cp's OEIS Frontend

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

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A055249 Triangle of partial row sums (prs) of triangle A055248 (prs of Pascal's triangle A007318).

Original entry on oeis.org

1, 3, 1, 8, 4, 1, 20, 12, 5, 1, 48, 32, 17, 6, 1, 112, 80, 49, 23, 7, 1, 256, 192, 129, 72, 30, 8, 1, 576, 448, 321, 201, 102, 38, 9, 1, 1280, 1024, 769, 522, 303, 140, 47, 10, 1, 2816, 2304, 1793, 1291, 825, 443, 187, 57, 11, 1, 6144, 5120, 4097, 3084, 2116, 1268, 630
Offset: 0

Views

Author

Wolfdieter Lang, May 26 2000

Keywords

Comments

In the language of the Shapiro et al. reference (given in A053121) such a lower triangular (ordinary) convolution array, considered as matrix, belongs to the Riordan-group. The G.f. for the row polynomials p(n,x) (increasing powers of x) is ((1-z)/(1-2*z)^2)/(1-x*z/(1-z)).
This is the second member of the family of Riordan-type matrices obtained from A007318(n,m) (Pascal's triangle read as lower triangular matrix) by repeated application of the prs-procedure.
The column sequences appear in A001792, A001787, A000337, A045618, A045889, A034009, A055250, A055251 for m=0..7.

Examples

			1;
3,1;
8,4,1;
20,12,5,1;
...
Fourth row polynomial (n=3): p(3,x)= 20+12*x+5*x^2+x^3

Links

Crossrefs

Cf. A007318, A055248, A008949. Row sums: A049611(n+1) = A055252(n, 0).

Programs

  • Mathematica
    a[n_, m_] := Binomial[n, m]*Hypergeometric2F1[2, m-n, m+1, -1]; Table[a[n, m], {n, 0, 10}, {m, 0, n}] // Flatten (* Jean-François Alcover, Mar 11 2014 *)

Formula

a(n, m) = Sum_{k=m,..,n} ( A055248(n, k) ), n >= m >= 0, a(n, m) := 0 if n
Column m recursion: a(n, m) = Sum_{j=m,..,(n-1)} ( a(j, m) ) + A055248(n, m), n >= m >= 0, a(n, m) := 0 if n
G.f. for column m: ((1-x)/(1-2*x)^2)*(x/(1-x))^m, m >= 0.
a(n, m) = binomial(n, m) * 2F1(2, m-n; m+1; -1) where 2F1 is the hypergeometric function. Jean-François Alcover, Mar 11 2014

A159573 Triangle read by rows, A055248 * (A005493 * 0^(n-k)).

Original entry on oeis.org

1, 2, 1, 4, 3, 3, 8, 7, 12, 10, 16, 15, 33, 50, 37, 32, 31, 78, 160, 222, 151, 64, 63, 171, 420, 814, 1057, 674, 128, 127, 360, 990, 2368, 4379, 5392, 3263, 256, 255, 741, 2190, 6031, 14043, 24938, 29367, 17007, 512, 511, 1506, 4660, 14134, 38656, 87620
Offset: 0

Views

Author

Gary W. Adamson, Apr 16 2009

Keywords

Comments

Row sums = A005493: (1, 3, 10, 37, 151, 674, 3263,...); = row sums of Aitken's array. As a property of eigentriangles, sum of n-th row terms = rightmost term of next row.

Examples

			First few rows of the triangle =
1;
2, 1;
4, 3, 3;
8, 7, 12, 10;
16, 15, 33, 50, 37;
32, 31, 78, 160, 222, 151;
64, 63, 171, 420, 814, 1057, 674;
128, 127, 360, 990, 2368, 4379, 5392, 3263;
256, 255, 741, 2190, 6031, 14043, 24938, 29367, 17007;
512, 511, 1506, 4660, 14134, 38656, 87620, 150098, 170070, 94828;
1024, 1023, 3039, 9680, 31376, 96338, 260164, 574288, 952392, 1043108, 562595;
...
Example: row 3 = (8, 7, 12, 10) = termwise products of (8, 7, 4, 1) and
(1, 1, 3, 10), where (8, 7, 12, 10) = row 3 of triangle A055248.

