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.

User: Yahia Kahloune

Yahia Kahloune's wiki page.

Yahia Kahloune has authored 12 sequences. Here are the ten most recent ones:

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

Views

Author

Yahia Kahloune, Feb 05 2014

Keywords

Comments

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

Examples

			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).

Links

Crossrefs

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.

Programs

  • Mathematica
    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 *)
  • PARI
    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

Formula

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).

Extensions

Edited by Andrew Howroyd, May 06 2020

A237202 Irregular triangle read by rows: T(n,k) = Sum_{i=0..k} (-1)^i * binomial(5*n+1,i) * binomial(k+5-i,5)^n, 0 <= k <= 5*(n-1).

Original entry on oeis.org

1, 1, 25, 100, 100, 25, 1, 1, 200, 5925, 52800, 182700, 273504, 182700, 52800, 5925, 200, 1, 1, 1275, 167475, 6021225, 84646275, 554083761, 1858142825, 3363309675, 3363309675, 1858142825, 554083761, 84646275
Offset: 1

Views

Author

Yahia Kahloune, Feb 05 2014

Keywords

Comments

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,5,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 5 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 08 2020

Examples

			T(n,0) = 1;
T(n,1) = 6^n - (5*n+1);
T(n,2) = 21^n - (5*n+1)*6^n + C(5*n+1,2);
T(n,3) = 56^n - (5*n+1)*21^n + C(5*n+1,2)*6^n - C(5*n+1,3) ;
T(n,4) = 126^n - (5*n+1)*56^n + C(5*n+1,2)*21^n - C(5*n+1,3)*6^n  + C(5*n+1,4).
Triangle T(n,k) begins:
1;
1, 25, 100, 100, 25, 1;
1, 200, 5925, 52800, 182700, 273504, 182700, 52800, 5925, 200, 1;
1, 1275, 167475, 6021225, 84646275, 554083761, 1858142825, 3363309675, 3363309675, 1858142825, 554083761, 84646275, 6021225, 167475, 125, 1;
1, 7750, 3882250, 447069750, 18746073375, 359033166276, 3575306548500, 20052364456500, 66640122159000, 135424590593500, 171219515211316, 135424590593500, 66640122159000, 20052364456500, 3575306548500, 359033166276, 18746073375, 447069750, 3882250, 7750, 1;
...
Example:
Sum_{i=1..n} C(4+i,5)^3 = C(n+5,16) + 200*C(n+6,16) + 5925*(n+7,16) + 52800*C(n+8,16) + 182700*C(n+9,16) + 273504*C(n+10,16) + 182700*C(n+11,16) + 52800*C(n+12,16) + 5925*C(n+13,16) + 200*C(n+14,16) + C(n+15,16).
C(n,5)^3 = C(n,15) + 200*C(n+1,15) + 5925*C(n+2,15) + 52800*C(n+3,15) + 182700*C(n+4,15) + 273504*C(n+5,15) + 182700*C(n+6,15) + 52800*C(n+7,15) + 5925*C(n+8,15) + 200*C(n+9,15) + C(n+10,15).

Links

Crossrefs

Columns k=2..5 are A151647, A151648, A151649, A151650.
Row sums are A014609.
Similar triangles for e=1..6: A173018 (or A008292), A154283, A174266, A236463, this sequence, A237252.
Sum_{i=1..n} binomial(4+i,5)^p for p=2..3 gives: A086025, A086026.
Cf. A087127, A087107, A087108, A087109, A087110, A087111, A181544.

Programs

  • Mathematica
    b[k_, 5, p_] := Sum[(-1)^i*Binomial[5*p+1, i]*Binomial[k-i, 5]^p /. k -> 5+i, {i, 0, k-5}]; row[p_] := Table[b[k, 5, p], {k, 5, 5*p}]; Table[row[p], {p, 1, 5}] // Flatten (* Jean-François Alcover, Feb 05 2014 *)
  • PARI
    T(n,k)={sum(i=0, k, (-1)^i*binomial(5*n+1, i)*binomial(k+5-i, 5)^n)} \\ Andrew Howroyd, May 08 2020

Formula

Sum_{i=1..n} binomial(4+i,5)^p = Sum{k=0..5*(p-1)} T(p,k) * binomial(n+5+k, 5*p+1).
binomial(n,5)^p = Sum_{k=0..5*(p-1)} T(p,k) * binomial(n+k, 5*p).

