A000580 -id:A000580 - cp's OEIS frontend

A085439 a(n) = Sum_{i=1..n} binomial(i+1,2)^4.

1, 82, 1378, 11378, 62003, 256484, 871140, 2550756, 6651381, 15802006, 34776742, 71791798, 140366759, 261917384, 469277384, 811379400, 1359360681, 2214396762, 3517606762, 5462416762, 8309813083, 12406965164, 18209748140, 26309748140, 37466388765, 52644875166

Offset: 1

André F. Labossière, Jul 03 2003

			a(15) = (2520*(15^9) +22680*(15^8) +79920*(15^7) +136080*(15^6) +107352*(15^5) +22680*(15^4) -10080*(15^3) +1728*15)/9! = 469277384.

G. C. Greubel, Table of n, a(n) for n = 1..5000
Feihu Liu, Guoce Xin, and Chen Zhang, Ehrhart Polynomials of Order Polytopes: Interpreting Combinatorial Sequences on the OEIS, arXiv:2412.18744 [math.CO], 2024. See p. 13.
Index entries for linear recurrences with constant coefficients, signature (10,-45,120,-210,252,-210,120,-45,10,-1).

Column k=4 of A334781.

Cf. A000292, A087127, A024166, A024166, A085438, A085440, A085441, A085442, A000332, A086020, A086021, A086022, A000389, A086023, A086024, A000579, A086025, A086026, A000580, A086027, A086028, A027555, A086029, A086030.

[(2520*n^9 +22680*n^8 +79920*n^7 +136080*n^6 +107352*n^5 +22680*n^4 -10080*n^3 +1728*n)/Factorial(9): n in [1..30]]; // G. C. Greubel, Nov 22 2017

Table[(2520*(n^9) + 22680*(n^8) + 79920*(n^7) + 136080*(n^6) + 107352*(n^5) + 22680*(n^4) - 10080*(n^3) + 1728*n)/9!, {n, 1, 50}] (* G. C. Greubel, Nov 22 2017 *)

Vec(x*(x^6+72*x^5+603*x^4+1168*x^3+603*x^2+72*x+1)/(x-1)^10 + O(x^100)) \\ Colin Barker, May 02 2014

a(n) = sum(i=1, n, binomial(i+1, 2)^4); \\ Michel Marcus, Nov 22 2017

a(n) = (2520*n^9 +22680*n^8 +79920*n^7 +136080*n^6 +107352*n^5 +22680*n^4 -10080*n^3 +1728*n)/9!.

G.f.: x*(x^6+72*x^5+603*x^4+1168*x^3+603*x^2+72*x+1) / (x-1)^10. - Colin Barker, May 02 2014

More terms from Colin Barker, May 02 2014

Typo in example fixed by Colin Barker, May 02 2014

A085440 a(n) = Sum_{i=1..n} binomial(i+1,2)^5.

Original entry on oeis.org

1, 244, 8020, 108020, 867395, 4951496, 22161864, 82628040, 267156165, 770440540, 2022773116, 4909947484, 11150268935, 23913084560, 48796284560, 95322158736, 179163294729, 325374464580, 572984364580, 981394464580, 1639143014731, 2675722491224, 4277290592600

Offset: 1

André F. Labossière, Jun 30 2003

Elisabeth Busser and Gilles Cohen, Neuro-Logies - "Chercher, jouer, trouver", La Recherche, April 1999, No. 319, page 97.

T. D. Noe, Table of n, a(n) for n = 1..1000
Feihu Liu, Guoce Xin, and Chen Zhang, Ehrhart Polynomials of Order Polytopes: Interpreting Combinatorial Sequences on the OEIS, arXiv:2412.18744 [math.CO], 2024. See p. 13.
Index entries for linear recurrences with constant coefficients, signature (12,-66,220,-495,792,-924,792,-495,220,-66,12,-1).

Column k=5 of A334781.

Cf. A000292, A087127, A024166, A024166, A085438, A085439, A085441, A085442, A000332, A086020, A086021, A086022, A000389, A086023, A086024, A000579, A086025, A086026, A000580, A086027, A086028, A027555, A086029, A086030.

[(113400*n^11 +1247400*n^10 +5544000*n^9 +12474000*n^8 +14196600*n^7 +6237000*n^6 -831600*n^5 +1108800*n^3 -172800*n )/Factorial(11): n in [1..30]]; // G. C. Greubel, Nov 22 2017

Table[(113400*n^11 +1247400*n^10 +5544000*n^9 +12474000*n^8 +14196600*n^7 +6237000*n^6 -831600*n^5 +1108800*n^3 -172800*n)/11!, {n,1,50}] (* G. C. Greubel, Nov 22 2017 *)

for(n=1,30, print1(sum(k=1,n, binomial(k+1,2)^5), ", ")) \\ G. C. Greubel, Nov 22 2017

a(n) = (113400*n^11 +1247400*n^10 +5544000*n^9 +12474000*n^8 +14196600*n^7 +6237000*n^6 -831600*n^5 +1108800*n^3 -172800*n)/11!.

G.f.: x*(x^8+232*x^7+5158*x^6+27664*x^5+47290*x^4+27664*x^3+5158*x^2+232*x+1) / (x-1)^12. - Colin Barker, May 02 2014

Formula edited by Colin Barker, May 02 2014

A085441 a(n) = Sum_{i=1..n} binomial(i+1,2)^6.

Original entry on oeis.org

1, 730, 47386, 1047386, 12438011, 98204132, 580094436, 2756876772, 11060642397, 38741283022, 121395233038, 346594833742, 914464085783, 2254559726408, 5240543726408, 11568062614344, 24395756421273, 49397866465794, 96443747465794, 182209868465794

Offset: 1

André F. Labossière, Jul 07 2003

			a(5) = C(7,3)*[191*106 + 450*(18*C(14,10) + 3851*C(13,10) + 61839*C(12,10) + 225352*C(11,10) + 225352*C(10,10))]/10010 = 12438011.

