 55, 55, 55, 50, 30, 9, 1, 89, 89, 89, 89, 89, 89, 88, 73, 38
Offset: 0

Views

Author

Gary W. Adamson, Mar 20 2005

Keywords

Comments

If the triangular factors A026729 and A000012 are commuted in the product, A004070 results.
Riordan array (1/(1-x-x^2), x*(1+x)). - Philippe Deléham, Mar 06 2013

Examples

			First few rows of the triangle are
1;
1, 1;
2, 2, 1;
3, 3, 3, 1;
5, 5, 5, 4, 1;
8, 8, 8, 8, 5, 1;
13, 13, 13, 13, 12, 6, 1;
21, 21, 21, 21, 21, 17, 7, 1;
...
Production array begins
1, 1
1, 1, 1
-1, -1, 1, 1
2, 2, -1, 1, 1
-5, -5, 2, -1, 1, 1
14, 14, -5, 2, -1, 1, 1
-42, -42, 14, -5, 2, -1, 1, 1
132, 132, -42, 14, -5, 2, -1, 1, 1
-429, -429, 132, -42, 14, -5, 2, -1, 1, 1
... which is based on A000108 or A168491. - _Philippe Deléham_, Mar 06 2013
		

Crossrefs

Cf. A001629 (row sums), A026729, A004070, A000071.

Programs

  • Maple
    A104726 := proc(n,k)
            add( binomial(j,n-j),j=k..n) ;
    end proc:
    seq(seq(A104726(n,k),k=0..n),n=0..10) ; # R. J. Mathar, Oct 30 2011

Formula

T(n,k) = sum_{j=k..n} binomial(j,n-j). - R. J. Mathar, Oct 30 2011
T(n,0) = T(n-1,0) + T(n-2,0), T(n,k) = T(n-1,k-1) + T(n-2,k-1) for k>0. - Philippe Deléham, Mar 06 2013
T(2*n,n) = A000045(2n+1) = A001519(n+1) = A122367(n). - Philippe Deléham, Mar 06 2013

A144955 Binomial transform of a modified version of A026729.

Original entry on oeis.org

1, 2, 0, 3, 1, 0, 4, 4, 1, 0, 5, 10, 6, 1, 0, 6, 20, 21, 8, 1, 0, 7, 35, 56, 36, 10, 1, 0, 8, 56, 126, 120, 55, 12, 1, 9, 84, 252, 330, 220, 78, 14, 1, 10, 120, 462, 792, 715, 364, 105, 16, 1, 11, 165, 792, 1716, 2002, 1365, 560, 136, 18, 1
Offset: 0

Views

Author

Gary W. Adamson, Sep 27 2008

Keywords

Comments

Row sums = A055588: (1, 2, 4, 9, 22, 56, 145,...).

Examples

			First few rows of the triangle =
1;
2, 0;
3, 1, 0;
4, 4, 1, 0;
5, 10, 6, 1, 0;
6, 20, 21, 8, 1, 0;
7, 35, 56, 36, 10, 1, 0;
8, 56, 126, 120, 55, 12, 1, 0;
...
		

Crossrefs

Formula

Triangle A026729 is shifted one place down, with a (1, 1, 0, 0, 0...) in the first column; getting:
1;
1, 0;
0, 1, 0;
0, 1, 1, 0;
0, 0, 2, 1, 0;
0, 0, 1, 3, 1, 0;
...
A144955 = A007318 * the modified triangle.

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

Original entry on oeis.org

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

Views

Author

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

Keywords

Comments

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

Examples

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

References

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

Crossrefs

Equals differences between consecutive terms of A102363. - David G. Williams (davidwilliams(AT)Paxway.com), Jan 23 2006
Row sums give A000079 (powers of 2).
Cf. A083093 (triangle read mod 3), A214292 (first differences of rows).
Partial sums of rows give triangle A008949.
The triangle of the antidiagonals is A011973.
Infinite matrix squared: A038207, cubed: A027465.
Cf. A101164. If rows are sorted we get A061554 or A107430.
Another version: A108044.
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
Cf. A115940 (pandigital binomial coefficients C(m,k) with k>1).
Cf. (simplex colorings) A325002 (oriented), [k==n+1] (chiral), A325003 (achiral), A325000 (k or fewer colors), A325009 (orthotope facets, orthoplex vertices), A325017 (orthoplex facets, orthotope vertices).
Triangles of generalized binomial coefficients (n,k)_m (or generalized Pascal triangles) for m = 2..12: A001263, A056939, A056940, A056941, A142465, A142467, A142468, A174109, A342889, A342890, A342891.

Programs

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

Formula

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

Extensions

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

A000931 Padovan sequence (or Padovan numbers): a(n) = a(n-2) + a(n-3) with a(0) = 1, a(1) = a(2) = 0.

Original entry on oeis.org

1, 0, 0, 1, 0, 1, 1, 1, 2, 2, 3, 4, 5, 7, 9, 12, 16, 21, 28, 37, 49, 65, 86, 114, 151, 200, 265, 351, 465, 616, 816, 1081, 1432, 1897, 2513, 3329, 4410, 5842, 7739, 10252, 13581, 17991, 23833, 31572, 41824, 55405, 73396, 97229, 128801, 170625
Offset: 0

