A237202 -id:A237202 - cp's OEIS frontend

A154283 Irregular triangle read by rows: T(n,k) = Sum_{i=0..k} (-1)^i * binomial(2n+1,i) binomial(k+2-i,2)^n, 0 <= k <= 2*(n-1).

1, 1, 4, 1, 1, 20, 48, 20, 1, 1, 72, 603, 1168, 603, 72, 1, 1, 232, 5158, 27664, 47290, 27664, 5158, 232, 1, 1, 716, 37257, 450048, 1822014, 2864328, 1822014, 450048, 37257, 716, 1, 1, 2172, 247236, 6030140, 49258935, 163809288, 242384856, 163809288, 49258935, 6030140, 247236, 2172, 1

Offset: 1

Roger L. Bagula, Jan 06 2009

From Yahia Kahloune, Jan 30 2014: (Start)
In general, let 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,2,n).
With these coefficients we can calculate: Sum_{i=1..n} binomial(i+e-1,e)^p = Sum_{k=0..e*(p-1)} b(k,e,p)*binomial(n+e+k,e*p+k).
For example, A085438(n) = Sum_{i=1..n} binomial(1+i,2)^3 = T(3,0)*binomial(2+n,7) + T(3,1)*binomial(3+n,7) + T(3,2)*binomial(4+n,7) + T(3,3)*binomial(5+n,7) + T(3,4)*binomial(6+n,7) = (1/5040)*(90*n^7 + 630*n^6 + 1638*n^5 + 1890*n^4 + 840*n^3 - 48*n).
(End)
T(n,k) is the number of permutations of 2 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

			Triangle begins:
  1;
  1,     4,       1;
  1,    20,      48,        20,           1;
  1,    72,     603,      1168,         603,           72,           1;
  1,   232,    5158,     27664,       47290,        27664,        5158,  232, 1;
  1,   716,   37257,    450048,     1822014,      2864328,     1822014, ...;
  1,  2172,  247236,   6030140,    49258935,    163809288,   242384856, ...;
  1,  6544, 1568215,  72338144,  1086859301,   6727188848, 19323413187, ...;
  1, 19664, 9703890, 811888600, 21147576440, 225167210712, ... ;
  ...
The T(2,1) = 4 permutations of 1122 with 1 descent are 1212, 1221, 2112, 2211. - _Andrew Howroyd_, May 15 2020

Andrew Howroyd, Table of n, a(n) for n = 1..1600 (rows 1..40)
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.
Helmut Prodinger, On Touchard's continued fraction and extensions: combinatorics-free, self-contained proofs , arXiv:1102.5186 [math.CO], 2011.

Columns k=0..9 are A000012, A061981, A151624, A151625, A151626, A151627, A151628, A151629, A151630, A151631.
Row sums are A000680.
Similar triangles for e=1..6: A173018 (or A008292), this sequence, A174266, A236463, A237202, A237252.
Cf. A000680, A005799, A085438, A334781.

[(&+[(-1)^j*Binomial(2*n+1,j)*Binomial(k-j+2,2)^n: j in [0..k]]): k in [0..2*n-2], n in [1..12]]; // G. C. Greubel, Jun 13 2022

A154283 := proc(n,k)
        (1-x)^(2*n+1)*add( (l*(l+1)/2)^n*x^(l-1),l=0..k+1) ;
        coeftayl(%,x=0,k) ;
end proc: # R. J. Mathar, Feb 01 2013

p[x_, n_]= (1-x)^(2*n+1)*Sum[(k*(k+1)/2)^n*x^k, {k, 0, Infinity}]/x;
Table[CoefficientList[FullSimplify[ExpandAll[p[x, n]]], x], {n,10}]//Flatten

T(n,k)={sum(i=0, k, (-1)^i*binomial(2*n+1, i)*binomial(k+2-i, 2)^n)} \\ Andrew Howroyd, May 09 2020

def A154283(n,k): return sum((-1)^j*binomial(2*n+1, j)*binomial(k-j+2, 2)^n for j in (0..k))
flatten([[A154283(n,k) for k in (0..2*n-2)] for n in (1..12)]) # G. C. Greubel, Jun 13 2022