T. D. Noe, Table of n, a(n) for n = 1..1000
Feihu Liu, Guoce Xin, and Chen Zhang, Ehrhart Polynomials of Order Polytopes: Interpreting Combinatorial Sequences on the OEIS, arXiv:2412.18744 [math.CO], 2024. See p. 13.
Index entries for linear recurrences with constant coefficients, signature (14,-91,364,-1001,2002,-3003,3432,-3003,2002,-1001,364,-91,14,-1).

Column k=6 of A334781.

Cf. A000292, A087127, A024166, A024166, A085438, A085439, A085440, A085442, A000332, A086020, A086021, A086022, A000389, A086023, A086024, A000579, A086025, A086026, A000580, A086027, A086028, A027555, A086029, A086030, A234253.

[(n/960960)*(6112 - 40040*n^2 + 78078*n^4 + 15015*n^5 + 19305*n^6 + 225225*n^7 + 335335*n^8 + 225225*n^9 + 80535*n^10 + 15015*n^11 + 1155*n^12): n in [1..30]]; // G. C. Greubel, Nov 22 2017

f:= sum(binomial(1+i,2)^6,i=1..n):
seq(f, n=1..30); # Robert Israel, Nov 22 2017

Table[Sum[Binomial[i+1,2]^6,{i,n}],{n,20}] (* or *) LinearRecurrence[ {14,-91,364,-1001,2002,-3003,3432,-3003,2002,-1001,364,-91,14,-1},{1,730,47386,1047386,12438011, 98204132,580094436, 2756876772,11060642397, 38741283022,121395233038, 346594833742, 914464085783, 2254559726408},20] (* Harvey P. Dale, Jun 05 2017 *)

for(n=1,30, print1(sum(k=1,n, binomial(k+1,2)^6), ", ")) \\ G. C. Greubel, Nov 22 2017

G.f.: x*(x^10 +716*x^9 +37257*x^8 +450048*x^7 +1822014*x^6 +2864328*x^5 +1822014*x^4 +450048*x^3 +37257*x^2 +716*x +1) / (x -1)^14. - Colin Barker, May 02 2014

a(n) = (n/960960)*(6112 - 40040*n^2 + 78078*n^4 + 15015*n^5 + 19305*n^6 + 225225*n^7 + 335335*n^8 + 225225*n^9 + 80535*n^10 + 15015*n^11 + 1155*n^12). - G. C. Greubel, Nov 22 2017

A119900 Triangle read by rows: T(n,k) is the number of binary words of length n with k strictly increasing runs, for 0<=k<=n.

Original entry on oeis.org

1, 0, 2, 0, 1, 3, 0, 0, 4, 4, 0, 0, 1, 10, 5, 0, 0, 0, 6, 20, 6, 0, 0, 0, 1, 21, 35, 7, 0, 0, 0, 0, 8, 56, 56, 8, 0, 0, 0, 0, 1, 36, 126, 84, 9, 0, 0, 0, 0, 0, 10, 120, 252, 120, 10, 0, 0, 0, 0, 0, 1, 55, 330, 462, 165, 11, 0, 0, 0, 0, 0, 0, 12, 220, 792, 792, 220, 12, 0, 0, 0, 0, 0, 0, 1, 78

Offset: 0

Emeric Deutsch, May 27 2006

Sum of terms in row n is 2^n (A000079). Sum of terms in column k is A001906(k+1) (the even-indexed Fibonacci numbers). Row n contains 1+floor(n/2) nonzero terms. Sum_{k=0..n} k*T(n,k) = (3n+1)*2^(n-2) = A066373(n+1) for n>=1.
Triangle T(n,k), 0<=k<=n, read by rows, given by [0,1/2,-1/2,0,0,0,0,0, 0,...] DELTA [2,-1/2,1/2,0,0,0,0,0,0,...] where DELTA is the operator defined in A084938. - Philippe Deléham, Dec 02 2008
From R. Bagula's comment in A053122 (cf. Damianou link), the columns of this array give the coefficients (mod sign) of the characteristic polynomials for the Cartan matrix of the root system A_n. - Tom Copeland, Oct 11 2014
Odd rows contain the Pascal triangle numbers A091042. See A034867 and A034839 for some relations to tan(x). - Tom Copeland, Oct 15 2014

			The binary word 1/0/01/01/1/1/01 has 7 strictly increasing runs.
T(5,3)=6 because we have 0/01/01, 01/0/01, 01/01/0, 01/1/01, 01/01/1 and 1/01/01 (the runs are separated by /).
Triangle starts:
  1;
  0,2;
  0,1,3;
  0,0,4,4;
  0,0,1,10,5;
  0,0,0,6,20,6;

G. C. Greubel, Table of n, a(n) for the first 50 rows, flattened
A. Collins et al., Binary words, n-color compositions and bisection of the Fibonacci numbers, Fib. Quarterly, 51 (2013), 130-136.
P. Damianou, On the characteristic polynomials of Cartan matrices and Chebyshev polynomials, arXiv:1110.6620 [math.RT], 2014.
R. Zielinski, Induction and Analogy in a Problem of Finite Sums arXiv:1608.04006 [math.GM], 2016.

Cf. A000079, A001906, A066373, A034867.
Cf. A098158. - Philippe Deléham, Dec 02 2008
Cf. A000292, A000389, A000580, A001263, A049310, A053122, A134264, A207543.