Keywords

Comments

Number of compositions of n into parts congruent to 2 mod 3 (offset -1). - Vladeta Jovovic, Feb 09 2005
a(n) is the number of compositions of n into parts that are odd and >= 3. Example: a(10)=3 counts 3+7, 5+5, 7+3. - David Callan, Jul 14 2006
Referred to as N0102 in R. K. Guy's "Anyone for Twopins?" - Rainer Rosenthal, Dec 05 2006
Zagier conjectures that a(n+3) is the maximum number of multiple zeta values of weight n > 1 which are linearly independent over the rationals. - Jonathan Sondow and Sergey Zlobin (sirg_zlobin(AT)mail.ru), Dec 20 2006
Starting with offset 6: (1, 1, 2, 2, 3, 4, 5, ...) = INVERT transform of A106510: (1, 1, -1, 0, 1, -1, 0, 1, -1, ...). - Gary W. Adamson, Oct 10 2008
Starting with offset 7, the sequence 1, 2, 2, 3, 4, 5, 7, 9, 12, 16, 21, 28, ... is called the Fibonacci quilt sequence by Catral et al., in Fib. Q. 2017. - N. J. A. Sloane, Dec 24 2021
Triangle A145462: right border = A000931 starting with offset 6. Row sums = Padovan sequence starting with offset 7. - Gary W. Adamson, Oct 10 2008
Starting with offset 3 = row sums of triangle A146973 and INVERT transform of [1, -1, 2, -2, 3, -3, ...]. - Gary W. Adamson, Nov 03 2008
a(n+5) corresponds to the diagonal sums of "triangle": 1; 1; 1,1; 1,1; 1,2,1; 1,2,1; 1,3,3,1; 1,3,3,1; 1,4,6,4,1; ..., rows of Pascal's triangle (A007318) repeated. - Philippe Deléham, Dec 12 2008
With offset 3: (1, 0, 1, 1, 1, 2, 2, ...) convolved with the tribonacci numbers prefaced with a "1": (1, 1, 1, 2, 4, 7, 13, ...) = the tribonacci numbers, A000073. (Cf. triangle A153462.) - Gary W. Adamson, Dec 27 2008
a(n) is also the number of strings of length (n-8) from an alphabet {A, B} with no more than one A or 2 B's consecutively. (E.g., n = 4: {ABAB,ABBA,BABA,BABB,BBAB} and a(4+8) = 5.) - Toby Gottfried, Mar 02 2010
p(n):=A000931(n+3), n >= 1, is the number of partitions of the numbers {1,2,3,...,n} into lists of length two or three containing neighboring numbers. The 'or' is inclusive. For n=0 one takes p(0)=1. For details see the W. Lang link. There the explicit formula for p(n) (analog of the Binet-de Moivre formula for Fibonacci numbers) is also given. Padovan sequences with different inputs are also considered there. - Wolfdieter Lang, Jun 15 2010
Equals the INVERTi transform of Fibonacci numbers prefaced with three 1's, i.e., (1 + x + x^2 + x^3 + x^4 + 2x^5 + 3x^6 + 5x^7 + 8x^8 + 13x^9 + ...). - Gary W. Adamson, Apr 01 2011
When run backwards gives (-1)^n*A050935(n).
a(n) is the top left entry of the n-th power of the 3 X 3 matrix [0, 0, 1; 1, 0, 1; 0, 1, 0] or of the 3 X 3 matrix [0, 1, 0; 0, 0, 1; 1, 1, 0]. - R. J. Mathar, Feb 03 2014
Figure 4 of Brauchart et al., 2014, shows a way to "visualize the Padovan sequence as cuboid spirals, where the dimensions of each cuboid made up by the previous ones are given by three consecutive numbers in the sequence". - N. J. A. Sloane, Mar 26 2014
a(n) is the number of closed walks from a vertex of a unidirectional triangle containing an opposing directed edge (arc) between the second and third vertices. Equivalently the (1,1) entry of A^n where the adjacency matrix of digraph is A=(0,1,0;0,0,1;1,1,0). - David Neil McGrath, Dec 19 2014
Number of compositions of n-3 (n >= 4) into 2's and 3's. Example: a(12)=5 because we have 333, 3222, 2322, 2232, and 2223. - Emeric Deutsch, Dec 28 2014
The Hoffman (2015) paper "offers significant evidence that the number of quantities needed to generate the weight-n multiple harmonic sums mod p is" a(n). - N. J. A. Sloane, Jun 24 2016
a(n) gives the number of compositions of n-5 into odd parts where the order of the 1's does not matter. For example, a(11)=4 counts the following compositions of 6: (5,1)=(1,5), (3,3), (3,1,1,1)=(1,3,1,1)=(1,1,3,1)=(1,1,1,3), (1,1,1,1,1,1). - Gregory L. Simay, Aug 04 2016
For n > 6, a(n) is the number of maximal matchings in the (n-5)-path graph, maximal independent vertex sets and minimal vertex covers in the (n-6)-path graph, and minimal edge covers in the (n-5)-pan graph and (n-3)-path graphs. - Eric W. Weisstein, Mar 30, Aug 03, and Aug 07 2017
From James Mitchell and Wilf A. Wilson, Jul 21 2017: (Start)
a(2n + 5) + 2n - 4, n > 2, is the number of maximal subsemigroups of the monoid of order-preserving mappings on a set with n elements.
a(n + 6) + n - 3, n > 3, is the number of maximal subsemigroups of the monoid of order-preserving or reversing mappings on a set with n elements.
(End)
Has the property that the largest of any four consecutive terms equals the sum of the two smallest. - N. J. A. Sloane, Aug 29 2017 [David Nacin points out that there are many sequences with this property, such as 1,1,1,2,1,1,1,2,1,1,1,2,... or 2,3,4,5,2,3,4,5,2,3,4,5,... or 2,2,1,3,3, 4,1,4, 5,5,1,6,6, 7,1,7, 8,8,1,9,9, 10,1,10, ... (spaces added for clarity), and a conjecture I made here in 2017 was simply wrong. I have deleted it. - N. J. A. Sloane, Oct 23 2018]
a(n) is also the number of maximal cliques in the (n+6)-path complement graph. - Eric W. Weisstein, Apr 12 2018
a(n+8) is the number of solus bitstrings of length n with no runs of 3 zeros. - Steven Finch, Mar 25 2020
Named after the architect Richard Padovan (b. 1935). - Amiram Eldar, Jun 08 2021
Shannon et al. (2006) credit a French architecture student Gérard Cordonnier with the discovery of these numbers.
For n >= 3, a(n) is the number of sequences of 0s and 1s of length (n-2) that begin with a 0, end with a 0, contain no two consecutive 0s, and contain no three consecutive 1s. - Yifan Xie, Oct 20 2022
For n >= 2, a(n+5) is the number of ways to tile the 1xn board with dominoes and squares (ie. size 1x1) such that are either none or one squares between dominoes, none or one squares at both ends of the board, and there is at least one domino. For example, for n=6, a(11)=4 since the tilings are |2|2, |22|, 2|2| and 222 (where 2 represents a domino and | a square). - Enrique Navarrete, Aug 31 2024

Examples

			G.f. = 1 + x^3 + x^5 + x^6 + x^7 + 2*x^8 + 2*x^9 + 3*x^10 + 4*x^11 + ...
		

References

  • A. T. Benjamin and J. J. Quinn, Proofs that really count: the art of combinatorial proof, M.A.A. 2003, p. 47, ex. 4.
  • Minerva Catral, Pari L. Ford, Pamela E. Harris, Steven J. Miller, Dawn Nelson, Zhao Pan, and Huanzhong Xu, Legal Decompositions Arising from Non-positive Linear Recurrences, Fib. Quart., 55:3 (2017), 252-275. [Note that there is an earlier version of this paper, with only five authors, on the arXiv in 2016. Note to editors: do not merge these two citations. - N. J. A. Sloane, Dec 24 2021]
  • Richard K. Guy, "Anyone for Twopins?" in D. A. Klarner, editor, The Mathematical Gardner. Prindle, Weber and Schmidt, Boston, 1981, pp. 10-11.
  • Silvia Heubach and Toufik Mansour, Combinatorics of Compositions and Words, CRC Press, 2010.
  • A. G. Shannon, P. G. Anderson and A. F. Horadam, Properties of Cordonnier, Perrin and Van der Laan numbers, International Journal of Mathematical Education in Science and Technology, Volume 37:7 (2006), 825-831. See P_n.
  • 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).
  • Ian Stewart, L'univers des nombres, "La sculpture et les nombres", pp. 19-20, Belin-Pour La Science, Paris, 2000.
  • Hans van der Laan, Het plastische getal. XV lessen over de grondslagen van de architectonische ordonnantie. Leiden, E.J. Brill, 1967.
  • Don Zagier, Values of zeta functions and their applications, in First European Congress of Mathematics (Paris, 1992), Vol. II, A. Joseph et al. (eds.), Birkhäuser, Basel, 1994, pp. 497-512.

Crossrefs

The following are basically all variants of the same sequence: A000931, A078027, A096231, A124745, A133034, A134816, A164001, A182097, A228361 and probably A020720. However, each one has its own special features and deserves its own entry.
Closely related to A001608.
Doubling every term gives A291289.