Crossrefs

Cf. A055248, A005493, A011971

Formula

Triangle read by rows, A055248 * (A005493 * 0^(n-k)); equivalent to the product of triangle A055248 and its own eigensequence (diagonalized with the rest zeros, as an infinite lower triangular matrix).

A193604 Augmentation of the triangle A055248. See Comments.

Original entry on oeis.org

1, 2, 1, 8, 8, 3, 64, 88, 62, 19, 1024, 1664, 1568, 896, 233, 32768, 58368, 64128, 49248, 23890, 5385, 2097152, 3932160, 4703232, 4249728, 2800488, 1179656, 233787, 268435456, 517996544, 649887744, 645547008, 507802304, 293645688, 108978862
Offset: 0

Views

Author

Clark Kimberling, Jul 31 2011

Keywords

Comments

For an introduction to the unary operation "augmentation" as applied to triangular arrays or sequences of polynomials, see A193091.

Examples

			First five rows of A193604:
1
2.....1
8.....8.....3
64....88....62....19
1024..1664..1568..896...233

Crossrefs

Cf. A055248, A193091.

Programs

  • Mathematica
    u[n_, k_] := Sum[Binomial[n, h], {h, 0, k}] (* A055248 *)
p[n_, k_] := u[n, n - k]
Table[p[n, k], {n, 0, 5}, {k, 0, n}]
m[n_] := Table[If[i <= j, p[n + 1 - i, j - i], 0], {i, n}, {j, n + 1}]
TableForm[m[4]]
w[0, 0] = 1; w[1, 0] = p[1, 0]; w[1, 1] = p[1, 1];
v[0] = w[0, 0]; v[1] = {w[1, 0], w[1, 1]};
v[n_] := v[n - 1].m[n]
TableForm[Table[v[n], {n, 0, 6}]] (* A193604 *)
Flatten[Table[v[n], {n, 0, 8}]]

A157603 Triangle read by rows: T(n,k) = 1 for k <= n/2, T(n,k)=A055248 otherwise.

Original entry on oeis.org

1, 1, 1, 1, 3, 1, 1, 1, 4, 1, 1, 1, 11, 5, 1, 1, 1, 1, 16, 6, 1, 1, 1, 1, 42, 22, 7, 1, 1, 1, 1, 1, 64, 29, 8, 1, 1, 1, 1, 1, 163, 93, 37, 9, 1, 1, 1, 1, 1, 1, 256, 130, 46, 10, 1, 1, 1, 1, 1, 1, 638, 386, 176, 56, 11, 1
Offset: 0

Views

Author

Roger L. Bagula and Gary W. Adamson, Mar 02 2009

Keywords

Examples

			{1},
{1, 1},
{1, 3, 1},
{1, 1, 4, 1},
{1, 1, 11, 5, 1},
{1, 1, 1, 16, 6, 1},
{1, 1, 1, 42, 22, 7, 1},
{1, 1, 1, 1, 64, 29, 8, 1},
{1, 1, 1, 1, 163, 93, 37, 9, 1},
{1, 1, 1, 1, 1, 256, 130, 46, 10, 1},
{1, 1, 1, 1, 1, 638, 386, 176, 56, 11, 1}

Programs

  • Mathematica
    t[n_, m_] = Sum[Binomial[n, m - k], {k, 0, m}];
a = Table[Table[If[t[n, m] <= Sum[t[n, m], {m, 0, n}]/( n + 1), 1, t[n, n - m]], {m, 0, n}], {n, 0, 10}];
Flatten[%]