/* triangle */ [[Binomial(n+1, 2*k-n): k in [0..n]]: n in [0..10]]; // G. C. Greubel, Oct 22 2017

T:=(n,k)->binomial(n+1,2*k-n): for n from 0 to 12 do seq(T(n,k),k=0..n) od; # yields sequence in triangular form

Table[Binomial[n + 1, 2 k - n], {n, 0, 12}, {k, 0, n}] // Flatten (* Michael De Vlieger, Aug 21 2016 *)

for(n=0,10, for(k=0,n, print1(binomial(n+1, 2*k-n), ", "))) \\ G. C. Greubel, Oct 22 2017

T(n,k) = binomial(n+1,2k-n).
G.f.: 1/(1 - 2*t*z - t*(1-t)*z^2).
T(n,k) = A034867(n,n-k)
From Tom Copeland, Sep 30 2011: (Start)
With K(x,t) = 1/{d/dx{x/[t-1+1/(1-x)]}} = [t-1+1/(1-x)]^2/{t-[x/(1-x)]^2}, the g.f. of A119900 = K(x*t,t)-t+1.
From formulas in A134264: K(x,t)d/dx is a generator for A001263. A refinement of A119900 to partition polynomials is given by umbralizing
  K(x,t) roughly as K(h.x,h_0) and precisely as in A134264 as
  W(x)= 1/{d/dx[f(x)]}=1/{d/dx[x/h(x)]}. (End)
T(n,k) = 2*T(n-1,k-1) + T(n-2,k-1) - T(n-2,k-2). - Philippe Deléham, Oct 02 2011
From Tom Copeland, Dec 07 2015: (Start)
An alternate o.g.f. is (1/(x*t)) {-1 + 1 / [1 - (1/t)[x*t/(1-x*t)]^2]} = Sum_{n>0} x^(2(n-1)+1) t^(n-1) / (1-t*x)^(2n) = x + 2t x^2 + (t+3t^2) x^3 + ... .
The n-th diagonal has elements binomial(2n+1+k,k), starting with k=0 for the first non-vanishing element, with o.g.f. (1-x)^(-2(n+1)). The first few subdiagonals are shifted versions of A000292, A000389, and A000580. Cf. A049310.
See A034867 for the matrix representation for the infinitesimal generator K(x,t) d/dx for the Narayana polynomials. (End)
From Peter Bala, Aug 17 2016: (Start)
Let S(k,n) = Sum_{i = 1..n} i^k. Calculations in Zielinski 2016 suggest the following identity holds involving the p-th row elements of this triangle:
Sum_{k = 0..p} T(p,k)*S(2*k + 1,n) = (n*(n + 1)/2)^(p+1).
For example, for row 6 we find S(7,n) + 21*S(9,n) + 35*S(11,n) + 7*S(13,n) = (n*(n + 1)/2)^7.
There appears to be a similar result for the even power sums S(2*k,n) involving A207543. (End)

Keyword tabl added by Philippe Deléham, Jan 26 2010

A087107 This table shows the coefficients of combinatorial formulas needed for generating the sequential sums of p-th powers of tetrahedral numbers. The p-th row (p>=1) contains a(i,p) for i=1 to 3p-2, where a(i,p) satisfies Sum_{i=1..n} C(i+2,3)^p = 4 C(n+3,4) * Sum_{i=1..3p-2} a(i,p) C(n-1,i-1)/(i+3).

Original entry on oeis.org

1, 1, 3, 3, 1, 1, 15, 69, 147, 162, 90, 20, 1, 63, 873, 5191, 16620, 31560, 36750, 25830, 10080, 1680, 1, 255, 9489, 130767, 919602, 3832650, 10238000, 18244380, 21990360, 17745000, 9198000, 2772000, 369600, 1, 1023, 97953, 2903071, 40317780

Offset: 1

André F. Labossière, Aug 11 2003

Let s_n denote the sequence (1, 4^n, 10^n, 20^n, ...) regarded as an infinite column vector, where 1, 4, 10, 20, ... is the sequence of tetrahedral numbers A000292. It appears that the n-th row of this table is determined by the matrix product P^(-1)s_n, where P denotes Pascal's triangle A007318. - Peter Bala, Nov 26 2017
From Peter Bala, Mar 11 2018: (Start)
The observation above is correct.
The table entries T(n,k) are the coefficients when expressing the polynomial C(x+3,3)^p of degree 3*p in terms of falling factorials: C(x+3,3)^p = Sum_{k = 0..3*p} T(p,k)*C(x,k). It follows that Sum_{i = 0..n-1} C(i+3,3)^p = Sum_{k = 0..3*p} T(p,k)*C(n,k+1).
The sum of the p-th powers of the tetrahedral numbers is also given by Sum_{i = 0..n-1} C(i+3,3)^p = Sum_{k = 3..3*p} A299041(p,k)*C(n+3,k+1) for p >= 1. (End)

			Row 3 contains 1,15,69,147,162,90,20, so Sum_{i=1..n} C(i+2,3)^3 = 4 * C(n+3,4) * [ a(1,3)/4 + a(2,3)*C(n-1,1)/5 + a(3,3)*C(n-1,2)/6 + ... + a(7,3)*C(n-1,6)/10 ] = 4 * C(n+3,4) * [ 1/4 + 15*C(n-1,1)/5 + 69*C(n-1,2)/6 + 147*C(n-1,3)/7 + 162*C(n-1,4)/8 + 90*C(n-1,5)/9 + 20*C(n-1,6)/10 ]. Cf. A086021 for more details.
From _Peter Bala_, Mar 11 2018: (Start)
Table begins
n=0 | 1
n=1 | 1  3   3    1
n=2 | 1 15  69  147   162    90    20
n=3 | 1 63 873 5191 16620 31560 36750 25830 10080 1680
...
Row 2: C(i+3,3)^2 = C(i,0) + 15*C(i,1) + 69*C(i,2) + 147*C(i,3) + 162*C(i,4) + 90*C(i,5) + 20*C(i,6). Hence, Sum_{i = 0..n-1} C(i+3,3)^2 =  C(n,1) + 15*C(n,2) + 69*C(n,3) + 147*C(n,4) + 162*C(n,5) + 90*C(n,6) + 20*C(n,7). (End)

G. C. Greubel, Table of n, a(n) for the first 50 rows, flattened
Dukes, C. D. White, Web Matrices: Structural Properties and Generating Combinatorial Identities, arXiv:1603.01589 [math.CO], 2016.

