A086022 -id:A086022 - cp's OEIS frontend

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

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

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

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

A087108 This table shows the coefficients of combinatorial formulas needed for generating the sequential sums of p-th powers of binomial coefficients C(n,4). The p-th row (p>=1) contains a(i,p) for i=1 to 4p-3, where a(i,p) satisfies Sum_{i=1..n} C(i+3,4)^p = 5 C(n+4,5) * Sum_{i=1..4p-3} a(i,p) C(n-1,i-1)/(i+4).

Original entry on oeis.org

1, 1, 4, 6, 4, 1, 1, 24, 176, 624, 1251, 1500, 1070, 420, 70, 1, 124, 3126, 33124, 191251, 681000, 1596120, 2543520, 2780820, 2058000, 987000, 277200, 34650, 1, 624, 49376, 1350624, 18308751, 146500500, 763418870, 2749648020, 7101675070, 13440210000

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+4,4)^p of degree 4*p in terms of falling factorials: C(x+4,4)^p = Sum_{k = 0..4*p} T(p,k)*C(x,k). It follows that Sum_{i = 0..n-1} C(i+4,4)^p = Sum_{k = 0..4*p} T(p,k)*C(n,k+1). (End)

			Row 3 contains 1,24,176,...,70, so Sum_{i=1..n} C(i+3,4)^3 = 5 * C(n+4,5) * [ a(1,3)/5 + a(2,3)*C(n-1,1)/6 + a(3,3)*C(n-1,2)/7 + ... + a(9,3)*C(n-1,8)/13 ] = 5 * C(n+4,5) * [ 1/5 + 24*C(n-1,1)/6 + 176*C(n-1,2)/7 + ... + 70*C(n-1,8)/13 ]. Cf. A086024 for more details.
From _Peter Bala_, Mar 11 2018: (Start)
Table begins
  n = 0 | 1
  n = 1 | 1   4    6     4      1
  n = 2 | 1  24  176   624   1251   1500    1070  420  70
  n = 3 | 1 124 3126 33124 191251 681000 1596120 ...
  ...
Row 2: C(i+4,4)^2 = C(i,0) + 24*C(i,1) + 176*C(i,2) + 624*C(i,3) + 1251*C(i,4) + 1500*C(i,5) + 1070*C(i,6) + 420*C(i,7) + 70*C(i,8). Hence, Sum_{i = 0..n-1} C(i+4,4)^2 =  C(n,1) + 24*C(n,2) + 176*C(n,3) + 624*C(n,4) + 1251*C(n,5) + 1500*C(n,6) + 1070*C(n,7) + 420*C(n,8) + 70*C(n,9) .(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, A087107, A000332, A086020, A086021, A086022, A000389, A086023, A086024, A087109, A000579, A086025, A086026, A087110, A000580, A086027, A086028, A087111, A027555, A086029, A086030.

Cf. A087127, A236463.

seq(seq(add( (-1)^(k-i)*binomial(k, i)*binomial(i+4, 4)^n, i = 0..k), k = 0..4*n), n = 0..6); # Peter Bala, Mar 11 2018

a[i_, p_] := Sum[Binomial[i - 1, 2*k - 2]*Binomial[i - 2*k + 5, i - 2*k + 1]^(p - 1) - Binomial[i - 1, 2*k - 1]*Binomial[i - 2*k + 4, 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, 4*p - 3}]//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 + 5, i - 2*k + 1)^(p - 1) - binomial(i - 1, 2*k - 1)*binomial(i - 2*k + 4, i - 2*k)^(p - 1))}; for(p=1,8, for(i=1, 4*p-3, 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+5, i-2*k+1)^(p-1) -C(i-1, 2*k-1)*C(i-2*k+4, 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+4,4)^n. Equivalently, let v_n denote the sequence (1, 5^n, 15^n, 35^n, ...) regarded as an infinite column vector, where 1, 5, 15, 35, ... is the sequence binomial(n+4,4) - see A000332. 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) = C(k+4,4)*T(n,k) + 4*C(k+3,4)*T(n,k-1) + 6*C(k+2,4)*T(n,k-2) + 4*C(k+1,4)*T(n,k-3) + C(k,4)*T(n,k-4) with boundary conditions T(n,0) = 1 for all n and T(n,k) = 0 for k > 4*n.
n-th row polynomial R(n,x) = (1 + x)^4 o (1 + x)^4 o ... o (1 + x)^4 (n factors), where o denotes the black diamond product of power series defined in Dukes and White.
R(n,x) = Sum_{i >= 0} binomial(i+4,4)^n*x^i/(1 + x)^(i+1).
R(n+1,x) = 1/4! * (1 + x)^4 * (d/dx)^4(x^4*R(n,x)).
(1 - x)^(4*n)*R(n,x/(1 - x)) appears to equal the n-th row polynomial of A236463. (End)