Extensions

Edited by Joerg Arndt, Sep 17 2013

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

Original entry on oeis.org

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

Views

Author

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

Keywords

Comments

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

Examples

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

References

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

Links

Crossrefs

Equals differences between consecutive terms of A102363. - David G. Williams (davidwilliams(AT)Paxway.com), Jan 23 2006
Row sums give A000079 (powers of 2).
Cf. A083093 (triangle read mod 3), A214292 (first differences of rows).
Partial sums of rows give triangle A008949.
The triangle of the antidiagonals is A011973.
Infinite matrix squared: A038207, cubed: A027465.
Cf. A101164. If rows are sorted we get A061554 or A107430.
Another version: A108044.
Cf. A008277, A132311, A132312, A052216, A052217, A052218, A052219, A052220, A052221, A052222, A052223, A144225, A202750, A211226, A047999, A026729, A052553, A051920, A193242.
Triangle sums (see the comments): A000079 (Row1); A000007 (Row2); A000045 (Kn11 & Kn21); A000071 (Kn12 & Kn22); A001924 (Kn13 & Kn23); A014162 (Kn14 & Kn24); A014166 (Kn15 & Kn25); A053739 (Kn16 & Kn26); A053295 (Kn17 & Kn27); A053296 (Kn18 & Kn28); A053308 (Kn19 & Kn29); A053309 (Kn110 & Kn210); A001519 (Kn3 & Kn4); A011782 (Fi1 & Fi2); A000930 (Ca1 & Ca2); A052544 (Ca3 & Ca4); A003269 (Gi1 & Gi2); A055988 (Gi3 & Gi4); A034943 (Ze1 & Ze2); A005251 (Ze3 & Ze4). - Johannes W. Meijer, Sep 22 2010
Fibonacci-Pascal triangles: A027926, A036355, A037027, A074829, A105809, A109906, A111006, A114197, A162741, A228074, A228196, A228576.
Cf. A137948, A245334.
Cf. A085478, A258993.
Cf. A115940 (pandigital binomial coefficients C(m,k) with k>1).
Cf. (simplex colorings) A325002 (oriented), [k==n+1] (chiral), A325003 (achiral), A325000 (k or fewer colors), A325009 (orthotope facets, orthoplex vertices), A325017 (orthoplex facets, orthotope vertices).
Triangles of generalized binomial coefficients (n,k)_m (or generalized Pascal triangles) for m = 2..12: A001263, A056939, A056940, A056941, A142465, A142467, A142468, A174109, A342889, A342890, A342891.

Programs

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

Formula

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

Extensions

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

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

Original entry on oeis.org

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

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Author

N. J. A. Sloane

Keywords

Comments

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

Examples

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

References

  • M. Abramowitz and I. A. Stegun, eds., Handbook of Mathematical Functions, National Bureau of Standards Applied Math. Series 55, 1964 (and various reprintings), p. 795.
  • N. J. A. Sloane, A Handbook of Integer Sequences, Academic Press, 1973 (includes this sequence).
  • N. J. A. Sloane and Simon Plouffe, The Encyclopedia of Integer Sequences, Academic Press, 1995 (includes this sequence).
  • A. M. Stepin and A. T. Tagi-Zade, Words with restrictions, pp. 67-74 of Kvant Selecta: Combinatorics I, Amer. Math. Soc., 2001 (G_n on p. 70).

Links

Crossrefs

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

Programs

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

Formula

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

A029653 Numbers in (2,1)-Pascal triangle (by row).

Original entry on oeis.org

1, 2, 1, 2, 3, 1, 2, 5, 4, 1, 2, 7, 9, 5, 1, 2, 9, 16, 14, 6, 1, 2, 11, 25, 30, 20, 7, 1, 2, 13, 36, 55, 50, 27, 8, 1, 2, 15, 49, 91, 105, 77, 35, 9, 1, 2, 17, 64, 140, 196, 182, 112, 44, 10, 1, 2, 19, 81, 204, 336, 378, 294, 156, 54, 11, 1, 2, 21, 100, 285
Offset: 0