T(n,k) = (-1) times coefficient of x^k in (x-1)^(2*n+1) * Sum_{k>=0} (k*(k+1)/2)^n *x^(k-1).
From Yahia Kahloune, Jan 29 2014: (Start)
Sum_{i=1..n} binomial(1+i,2)^p = Sum_{k=0..2*p-2} T(p,k)*binomial(n+2+k,2*p+1).
binomial(n,2)^p = Sum_{k=0..2*p-2} T(p,k)*binomial(n+k,2*p). (End)
From Peter Bala, Dec 21 2019: (Start)
E.g.f. as a continued fraction: (1-x)/(1-x + ( 1-exp((1-x)^2*t))*x/(1-x + (1-exp(2*(1-x)^2*t))*x/(1-x + (1-exp(3*(1-x)^2*t))*x/(1-x + ... )))) =  1 + x*t + x*(x^2 + 4*x + 1)*t^2/2! + x*(x^4 + 20*x^3 + 48*x^2 + 20*x + 1)*t^3/3! + ... (use Prodinger equation 1.1).
The sequence of alternating row sums (unsigned) [1, 1, 2, 10, 104, 1816,...] appears to be A005799. (End)

Edited by N. J. A. Sloane, Jan 30 2014 following suggestions from Yahia Kahloune (among other things, the signs of all terms have been reversed).

Edited by Andrew Howroyd, May 09 2020

A174266 Irregular triangle read by rows: T(n,k) = Sum_{i=0..k} (-1)^i * binomial(3n+1,i) binomial(k+3-i,3)^n, 0 <= k <= 3*(n-1).

Original entry on oeis.org

1, 1, 9, 9, 1, 1, 54, 405, 760, 405, 54, 1, 1, 243, 6750, 49682, 128124, 128124, 49682, 6750, 243, 1, 1, 1008, 83736, 1722320, 12750255, 40241088, 58571184, 40241088, 12750255, 1722320, 83736, 1008, 1, 1, 4077, 922347, 45699447, 789300477, 5904797049, 21475242671, 40396577931, 40396577931, 21475242671, 5904797049, 789300477, 45699447, 922347, 4077, 1

Offset: 1

Roger L. Bagula, Mar 14 2010

From Yahia Kahloune, Jan 30 2014: (Start)
In general, let 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,3,n).
With these coefficients we can calculate: Sum_{i=1..n} binomial(i+e-1,e)^p = Sum_{i=0..e*(p-1)} b(i,e,p)*binomial(n+e+i,e*p+1).
For example, A086020(n) = Sum_{i=1..n} binomial(2+i, 3)^2 = T(2,0)*binomial(n+3, 7) + T(2,1)*binomial(n+4,7) + T(2,2)*binomial(n+5,7) + T(2,3)*binomial(n+6,7) = (1/5040)*(20*n^7 + 210*n^6 + 854*n^5 + 1680*n^4 + 1610*n^3 + 630*n^2 + 36*n). (End)
T(n,k) is the number of permutations of 3 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

			Triangle begins:
  1;
  1,      9,         9,            1;
  1,     54,       405,          760,            405,       54,        1;
  1,    243,      6750,        49682,         128124,   128124,    49682, ... ;
  1,   1008,     83736,      1722320,       12750255, 40241088, 58571184, ... ;
  1,   4077,    922347,     45699447,      789300477, ... ;
  1,  16362,   9639783,   1063783164,    38464072830, ... ;
  1,  65511,  98361900,  23119658500,  1641724670475, ... ;
  1, 262116, 992660346, 484099087156, 64856779908606, ... ;
...
The T(2,1) = 9 permutations of 111222 with 1 descent are: 112221, 112212, 112122, 122211, 122112, 121122, 222111, 221112, 211122. - _Andrew Howroyd_, May 07 2020

Andrew Howroyd, Table of n, a(n) for n = 1..1335
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=0..9 are A000012, A289254, A151632, A151633, A151634, A151635, A151636, A151637, A151638, A151639.
Row sums are A014606.
Similar triangles for e=1..6: A173018 (or A008292), A154283, this sequence, A236463, A237202, A237252.
Cf. A060187, A174264.