Extensions

Edited by Andrew Howroyd, May 08 2020

A236463 Irregular triangle read by rows: T(n,k) = Sum_{i=0..k} (-1)^i * binomial(4*n+1,i) * binomial(k+4-i,4)^n, 0 <= k <= 4*(n-1).

Original entry on oeis.org

1, 1, 16, 36, 16, 1, 1, 112, 1828, 8464, 13840, 8464, 1828, 112, 1, 1, 608, 40136, 724320, 4961755, 15018688, 21571984, 15018688, 4961755, 724320, 40136, 608, 1, 1, 3104, 693960, 37229920, 733059110, 6501577152, 29066972368, 69830127680, 93200908410, 69830127680
Offset: 1

Views

Author

Yahia Kahloune, Feb 01 2014

Keywords

Comments

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,4,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 4 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 08 2020

Examples

			T(n,0) = 1;
T(n,1) = 5^n - (4*n+1);
T(n,2) = 15^n - (4*n+1)*5^n + C(4*n+1,2);
T(n,3) = 35^n - (4*n+1)*15^n + C(4*n+1,2)*5^n - C(4*n+1,3);
T(n,4) = 70^n - (4*n+1)*35^n + C(4*n+1,2)*15^n - C(4*n+1,3)*5^n + C(4*n+1,4).
Triangle T(n,k) begins:
1,
1, 16, 36, 16, 1;
1, 112, 1828, 8464, 13840, 8464, 1828, 112, 1;
1, 608, 40136, 724320, 4961755, 15018688, 21571984, 15018688, 4961755, 724320, 40136, 608, 1;
1, 3104, 693960, 37229920, 733059110, 6501577152, 29066972368, 69830127680, 93200908410, 69830127680, 29066972368, 6501577152, 733059110, 37229920, 693960, 3104, 1;
1, 15600, 11000300, 1558185200, 75073622025, 1585757994496, 16938467955200, 99825129369600, 342907451401150, 710228619472800, 903546399077256, 710228619472800, 342907451401150, 99825129369600, 16938467955200, 1585757994496, 75073622025, 1558185200, 11000300, 15600, 1;
  ...
Example:
Sum_{i=1..n} C(3+i,4)^3 = C(n+4,13) + 112*C(n+5,13) + 1828*C(n+6,13) + 8464*C(n+7,13) + 13840*C(n+8,13) + 8464*C(n+9,13) + 1828*C(n+10,13) + 112*C(n+11,13) + C(+12,13).
C(n,4)^3 = C(n,12) + 112*C(n+1,12) + 1828*C(n+2,12) + 8464*C(n+3,12) + 13840*C(n+4,12) + 8464*C(n+5,12) + 1828*C(n+6,12) + 112*C(n+7,12) + C(n+8,12).

Links

Crossrefs

Columns k=2..8 are A151640, A151641, A151642, A151643, A151644, A151645, A151646.
Row sums are A014608.
Similar triangles for e=1..6: A173018 (or A008292), A154283, A174266, this sequence, A237202, A237252.
Sum_{i=1..n} binomial(3+i,4)^p for p=2..3 gives: A086023, A086024.
Cf. A087127, A087107, A087108, A087109, A087110, A087111, A181544.

Programs

  • Mathematica
    b[k_, 4, p_] := Sum[(-1)^i*Binomial[4*p+1, i]*Binomial[k-i, 4]^p /. k -> 4+i, {i, 0, k-4}]; row[p_] := Table[b[k, 4, p], {k, 4, 4*p}]; Table[row[p], {p, 1, 6}] // Flatten (* Jean-François Alcover, Feb 05 2014 *)
  • PARI
    T(n,k)={sum(i=0, k, (-1)^i*binomial(4*n+1, i)*binomial(k+4-i, 4)^n)} \\ Andrew Howroyd, May 08 2020

Formula

Sum_{i=1..n} binomial(3+i,4)^p = Sum{k=0..4*(p-1)} T(p,k) * binomial(n+4+k, 4*p+1).
binomial(n,4)^p = Sum_{k=0..4*(p-1)} T(p,k) * binomial(n+k, 4*p).