Programs

  • GAP
    a:=[1,0,0];; for n in [4..50] do a[n]:=a[n-2]+a[n-3]; od; a; # G. C. Greubel, Dec 30 2019
    
  • Haskell
    a000931 n = a000931_list !! n
    a000931_list = 1 : 0 : 0 : zipWith (+) a000931_list (tail a000931_list)
    -- Reinhard Zumkeller, Feb 10 2011
    
  • Magma
    I:=[1,0,0]; [n le 3 select I[n] else Self(n-2) + Self(n-3): n in [1..60]]; // Vincenzo Librandi, Jul 21 2015
    
  • Maple
    A000931 := proc(n) option remember; if n = 0 then 1 elif n <= 2 then 0 else procname(n-2)+procname(n-3); fi; end;
    A000931:=-(1+z)/(-1+z^2+z^3); # Simon Plouffe in his 1992 dissertation; gives sequence without five leading terms
    a[0]:=1; a[1]:=0; a[2]:=0; for n from 3 to 50 do a[n]:=a[n-2]+a[n-3]; end do; # Francesco Daddi, Aug 04 2011
  • Mathematica
    CoefficientList[Series[(1-x^2)/(1-x^2-x^3), {x, 0, 50}], x]
    a[0]=1; a[1]=a[2]=0; a[n_]:= a[n]= a[n-2] + a[n-3]; Table[a[n], {n, 0, 50}] (* Robert G. Wilson v, May 04 2006 *)
    LinearRecurrence[{0,1,1}, {1,0,0}, 50] (* Harvey P. Dale, Jan 10 2012 *)
    Table[RootSum[-1 -# +#^3 &, 5#^n -6#^(n+1) +4#^(n+2) &]/23, {n,0,50}] (* Eric W. Weisstein, Nov 09 2017 *)
  • PARI
    Vec((1-x^2)/(1-x^2-x^3) + O(x^50)) \\ Charles R Greathouse IV, Feb 11 2011
    
  • PARI
    {a(n) = if( n<0, polcoeff(1/(1+x-x^3) + x * O(x^-n), -n), polcoeff( (1 - x^2)/(1-x^2-x^3) + x * O(x^n), n))}; /* Michael Somos, Sep 18 2012 */
    
  • Python
    def aupton(nn):
        alst = [1, 0, 0]
        for n in range(3, nn+1): alst.append(alst[n-2]+alst[n-3])
        return alst
    print(aupton(49)) # Michael S. Branicky, Mar 28 2022
  • Sage
    def A000931_list(prec):
        P. = PowerSeriesRing(ZZ, prec)
        return P( (1-x^2)/(1-x^2-x^3) ).list()
    A000931_list(50) # G. C. Greubel, Dec 30 2019
    

Formula

G.f.: (1-x^2)/(1-x^2-x^3).
a(n) is asymptotic to r^n / (2*r+3) where r = 1.3247179572447... = A060006, the real root of x^3 = x + 1. - Philippe Deléham, Jan 13 2004
a(n)^2 + a(n+2)^2 + a(n+6)^2 = a(n+1)^2 + a(n+3)^2 + a(n+4)^2 + a(n+5)^2 (Barniville, Question 16884, Ed. Times 1911).
a(n+5) = a(0) + a(1) + ... + a(n).
a(n) = central and lower right terms in the (n-3)-th power of the 3 X 3 matrix M = [0 1 0 / 0 0 1 / 1 1 0]. E.g., a(13) = 7. M^10 = [3 5 4 / 4 7 5 / 5 9 7]. - Gary W. Adamson, Feb 01 2004
G.f.: 1/(1 - x^3 - x^5 - x^7 - x^9 - ...). - Jon Perry, Jul 04 2004
a(n+4) = Sum_{k=0..floor((n-1)/2)} binomial(floor((n+k-2)/3), k). - Paul Barry, Jul 06 2004
a(n+3) = Sum_{k=0..floor(n/2)} binomial(k, n-2k). - Paul Barry, Sep 17 2004, corrected by Greg Dresden and Zi Ye, Jul 06 2021
a(n+3) is diagonal sum of A026729 (as a number triangle), with formula a(n+3) = Sum_{k=0..floor(n/2)} Sum_{i=0..n-k} (-1)^(n-k+i)*binomial(n-k, i)*binomial(i+k, i-k). - Paul Barry, Sep 23 2004
a(n) = a(n-1) + a(n-5) = A003520(n-4) + A003520(n-13) = A003520(n-3) - A003520(n-9). - Henry Bottomley, Jan 30 2005
a(n+3) = Sum_{k=0..floor(n/2)} binomial((n-k)/2, k)(1+(-1)^(n-k))/2. - Paul Barry, Sep 09 2005
The sequence 1/(1-x^2-x^3) (a(n+3)) is given by the diagonal sums of the Riordan array (1/(1-x^3), x/(1-x^3)). The row sums are A000930. - Paul Barry, Feb 25 2005
a(n) = A023434(n-7) + 1 for n >= 7. - David Callan, Jul 14 2006
a(n+5) corresponds to the diagonal sums of A030528. The binomial transform of a(n+5) is A052921. a(n+5) = Sum_{k=0..floor(n/2)} Sum_{k=0..n} (-1)^(n-k+i)*binomial(n-k, i)binomial(i+k+1, 2k+1). - Paul Barry, Jun 21 2004
r^(n-1) = (1/r)*a(n) + r*a(n+1) + a(n+2), where r = 1.32471... is the real root of x^3 - x - 1 = 0. Example: r^8 = (1/r)*a(9) + r*a(10) + a(11) = (1/r)*2 + r*3 + 4 = 9.483909... - Gary W. Adamson, Oct 22 2006
a(n) = (r^n)/(2r+3) + (s^n)/(2s+3) + (t^n)/(2t+3) where r, s, t are the three roots of x^3-x-1. - Keith Schneider (schneidk(AT)email.unc.edu), Sep 07 2007
a(n) = -k*a(n-1) + a(n-2) + (k+1)a(n-2) + k*a(n-4), n > 3, for any value of k. - Gary Detlefs, Sep 13 2010
From Francesco Daddi, Aug 04 2011: (Start)
a(0) + a(2) + a(4) + a(6) + ... + a(2*n) = a(2*n+3).
a(0) + a(3) + a(6) + a(9) + ... + a(3*n) = a(3*n+2)+1.
a(0) + a(5) + a(10) + a(15) + ... + a(5*n) = a(5*n+1)+1.
a(0) + a(7) + a(14) + a(21) + ... + a(7*n) = (a(7*n) + a(7*n+1) + 1)/2. (End)
a(n+3) = Sum_{k=0..floor((n+1)/2)} binomial((n+k)/3,k), where binomial((n+k)/3,k)=0 for noninteger (n+k)/3. - Nikita Gogin, Dec 07 2012
a(n) = A182097(n-3) for n > 2. - Jonathan Sondow, Mar 14 2014
a(n) = the k-th difference of a(n+5k) - a(n+5k-1), k>=1. For example, a(10)=3 => a(15)-a(14) => 2nd difference of a(20)-a(19) => 3rd difference of a(25)-a(24)... - Bob Selcoe, Mar 18 2014
Construct the power matrix T(n,j) = [A^*j]*[S^*(j-1)] where A=(0,0,1,0,1,0,1,...) and S=(0,1,0,0,...) or A063524. [* is convolution operation] Define S^*0=I with I=(1,0,0,...). Then a(n) = Sum_{j=1...n} T(n,j). - David Neil McGrath, Dec 19 2014
If x=a(n), y=a(n+1), z=a(n+2), then x^3 + 2*y*x^2 - z^2*x - 3*y*z*x + y^2*x + y^3 - y^2*z + z^3 = 1. - Alexander Samokrutov, Jul 20 2015
For the sequence shifted by 6 terms, a(n) = Sum_{k=ceiling(n/3)..ceiling(n/2)} binomial(k+1,3*k-n) [Doslic-Zubac]. - N. J. A. Sloane, Apr 23 2017
From Joseph M. Shunia, Jan 21 2020: (Start)
a(2n) = 2*a(n-1)*a(n) + a(n)^2 + a(n+1)^2, for n > 8.
a(2n-1) = 2*a(n)*a(n+1) + a(n-1)^2, for n > 8.
a(2n+1) = 2*a(n+1)*a(n+2) + a(n)^2, for n > 7. (End)
0*a(0) + 1*a(1) + 2*a(2) + ... + n*a(n) = n*a(n+5) - a(n+9) + 2. - Greg Dresden and Zi Ye, Jul 02 2021
From Greg Dresden and Zi Ye, Jul 06 2021: (Start)
2*a(n) = a(n+2) + a(n-5) for n >= 5.
3*a(n) = a(n+4) - a(n-9) for n >= 9.
4*a(n) = a(n+5) - a(n-9) for n >= 9. (End)