(* First program *)
p[n_, x_]:= p[n,x]= (1-x)^(3*n+1)*Sum[(Binomial[k+1, 3])^n*x^k, {k, 0, Infinity}]/x^2;
Table[CoefficientList[p[x, n], x], {n,10}]//Flatten (* corrected by G. C. Greubel, Mar 26 2022 *)
(* Second program *)
T[n_, k_]:= T[n, k]= Sum[(-1)^(k-j+1)*Binomial[3*n+1, k-j+1]*(j*(j^2-1)/2)^n, {j, 0, k+1}]/(3^n);
Table[T[n, k], {n,10}, {k,3*n-2}]//Flatten (* G. C. Greubel, Mar 26 2022 *)

T(n,k)={sum(i=0, k, (-1)^i*binomial(3*n+1, i)*binomial(k+3-i, 3)^n)} \\ Andrew Howroyd, May 06 2020

@CachedFunction
def T(n, k): return (1/3^n)*sum( (-1)^(k-j+1)*binomial(3*n+1, k-j+1)*(j*(j^2-1)/2)^n for j in (0..k+1) )
flatten([[T(n, k) for k in (1..3*n-2)] for n in (1..10)]) # G. C. Greubel, Mar 26 2022

T(n,k) = [x^k] (1-x)^(3*n+1)*(Sum_{k>=0} (k*(k+1)*(k-1)/2)^n*x^k)/(3^n*x^2).
T(n,k) = T(n, 3*n-k).
From Yahia Kahloune, Jan 30 2014: (Start)
Sum_{i=1..n} binomial(2+i,3)^p = Sum_{i=0..3*p-3} T(p,i)*binomial(n+3+i,3*p+1).
binomial(n,3)^p = Sum_{i=0..3*p-3} T(p,i)*binomial(n+i,3*p). (End)
From Sergii Voloshyn, Dec 18 2024: (Start)
Let E be the operator (x^2)D*(1/x)*D*(x^2)*D, where D denotes the derivative operator d/dx. Then (1/6^n)* E^n(x^2/(1 - x)^4) = (row n generating polynomial)/(1 - x)^(3*n+4) = Sum_{i=0..k} (-1)^i * binomial(3*n+1,i) * binomial(k+3-i,3)^n.
For example, when n = 3 we have 1/216*E^3(x^2/(1 - x)^4) = x^2 (1 + 243x + 6750x^2 + 49682x^3 + 128124x^4 + 128124x^5 + 49682x^6 + 6750x^7 + 243x^8 + x^9)/(1 - x)^13. (End)

Edited by Andrew Howroyd, May 06 2020

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

Yahia Kahloune, Feb 01 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,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

			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).

Vincenzo Librandi, Table of n, a(n) for n = 1..780
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..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.

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 *)

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

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).

a(36) corrected by Vincenzo Librandi, Feb 14 2014

Edited by Andrew Howroyd, May 08 2020

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

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

A151647 Number of permutations of 5 indistinguishable copies of 1..n with exactly 2 adjacent element pairs in decreasing order.

Original entry on oeis.org

0, 100, 5925, 167475, 3882250, 84320250, 1791011475, 37753995925, 793816473600, 16676797204500, 350257183908625, 7355694727665975, 154471515733316550, 3243914368665860350, 68122282848892857375, 1430568461732082827625, 30041941039388979651100

Offset: 1

R. H. Hardin, May 29 2009

Andrew Howroyd, Table of n, a(n) for n = 1..500
Index entries for linear recurrences with constant coefficients, signature (36,-390,1720,-3165,2556,-756).

Column k=2 of A237202.