Cf. A000292, A024166, A087127, A024166, A085438, A085439, A085440, A085441, A085442, A000332, A086020, A086021, A086022, A087108, A000389, A086023, A086024, A087109, A000579, A086025, A086026, A087110, A000580, A086027, A086028, A087111, A027555, A086029, A086030.

Cf. A087127, A087110, A174266, A299041.

seq(seq(add( (-1)^(k-i)*binomial(k, i)*binomial(i+3, 3)^n, i= 0..k), k = 0..3*n), n = 0..8); # Peter Bala, Mar 11 2018

a[i_, p_] := Sum[Binomial[i - 1, 2*k - 2]*Binomial[i - 2*k + 4, i - 2*k + 1]^(p - 1) - Binomial[i - 1, 2*k - 1]*Binomial[i - 2*k + 3, i - 2*k]^(p - 1), {k, 1, (2*i + 1 + (-1)^(i - 1))/4}]; Table[If[p == 1, 1, a[i, p]], {p, 1, 10}, {i, 1, 3*p - 2}]//Flatten (* G. C. Greubel, Nov 23 2017 *)

{a(i, p) = sum(k=1, (2*i + 1 + (-1)^(i - 1))/4, binomial(i - 1, 2*k - 2)*binomial(i - 2*k + 4, i - 2*k + 1)^(p - 1) - binomial(i - 1, 2*k - 1)*binomial(i - 2*k + 3, i - 2*k)^(p - 1))}; for(p=1,8, for(i=1, 3*p-2, print1(if(p==1,1,a(i,p)), ", "))) \\ G. C. Greubel, Nov 23 2017

a(i, p) = Sum_{k=1..[2*i+1+(-1)^(i-1)]/4} [ C(i-1, 2*k-2)*C(i-2*k+4, i-2*k+1)^(p-1) -C(i-1, 2*k-1)*C(i-2*k+3, i-2*k)^(p-1) ].
From Peter Bala, Nov 26 2017: (Start)
Conjectural formula for table entries: T(n,k) = Sum_{j = 0..k} (-1)^(k+j)*binomial(k,j)*binomial(j+3,3)^n.
Conjecturally, the n-th row polynomial R(n,x) = 1/(1 + x)*Sum_{i >= 0} binomial(i+3,3)^n *(x/(1 + x))^n. (End)
From Peter Bala, Mar 11 2018: (Start)
The conjectures above are correct.
The following remarks assume the row and column indices start at 0.
T(n+1,k) = C(k+3,3)*T(n,k) + 3*C(k+2,3)*T(n,k-1) + 3*C(k+1,3)*T(n,k-2) + C(k,3)*T(n,k-3) with boundary conditions T(n,0) = 1 for all n and T(n,k) = 0 for k > 3*n.
Sum_{k = 0..3*n} T(n,k)*binomial(x,k) = (binomial(x+3,3))^n.
x^3*R(n,x) = (1 + x)^3 * the n-th row polynomial of A299041.
R(n+1,x) = 1/3!*(1 + x)^3*(d/dx)^3 (x^3*R(n,x)).
(1 - x)^(3*n)*R(n,x/(1 - x)) gives the n-th row polynomial of A174266.
R(n,x) = (1 + x)^3 o (1 + x)^3 o ... o (1 + x)^3 (n factors), where o denotes the black diamond product of power series defined in Dukes and White. Note the polynomial x^3 o ... o x^3 (n factors) is the n-th row polynomial of A299041. (End)

Edited by Dean Hickerson, Aug 16 2003

A110813 A triangle of pyramidal numbers.

Original entry on oeis.org

1, 3, 1, 5, 4, 1, 7, 9, 5, 1, 9, 16, 14, 6, 1, 11, 25, 30, 20, 7, 1, 13, 36, 55, 50, 27, 8, 1, 15, 49, 91, 105, 77, 35, 9, 1, 17, 64, 140, 196, 182, 112, 44, 10, 1, 19, 81, 204, 336, 378, 294, 156, 54, 11, 1, 21, 100, 285, 540, 714, 672, 450, 210, 65, 12, 1, 23, 121, 385, 825

Offset: 0

Paul Barry, Aug 05 2005

Triangle A029653 less first column. In general, the product (1/(1-x),x/(1-x))*(1+m*x,x) yields the Riordan array ((1+(m-1)x)/(1-x)^2,x/(1-x)) with general term T(n,k)=(m*n-(m-1)*k+1)*C(n+1,k+1)/(n+1). This is the reversal of the (1,m)-Pascal triangle, less its first column. - Paul Barry, Mar 01 2006
The column sequences give, for k=0..10: A005408 (odd numbers), A000290 (squares), A000330, A002415, A005585, A040977, A050486, A053347, A054333, A054334, A057788.
Linked to Chebyshev polynomials by the fact that this triangle with interpolated zeros in the rows and columns is a scaled version of A053120.
Row sums are A033484. Diagonal sums are A001911(n+1) or F(n+4)-2. Factors as (1/(1-x),x/(1-x))*(1+2x,x). Inverse is A110814 or (-1)^(n-k)*A104709.
This triangle is a subtriangle of the [2,1] Pascal triangle A029653 (omit there the first column).
Subtriangle of triangles in A029653, A131084, A208510. - Philippe Deléham, Mar 02 2012
This is the iterated partial sums triangle of A005408 (odd numbers). Such iterated partial sums of arithmetic progression sequences have been considered by Narayana Pandit (see the Mar 20 2015 comment on A000580 where the MacTutor History of Mathematics archive link and the Gottwald et al. reference, p. 338, are given). - Wolfdieter Lang, Mar 23 2015

			The number triangle T(n, k) begins
n\k  0   1   2   3    4    5    6   7   8  9 10 11
0:   1
1:   3   1
2:   5   4   1
3:   7   9   5   1
4:   9  16  14   6    1
5:  11  25  30  20    7    1
6:  13  36  55  50   27    8    1
7:  15  49  91 105   77   35    9   1
8:  17  64 140 196  182  112   44  10   1
9:  19  81 204 336  378  294  156  54  11  1
10: 21 100 285 540  714  672  450 210  65 12  1
11: 23 121 385 825 1254 1386 1122 660 275 77 13  1
... reformatted by _Wolfdieter Lang_, Mar 23 2015
As a number square S(n, k) = T(n+k, k), rows begin
  1,   1,   1,   1,   1,   1, ...
  3,   4,   5,   6,   7,   8, ...
  5,   9,  14,  20,  27,  35, ...
  7,  16,  30,  50,  77, 112, ...
  9,  25,  55, 105, 182, 294, ...