Extensions

a(36) corrected by Vincenzo Librandi, Feb 14 2014
Edited by Andrew Howroyd, May 08 2020

A234253 a(n) = Sum_{i=1..n} C(7+i,8)^2.

Original entry on oeis.org

1, 82, 2107, 29332, 274357, 1930726, 10948735, 52357960, 217994860, 808970960, 2723733524, 8436372248, 24304813148, 65712993248, 167965846148, 408373664744, 949291256585, 2119095737210, 4559798912835, 9488531918460, 19148848609485, 37571357310510
Offset: 1

Views

Author

Yahia Kahloune, Dec 22 2013

Keywords

Links

Crossrefs

Cf. A087127, A086023, A086025, A086027, A086029.

Programs

  • Maple
    A234253:=n->add(binomial(7+i,8)^2, i=1..n); seq(A234253(n), n=1..30); # Wesley Ivan Hurt, Dec 23 2013
  • Mathematica
    Table[Sum[Binomial[7 + i, 8]^2, {i, n}], {n, 30}] (* Wesley Ivan Hurt, Dec 23 2013 *)
CoefficientList[Series[(x^8 + 64 x^7 + 784 x^6 + 3136 x^5 + 4900 x^4 + 3136 x^3 + 784 x^2 + 64 x + 1)/(x - 1)^18, {x, 0, 40}], x] (* Vincenzo Librandi, May 06 2014 *)
  • PARI
    Vec(x*(x^8 +64*x^7 +784*x^6 +3136*x^5 +4900*x^4 +3136*x^3 +784*x^2 +64*x +1)/(x-1)^18 + O(x^100)) \\ Colin Barker, May 02 2014

Formula

G.f.: x*(x^8 +64*x^7 +784*x^6 +3136*x^5 +4900*x^4 +3136*x^3 +784*x^2 +64*x +1) / (x-1)^18. - Colin Barker, May 02 2014

Extensions

One term corrected and more terms from Colin Barker, May 02 2014

A229702 Expansion of 1/((1-x)^4*(1-6x)).

Original entry on oeis.org

1, 10, 70, 440, 2675, 16106, 96720, 580440, 3482805, 20897050, 125382586, 752295880, 4513775735, 27082654970, 162495930500, 974975583816, 5849853503865, 35099121024330, 210594726147310, 1263568356885400, 7581410141314171
Offset: 0

Views

Author

Yahia Kahloune, Sep 27 2013

Keywords

Comments

This sequence was chosen to illustrate a way to match generating functions and closed-form solutions.
The general term associated with the generating function
1/((1-s*x)^4*(1-r*x)) with r>s>=1 is a(n) = [ r^(n+4) - s^(n+1)*(s^3 + s^2*(r-s)*binomial(n+4,1) + s*(r-s)^2*binomial(n+4,2)+(r-s)^3*binomial(n+4,3))]/(r-s)^4.

Examples

			a(3) = (6^8 - (125*3^3  + 1200*3^2 + 3805*3 + 4026))/3750 = 440.

Links

Crossrefs

Cf. A002663, A097786, A097788, A097790.

Formula

a(n) = (6^(n+4) - (1 + 5*C(n+4,1) + 25*C(n+4,2) + 125*C(n+4,3)))/625 = (6^(n+5) - (125*n^3 + 1200*n^2 + 3805*n + 4026))/3750.

A229611 Expansion of 1/((1-x)^3*(1-11x)).

Original entry on oeis.org

1, 14, 160, 1770, 19485, 214356, 2357944, 25937420, 285311665, 3138428370, 34522712136, 379749833574, 4177248169405, 45949729863560, 505447028499280, 5559917313492216, 61159090448414529, 672749994932559990, 7400249944258160080, 81402749386839761090
Offset: 0

Views

Author

Yahia Kahloune, Sep 26 2013

Keywords

Comments

This sequence was chosen to illustrate a method of matching generating functions and closed-form solutions: The general term associated with the generating function 1/((1-s*x)^3*(1-r*x)) with r>s>=1 is a(n) = [r^(n+3) - s^(n+1)*(s^2 + (r-s)*s*binomial(n+3,1) +(r-s)^2*binomial(n+3,2))] / (r-s)^3 .