Extensions

Edited by Charles R Greathouse IV, Mar 17 2010
Deleted certain dangerous or potentially dangerous links. - N. J. A. Sloane, Jan 30 2021

A085478 Triangle read by rows: T(n, k) = binomial(n + k, 2*k).

Original entry on oeis.org

1, 1, 1, 1, 3, 1, 1, 6, 5, 1, 1, 10, 15, 7, 1, 1, 15, 35, 28, 9, 1, 1, 21, 70, 84, 45, 11, 1, 1, 28, 126, 210, 165, 66, 13, 1, 1, 36, 210, 462, 495, 286, 91, 15, 1, 1, 45, 330, 924, 1287, 1001, 455, 120, 17, 1, 1, 55, 495, 1716, 3003, 3003, 1820, 680, 153, 19, 1
Offset: 0

Author

Philippe Deléham, Aug 14 2003

Keywords

Comments

Coefficient array for Morgan-Voyce polynomial b(n,x). A053122 (unsigned) is the coefficient array for B(n,x). Reversal of A054142. - Paul Barry, Jan 19 2004
This triangle is formed from even-numbered rows of triangle A011973 read in reverse order. - Philippe Deléham, Feb 16 2004
T(n,k) is the number of nondecreasing Dyck paths of semilength n+1, having k+1 peaks. T(n,k) is the number of nondecreasing Dyck paths of semilength n+1, having k peaks at height >= 2. T(n,k) is the number of directed column-convex polyominoes of area n+1, having k+1 columns. - Emeric Deutsch, May 31 2004
Riordan array (1/(1-x), x/(1-x)^2). - Paul Barry, May 09 2005
The triangular matrix a(n,k) = (-1)^(n+k)*T(n,k) is the matrix inverse of A039599. - Philippe Deléham, May 26 2005
The n-th row gives absolute values of coefficients of reciprocal of g.f. of bottom-line of n-wave sequence. - Floor van Lamoen (fvlamoen(AT)planet.nl), Sep 24 2006
Unsigned version of A129818. - Philippe Deléham, Oct 25 2007
T(n, k) is also the number of idempotent order-preserving full transformations (of an n-chain) of height k >=1 (height(alpha) = |Im(alpha)|) and of waist n (waist(alpha) = max(Im(alpha))). - Abdullahi Umar, Oct 02 2008
A085478 is jointly generated with A078812 as a triangular array of coefficients of polynomials u(n,x): initially, u(1,x) = v(1,x) = 1; for n>1, u(n,x) = u(n-1,x)+x*v(n-1)x and v(n,x) = u(n-1,x)+(x+1)*v(n-1,x). See the Mathematica section. - Clark Kimberling, Feb 25 2012
Per Kimberling's recursion relations, see A102426. - Tom Copeland, Jan 19 2016
Subtriangle of the triangle given by (0, 1, 0, 1, 0, 0, 0, 0, 0, 0, 0, ...) DELTA (1, 0, 1, -1, 0, 0, 0, 0, 0, 0, 0, ...) where DELTA is the operator defined in A084938. - Philippe Deléham, Mar 26 2012
T(n,k) is also the number of compositions (ordered partitions) of 2*n+1 into 2*k+1 parts which are all odd. Proof: The o.g.f. of column k, x^k/(1-x)^(2*k+1) for k >= 0, is the o.g.f. of the odd-indexed members of the sequence with o.g.f. (x/(1-x^2))^(2*k+1) (bisection, odd part). Thus T(n,k) is obtained from the sum of the multinomial numbers A048996 for the partitions of 2*n+1 into 2*k+1 parts, all of which are odd. E.g., T(3,1) = 3 + 3 from the numbers for the partitions [1,1,5] and [1,3,3], namely 3!/(2!*1!) and 3!/(1!*2!), respectively. The number triangle with the number of these partitions as entries is A152157. - Wolfdieter Lang, Jul 09 2012
The matrix elements of the inverse are T^(-1)(n,k) = (-1)^(n+k)*A039599(n,k). - R. J. Mathar, Mar 12 2013
T(n,k) = A258993(n+1,k) for k = 0..n-1. - Reinhard Zumkeller, Jun 22 2015
The n-th row polynomial in descending powers of x is the n-th Taylor polynomial of the algebraic function F(x)*G(x)^n about 0, where F(x) = (1 + sqrt(1 + 4*x))/(2*sqrt(1 + 4*x)) and G(x) = ((1 + sqrt(1 + 4*x))/2)^2. For example, for n = 4, (1 + sqrt(1 + 4*x))/(2*sqrt(1 + 4*x)) * ((1 + sqrt(1 + 4*x))/2)^8 = (x^4 + 10*x^3 + 15*x^2 + 7*x + 1) + O(x^5). - Peter Bala, Feb 23 2018
Row n also gives the coefficients of the characteristc polynomial of the tridiagonal n X n matrix M_n given in A332602: Phi(n, x) := Det(M_n - x*1_n) = Sum_{k=0..n} T(n, k)*(-x)^k, for n >= 0, with Phi(0, x) := 1. - Wolfdieter Lang, Mar 25 2020
It appears that the largest root of the n-th degree polynomial is equal to the sum of the distinct diagonals of a (2*n+1)-gon including the edge, 1. The largest root of x^3 - 6*x^2 + 5*x - 1 is 5.048917... = the sum of (1 + 1.80193... + 2.24697...). Alternatively, the largest root of the n-th degree polynomial is equal to the square of sigma(2*n+1). Check: 5.048917... is the square of sigma(7), 2.24697.... Given N = 2*n+1, sigma(N) (N odd) can be defined as 1/(2*sin(Pi/(2*N))). Relating to the 9-gon, the largest root of x^4 - 10*x^3 + 15*x^2 - 7*x + 1 is 8.290859..., = the sum of (1 + 1.879385... + 2.532088... + 2.879385...), and is the square of sigma(9), 2.879385... Refer to A231187 for a further clarification of sigma(7). - Gary W. Adamson, Jun 28 2022
For n >=1, the n-th row is given by the coefficients of the minimal polynomial of -4*sin(Pi/(4*n + 2))^2. - Eric W. Weisstein, Jul 12 2023
Denoting this lower triangular array by L, then L * diag(binomial(2*k,k)^2) * transpose(L) is the LDU factorization of A143007, the square array of crystal ball sequences for the A_n X A_n lattices. - Peter Bala, Feb 06 2024
T(n, k) is the number of occurrences of the periodic substring (01)^k in the periodic string (01)^n (see Proposition 4.7 at page 7 in Fang). - Stefano Spezia, Jun 09 2024