[(&+[(-1)^j*Binomial(5*n+1, j)*Binomial(7-j, 5)^n: j in [0..2]]): n in [1..30]]; // G. C. Greubel, Sep 12 2022

LinearRecurrence[{36,-390,1720,-3165,2556,-756},{0,100,5925,167475,3882250, 84320250}, 30] (* Harvey P. Dale, Nov 01 2021 *)

a(n) = {21^n -(5*n+1)*6^n +5*n*(5*n+1)/2} \\ Andrew Howroyd, May 06 2020

concat(0, Vec(25*x^2*(4 + 93*x - 273*x^2 - 324*x^3) / ((1 - x)^3*(1 - 6*x)^2*(1 - 21*x)) + O(x^20))) \\ Colin Barker, Jul 18 2020

def A151647(n): return sum((-1)^j*binomial(5*n+1, j)*binomial(7-j, 5)^n for j in (0..2))
[A151647(n) for n in (1..30)] # G. C. Greubel, Sep 12 2022

a(n) = 21^n - (5*n + 1)*6^n + 5*n*(5*n + 1)/2. - Andrew Howroyd, May 06 2020
From Colin Barker, Jul 18 2020: (Start)
G.f.: 25*x^2*(4 + 93*x - 273*x^2 - 324*x^3)/((1 - x)^3*(1 - 6*x)^2*(1 - 21*x)).
a(n) = 36*a(n-1) - 390*a(n-2) + 1720*a(n-3) - 3165*a(n-4) + 2556*a(n-5) - 756*a(n-6) for n>6. (End)
From G. C. Greubel, Sep 12 2022: (Start)
a(n) = Sum_{j=0..2} (-1)^j*binomial(5*n+1, j)*binomial(7-j, 5)^n.
E.g.f.: exp(21*x) - (1 + 30*x)*exp(6*x) + (5/2)*x*(6 + 5*x)*exp(x). (End)

Terms a(8) and beyond from Andrew Howroyd, May 06 2020

A151648 Number of permutations of 5 indistinguishable copies of 1..n with exactly 3 adjacent element pairs in decreasing order.

Original entry on oeis.org

0, 100, 52800, 6021225, 447069750, 28203920250, 1662432014600, 95167951614675, 5379642996402350, 302454892260579500, 16965492408059468000, 950717312310731109725, 53255101624187593866550, 2982626301173304400020350, 167034793385579317725373000, 9354122482050520106734846375

Offset: 1

R. H. Hardin, May 29 2009

nonn

Andrew Howroyd, Table of n, a(n) for n = 1..200
Index entries for linear recurrences with constant coefficients, signature (120,-5135,105130,-1138875,6873212,-23367945,44115210,-45855180,24607800,-5334336).

Column k=3 of A237202.

[(&+[(-1)^j*Binomial(5*n+1, j)*Binomial(8-j, 5)^n: j in [0..3]]): n in [1..30]]; // G. C. Greubel, Sep 12 2022

Table[Sum[(-1)^j*Binomial[5*n+1, j]*Binomial[8-j, 5]^n, {j,0,3}], {n, 30}] (* G. C. Greubel, Sep 12 2022 *)

a(n) = {56^n - (5*n + 1)*21^n + binomial(5*n+1, 2)*6^n - binomial(5*n+1, 3)} \\ Andrew Howroyd, May 07 2020

def A151648(n): return sum((-1)^j*binomial(5*n+1, j)*binomial(8-j, 5)^n for j in (0..3))
[A151648(n) for n in (1..30)] # G. C. Greubel, Sep 12 2022

a(n) = 56^n - (5*n + 1)*21^n + binomial(5*n+1, 2)*6^n - binomial(5*n+1, 3). - Andrew Howroyd, May 07 2020
From G. C. Greubel, Sep 12 2022: (Start)
a(n) = Sum_{j=0..3} (-1)^j*binomial(5*n+1, j)*binomial(8-j, 5)^n.
G.f.: 25*x^2*(4 + 1632*x + 7949*x^2 - 594490*x^3 + 1502565*x^4 + 3945816*x^5 - 13945932*x^6 - 4667544*x^7)/( Product_{j=0.3} (1 - binomial(j+5, 5)*x)^(4-j) ).
E.g.f.: exp(56*x) - (1 + 105*x)*exp(21*x) + 90*x*(1 + 5*x)*exp(6*x) - (5/6)*x*(24 + 75*x + 25*x^2)*exp(x). (End)

Terms a(8) and beyond from Andrew Howroyd, May 07 2020

A151649 Number of permutations of 5 indistinguishable copies of 1..n with exactly 4 adjacent element pairs in decreasing order.