G. C. Greubel, Table of n, a(n) for the first 50 rows, flattened

Cf. A000290, A000330, A002415, A005408, A005585, A029655, A040977, A050486, A053347, A054333, A054334, A057788.

Table[2*Binomial[n + 1, k + 1] - Binomial[n, k], {n, 0, 10}, {k, 0, n}] // Flatten (* G. C. Greubel, Oct 19 2017 *)

for(n=0,10, for(k=0,n, print1(2*binomial(n+1, k+1) - binomial(n,k), ", "))) \\ G. C. Greubel, Oct 19 2017

Number triangle T(n, k) = C(n, k)*(2n-k+1)/(k+1) = 2*C(n+1, k+1) - C(n, k); Riordan array ((1+x)/(1-x)^2, x/(1-x)); As a number square read by antidiagonals, T(n, k)=C(n+k, k)(2n+k+1)/(k+1).
Equals A007318 * an infinite bidiagonal matrix with 1's in the main diagonal and 2's in the subdiagonal. - Gary W. Adamson, Dec 01 2007
Binomial transform of an infinite lower triangular matrix with all 1's in the main diagonal, all 2's in the subdiagonal and the rest zeros. - Gary W. Adamson, Dec 12 2007
T(n,k) = 2*T(n-1,k) + T(n-1,k-1) - T(n-2,k) - T(n-2,k-1), T(0,0)=T(1,1)=1, T(1,0)=3, T(n,k)=0 if k<0 or if k>n. - Philippe Deléham, Nov 30 2013
exp(x) * e.g.f. for row n = e.g.f. for diagonal n. For example, for n = 3 we have exp(x)*(7 + 9*x + 5*x^2/2! + x^3/3!) = 7 + 16*x + 30*x^2/2! + 50*x^3/3! + 77*x^4/4! + .... The same property holds more generally for Riordan arrays of the form ( f(x), x/(1 - x) ). - Peter Bala, Dec 21 2014
T(n, k) = ps(1, 2; k, n-k) with ps(a, d; k, n) = sum(ps(a, d; k-1, j), j=0..n) and input ps(a, d; 0, j) = a + d*j. See the iterated partial sums comment from Mar 23 2015 above. - Wolfdieter Lang, Mar 23 2015
From Franck Maminirina Ramaharo, May 21 2018: (Start)
T(n,k) = coefficients in the expansion of ((x + 2)*(x + 1)^n - 2)/x.
T(n,k) = A135278(n,k) + A135278(n-1,k).
T(n,k) = A097207(n,n-k).
G.f.: (y + 1)/((y - 1)*(x*y + y - 1)).
E.g.f.: ((x + 2)*exp(x*y + y) - 2*exp(y))/x.
(End)

A087111 This table shows the coefficients of combinatorial formulas needed for generating the sequential sums of p-th powers of binomial coefficients C(n,7). The p-th row (p>=1) contains a(i,p) for i=1 to 7p-6, where a(i,p) satisfies Sum_{i=1..n} C(i+6,7)^p = 8 C(n+7,8) * Sum_{i=1..7p-6} a(i,p) C(n-1,i-1)/(i+7).

Original entry on oeis.org

1, 1, 7, 21, 35, 35, 21, 7, 1, 1, 63, 1169, 10703, 58821, 214123, 545629, 1004307, 1356194, 1347318, 974862, 500346, 172788, 36036, 3432, 1, 511, 45633, 1589567, 29302889, 333924087, 2577462937, 14287393351, 59159005164, 188008120188

Offset: 1

André F. Labossière, Aug 11 2003

From Peter Bala, Mar 11 2018: (Start)

The table entries T(n,k) are the coefficients when expressing the polynomial C(x+7,7)^p of degree 7*p in terms of falling factorials: C(x+7,7)^p = Sum_{k = 0..7*p} T(p,k)*C(x,k). It follows that Sum_{i = 0..n-1} C(i+7,7)^p = Sum_{k = 0..7*p} T(p,k)*C(n,k+1). (End)

			Row 3 contains 1,63,1169,...,3432, so Sum_{i=1..n} C(i+6,7)^3 = 8 * C(n+7,8) * [ a(1,3)/8 + a(2,3)*C(n-1,1)/9 + a(3,3)*C(n-1,2)/10 + ... + a(15,3)*C(n-1,14)/22 ] = 8 * C(n+7,8) * [ 1/8 + 63*C(n-1,1)/9 + 1169*C(n-1,2)/10 + ... + 3432*C(n-1,14)/22 ]. Cf. A086030 for more details.

G. C. Greubel, Table of n, a(n) for the first 40 rows, flattebed
Dukes, C. D. White, Web Matrices: Structural Properties and Generating Combinatorial Identities, arXiv:1603.01589 [math.CO], 2016.