Examples

			a(3) = (11^6 - (50*3^2+260*3 + 331))/1000 = 1770 .

Links

Crossrefs

Cf. A002662, A052150, A052161, A052244, A052262.

Programs

  • Magma
    [(11^(n+3) - (50*n^2 + 260*n + 331))/1000: n in [0..25]]; // Vincenzo Librandi, Sep 27 2013
  • Mathematica
    CoefficientList[Series[1/((1 - x)^3 (1 - 11 x)), {x, 0, 20}], x] (* Vincenzo Librandi, Sep 27 2013 *)
LinearRecurrence[{14,-36,34,-11},{1,14,160,1770},30] (* Harvey P. Dale, Apr 09 2016 *)

Formula

a(n) = (11^(n+3) - (1 + 10*C(n+3,1) + 100*C(n+3,2)))/1000 = (11^(n+3) - (50*n^2 + 260*n + 331))/1000.
a(n) = 14*a(n-1) -36*a(n-2) +34*a(n-3) -11*a(n-4). - Vincenzo Librandi, Sep 27 2013

A229463 Expansion of g.f. 1/((1-x)^2*(1-26*x)).

Original entry on oeis.org

1, 28, 731, 19010, 494265, 12850896, 334123303, 8687205886, 225867353045, 5872551179180, 152686330658691, 3969844597125978, 103215959525275441, 2683614947657161480, 69773988639086198495, 1814123704616241160886, 47167216320022270183053, 1226347624320579024759396
Offset: 0

Views

Author

Yahia Kahloune, Sep 24 2013

Keywords

Comments

This sequence was chosen to illustrate a method of solution.

Examples

			a(3) = (26^5 - 25*3 - 51)/625 = 19010.

Links

Crossrefs

Cf. A000295, A000340, A014825, A014827, A014936.

Programs

  • PARI
    my(x='x+O('x^18)); Vec(1/((1-26*x)*(1-x)^2)) \\ Elmo R. Oliveira, May 24 2025

Formula

a(n) = (26^(n+2) - 25*n - 51)/625.
In general, for the expansion of 1/((1-s*x)^2*(1-r*x)) with r>s>=1 we have the formula: a(n) = (r^(n+2)- s^(n+1)*((r-s)*n +(2*r-s)))/(r-s)^2.
From Elmo R. Oliveira, May 24 2025: (Start)
E.g.f.: exp(x)*(-51 - 25*x + 676*exp(25*x))/625.
a(n) = 28*a(n-1) - 53*a(n-2) + 26*a(n-3). (End)

A227209 Expansion of 1/((1-x)^2*(1-2*x)*(1-4*x)).

Original entry on oeis.org

1, 8, 43, 198, 849, 3516, 14311, 57746, 231997, 930024, 3724179, 14904894, 59635945, 238576532, 954371647, 3817617642, 15270732693, 61083455040, 244334868715, 977341571990, 3909370482241, 15637490317548
Offset: 0

Views

Author

Yahia Kahloune, Sep 19 2013

Keywords

Comments

This sequence was chosen to illustrate a method of solution.
In general, for the expansion of 1/((1-t*x)^2*(1-s*x)*(1-r*x)) with r>s>t we have the formula: a(n) = ( K*r^(n+3) + L*s^(n+3) + M*t^(n+3) + N*t^(n+3) )/D where K,L,M,N,D have the following values:
K = (s-t)^2;
L = -(r-t)^2;
M = (r-s)*(r+s-2*t);
N = (r-t)*(s-t)*(r-s)*(n+3);
D = (r-s)*(r-t)^2*(s-t)^2.
Directly using formula we get a(n) = ( 4^(n+3) - 9*2^(n+3) + 8 + 6*(n+3) )/18. After transformation we obtain previous formula.

Links

Crossrefs

Cf. A229026.
Partial sums of A171477.