Original entry on oeis.org

0, 25, 182700, 84646275, 18746073375, 3085105337250, 443146794326775, 59593466814021450, 7756190980563441400, 993121304532091347375, 126129019383244869440750, 15954027727693152179563525, 2014001281330236936094898325, 253995299147567448467485168400

Offset: 1

R. H. Hardin, May 29 2009

nonn

Andrew Howroyd, Table of n, a(n) for n = 1..200
Index entries for linear recurrences with constant coefficients, signature (330,-42640,2885620,-114569650,2830998568, -44959643760,466859252380,-3196112991765,14413138051950,-42392995156008, 80058527221680,-95391910308240,69101028551520,-27758817676800,4742523426816).

Column k=4 of A237202.

[(&+[(-1)^j*Binomial(5*n+1, j)*Binomial(9-j, 5)^n: j in [0..4]]): n in [1..30]]; // G. C. Greubel, Sep 12 2022

a := n -> 126^n - 56^n + 5*(5*n - 2 - 2^(n + 2)*3^n)*(25*n^3 - n)/24 + 5*7^n*n*(5*3^n*n + 3^n - 2^(3*n + 1))/2: seq(a(n), n = 1..14); # Peter Luschny, Sep 15 2022

Table[Sum[(-1)^j*Binomial[5*n+1, j]*Binomial[9-j, 5]^n, {j,0,4}], {n, 30}] (* G. C. Greubel, Sep 12 2022 *)

def A151649(n): return sum((-1)^j*binomial(5*n+1, j)*binomial(9-j, 5)^n for j in (0..4))
[A151649(n) for n in (1..30)] # G. C. Greubel, Sep 12 2022

From G. C. Greubel, Sep 12 2022: (Start)
a(n) = Sum_{j=0..4} (-1)^j*binomial(5*n+1, j)*binomial(9-j, 5)^n.
G.f.: 25*x^2*(1 + 6978*x + 1016851*x^2 - 58760395*x^3 - 644809730*x^4 + 39948710783*x^5 - 333333302706*x^6 - 615347004762*x^7 + 9446377773420*x^8 - 24901972278120*x^9 + 4642437947616*x^10 + 51610957036128*x^11 + 7377258663936*x^12)/( Product_{j=0..4} (1 - binomial(j+5, 5)*x)^(5-j) ).
E.g.f.: exp(126*x) - (1 + 280*x)*exp(56*x) + (315/2)*x*(2 + 35*x)*exp(21*x) - 30*x*(4 + 75*x + 150*x^2)*exp(6*x) + (5/24)*x*(72 + 720*x + 700*x^2 + 125*x^3)*exp(x). (End)

Terms a(7) and beyond from Andrew Howroyd, May 06 2020

A151650 Number of permutations of 5 indistinguishable copies of 1..n with exactly 5 adjacent element pairs in decreasing order.

Original entry on oeis.org

0, 1, 273504, 554083761, 359033166276, 146006641259682, 47460662479108620, 13737399319828223622, 3735696667164317656002, 981723633332192745554763, 253168802147494901125791536, 64610303181638008483181729583, 16394452997496205694559810296928

Offset: 1

R. H. Hardin, May 29 2009

nonn

Andrew Howroyd, Table of n, a(n) for n = 1..200
Index entries for linear recurrences with constant coefficients, signature (792,-260325,47435950,-5384570670,405212887824, -20987115153218,766141819691700,-20020244758559925,378098953185430160, -5187747941307534441,51805847315420838702,-376119699869936994300, 1976772497606552079000,-7462030652722893940560,20005194459069943157664, -37570195548784110639168,48620286446529148444800,-42255660508281014054400, 23493226909575103672320,-7540975791513246007296,1062524964753334075392).