Views

Author

Mohammad K. Azarian

Keywords

Comments

Reverse of A029635. Row sums are A003945. Diagonal sums are Fibonacci(n+2) = Sum_{k=0..floor(n/2)} (2n-3k)*C(n-k,n-2k)/(n-k). - Paul Barry, Jan 30 2005
Riordan array ((1+x)/(1-x), x/(1-x)). The signed triangle (-1)^(n-k)T(n,k) or ((1-x)/(1+x), x/(1+x)) is the inverse of A055248. Row sums are A003945. Diagonal sums are F(n+2). - Paul Barry, Feb 03 2005
Row sums = A003945: (1, 3, 6, 12, 24, 48, 96, ...) = (1, 3, 7, 15, 31, 63, 127, ...) - (0, 0, 1, 3, 7, 15, 31, ...); where (1, 3, 7, 15, ...) = A000225. - Gary W. Adamson, Apr 22 2007
Triangle T(n,k), read by rows, given by (2,-1,0,0,0,0,0,0,0,...) DELTA (1,0,0,0,0,0,0,0,0,...) where DELTA is the operator defined in A084938. - Philippe Deléham, Nov 17 2011
A029653 is jointly generated with A208510 as an array of coefficients of polynomials v(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*v(n-1,x)+1. See the Mathematica section. - Clark Kimberling, Feb 28 2012
For a closed-form formula for arbitrary left and right borders of Pascal like triangle, see A228196. - Boris Putievskiy, Aug 18 2013
For a closed-form formula for generalized Pascal's triangle, see A228576. - Boris Putievskiy, Sep 04 2013
The n-th row polynomial is (2 + x)*(1 + x)^(n-1) for n >= 1. More generally, the n-th row polynomial of the Riordan array ( (1-a*x)/(1-b*x), x/(1-b*x) ) is (b - a + x)*(b + x)^(n-1) for n >= 1. - Peter Bala, Feb 25 2018

Examples

			The triangle T(n,k) begins:
n\k 0  1  2   3   4   5   6   7  8  9 10 ...
0:  1
1:  2  1
2:  2  3  1
3:  2  5  4   1
4:  2  7  9   5   1
5:  2  9 16  14   6   1
6:  2 11 25  30  20   7   1
7:  2 13 36  55  50  27   8   1
8:  2 15 49  91 105  77  35   9  1
9:  2 17 64 140 196 182 112  44 10  1
10: 2 19 81 204 336 378 294 156 54 11  1
... Reformatted. - _Wolfdieter Lang_, Jan 09 2015
With the array M(k) as defined in the Formula section, the infinite product M(0)*M(1)*M(2)*... begins
/1        \/1         \/1        \      /1        \
|2 1      ||0 1       ||0 1      |      |2 1      |
|2 1 1    ||0 2 1     ||0 0 1    |... = |2 3 1    |
|2 1 1 1  ||0 2 1 1   ||0 0 2 1  |      |2 5 4 1  |
|2 1 1 1 1||0 2 1 1 1 ||0 0 2 1 1|      |2 7 9 5 1|
|...      ||...       ||...      |      |...      |
- _Peter Bala_, Dec 27 2014

References

  • Boris A. Bondarenko, Generalized Pascal Triangles and Pyramids (in Russian), FAN, Tashkent, 1990, ISBN 5-648-00738-8.

Links

Crossrefs

(d, 1) Pascal triangles: A007318(d=1), A093560(3), A093561(4), A093562(5), A093563(6), A093564(7), A093565(8), A093644(9), A093645(10).
Cf. A003945, A208510, A228196, A228576.
Cf. A078812, A106195.

Programs

  • Haskell
    a029653 n k = a029653_tabl !! n !! k
a029653_row n = a029653_tabl !! n
a029653_tabl = [1] : iterate
               (\xs -> zipWith (+) ([0] ++ xs) (xs ++ [0])) [2, 1]
-- Reinhard Zumkeller, Dec 16 2013
  • Maple
    A029653 :=  proc(n,k)
if n = 0 then
  1;
else
  binomial(n-1, k)+binomial(n, k)
fi
end proc: # R. J. Mathar, Jun 30 2013
  • Mathematica
    u[1, x_] := 1; v[1, x_] := 1; z = 16;
u[n_, x_] := u[n - 1, x] + x*v[n - 1, x];
v[n_, x_] := u[n - 1, x] + x*v[n - 1, x] + 1;
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[%]  (* A208510 *)
Table[Expand[v[n, x]], {n, 1, z}]
cv = Table[CoefficientList[v[n, x], x], {n, 1, z}];
TableForm[cv]
Flatten[%]  (* A029653 *)
(* Clark Kimberling, Feb 28 2012 *)
  • Python
    from sympy import Poly
from sympy.abc import x
def u(n, x): return 1 if n==1 else u(n - 1, x) + x*v(n - 1, x)
def v(n, x): return 1 if n==1 else u(n - 1, x) + x*v(n - 1, x) + 1
def a(n): return Poly(v(n, x), x).all_coeffs()[::-1]
for n in range(1, 13): print(a(n)) # Indranil Ghosh, May 27 2017
  • Python
    from math import comb, isqrt
def A029653(n): return comb(r:=(m:=isqrt(k:=n+1<<1))-(k<=m*(m+1)),a:=n-comb(r+1,2))*((r<<1)-a)//r if n else 1 # Chai Wah Wu, Nov 12 2024