Edited by Dean Hickerson, Aug 16 2003

A087109 This table shows the coefficients of combinatorial formulas needed for generating the sequential sums of p-th powers of binomial coefficients C(n,5). The p-th row (p>=1) contains a(i,p) for i=1 to 5p-4, where a(i,p) satisfies Sum_{i=1..n} C(i+4,5)^p = 6 C(n+5,6) * Sum_{i=1..5p-4} a(i,p) C(n-1,i-1)/(i+5).

Original entry on oeis.org

1, 1, 5, 10, 10, 5, 1, 1, 35, 370, 1920, 5835, 11253, 14240, 11830, 6230, 1890, 252, 1, 215, 8830, 148480, 1352615, 7665757, 29224020, 78518790, 152794740, 218270220, 229279512, 175227360, 94864770, 34504470, 7567560, 756756, 1, 1295, 191890

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+5,5)^p of degree 5*p in terms of falling factorials: C(x+5,5)^p = Sum_{k = 0..5*p} T(p,k)*C(x,k). It follows that Sum_{i = 0..n-1} C(i+5,5)^p = Sum_{k = 0..5*p} T(p,k)*C(n,k+1). (End)

			Row 3 contains 1,35,370,...,252, so Sum_{i=1..n} C(i+4,5)^3 = 6 * C(n+5,6) * [ a(1,3)/6 + a(2,3)*C(n-1,1)/7 + a(3,3)*C(n-1,2)/8 + ... + a(11,3)*C(n-1,10)/16 ] = 6 * C(n+5,6) * [ 1/6 + 35*C(n-1,1)/7 + 370*C(n-1,2)/8 + ... + 252*C(n-1,10)/16 ]. Cf. A086026 for more details.
From _Peter Bala_, Mar 11 2018: (Start)
Table begins
1
1  5  10   10    5     1
1 35 370 1920 5835 11253 14240 11830 6230 1890 252
...
Row 2: C(i+5,5)^2 = C(i,0) + 35*C(i,1) + 370*C(i,2) + 1920*C(i,3) + 5835*C(i,4) + 11253*C(i,5) + 14240*C(i,6) + 11830*C(i,7) + 6230*C(i,8) + 1890*C(i,9) + 252*C(i,10). Hence, Sum_{i = 0..n-1} C(i+5,5)^2 = C(n,1) + 35*C(n,2) + 370*C(n,3) + 1920*C(n,4) + 5835*C(n,5) + 11253*C(n,6) + 14240*C(n,7) + 11830*C(n,8) + 6230*C(n,9) + 1890*C(n,10) + 252*C(n,11). (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, A087107, A000332, A086020, A086021, A086022, A087108, A000389, A086023, A086024, A000579, A086025, A086026, A087110, A000580, A086027, A086028, A087111, A027555, A086029, A086030.

Cf. A087127, A237202.

seq(seq(add( (-1)^(k-i)*binomial(k, i)*binomial(i+5, 5)^n, i = 0..k), k = 0..5*n), n = 0..5); # Peter Bala, Mar 11 2018