Cf. A000292, A024166, A087127, A024166, A085438, A085439, A085440, A085441, A085442, A087107, A000332, A086020, A086021, A086022, A087108, A000389, A086023, A086024, A087109, A000579, A086025, A086026, A087110, A000580, A086027, A086028, A027555, A086029, A086030, A087127.

seq(seq(add( (-1)^(k-i)*binomial(k, i)*binomial(i+7, 7)^n, i = 0..k), k = 0..7*n), n = 0..4); # Peter Bala, Mar 11 2018

a[i_, p_] := Sum[Binomial[i - 1, 2*k - 2]*Binomial[i - 2*k + 8, i - 2*k + 1]^(p - 1) - Binomial[i - 1, 2*k - 1]*Binomial[i - 2*k + 7, i - 2*k]^(p - 1), {k, 1, (2*i + 1 + (-1)^(i - 1))/4}]; Table[If[p == 1, 1, a[i, p]], {p, 1, 10}, {i, 1, 7*p - 6}]//Flatten (* G. C. Greubel, Nov 23 2017 *)

{a(i, p) = sum(k=1, (2*i + 1 + (-1)^(i - 1))/4, binomial(i - 1, 2*k - 2)*binomial(i - 2*k + 8, i - 2*k + 1)^(p - 1) - binomial(i - 1, 2*k - 1)*binomial(i - 2*k + 7, i - 2*k)^(p - 1))}; for(p=1,8, for(i=1, 7*p-6, print1(if(p==1,1,a(i,p)), ", "))) \\ G. C. Greubel, Nov 23 2017

a(i, p) = Sum_{k=1..[2*i+1+(-1)^(i-1)]/4} [ C(i-1, 2*k-2)*C(i-2*k+8, i-2*k+1)^(p-1) -C(i-1, 2*k-1)*C(i-2*k+7, i-2*k)^(p-1) ]
From Peter Bala, Mar 11 2018: (Start)
The following remarks assume the row and column indices start at 0.
T(n,k) = Sum_{i = 0..k} (-1)^(k-i)*binomial(k,i) * binomial(i+7,7)^n. Equivalently, let v_n denote the sequence (1, 8^n, 36^n, 120^n, ...) regarded as an infinite column vector, where 1, 8, 36, 120, ... is the sequence binomial(n+7,7) - see A000580. Then the n-th row of this table is determined by the matrix product P^(-1)*v_n, where P denotes Pascal's triangle A007318.
Recurrence: T(n+1,k) = Sum_{i = 0..7} C(7,i)*C(k+7-i,7)*T(n,k-i) with boundary conditions T(n,0) = 1 for all n and T(n,k) = 0 for k > 7*n.
n-th row polynomial R(n,x) = (1 + x)^7 o (1 + x)^7 o ... o (1 + x)^7 (n factors), where o denotes the black diamond product of power series defined in Dukes and White.
R(n+1,x) = 1/7!*(1 + x)^7 * (d/dx)^7(x^7*R(n,x)).
R(n,x) = Sum_{i >= 0} binomial(i+7,7)^n*x^i/(1 + x)^(i+1).
(End)

Edited by Dean Hickerson, Aug 16 2003

A128908 Riordan array (1, x/(1-x)^2).

Original entry on oeis.org

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

Offset: 0

Philippe Deléham, Apr 22 2007

Triangle T(n,k), 0 <= k <= n, read by rows given by [0,2,-1/2,1/2,0,0,0,0,0,...] DELTA [1,0,0,0,0,0,0,0,...] where DELTA is the operator defined in A084938.
Row sums give A088305. - Philippe Deléham, Nov 21 2007
Column k is C(n,2k-1) for k > 0. - Philippe Deléham, Jan 20 2012
From R. Bagula's comment in A053122 (cf. Damianou link p. 10), this array gives the coefficients (mod sign) of the characteristic polynomials for the Cartan matrix of the root system A_n. - Tom Copeland, Oct 11 2014
T is the convolution triangle of the positive integers (see A357368). - Peter Luschny, Oct 19 2022

			The triangle T(n,k) begins:
   n\k  0    1    2    3    4    5    6    7    8    9   10
   0:   1
   1:   0    1
   2:   0    2    1
   3:   0    3    4    1
   4:   0    4   10    6    1
   5:   0    5   20   21    8    1
   6:   0    6   35   56   36   10    1
   7:   0    7   56  126  120   55   12    1
   8:   0    8   84  252  330  220   78   14    1
   9:   0    9  120  462  792  715  364  105   16    1
  10:   0   10  165  792 1716 2002 1365  560  136   18    1
  ... reformatted by _Wolfdieter Lang_, Jul 31 2017
From _Peter Luschny_, Mar 06 2022: (Start)
The sequence can also be seen as a square array read by upwards antidiagonals.
   1, 1,   1,    1,    1,     1,     1,      1,      1, ...  A000012
   0, 2,   4,    6,    8,    10,    12,     14,     16, ...  A005843
   0, 3,  10,   21,   36,    55,    78,    105,    136, ...  A014105
   0, 4,  20,   56,  120,   220,   364,    560,    816, ...  A002492
   0, 5,  35,  126,  330,   715,  1365,   2380,   3876, ... (A053126)
   0, 6,  56,  252,  792,  2002,  4368,   8568,  15504, ... (A053127)
   0, 7,  84,  462, 1716,  5005, 12376,  27132,  54264, ... (A053128)
   0, 8, 120,  792, 3432, 11440, 31824,  77520, 170544, ... (A053129)
   0, 9, 165, 1287, 6435, 24310, 75582, 203490, 490314, ... (A053130)
    A27,A292, A389, A580,  A582, A1288, A10966, A10968, A165817       (End)

G. C. Greubel, Table of n, a(n) for the first 100 rows, flattened
P. Damianou, On the characteristic polynomials of Cartan matrices and Chebyshev polynomials, arXiv:1110.6620 [math.RT], 2014.

Cf. A002450, A007318, A034008, A053122, A078812, A084938, A088305.
Cf. Columns : A000007, A000027, A000292, A000389, A000580, A000582, A001288, A010966 ..(+2).. A011000, A017713 ..(+2).. A017763.
Cf. A000007, A001352, A008574, A054888, A084099, A084103, A163810, A357368.
Cf. A165817 (the main diagonal of the array).