Formula

T(n, k) = C(n-2, k-1) + C(n-2, k) + C(n-1, k-1) + C(n-1, k) except for n=0.
G.f.: (1 + x + y + xy)/(1 - y - xy). - Ralf Stephan, May 17 2004
T(n, k) = (2n-k)*binomial(n, n-k)/n, n, k > 0. - Paul Barry, Jan 30 2005
Sum_{k=0..n} T(n, k)*x^k gives A003945-A003954 for x = 1, 2, 3, 4, 5, 6, 7, 8, 9, 10. - Philippe Deléham, Jul 10 2005
T(n, k) = C(n-1, k) + C(n, k). - Philippe Deléham, Jul 10 2005
Equals A097806 * A007318, i.e., the pairwise operator * Pascal's Triangle as infinite lower triangular matrices. - Gary W. Adamson, Apr 22 2007
From Peter Bala, Dec 27 2014: (Start)
exp(x) * e.g.f. for row n = e.g.f. for diagonal n. For example, for n = 3 we have exp(x)*(2 + 5*x + 4*x^2/2! + x^3/3!) = 2 + 7*x + 16*x^2/2! + 30*x^3/3! + 50*x^4/4! + .... The same property holds more generally for Riordan arrays of the form ( f(x), x/(1 - x) ).
Let M denote the lower unit triangular array with 1's on the main diagonal and 1's everywhere else below the main diagonal except for the first column which consists of the sequence [1,2,2,2,...]. For k = 0,1,2,... define M(k) to be the lower unit triangular block array
/I_k 0\
\ 0 M/ having the k X k identity matrix I_k as the upper left block; in particular, M(0) = M. Then the present triangle equals the infinite product M(0)*M(1)*M(2)*... (which is clearly well-defined). See the Example section. (End)

Extensions

More terms from James Sellers

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

Original entry on oeis.org

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

Views

Author

N. J. A. Sloane

Keywords

Comments

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

Examples

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

References

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

Links

Crossrefs

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

Programs

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

Formula

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

Extensions

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

A002662 a(n) = 2^n - 1 - n*(n+1)/2.

Original entry on oeis.org

0, 0, 0, 1, 5, 16, 42, 99, 219, 466, 968, 1981, 4017, 8100, 16278, 32647, 65399, 130918, 261972, 524097, 1048365, 2096920, 4194050, 8388331, 16776915, 33554106, 67108512, 134217349, 268435049, 536870476, 1073741358, 2147483151, 4294966767, 8589934030
Offset: 0

Views

Author

N. J. A. Sloane

Keywords

Comments

Number of subsets with at least 3 elements of an n-element set.
For n>4, number of simple rank-(n-1) matroids over S_n.
Number of non-interval subsets of {1,2,3,...,n} (cf. A000124). - Jose Luis Arregui (arregui(AT)unizar.es), Jun 27 2006
The partial sums of the second diagonal of A008292 or third column of A123125. - Tom Copeland, Sep 09 2008
a(n) is the number of binary sequences of length n having at least three 0's. - Geoffrey Critzer, Feb 11 2009
Starting with "1" = eigensequence of a triangle with the tetrahedral numbers (1, 4, 10, 20, ...) as the left border and the rest 1's. - Gary W. Adamson, Jul 24 2010
a(n) is also the number of crossing set partitions of [n+1] with two blocks. - Peter Luschny, Apr 29 2011
The Kn24 sums, see A180662, of triangle A065941 equal the terms (doubled) of this sequence minus the three leading zeros. - Johannes W. Meijer, Aug 14 2011
From L. Edson Jeffery, Dec 28 2011: (Start)
Nonzero terms of this sequence can be found from the row sums of the fourth sub-triangle extracted from Pascal's triangle as indicated below by braces:
1;
1, 1;
1, 2, 1;
{1}, 3, 3, 1;
{1, 4}, 6, 4, 1;
{1, 5, 10}, 10, 5, 1;
{1, 6, 15, 20}, 15, 6, 1;
... (End)
Partial sums of A000295 (Eulerian Numbers, Column 2).
Second differences equal 2^(n-2) - 1, for n >= 4. - Richard R. Forberg, Jul 11 2013
Starting (0, 0, 1, 5, 16, ...) is the binomial transform of (0, 0, 1, 2, 2, 2, ...). - Gary W. Adamson, Jul 27 2015
a(n - 1) is the rank of the divisor class group of the moduli space of stable rational curves with n marked points, see Keel p. 550. - Harry Richman, Aug 10 2024