a[i_, p_] := Sum[Binomial[i - 1, 2*k - 2]*Binomial[i - 2*k + 6, i - 2*k + 1]^(p - 1) - Binomial[i - 1, 2*k - 1]*Binomial[i - 2*k + 5, 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, 5*p - 4}]//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 + 6, i - 2*k + 1)^(p - 1) - binomial(i - 1, 2*k - 1)*binomial(i - 2*k + 5, i - 2*k)^(p - 1))}; for(p=1,8, for(i=1, 5*p-4, 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+6, i-2*k+1)^(p-1) -C(i-1, 2*k-1)*C(i-2*k+5, 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+5,5)^n. Equivalently, let v_n denote the sequence (1, 6^n, 21^n, 56^n, ...) regarded as an infinite column vector, where 1, 6, 21, 56, ... is the sequence binomial(n+5,5) - see A000389. 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..5} C(5,i)*C(k+5-i,5)*T(n,k-i) with boundary conditions T(n,0) = 1 for all n and T(n,k) = 0 for k > 5*n.
n-th row polynomial R(n,x) = (1 + x)^5 o (1 + x)^5 o ... o (1 + x)^5 (n factors), where o denotes the black diamond product of power series defined in Dukes and White.
R(n+1,x) = 1/5!*(1 + x)^5 * (d/dx)^5(x^5*R(n,x)).
R(n,x) = Sum_{i >= 0} binomial(i+5,5)^n*x^i/(1 + x)^(i+1).
(1 - x)^(5*n)*R(n,x/(1 - x)) appears to equal the n-th row polynomial of A237202. (End)

Edited by Dean Hickerson, Aug 16 2003

A087110 This table shows the coefficients of combinatorial formulas needed for generating the sequential sums of p-th powers of binomial coefficients C(n,6). The p-th row (p>=1) contains a(i,p) for i=1 to 6p-5, where a(i,p) satisfies Sum_{i=1..n} C(i+5,6)^p = 7 C(n+6,7) * Sum_{i=1..6p-5} a(i,p) C(n-1,i-1)/(i+6).

Original entry on oeis.org

1, 1, 6, 15, 20, 15, 6, 1, 1, 48, 687, 4850, 20385, 55908, 104959, 137886, 127050, 80640, 33642, 8316, 924, 1, 342, 21267, 527876, 7020525, 58015362, 324610399, 1297791264, 3839203452, 8595153000, 14760228672, 19560928464, 19987430694

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+6,6)^p of degree 6*p in terms of falling factorials: C(x+6,6)^p = Sum_{k = 0..6*p} T(p,k)*C(x,k). It follows that Sum_{i = 0..n-1} C(i+6,6)^p = Sum_{k = 0..6*p} T(p,k)*C(n,k+1). (End)

			Row 3 contains 1,48,687,...,924, so Sum_{i=1..n} C(i+5,6)^3 = 7 * C(n+6,7) * [ a(1,3)/7 + a(2,3)*C(n-1,1)/8 + a(3,3)*C(n-1,2)/9 + ... + a(13,3)*C(n-1,12)/19 ] = 7 * C(n+6,7) * [ 1/7 + 48*C(n-1,1)/8 + 687*C(n-1,2)/9 + ... + 924*C(n-1,12)/19 ]. Cf. A086028 for more details.

G. C. Greubel, Table of n, a(n) for the first 40 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, A087107, A000332, A086020, A086021, A086022, A087108, A000389, A086023, A086024, A087109, A000579, A086025, A086026, A000580, A086027, A086028, A087111, A027555, A086029, A086030.

Cf. A087127, A237252.

seq(seq(add( (-1)^(k-i)*binomial(k, i)*binomial(i+6, 6)^n, i = 0..k), k = 0..6*n), n = 0..5); # Peter Bala, Mar 11 2018

a[i_, p_] := Sum[Binomial[i - 1, 2*k - 2]*Binomial[i - 2*k + 7, i - 2*k + 1]^(p - 1) - Binomial[i - 1, 2*k - 1]*Binomial[i - 2*k + 6, 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, 6*p - 5}]//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 + 7, i - 2*k + 1)^(p - 1) - binomial(i - 1, 2*k - 1)*binomial(i - 2*k + 6, i - 2*k)^(p - 1))}; for(p=1,8, for(i=1, 6*p-5, 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+7, i-2*k+1)^(p-1) -C(i-1, 2*k-1)*C(i-2*k+6, 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+6,6)^n. Equivalently, let v_n denote the sequence (1, 7^n, 28^n, 84^n, ...) regarded as an infinite column vector, where 1, 7, 28, 84, ... is the sequence binomial(n+6,6) - see A000579. 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..6} C(6,i)*C(k+6-i,6)*T(n,k-i) with boundary conditions T(n,0) = 1 for all n and T(n,k) = 0 for k > 6*n.
n-th row polynomial R(n,x) = (1 + x)^6 o (1 + x)^6 o ... o (1 + x)^6 (n factors), where o denotes the black diamond product of power series defined in Dukes and White.
R(n+1,x) = 1/6!*(1 + x)^6 * (d/dx)^6(x^6*R(n,x)).
R(n,x) = Sum_{i >= 0} binomial(i+6,6)^n*x^i/(1 + x)^(i+1).
(1 - x)^(6*n)*R(n,x/(1 - x)) appears to equal the n-th row polynomial of A237252. (End)