Column k=5 of A237202.

[(&+[(-1)^j*Binomial(5*n+1, j)*Binomial(10-j, 5)^n: j in [0..5]]): n in [1..30]]; // G. C. Greubel, Sep 12 2022

Table[Sum[(-1)^j*Binomial[5*n+1, j]*Binomial[10-j, 5]^n, {j,0,5}], {n, 30}] (* G. C. Greubel, Sep 12 2022 *)

def A151650(n): return sum((-1)^j*binomial(5*n+1, j)*binomial(10-j, 5)^n for j in (0..5))
[A151650(n) for n in (1..30)] # G. C. Greubel, Sep 12 2022

From G. C. Greubel, Sep 12 2022: (Start)
a(n) = Sum_{j=0..5} (-1)^j*binomial(5*n+1, j)*binomial(10-j, 5)^n.
G.f.: x^2*(1 + 272712*x + 337728918*x^2 - 8648679586*x^3 - 7074308846715*x^4 + 477518433272082*x^5 - 648991861481176*x^6 - 633499131060575994*x^7 + 16047431893076948868*x^8 - 110517528944798318070*x^9 - 876288066908074857792*x^10 + 15640555372102290699216*x^11 - 63236662801921291034016*x^12 + 68632685088119444662272*x^13 + 271292956749261231644160*x^14 - 1029875415092135380492416*x^15 + 783701219351671172149248*x^16 + 907411282152901890158592*x^17 + 66939072779460046749696*x^18)/( Product_{j=0..5} (1 - binomial(j+5, 5)*x)^(6-j) ).
E.g.f.: exp(252*x) - (1 + 630*x)*exp(126*x) + 280*x*(3 + 140*x)*exp(56*x) - (105/2)*x*(8 + 525*x + 3675*x^2)*exp(21*x) + 90*x*(1 + 60*x + 350*x^2 + 375*x^3)*exp(6*x) - (1/24)*x*(144 + 5400*x + 12000*x^2 + 5625*x^3 + 625*x^4)*exp(x). (End)

Terms a(7) and beyond from Andrew Howroyd, May 06 2020

cp's OEIS Frontend

A154283 Irregular triangle read by rows: T(n,k) = Sum_{i=0..k} (-1)^i * binomial(2*n+1,i) * binomial(k+2-i,2)^n, 0 <= k <= 2*(n-1).

Views

Author

Keywords

Comments

Examples

Links

Crossrefs

Programs

Magma

Maple

Mathematica

PARI

SageMath

Formula

Extensions

A174266 Irregular triangle read by rows: T(n,k) = Sum_{i=0..k} (-1)^i * binomial(3*n+1,i) * binomial(k+3-i,3)^n, 0 <= k <= 3*(n-1).

Views

Author

Keywords

Comments

Examples

Links

Crossrefs

Programs

Mathematica

PARI

Sage

Formula

Extensions

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).

Views

Author

Keywords

Comments

Examples

Links

Crossrefs

Programs

Mathematica

PARI

Formula

Extensions

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).

Views

Author

Keywords

Comments

Examples

Links

Crossrefs

Programs

Mathematica

PARI

Formula

Extensions

Views

Author

Keywords

Comments

Examples

Links

Crossrefs

Programs

Maple

Mathematica

PARI

Formula

Extensions

A151647 Number of permutations of 5 indistinguishable copies of 1..n with exactly 2 adjacent element pairs in decreasing order.

Views

Author

Keywords

Links

Crossrefs

Programs

Magma

Mathematica

PARI

A154283 Irregular triangle read by rows: T(n,k) = Sum_{i=0..k} (-1)^i * binomial(2n+1,i) binomial(k+2-i,2)^n, 0 <= k <= 2*(n-1).

A174266 Irregular triangle read by rows: T(n,k) = Sum_{i=0..k} (-1)^i * binomial(3n+1,i) binomial(k+3-i,3)^n, 0 <= k <= 3*(n-1).

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

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).