Examples

			a(4) = 5 is the number of crossing set partitions of {1,2,..,5}, card{13|245, 14|235, 24|135, 25|134, 35|124}. - _Peter Luschny_, Apr 29 2011

References

  • N. J. A. Sloane, A Handbook of Integer Sequences, Academic Press, 1973 (includes this sequence).
  • N. J. A. Sloane and Simon Plouffe, The Encyclopedia of Integer Sequences, Academic Press, 1995 (includes this sequence).

Links

Crossrefs

a(n) = A055248(n,3).
First differences are A000295.
Cf. A000079, A000124, A000217, A000225, A000247, A002663, A002664, A035038-A035042, A007318, A342352.
Cf. also A000290, A001045.

Programs

Formula

G.f.: x^3/((1-2*x)*(1-x)^3).
a(n) = Sum_{k=0..n} binomial(n,k+3) = Sum_{k=3..n} binomial(n,k). - Paul Barry, Jul 30 2004
a(n+1) = 2*a(n) + binomial(n,2). - Paul Barry, Aug 23 2004
(1, 5, 16, 42, 99, ...) = binomial transform of (1, 4, 7, 8, 8, 8, ...). - Gary W. Adamson, Sep 30 2007
E.g.f.: exp(x)*(exp(x)-x^2/2-x-1). - Geoffrey Critzer, Feb 11 2009
a(n) = n - 2 + 3*a(n-1) - 2*a(n-2), for n >= 2. - Richard R. Forberg, Jul 11 2013
For n>1, a(n) = (1/4)*Sum_{k=1..n-2} 2^k*(n-k-1)*(n-k). For example, (1/4)*(2^1*(4*5) + 2^2*(3*4) + 2^3*(2*3) + 2^4*(1*2)) = 168/4 = 42. - J. M. Bergot, May 27 2014 [edited by Danny Rorabaugh, Apr 19 2015]
Convolution of A001045 and (A000290 shifted by one place). - Oboifeng Dira, Aug 16 2016
a(n) = Sum_{k=1..n-2} Sum_{i=1..n} (n-k-1) * C(k,i). - Wesley Ivan Hurt, Sep 19 2017
a(n) = 5*a(n-1) - 9*a(n-2) + 7*a(n-3) - 2*a(n-4) for n > 3. - Chai Wah Wu, Apr 03 2021
a(n) = a(n-1) + 1 + A000247(n-1). - Harry Richman, Aug 13 2024

A143494 Triangle read by rows: 2-Stirling numbers of the second kind.

Original entry on oeis.org

1, 2, 1, 4, 5, 1, 8, 19, 9, 1, 16, 65, 55, 14, 1, 32, 211, 285, 125, 20, 1, 64, 665, 1351, 910, 245, 27, 1, 128, 2059, 6069, 5901, 2380, 434, 35, 1, 256, 6305, 26335, 35574, 20181, 5418, 714, 44, 1, 512, 19171, 111645, 204205, 156660, 58107, 11130, 1110, 54, 1
Offset: 2