Edited by Dean Hickerson, Aug 16 2003

A299041 Irregular table: T(n,k) equals the number of alignments of length k of n strings each of length 3.

Original entry on oeis.org

1, 1, 12, 30, 20, 1, 60, 690, 2940, 5670, 5040, 1680, 1, 252, 8730, 103820, 581700, 1767360, 3087000, 3099600, 1663200, 369600, 1, 1020, 94890, 2615340, 32186070, 214628400, 859992000, 2189325600, 3628409400, 3903900000, 2630628000, 1009008000, 168168000, 1, 4092, 979530, 58061420, 1411122300

Offset: 1

Peter Bala, Feb 02 2018

An alignment of n strings of various lengths is a way of inserting blank characters into the n strings so that the resulting strings all have the same length. We don't allow insertion of a blank character into the same position in each of the n strings.
In this case, let s_1,...,s_n be n strings each of length 3 over an alphabet A. Let - be a gap symbol not in A and let A' = union of A and {-}. An alignment of the n strings is an n-tuple (s_1',...,s_n') of strings each of length >= 3 over the alphabet A' such that
(a) the strings s_i', 1 <= i <= n, have the same length. This common length is called the length of the alignment.
(b) deleting the gap symbols from s_i' yields the string s_i for 1 <= i <= n
(c) there is no value j such that all the strings s_i', 1 <= i <= n have a gap symbol at position j.
By writing the strings s_i' one under another we can consider an alignment of n strings as an n X L matrix, where L, the length of the alignment, ranges from a minimum value of 3 to a maximum value of 3*n. Each row of the matrix has 3 characters from the alphabet A and (L - 3) gap characters.
For example,
  s_1' = ABC------
  s_2' = ---DEF---
  s_3' = ------GHI
is an alignment (of maximum length L = 9) of three strings s_1 = ABC, s_2 = DEF and s_3 = GHI each of length 3.
For the number of alignments of length k of n strings of length 1 (resp. 2) see A131689 (resp. A122193).

			Table begins
n\k| 3   4     5       6      7      8        9      10
- - - - - - - - - - - - - - - - - - - - - - - - - - - -
  1| 1
  2| 1  12    30      20
  3| 1  60   690    2940   5670    5040    1680
  4| 1 252  8730  103820 581700 1767360 3087000 3099600 ...
...
T(2,5) = 30: An alignment of length 5 will have two gap symbols on each line. There are C(5,2) = 10 ways of choosing the 2 positions to insert the gap symbols in the first string. The second string in the alignment must then have nongap symbols at these two positions leaving three positions in which to insert the remaining 1 nongap symbol, giving in total 10 x 3 = 30 possible alignments of 2 strings of 3 characters. Some examples are
  ABC--   ABC--   ABC--
  D--EF   -D-EF   --DEF
Row 2: Sum_{i = 3..n-1} C(i,3)^2 = C(n,4) + 12*C(n,5) + 30*C(n,6) + 20*C(n,7).
Row 3: Sum_{i = 3..n-1} C(i,3)^3 = C(n,4) + 60*C(n,5) + 690*C(n,6) + 2940*C(n,7) + 5670*C(n,8)+ 5040*C(n,9)+ 1680*C(n,10).
exp( Sum_{n >= 1} R(n,2)*x^n/n ) = (1 + x + 153*x^2 + 128793*x^3 + 319155321*x^4 + 1744213657689*x^5 + ....)^8
exp( Sum_{n >= 1} R(n,3)*x^n/n ) = (1 + x + 424*x^2 + 998584*x^3 + 6925040260*x^4 + 105920615923684*x^5 + ....)^27.