Examples

			Triangle begins as:
  1;
  1    1;
  1    3    1;
  1    6    5    1;
  1   10   15    7    1;
  1   15   35   28    9    1;
  1   21   70   84   45   11    1;
  1   28  126  210  165   66   13    1;
  1   36  210  462  495  286   91   15    1;
  1   45  330  924 1287 1001  455  120   17    1;
  1   55  495 1716 3003 3003 1820  680  153   19    1;
...
From _Philippe Deléham_, Mar 26 2012: (Start)
(0, 1, 0, 1, 0, 0, 0, ...) DELTA (1, 0, 1, -1, 0, 0, 0, ...) begins:
  1
  0, 1
  0, 1,  1
  0, 1,  3,   1
  0, 1,  6,   5,   1
  0, 1, 10,  15,   7,   1
  0, 1, 15,  35,  28,   9,  1
  0, 1, 21,  70,  84,  45, 11,  1
  0, 1, 28, 126, 210, 165, 66, 13, 1. (End)
		

Crossrefs

Programs

  • GAP
    Flat(List([0..12], n-> List([0..n], k-> Binomial(n+k, 2*k) ))); # G. C. Greubel, Aug 01 2019
  • Haskell
    a085478 n k = a085478_tabl !! n !! k
    a085478_row n = a085478_tabl !! n
    a085478_tabl = zipWith (zipWith a007318) a051162_tabl a025581_tabl
    -- Reinhard Zumkeller, Jun 22 2015
    
  • Magma
    [Binomial(n+k, 2*k): k in [0..n], n in [0..12]]; // G. C. Greubel, Aug 01 2019
    
  • Maple
    T := (n,k) -> binomial(n+k,2*k): seq(seq(T(n,k), k=0..n), n=0..11);
  • Mathematica
    (* First program *)
    u[1, x_]:= 1; v[1, x_]:= 1; z = 13;
    u[n_, x_]:= u[n-1, x] + x*v[n-1, x];
    v[n_, x_]:= u[n-1, x] + (x+1)*v[n-1, x];
    Table[Expand[u[n, x]], {n, 1, z/2}]
    Table[Expand[v[n, x]], {n, 1, z/2}]
    cu = Table[CoefficientList[u[n, x], x], {n, 1, z}];
    TableForm[cu]
    Flatten[%]   (* A085478 *)
    Table[Expand[v[n, x]], {n, 1, z}]
    cv = Table[CoefficientList[v[n, x], x], {n, 1, z}];
    TableForm[cv]
    Flatten[%]   (* A078812 *) (*Clark Kimberling, Feb 25 2012 *)
    (* Second program *)
    Table[Binomial[n + k, 2 k], {n, 0, 12}, {k, 0, n}] // Flatten (* G. C. Greubel, Aug 01 2019 *)
    CoefficientList[Table[Fibonacci[2 n + 1, Sqrt[x]], {n, 0, 10}], x] // Flatten (* Eric W. Weisstein, Jul 03 2023 *)
    Join[{{1}}, CoefficientList[Table[MinimalPolynomial[-4 Sin[Pi/(4 n + 2)]^2, x], {n, 20}], x]] (* Eric W. Weisstein, Jul 12 2023 *)
  • PARI
    T(n,k) = binomial(n+k,n-k)
    
  • Sage
    [[binomial(n+k,2*k) for k in (0..n)] for n in (0..12)] # G. C. Greubel, Aug 01 2019
    

Formula

T(n, k) = (n+k)!/((n-k)!*(2*k)!).
G.f.: (1-z)/((1-z)^2-tz). - Emeric Deutsch, May 31 2004
Row sums are A001519 (Fibonacci(2n+1)). Diagonal sums are A011782. Binomial transform of A026729 (product of lower triangular matrices). - Paul Barry, Jun 21 2004
T(n, 0) = 1, T(n, k) = 0 if n=0} T(n-1-j, k-1)*(j+1). T(0, 0) = 1, T(0, k) = 0 if k>0; T(n, k) = T(n-1, k-1) + T(n-1, k) + Sum_{j>=0} (-1)^j*T(n-1, k+j)*A000108(j). For the column k, g.f.: Sum_{n>=0} T(n, k)*x^n = (x^k) / (1-x)^(2*k+1). - Philippe Deléham, Feb 15 2004
Sum_{k=0..n} T(n,k)*x^(2*k) = A000012(n), A001519(n+1), A001653(n), A078922(n+1), A007805(n), A097835(n), A097315(n), A097838(n), A078988(n), A097841(n), A097727(n), A097843(n), A097730(n), A098244(n), A097733(n), A098247(n), A097736(n), A098250(n), A097739(n), A098253(n), A097742(n), A098256(n), A097767(n), A098259(n), A097770(n), A098262(n), A097773(n), A098292(n), A097776(n) for x=0,1,2,...,27,28 respectively. - Philippe Deléham, Dec 31 2007
T(2*n,n) = A005809(n). - Philippe Deléham, Sep 17 2009
A183160(n) = Sum_{k=0..n} T(n,k)*T(n,n-k). - Paul D. Hanna, Dec 27 2010
T(n,k) = 2*T(n-1,k) + T(n-1,k-1) - T(n-2,k). - Philippe Deléham, Feb 06 2012
O.g.f. for column k: x^k/(1-x)^(2*k+1), k >= 0. [See the o.g.f. of the triangle above, and a comment on compositions. - Wolfdieter Lang, Jul 09 2012]
E.g.f.: (2/sqrt(x + 4))*sinh((1/2)*t*sqrt(x + 4))*cosh((1/2)*t*sqrt(x)) = t + (1 + x)*t^3/3! + (1 + 3*x + x^2)*t^5/5! + (1 + 6*x + 5*x^2 + x^3)*t^7/7! + .... Cf. A091042. - Peter Bala, Jul 29 2013
T(n, k) = A065941(n+3*k, 4*k) = A108299(n+3*k, 4*k) = A194005(n+3*k, 4*k). - Johannes W. Meijer, Sep 05 2013
Sum_{k=0..n} (-1)^k*T(n,k)*A000108(k) = A000007(n) for n >= 0. - Werner Schulte, Jul 12 2017
Sum_{k=0..floor(n/2)} T(n-k,k)*A000108(k) = A001006(n) for n >= 0. - Werner Schulte, Jul 12 2017
From Peter Bala, Jun 26 2025: (Start)
The n-th row polynomial b(n, x) = (-1)^n * U(2*n, (i/2)*sqrt(x)), where U(n,x) is the n-th Chebyshev polynomial of the second kind.
b(n, x) = (-1)^n * Dir(n, -1 - x/2), where Dir(n, x) is the n-th row polynomial of the triangle A244419.
b(n, -1 - x) is the n-th row polynomial of A098493. (End)

A109466 Riordan array (1, x(1-x)).