# Computing the rows of the array representation:
S := proc(n,k) option remember;
if n = k then 1 elif k < 0 or k > n then 0 else
S(n-1, k-1) + 2*S(n-1, k) - S(n-2, k) fi end:
Arow := (n, len) -> seq(S(n+k-1, k-1), k = 0..len-1):
for n from 0 to 8 do Arow(n, 9) od; # Peter Luschny, Mar 06 2022
# Uses function PMatrix from A357368.
PMatrix(10, n -> n); # Peter Luschny, Oct 19 2022

With[{nmax = 10}, CoefficientList[CoefficientList[Series[(1 - x)^2/(1 - (2 + y)*x + x^2), {x, 0, nmax}, {y, 0, nmax}], x], y]] // Flatten (* G. C. Greubel, Nov 22 2017 *)

for(n=0,10, for(k=0,n, print1(if(n==0 && k==0, 1, if(k==0, 0, binomial(n+k-1,2*k-1))), ", "))) \\ G. C. Greubel, Nov 22 2017

from functools import cache
@cache
def A128908(n, k):
    if n == k: return 1
    if (k <= 0 or k > n): return 0
    return A128908(n-1, k-1) + 2*A128908(n-1, k) - A128908(n-2, k)
for n in range(10):
    print([A128908(n, k) for k in range(n+1)]) # Peter Luschny, Mar 07 2022

@cached_function
def T(k,n):
    if k==n: return 1
    if k==0: return 0
    return sum(i*T(k-1,n-i) for i in (1..n-k+1))
A128908 = lambda n,k: T(k,n)
for n in (0..10): print([A128908(n,k) for k in (0..n)]) # Peter Luschny, Mar 12 2016

T(n,0) = 0^n, T(n,k) = binomial(n+k-1, 2k-1) for k >= 1.
Sum_{k=0..n} T(n,k)*2^(n-k) = A002450(n) = (4^n-1)/3 for n>=1. - Philippe Deléham, Oct 19 2008
G.f.: (1-x)^2/(1-(2+y)*x+x^2). - Philippe Deléham, Jan 20 2012
Sum_{k=0..n} T(n,k)*x^k = (-1)^n*A001352(n), (-1)^(n+1)*A054888(n+1), (-1)^n*A008574(n), (-1)^n*A084103(n), (-1)^n*A084099(n), A163810(n), A000007(n), A088305(n) for x = -6, -5, -4, -3, -2, -1, 0, 1 respectively. - Philippe Deléham, Jan 20 2012
Riordan array (1, x/(1-x)^2). - Philippe Deléham, Jan 20 2012

A237252 Irregular triangle read by rows: T(n,k) = Sum_{i=0..k} (-1)^i * binomial(6n+1,i) binomial(k+6-i,6)^n, 0 <= k <= 6*(n-1).

Original entry on oeis.org

1, 1, 36, 225, 400, 225, 36, 1, 1, 324, 15606, 233300, 1424925, 4050864, 5703096, 4050864, 1424925, 233300, 15606, 324, 1, 1, 2376, 554931, 35138736, 879018750, 10490842656, 66555527346, 239677178256, 509723668476, 654019630000, 509723668476, 239677178256, 66555527346, 10490842656, 879018750, 35138736, 554931, 2376, 1

Offset: 1

Yahia Kahloune, Feb 05 2014

In general, define b(k,e,p) = Sum_{i=0..k} (-1)^i*binomial(e*p+1,i)*binomial(k+e-i,e)^p. Then T(n,k) = b(k,6,n).
Using these coefficients we can obtain formulas for binomial(n,e)^p and for Sum_{i=1..n} binomial(e-1+i,e)^p.
In particular:
  binomial(n, e)^p = Sum_{k=0..e*(p-1)} b(k,e p) * binomial(n+k, e*p).
  Sum_{i=1..n} binomial(e-1+i, e)^p = Sum_{k=0..e*(p-1)} b(k,e,p) * binomial(n+e+k, e*p+1).
T(n,k) is the number of permutations of 6 indistinguishable copies of 1..n with exactly k descents. A descent is a pair of adjacent elements with the second element less than the first. - Andrew Howroyd, May 06 2020

			For example :
  T(n,0) = 1;
  T(n,1) = 7^n - (6*n+1);
  T(n,2) = 28^n - (6*n+1)*7^n + C(6*n+1,2);
  T(n,3) = 84^n - (6*n+1)*28^n + C(6*n+1,2)*7^n + C(6*n+1,3);
  T(n,4) = 210^n - (6*n+1)*84^n + C(6*n+1,2)*28^n - C(6*n+1,3)*7^n + C(6*n+1,4).
Triangle T(n,k) begins:
 1;
 1, 36, 225, 400, 225, 36, 1;
 1, 324, 15606, 233300, 1424925, 4050864, 5703096, 4050864, 1424925, 233300, 15606, 324, 1;
 1, 2376, 554931, 35138736, 879018750, 10490842656, 66555527346, 239677178256, 509723668476, 654019630000, 509723668476, 239677178256, 66555527346, 10490842656, 879018750, 35138736, 554931, 2376, 1;
 1, 16776, 16689816, 3656408776, 286691702976, 10255094095176, 192698692565176, 2080037792142216, 13690633212385551, 57229721552316976, 156200093827061616, 283397584598631216, 345271537321293856, 283397584598631216, 156200093827061616, 57229721552316976,13690633212385551, 2080037792142216, 192698692565176, 10255094095176, 286691702976, 3656408776, 16689816, 16776, 1;
...
Example:
Sum_{i=1..n} C(5+i,6)^2 = A086027(n) = C(n+6,13) + 36*C(n+7,13) + 225*C(n+8,13) + 400*C(n+9,13) + 225*C(n+10,13) + 36*C(n+11,13) + C(n+12,13).
binomial(n,6)^2 = C(n,12) + 36*C(n+1,12) + 225*C(n+2,12) + 400*C(n+3,12) + 225*C(n+4,12) + 36*C(n+5,12) + C(n+6,12).