P. Bala, Notes on A299041
M. Dukes, C. D. White, Web Matrices: Structural Properties and Generating Combinatorial Identities, arXiv:1603.01589 [math.CO], 2016.
J. Engbers and C. Stocker, Two Combinatorial Proofs of Identities Involving Sums of Powers of Binomial Coefficients, Integers 16 (2016), #A58.
J. B. Slowinski, The Number of Multiple Alignments, Molecular Phylogenetics and Evolution 10:2 (1998), 264-266. doi:10.1006/mpev.1998.0522

Row sums A062208. Cf. A014606, A019538, A078741, A087127, A122193, A131689, A174266.

Cf. A086020, A086021, A086022.

seq(seq(add( (-1)^(k-i) *binomial(k,i)*binomial(i,3)^n, i = 0..k ), k = 3..3*n), n = 1..6);

nmax = 6; T[n_, k_] := Sum[(-1)^(k-i) Binomial[k, i] Binomial[i, 3]^n, {i, 0, k}]; Table[T[n, k], {n, 1, nmax}, {k, 3, 3n}] // Flatten (* Jean-François Alcover, Feb 20 2018 *)

T(n,k) = Sum_{i = 0..k} (-1)^(k-i)*binomial(k,i)*binomial(i,3)^n.
T(n,3) = 1; T(n,3*n) = (3*n)!/6^n = A014606(n)
T(n,k) = binomial(k,3)*( T(n-1,k) + 3*T(n-1,k-1) + 3*T(n-1,k-2) + T(n-1,k-3) ) for 3 <= k <= 3*n with boundary conditions T(n,3) = 1 for n >= 1 and T(n,k) = 0 if (k < 3) or (k > 3*n).
Double e.g.f.: exp(-x)*Sum_{n >= 0} exp(binomial(n,3)*y)*x^n/n! = 1 + (x^3/3!)*y + (x^3/3! + 12*x^4/4! + 30*x^5/5! + 20*x^6/6!)*y^2/2! + ....
n-th row polynomial R(n,x) = Sum_{i >= 3} binomial(i,3)^n*x^i/(1 + x)^(i+1) for n >= 1.
1/(1 - x)*R(n,x/(1 - x)) = Sum_{i >= 3} binomial(i,3)^n*x^i for n >= 1.
R(n,x) = x^3 o x^3 o ... o x^3 (n factors), where o is the black diamond product of power series defined in Dukes and White.
R(n,x) = coefficient of (z_1)^3*...*(z_n)^3 in the expansion of the rational function 1/(1 + x - x*(1 + z_1)*...*(1 + z_n)).
The polynomials Sum_{k = 3..3*n} T(n,k)*x^(k-3)*(1 - x)^(3*n-k) are the row polynomials of A174266.
Sum_{i = 3..n-1} binomial(i,3)^m = Sum_{k = 3..3*m} T(m,k)*binomial(n,k+1) for m >= 1. See Examples below.
x^3*R(n,-1 - x) = (-1)^n*(1 + x)^3*R(n,x).
R(n+1,x) = 1/3!*x^3*(d/dx)^3 ((1 + x)^3*R(n,x)) for n >= 1.
The zeros of R(n,x) belong to the interval [-1, 0].
Row sums R(n,1) = A062208(n); alternating row sums R(n,-1) = (-1)^n.
For k a nonzero integer, the power series A(k,x) := exp( Sum_{n >= 1} 1/k^3*R(n,k)*x^n/n ) appear to have integer coefficients. See the Example section.
Sum_{k = 3..3*n} T(n,k)*binomial(x,k) = ( binomial(x,3) )^n. Equivalently, Sum_{k = 3..3*n} (-1)^(n+k)*T(n,k)*binomial(x+k,k) = ( binomial(x+3,3) )^n. Cf. the Worpitzky-type identity Sum_{k = 1..n} A019538(n,k)* binomial(x,k) = x^n.
Sum_{k = 3..3*n} T(n,k)*binomial(x,k-3) = -binomial(x,3)^n + 3*binomial(x+1,3)^n - 3*binomial(x+2,3)^n + binomial(x+3,3)^n. These polynomials have their zeros on the vertical line Re x = -1/2 in the complex plane.

cp's OEIS Frontend

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

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

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