Original entry on oeis.org

1, 0, 1, 0, -1, 1, 0, 0, -2, 1, 0, 0, 1, -3, 1, 0, 0, 0, 3, -4, 1, 0, 0, 0, -1, 6, -5, 1, 0, 0, 0, 0, -4, 10, -6, 1, 0, 0, 0, 0, 1, -10, 15, -7, 1, 0, 0, 0, 0, 0, 5, -20, 21, -8, 1, 0, 0, 0, 0, 0, -1, 15, -35, 28, -9, 1, 0, 0, 0, 0, 0, 0, -6, 35, -56, 36, -10, 1, 0, 0, 0, 0, 0, 0, 1, -21, 70, -84, 45, -11, 1, 0, 0, 0, 0
Offset: 0

Author

Philippe Deléham, Aug 28 2005

Keywords

Comments

Inverse is Riordan array (1, xc(x)) (A106566).
Triangle T(n,k), 0 <= k <= n, read by rows, given by [0, -1, 1, 0, 0, 0, 0, 0, 0, ...] DELTA [1, 0, 0, 0, 0, 0, 0, 0, ...] where DELTA is the operator defined in A084938.
Modulo 2, this sequence gives A106344. - Philippe Deléham, Dec 18 2008
Coefficient array of the polynomials Chebyshev_U(n, sqrt(x)/2)*(sqrt(x))^n. - Paul Barry, Sep 28 2009

Examples

			Rows begin:
  1;
  0,  1;
  0, -1,  1;
  0,  0, -2,  1;
  0,  0,  1, -3,  1;
  0,  0,  0,  3, -4,   1;
  0,  0,  0, -1,  6,  -5,   1;
  0,  0,  0,  0, -4,  10,  -6,   1;
  0,  0,  0,  0,  1, -10,  15,  -7,  1;
  0,  0,  0,  0,  0,   5, -20,  21, -8,  1;
  0,  0,  0,  0,  0,  -1,  15, -35, 28, -9, 1;
From _Paul Barry_, Sep 28 2009: (Start)
Production array is
  0,    1,
  0,   -1,    1,
  0,   -1,   -1,   1,
  0,   -2,   -1,  -1,   1,
  0,   -5,   -2,  -1,  -1,  1,
  0,  -14,   -5,  -2,  -1, -1,  1,
  0,  -42,  -14,  -5,  -2, -1, -1,  1,
  0, -132,  -42, -14,  -5, -2, -1, -1,  1,
  0, -429, -132, -42, -14, -5, -2, -1, -1, 1 (End)
		

Crossrefs

Cf. A026729 (unsigned version), A000108, A030528, A124644.

Programs

  • Magma
    /* As triangle */ [[(-1)^(n-k)*Binomial(k, n-k): k in [0..n]]: n in [0.. 15]]; // Vincenzo Librandi, Jan 14 2016
  • Mathematica
    (* The function RiordanArray is defined in A256893. *)
    RiordanArray[1&, #(1-#)&, 13] // Flatten (* Jean-François Alcover, Jul 16 2019 *)

Formula

Number triangle T(n, k) = (-1)^(n-k)*binomial(k, n-k).
T(n, k)*2^(n-k) = A110509(n, k); T(n, k)*3^(n-k) = A110517(n, k).
Sum_{k=0..n} T(n,k)*A000108(k)=1. - Philippe Deléham, Jun 11 2007
From Philippe Deléham, Oct 30 2008: (Start)
Sum_{k=0..n} T(n,k)*A144706(k) = A082505(n+1).
Sum_{k=0..n} T(n,k)*A002450(k) = A100335(n).
Sum_{k=0..n} T(n,k)*A001906(k) = A100334(n).
Sum_{k=0..n} T(n,k)*A015565(k) = A099322(n).
Sum_{k=0..n} T(n,k)*A003462(k) = A106233(n). (End)
Sum_{k=0..n} T(n,k)*x^(n-k) = A053404(n), A015447(n), A015446(n), A015445(n), A015443(n), A015442(n), A015441(n), A015440(n), A006131(n), A006130(n), A001045(n+1), A000045(n+1), A000012(n), A010892(n), A107920(n+1), A106852(n), A106853(n), A106854(n), A145934(n), A145976(n), A145978(n), A146078(n), A146080(n), A146083(n), A146084(n) for x = -12,-11,-10,-9,-8,-7,-6,-5,-4,-3,-2,-1,0,1,2,3,4,5,6,7,8,9,10,11,12 respectively. - Philippe Deléham, Oct 27 2008
Sum_{k=0..n} T(n,k)*x^k = A000007(n), A010892(n), A099087(n), A057083(n), A001787(n+1), A030191(n), A030192(n), A030240(n), A057084(n), A057085(n+1), A057086(n) for x = 0,1,2,3,4,5,6,7,8,9,10 respectively. - Philippe Deléham, Oct 28 2008
G.f.: 1/(1-y*x+y*x^2). - Philippe Deléham, Dec 15 2011
T(n,k) = T(n-1,k-1) - T(n-2,k-1), T(n,0) = 0^n. - Philippe Deléham, Feb 15 2012
Sum_{k=0..n} T(n,k)*x^(n-k) = F(n+1,-x) where F(n,x)is the n-th Fibonacci polynomial in x defined in A011973. - Philippe Deléham, Feb 22 2013
Sum_{k=0..n} T(n,k)^2 = A051286(n). - Philippe Deléham, Feb 26 2013
Sum_{k=0..n} T(n,k)*T(n+1,k) = -A110320(n). - Philippe Deléham, Feb 26 2013
For T(0,0) = 0, the signed triangle below has the o.g.f. G(x,t) = [t*x(1-x)]/[1-t*x(1-x)] = L[t*Cinv(x)] where L(x) = x/(1-x) and Cinv(x)=x(1-x) with the inverses Linv(x) = x/(1+x) and C(x)= [1-sqrt(1-4*x)]/2, an o.g.f. for the shifted Catalan numbers A000108, so the inverse o.g.f. is Ginv(x,t) = C[Linv(x)/t] = [1-sqrt[1-4*x/(t(1+x))]]/2 (cf. A124644 and A030528). - Tom Copeland, Jan 19 2016

A030528 Triangle read by rows: a(n,k) = binomial(k,n-k).

Original entry on oeis.org

1, 1, 1, 0, 2, 1, 0, 1, 3, 1, 0, 0, 3, 4, 1, 0, 0, 1, 6, 5, 1, 0, 0, 0, 4, 10, 6, 1, 0, 0, 0, 1, 10, 15, 7, 1, 0, 0, 0, 0, 5, 20, 21, 8, 1, 0, 0, 0, 0, 1, 15, 35, 28, 9, 1, 0, 0, 0, 0, 0, 6, 35, 56, 36, 10, 1, 0, 0, 0, 0, 0, 1, 21, 70, 84, 45, 11, 1, 0, 0, 0, 0, 0, 0, 7, 56, 126, 120, 55, 12, 1
Offset: 1