Programs

  • Mathematica
    nn = 25; CoefficientList[Series[1/((1 - x)^2*(1 - 2 x)*(1 - 4 x)), {x, 0, nn}], x] (* T. D. Noe, Sep 19 2013 *)

Formula

G.f.: 1/((1-x)^2*(1-2*x)*(1-4*x)).
a(n) = ( 4^(n+3) - 9*2^(n+3) + 6*n + 26 )/18.
E.g.f.: exp(x)*(13 - 36*exp(x) + 32*exp(3*x) + 3*x)/9. - Stefano Spezia, Feb 23 2025

A229026 Expansion of 1/((1-x)*((1-5*x)^2)*(1-8*x)).

Original entry on oeis.org

1, 19, 238, 2490, 23631, 211509, 1823908, 15348100, 127057261, 1040261799, 8453319978, 68343722910, 550640774491, 4426107030889, 35521389816448, 284771933350920, 2281370275767321, 18267889925254779, 146232526369201318, 1170331087647336130, 9365122293936867751
Offset: 0

Views

Author

Yahia Kahloune, Sep 18 2013

Keywords

Comments

This sequence was chosen to illustrate a method of solution.

Links

Formula

a(n) = (2*8^(n+4) - (84*n+287)*5^(n+2) - 9)/1008.
In general, for the expansion of 1/((1-t*x)*((1-s*x)^2)*(1-r*x)) with r > s > t, we have the formula: a(n) = (K*r^(n+3) + L*s^(n+3) + M*s^(n+2) + N*t^(n+3))/D, where K, L, M, N, D have the following values:
K = (s-t)^2;
L = (r-t)*(r-2*s+t);
M = -(r-s)*(r-t)*(s-t)*(n+3);
N = -(r-s)^2;
D = (r-t)*((s-t)^2)*((r-s)^2).
Directly using formula we get: a(n) = (16*8^(n+3) - 7*5^(n+3) - 84*(n+3)*5^(n+2) - 9)/1008. After transformation we obtain previous formula.

A229025 Expansion of 1/((1-2x)(1-4x)(1-5x)(1-7x)(1-8x)).

Original entry on oeis.org

1, 26, 417, 5334, 59829, 616602, 5996089, 55931678, 505925277, 4470884418, 38804443041, 332070466182, 2809908472645, 23562316644074, 196128590350473, 1622684846427246, 13358308842842733, 109510501632625170, 894623468678425585, 7286920685445869270
Offset: 0

Views

Author

Yahia Kahloune, Sep 11 2013

Keywords

Comments

This sequence was chosen to illustrate a method of solution.

Links

Programs

  • Mathematica
    nn = 20; CoefficientList[Series[1/((1 - 2*x) (1 - 4*x) (1 - 5*x) (1 - 7*x) (1 - 8*x)), {x, 0, nn}], x] (* T. D. Noe, Sep 12 2013 *)

Formula

a(n) = (5*8^(n+4) - 12*7^(n+4) + 20*5^(n+4) - 15*4^(n+4) +2*2^(n+4))/360.
In general, for the expansion of 1/((1-r*x)(1-s*x)(1-t*x)(1-u*x)(1-v*x)) with v > u > t > s > r , we have the formula
a(n) = (K*v^(n+4) - L*u^(n+4) + M*t^(n+4) - N*s^(n+4) + P*r^(n+4)) / (K*L*M*N*P)^(1/3) where K,L,M,N,P have the following values:
K = (u-t)*(u-s)*(u-r)*(t-s)*(t-r)*(s-r);
L = (v-t)*(v-s)*(v-r)*(t-s)*(t-r)*(s-r);
M = (v-u)*(v-s)*(v-r)*(u-s)*(u-r)*(s-r);
N = (v-u)*(v-t)*(v-r)*(u-t)*(u-r)*(t-r);
P = (v-u)*(v-t)*(v-s)*(u-t)*(u-s)*(t-s).
Directly using the formula we obtain a(n) = (180*8^(n+4) - 432*7^(n+4) + 720*5^(n+4) - 540*4^(n+4) + 72*2^(n+4))/12960 simplifies after by 36.

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