Views

Author

Peter Bala, Aug 20 2008

Keywords

Comments

This is the case r = 2 of the r-Stirling numbers of the second kind. The 2-Stirling numbers of the second kind give the number of ways of partitioning the set {1,2,...,n} into k nonempty disjoint subsets with the restriction that the elements 1 and 2 belong to distinct subsets.
More generally, the r-Stirling numbers of the second kind give the number of ways of partitioning the set {1,2,...,n} into k nonempty disjoint subsets with the restriction that the numbers 1, 2, ..., r belong to distinct subsets. The case r = 1 gives the usual Stirling numbers of the second kind A008277; for other cases see A143495 (r = 3) and A143496 (r = 4).
The lower unitriangular array of r-Stirling numbers of the second kind equals the matrix product P^(r-1) * S (with suitable offsets in the row and column indexing), where P is Pascal's triangle, A007318 and S is the array of Stirling numbers of the second kind, A008277.
For the definition of and entries relating to the corresponding r-Stirling numbers of the first kind see A143491. For entries on r-Lah numbers refer to A143497. The theory of r-Stirling numbers of both kinds is developed in [Broder].
From Peter Bala, Sep 19 2008: (Start)
Let D be the derivative operator d/dx and E the Euler operator x*d/dx. Then x^(-2)*E^n*x^2 = Sum_{k = 0..n} T(n+2,k+2)*x^k*D^k.
The row generating polynomials R_n(x) := Sum_{k= 2..n} T(n,k)*x^k satisfy the recurrence R_(n+1)(x) = x*R_n(x) + x*d/dx(R_n(x)) with R_2(x) = x^2. It follows that the polynomials R_n(x) have only real zeros (apply Corollary 1.2. of [Liu and Wang]).
Relation with the 2-Eulerian numbers E_2(n,j) := A144696(n,j): T(n,k) = 2!/k!*Sum_ {j = n-k..n-2} E_2(n,j)*binomial(j,n-k) for n >= k >= 2. (End)
From Wolfdieter Lang, Sep 29 2011: (Start)
T(n,k) = S(n,k,2), n>=k>=2, in Mikhailov's first paper, eq.(28) or (A3). E.g.f. column no. k from (A20) with k->2, r->k. Therefore, with offset [0,0], this triangle is the Sheffer triangle (exp(2*x),exp(x)-1) with e.g.f. of column no. m>=0: exp(2*x)*((exp(x)-1)^m)/m!. See one of the formulas given below. For Sheffer matrices see the W. Lang link under A006232 with the S. Roman reference, also found in A132393. (End)

Examples

			Triangle begins
  n\k|...2....3....4....5....6....7
  =================================
  2..|...1
  3..|...2....1
  4..|...4....5....1
  5..|...8...19....9....1
  6..|..16...65...55...14....1
  7..|..32..211..285..125...20....1
  ...
T(4,3) = 5. The set {1,2,3,4} can be partitioned into three subsets such that 1 and 2 belong to different subsets in 5 ways: {{1}{2}{3,4}}, {{1}{3}{2,4}}, {{1}{4}{2,3}}, {{2}{3}{1,4}} and {{2}{4}{1,3}}; the remaining possibility {{1,2}{3}{4}} is not allowed.
From _Peter Bala_, Feb 23 2025: (Start)
The array factorizes as
/ 1               \       /1             \ /1             \ /1            \
| 2    1           |     | 2   1          ||0  1           ||0  1          |
| 4    5   1       |  =  | 4   3   1      ||0  2   1       ||0  0  1       | ...
| 8   19   9   1   |     | 8   7   4  1   ||0  4   3  1    ||0  0  2  1    |
|16   65  55  14  1|     |16  15  11  6  1||0  8   7  4  1 ||0  0  4  3  1 |
|...               |     |...             ||...            ||...           |
where, in the infinite product on the right-hand side, the first array is the Riordan array (1/(1 - 2*x), x/(1 - x)). See A055248. (End)

Links

Crossrefs

A001047 (column 3), A005493 (row sums), A008277, A016269 (column 4), A025211 (column 5), A049444 (matrix inverse), A074051 (alt. row sums).
Cf. A055248, A143491, A143495, A143496, A143497.