Keywords

Comments

A convolution triangle of numbers obtained from A019590.
a(n,m) := s1(-1; n,m), a member of a sequence of triangles including s1(0; n,m)= A023531(n,m) (unit matrix) and s1(2; n,m)= A007318(n-1,m-1) (Pascal's triangle).
The signed triangular matrix a(n,m)*(-1)^(n-m) is the inverse matrix of the triangular Catalan convolution matrix A033184(n+1,m+1), n >= m >= 0, with A033184(n,m) := 0 if n
Riordan array (1+x, x(1+x)). The signed triangle is the Riordan array (1-x,x(1-x)), inverse to (c(x),xc(x)) with c(x) g.f. for A000108. - Paul Barry, Feb 02 2005 [with offset 0]
Also, a(n,k)=number of compositions of n into k parts of 1's and 2's. Example: a(6,4)=6 because we have 2211, 2121, 2112, 1221, 1212 and 1122. - Emeric Deutsch, Apr 05 2005 [see MacMahon and Riordan. - Wolfdieter Lang, Jul 27 2023]
Subtriangle of A026729. - Philippe Deléham, Aug 31 2006
a(n,k) is the number of length n-1 binary sequences having no two consecutive 0's with exactly k-1 1's. Example: a(6,4)=6 because we have 01011, 01101, 01110, 10101, 10110, 11010. - Geoffrey Critzer, Jul 22 2013
Mirrored, shifted Fibonacci polynomials of A011973. The polynomials (illustrated below) of this entry have the property that p(n,t) = t * [p(n-1,t) + p(n-2,t)]. The additive properties of Pascal's triangle (A007318) are reflected in those of these polynomials, as can be seen in the Example Section below and also when the o.g.f. G(x,t) below is expanded as the series x*(1+x) + t * [x*(1+x)]^2 + t^2 * [x*(1+x)]^3 + ... . See also A053122 for a relation to Cartan matrices. - Tom Copeland, Nov 04 2014
Rows of this entry appear as columns of an array for an infinitesimal generator presented in the Copeland link. - Tom Copeland, Dec 23 2015
For n >= 2, the n-th row is also the coefficients of the vertex cover polynomial of the (n-1)-path graph P_{n-1}. - Eric W. Weisstein, Apr 10 2017
With an additional initial matrix element a_(0,0) = 1 and column of zeros a_(n,0) = 0 for n > 0, these are antidiagonals read from bottom to top of the numerical coefficients of the Maurer-Cartan form matrix of the Leibniz group L^(n)(1,1) presented on p. 9 of the Olver paper, which is generated as exp[c. * M] with (c.)^n = c_n and M the Lie infinitesimal generator A218272. Cf. A011973. And A169803. - Tom Copeland, Jul 02 2018

Examples

			Triangle starts:
  [ 1]  1
  [ 2]  1   1
  [ 3]  0   2   1
  [ 4]  0   1   3   1
  [ 5]  0   0   3   4   1
  [ 6]  0   0   1   6   5   1
  [ 7]  0   0   0   4  10   6   1
  [ 8]  0   0   0   1  10  15   7   1
  [ 9]  0   0   0   0   5  20  21   8   1
  [10]  0   0   0   0   1  15  35  28   9   1
  [11]  0   0   0   0   0   6  35  56  36  10   1
  [12]  0   0   0   0   0   1  21  70  84  45  11   1
  [13]  0   0   0   0   0   0   7  56 126 120  55  12   1
  ...
From _Tom Copeland_, Nov 04 2014: (Start)
For quick comparison to other polynomials:
  p(1,t) = 1
  p(2,t) = 1 + 1 t
  p(3,t) = 0 + 2 t + 1 t^2
  p(4,t) = 0 + 1 t + 3 t^2 + 1 t^3
  p(5,t) = 0 + 0   + 3 t^2 + 4 t^3 +  1 t^4
  p(6,t) = 0 + 0   + 1 t^2 + 6 t^3 +  5 t^4 +  1 t^5
  p(7,t) = 0 + 0   + 0     + 4 t^3 + 10 t^4 +  6 t^5 + 1 t^6
  p(8,t) = 0 + 0   + 0     + 1 t^3 + 10 t^4 + 15 t^5 + 7 t^6 + 1 t^7
  ...
Reading along columns gives rows for Pascal's triangle. (End)
		

References

  • P. A. MacMahon, Combinatory Analysis, Two volumes (bound as one), Chelsea Publishing Company, New York, 1960, Vol. I, nr. 124, p. 151.
  • John Riordan, An Introduction to Combinatorial Analysis, John Wiley & Sons, London, 1958. eq. (35), p.124, 11. p. 154.

Crossrefs

Row sums A000045(n+1) (Fibonacci). a(n, 1)= A019590(n) (Fermat's last theorem). Cf. A049403.

Programs

  • Magma
    /* As triangle */ [[Binomial(k, n-k): k in [1..n]]: n in [1.. 15]]; // Vincenzo Librandi, Nov 05 2014
  • Maple
    for n from 1 to 12 do seq(binomial(k,n-k),k=1..n) od; # yields sequence in triangular form - Emeric Deutsch, Apr 05 2005
  • Mathematica
    nn=10;CoefficientList[Series[(1+x)/(1-y x - y x^2),{x,0,nn}],{x,y}]//Grid (* Geoffrey Critzer, Jul 22 2013 *)
    Table[Binomial[k, n - k], {n, 13}, {k, n}] // Flatten (* Michael De Vlieger, Dec 23 2015 *)
    CoefficientList[Table[x^(n/2 - 1) Fibonacci[n + 1, Sqrt[x]], {n, 10}],
       x] // Flatten (* Eric W. Weisstein, Apr 10 2017 *)

Formula

a(n, m) = 2*(2*m-n+1)*a(n-1, m)/n + m*a(n-1, m-1)/n, n >= m >= 1; a(n, m) := 0, n
G.f. for m-th column: (x*(1+x))^m.
As a number triangle with offset 0, this is T(n, k) = Sum_{i=0..n} (-1)^(n+i)*binomial(n, i)*binomial(i+k+1, 2k+1). The antidiagonal sums give the Padovan sequence A000931(n+5). Inverse binomial transform of A078812 (product of lower triangular matrices). - Paul Barry, Jun 21 2004
G.f.: (1 + x)/(1 - y*x - y*x^2). - Geoffrey Critzer, Jul 22 2013 [offset 0] [with offset 1: g.f. of row polynomials in y: x*(1+x)*y/(1 - x*(1+x)*y). - Wolfdieter Lang, Jul 27 2023]
From Tom Copeland, Nov 04 2014: (Start)
O.g.f: G(x,t) = x*(1+x) / [1 - t*x*(1+x)] = -P[Cinv(-x),t], where P(x,t)= x / (1 + t*x) and Cinv(x)= x*(1-x) are the compositional inverses in x of Pinv(x,t) = -P(-x,t) = x / (1 - t*x) and C(x) = [1-sqrt(1-4*x)]/2, an o.g.f. for the shifted Catalan numbers A000108.
Therefore, Ginv(x,t) = -C[Pinv(-x,t)] = {-1 + sqrt[1 + 4*x/(1+t*x)]}/2, which is -A124644(-x,t).
This places this array in a family of arrays related by composition of P and C and their inverses and interpolation by t, such as A091867 and A104597, and associated to the Catalan, Motzkin, Fine, and Fibonacci numbers. Cf. A104597 (polynomials shifted in t) A125145, A146559, A057078, A000045, A155020, A125145, A039717, A001792, A057862, A011973, A115139. (End)

Extensions

More terms from Emeric Deutsch, Apr 05 2005

A005712 Coefficient of x^4 in expansion of (1+x+x^2)^n.

Original entry on oeis.org

1, 6, 19, 45, 90, 161, 266, 414, 615, 880, 1221, 1651, 2184, 2835, 3620, 4556, 5661, 6954, 8455, 10185, 12166, 14421, 16974, 19850, 23075, 26676, 30681, 35119, 40020, 45415, 51336, 57816, 64889, 72590, 80955, 90021, 99826, 110409, 121810, 134070
Offset: 2

Keywords

Comments

a(n) = A111808(n,4) for n>3. - Reinhard Zumkeller, Aug 17 2005
If a 2-set Y and 2-set Z, having one element in common, are subsets of an n-set X then a(n-3) is the number of 5-subsets of X intersecting both Y and Z. - Milan Janjic, Oct 03 2007
Antidiagonal sums of the convolution array A213781. - Clark Kimberling, Jun 22 2012

References

  • L. Comtet, Advanced Combinatorics, Reidel, 1974, p. 78.
  • N. J. A. Sloane and Simon Plouffe, The Encyclopedia of Integer Sequences, Academic Press, 1995 (includes this sequence).