G. C. Greubel, Table of n, a(n) for the first 25 rows, flattened
Feihu Liu, Guoce Xin, and Chen Zhang, Ehrhart Polynomials of Order Polytopes: Interpreting Combinatorial Sequences on the OEIS, arXiv:2412.18744 [math.CO], 2024. See p. 12.

Columns k=2..6 are A151651, A151652, A151653, A151654, A151655.
Row sums are A248814.
Similar triangles for e=1..5: A173018 (or A008292), A154283, A174266, A236463, A237202.
Sum_{i=1..n} binomial(5+i,6)^p for p=1..3 gives: A000580, A086027, A086028.
Cf. A087127, A086023, A086024, A086025, A087107, A087108, A087109, A087110, A087111, A181544.

b[k_, 6, p_] := Sum[(-1)^i*Binomial[6*p+1, i]*Binomial[k-i, 6]^p /. k -> 6+i, {i, 0, k-6}]; row[p_] := Table[b[k, 6, p], {k, 6, 6*p}]; Table[row[p], {p, 1, 5}] // Flatten (* Jean-François Alcover, Feb 05 2014 *)

T(n,k)={sum(i=0, k, (-1)^i*binomial(6*n+1, i)*binomial(k+6-i, 6)^n)} \\ Andrew Howroyd, May 06 2020

Sum_{i=1..n} binomial(5+i,6)^p = Sum{k=0..6*(p-1)} T(p,k) * binomial(n+6+k, 6*p+1).

binomial(n,6)^p = Sum_{k=0..6*(p-1)} T(p,k) * binomial(n+k, 6*p).

Edited by Andrew Howroyd, May 06 2020

A242023 Decimal expansion of Sum_{n >= 1} (-1)^(n + 1)24/(n(n + 1)(n + 2)(n + 3)).

Original entry on oeis.org

8, 4, 7, 3, 7, 6, 4, 4, 4, 5, 8, 4, 9, 1, 6, 5, 6, 8, 0, 1, 8, 0, 9, 4, 5, 5, 3, 3, 2, 8, 3, 1, 6, 8, 4, 5, 0, 8, 2, 6, 7, 0, 9, 6, 6, 1, 9, 4, 8, 3, 4, 7, 9, 8, 5, 2, 8, 4, 2, 6, 9, 7, 0, 4, 5, 5, 2, 6, 2, 5, 6, 9, 6, 9

Offset: 0

Richard R. Forberg, Aug 11 2014

Sum of terms of the inverse of Binomial(n,4) or A000332, for n>=4, with alternating signs.
In general the sums of Binomial coefficients of this form appear to have the form  m*log(2) - r, where m is an integer and r is rational as below:
For Binomial(n,1):  m = 1, r = 0. See A002162.
For Binomial(n,2):  m = 4, r = 2. See A000217.
For Binomial(n,3):  m = 12 r = 15/2. See A000292.
For Binomial(n,4):  m = 32, r = 64/3. See A000332.
For Binomial(n,5):  m = 80, r = 655/12. See A000389.
For Binomial(n,6):  m = 192, r = 661/5. See A000579.
For Binomial(n,7):  m = 448, r = 9289/30. See A000580.
For Binomial(n,8):  m = 1024, r = 74432/105. See A000581.
This is generalized as follows:
m grows as A001787(k) = k*2^(k-1) for Binomial(n,k).
r * (k-1)! produces the integer sequence: a(k) = 0, 2, 15, 128, 1310, 15864, 222936, 3572736,  where a(k+1)/a(k) approaches 2*k for large k.
Results are precise to 100 digits or more using Mathematica.

			0.8473764445849165680180945...

G. C. Greubel, Table of n, a(n) for n = 0..5000

Cf. A000332, A000217, A000292, A242024, A002162, A001787.

[32*Log(2) - 64/3]; // G. C. Greubel, Nov 23 2017

Sum[N[(-1)^(n + 1)*24/(n*(n + 1)*(n + 2)*(n + 3)), 150], {n, 1, Infinity}]
RealDigits[32*Log[2] - 64/3, 10, 50][[1]] (* G. C. Greubel, Nov 23 2017 *)

32*log(2) - 64/3 \\ Michel Marcus, Aug 13 2014

sumalt(n=1, (-1)^(n + 1)*24/(n*(n + 1)*(n + 2)*(n + 3))) \\ Michel Marcus, Aug 14 2014

Equals 32*log(2) - 64/3.

Equals 32*(A259284-1). - R. J. Mathar, Jun 30 2021

cp's OEIS Frontend

A085439 a(n) = Sum_{i=1..n} binomial(i+1,2)^4.

Views

Author

Keywords

Examples

Links

Crossrefs

Programs

Magma

Mathematica

PARI

PARI

Formula

Extensions

A085440 a(n) = Sum_{i=1..n} binomial(i+1,2)^5.

Views

Author

Keywords

References

Links

Crossrefs

Programs

Magma

Mathematica

PARI

Formula

Extensions

A085441 a(n) = Sum_{i=1..n} binomial(i+1,2)^6.

Views

Author

Keywords

Examples

Links

Crossrefs

Programs

Magma

Maple

Mathematica

PARI

Formula

A119900 Triangle read by rows: T(n,k) is the number of binary words of length n with k strictly increasing runs, for 0<=k<=n.

Views

Author

Keywords

Comments

Examples

Links

Crossrefs

Programs

Magma

Maple

Mathematica

PARI

Formula

Extensions

Views

Author

Keywords

Comments

Examples

Links

Crossrefs

Programs

Maple

Mathematica

PARI

Formula

Extensions

A110813 A triangle of pyramidal numbers.

Views

Author

Keywords

Comments

Examples

Links

Crossrefs

Programs

Mathematica

PARI

A237252 Irregular triangle read by rows: T(n,k) = Sum_{i=0..k} (-1)^i * binomial(6n+1,i) binomial(k+6-i,6)^n, 0 <= k <= 6*(n-1).

A242023 Decimal expansion of Sum_{n >= 1} (-1)^(n + 1)24/(n(n + 1)(n + 2)(n + 3)).