Programs

  • Maple
    with combinat: T := (n, k) -> (1/(k-2)!)*add ((-1)^(k-i)*binomial(k-2,i)*(i+2)^(n-2),i = 0..k-2): for n from 2 to 11 do seq(T(n, k), k = 2..n) end do;
  • Mathematica
    t[n_, k_] := StirlingS2[n, k] - StirlingS2[n-1, k]; Flatten[ Table[ t[n, k], {n, 2, 11}, {k, 2, n}]] (* Jean-François Alcover, Dec 02 2011 *)
  • Sage
    @CachedFunction
def stirling2r(n, k, r) :
    if n < r: return 0
    if n == r: return 1 if k == r else 0
    return stirling2r(n-1,k-1,r) + k*stirling2r(n-1,k,r)
A143494 = lambda n,k: stirling2r(n, k, 2)
for n in (2..6):
    [A143494(n, k) for k in (2..n)] # Peter Luschny, Nov 19 2012

Formula

T(n+2,k+2) = (1/k!)*Sum_{i = 0..k} (-1)^(k-i)*C(k,i)*(i+2)^n, n,k >= 0.
T(n,k) = Stirling2(n,k) - Stirling2(n-1,k) for n, k >= 2.
Recurrence relation: T(n,k) = T(n-1,k-1) + k*T(n-1,k) for n > 2, with boundary conditions T(n,1) = T(1,n) = 0 for all n, T(2,2) = 1 and T(2,k) = 0 for k > 2. Special cases: T(n,2) = 2^(n-2); T(n,3) = 3^(n-2) - 2^(n-2).
As a sum of monomial functions of degree m: T(n+m,n) = Sum_{2 <= i_1 <= ... <= i_m <= n} (i_1*i_2*...*i_m). For example, T(6,4) = Sum_{2 <= i <= j <= 4} (i*j) = 2*2 + 2*3 + 2*4 + 3*3 + 3*4 + 4*4 = 55.
E.g.f. column k+2 (with offset 2): 1/k!*exp(2*x)*(exp(x) - 1)^k.
O.g.f. k-th column: Sum_{n >= k} T(n,k)*x^n = x^k/((1-2*x)*(1-3*x)*...*(1-k*x)).
E.g.f.: exp(2*t + x*(exp(t) - 1)) = Sum_{n >= 0} Sum_{k = 0..n} T(n+2,k+2) *x^k*t^n/n! = Sum_{n >= 0} B_n(2;x)*t^n/n! = 1 + (2 + x)*t/1! + (4 + 5*x + x^2)*t^2/2! + ..., where the row polynomial B_n(2;x) := Sum_{k = 0..n} T(n+2,k+2)*x^k denotes the 2-Bell polynomial.
Dobinski-type identities: Row polynomial B_n(2;x) = exp(-x)*Sum_{i >= 0} (i + 2)^n*x^i/i!. Sum_{k = 0..n} k!*T(n+2,k+2)*x^k = Sum_{i >= 0} (i + 2)^n*x^i/(1 + x)^(i+1).
The T(n,k) are the connection coefficients between falling factorials and the shifted monomials (x + 2)^(n-2). For example, from row 4 we have 4 + 5*x + x*(x - 1) = (x + 2)^2, while from row 5 we have 8 + 19*x + 9*x*(x - 1) + x*(x - 1)*(x - 2) = (x + 2)^3.
The row sums of the array are the 2-Bell numbers, B_n(2;1), equal to A005493(n-2). The alternating row sums are the complementary 2-Bell numbers, B_n(2;-1), equal to (-1)^n*A074051(n-2).
This array is the matrix product P * S, where P denotes the Pascal triangle, A007318 and S denotes the lower triangular array of Stirling numbers of the second kind, A008277 (apply Theorem 10 of [Neuwirth]).
Also, this array equals the transpose of the upper triangular array A126351. The inverse array is A049444, the signed 2-Stirling numbers of the first kind. See A143491 for the unsigned version of the inverse.
Let f(x) = exp(exp(x)). Then for n >= 1, the row polynomials R(n,x) are given by R(n+2,exp(x)) = 1/f(x)*(d/dx)^n(exp(2*x)*f(x)). Similar formulas hold for A008277, A039755, A105794, A111577 and A154537. - Peter Bala, Mar 01 2012
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