Crossrefs

a(n)= A027907(n, 4), n >= 2 (fifth column of trinomial coefficients).

Programs

  • Magma
    I:=[1, 6, 19, 45, 90]; [n le 5 select I[n] else 5*Self(n-1)-10*Self(n-2)+10*Self(n-3)-5*Self(n-4)+Self(n-5): n in [1..40]]; Vincenzo Librandi, Jun 16 2012
    
  • Maple
    seq(binomial(n+2,n-2) + binomial(n+1,n-2) - binomial(n,n-2), n=2..50); # Zerinvary Lajos, May 16 2006
    A005712:=(-1-z+z**2)/(z-1)**5; # Conjectured (correctly) by Simon Plouffe in his 1992 dissertation.
    A005712 := n -> GegenbauerC(`if`(4A005712(n)), n=2..20); # Peter Luschny, May 10 2016
  • Mathematica
    CoefficientList[Series[(1+x-x^2)/(1-x)^5,{x,0,40}],x] (* Vincenzo Librandi, Jun 16 2012 *)
    LinearRecurrence[{5,-10,10,-5,1},{1,6,19,45,90},40] (* Harvey P. Dale, Apr 30 2015 *)
  • PARI
    Vec((x^2)*(1+x-x^2)/(1-x)^5+O(x^99)) \\ Charles R Greathouse IV, Sep 23 2012

Formula

G.f.: (x^2)*(1+x-x^2)/(1-x)^5.
a(n) = binomial(n+2,n-2) + binomial(n+1,n-2) - binomial(n,n-2). - Zerinvary Lajos, May 16 2006
a(n) = 5*a(n-1) - 10*a(n-2) + 10*a(n-3) - 5*a(n-4) + a(n-5). Vincenzo Librandi, Jun 16 2012
a(n) = binomial(n,2) + 3*binomial(n,3) + binomial(n,4) (see our comment in A026729). - Vladimir Shevelev and Peter J. C. Moses, Jun 22 2012
a(n) = GegenbauerC(N, -n, -1/2) where N = 4 if 4Peter Luschny, May 10 2016
E.g.f.: exp(x)*x^2*(12 + 12*x + x^2)/24. - Stefano Spezia, Jul 09 2023

Extensions

More terms from Vladeta Jovovic, Oct 02 2000

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

Original entry on oeis.org

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

Keywords

Comments

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

Examples

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

References

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

Crossrefs

Cf. A007799. As a triangle, mirror A052509.
Rows converge to powers of two (A000079). Subdiagonals include A000225, A000295, A002662, A002663, A002664, A035038, A035039, A035040, A035041, A035042. Antidiagonal sums are A000071.

Programs

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

Formula

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

Extensions

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

A063967 Triangle read by rows, T(n,k) = T(n-1,k) + T(n-2,k) + T(n-1,k-1) + T(n-2,k-1) and T(0,0) = 1.

Original entry on oeis.org

1, 1, 1, 2, 3, 1, 3, 7, 5, 1, 5, 15, 16, 7, 1, 8, 30, 43, 29, 9, 1, 13, 58, 104, 95, 46, 11, 1, 21, 109, 235, 271, 179, 67, 13, 1, 34, 201, 506, 705, 591, 303, 92, 15, 1, 55, 365, 1051, 1717, 1746, 1140, 475, 121, 17, 1, 89, 655, 2123, 3979, 4759, 3780, 2010, 703, 154, 19, 1
Offset: 0

Author

Henry Bottomley, Sep 05 2001

Keywords

Examples

			T(3,1) = T(2,1) + T(1,1) + T(2,0) + T(1,0) = 3 + 1 + 2 + 1 = 7.
Triangle begins:
   1,
   1,   1,
   2,   3,   1,
   3,   7,   5,   1,
   5,  15,  16,   7,   1,
   8,  30,  43,  29,   9,   1,
  13,  58, 104,  95,  46,  11,  1,
  21, 109, 235, 271, 179,  67, 13,  1,
  34, 201, 506, 705, 591, 303, 92, 15, 1
		

Crossrefs

Row sums are A002605.
Columns include: A000045(n+1), A023610(n-1).
Main diagonal: A000012, a(n, n-1) = A005408(n-1).
Matrix inverse: A091698, matrix square: A091700.
Cf. A321620.
Sum_{k=0..n} x^k*T(n,k) is (-1)^n*A057086(n) (x=-11), (-1)^n*A057085(n+1) (x=-10), (-1)^n*A057084(n) (x=-9), (-1)^n*A030240(n) (x=-8), (-1)^n*A030192(n) (x=-7), (-1)^n*A030191(n) (x=-6), (-1)^n*A001787(n+1) (x=-5), A000748(n) (x=-4), A108520(n) (x=-3), A049347(n) (x=-2), A000007(n) (x=-1), A000045(n) (x=0), A002605(n) (x=1), A030195(n+1) (x=2), A057087(n) (x=3), A057088(n) (x=4), A057089(n) (x=5), A057090(n) (x=6), A057091(n) (x=7), A057092(n) (x=8), A057093(n) (x=9). - Philippe Deléham, Nov 03 2006

Programs

  • Haskell
    a063967_tabl = [1] : [1,1] : f [1] [1,1] where
       f us vs = ws : f vs ws where
         ws = zipWith (+) ([0] ++ us ++ [0]) $
              zipWith (+) (us ++ [0,0]) $ zipWith (+) ([0] ++ vs) (vs ++ [0])
    -- Reinhard Zumkeller, Apr 17 2013
  • Mathematica
    T[n_, k_] := Sum[Binomial[j, n - j]*Binomial[j, k], {j, 0, n}]; Table[T[n, k], {n, 0, 10}, {k, 0, n}] // Flatten (* Jean-François Alcover, Oct 11 2017, after Paul Barry *)
    (* Function RiordanSquare defined in A321620. *)
    RiordanSquare[1/(1 - x - x^2), 11] // Flatten (* Peter Luschny, Nov 27 2018 *)

Formula

G.f.: 1/(1-x*(1+x)*(1+y)). - Vladeta Jovovic, Oct 11 2003
Riordan array (1/(1-x-x^2), x(1+x)/(1-x-x^2)). The inverse of the signed version (1/(1+x-x^2),x(1-x)/(1+x-x^2)) is abs(A091698). - Paul Barry, Jun 10 2005
T(n, k) = Sum_{j=0..n} C(j, n-j)C(j, k). - Paul Barry, Nov 09 2005
Diagonal sums are A002478. - Paul Barry, Nov 09 2005
A026729*A007318 as infinite lower triangular matrices. - Philippe Deléham, Dec 11 2008
Central coefficients T(2*n,n) are A137644. - Paul Barry, Apr 15 2010
Product of Riordan arrays (1, x(1+x))*(1/(1-x), x/(1-x)), that is, A026729*A007318. - Paul Barry, Mar 14 2011
Triangle T(n,k), read by rows, given by (1,1,-1,0,0,0,0,0,0,0,...) DELTA (1,0,0,0,0,0,0,0,0,0,...) where DELTA is the operator defined in A084938. - Philippe Deléham, Nov 12 2011
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