A154855 Triangle of coefficients of p(x,n) = (1/3)*(1-x)^(n+1)*Sum_{m >= 0} ((5*m+4)^n - (5*m+1)^n)*x^m, read by rows.

0, 1, -1, 5, 0, -5, 21, 87, -87, -21, 85, 1330, 0, -1330, -85, 341, 15045, 28160, -28160, -15045, -341, 1365, 152040, 816825, 0, -816825, -152040, -1365, 5461, 1457323, 16786931, 21064365, -21064365, -16786931, -1457323, -5461, 21845, 13592430, 297161830, 939811670, 0, -939811670, -297161830, -13592430, -21845

Offset

0,4

Comments

Row sums are zero.

Links

G. C. Greubel, Rows n = 0..50 of the triangle, flattened

Formula

Rows are coefficients of p(x,n) = (1/3)*(1-x)^(n+1)*Sum_{m >= 0} ((5*m+4)^n - (5*m+1)^n)*x^m.

Example

Triangle begins as:
     0;
     1,      -1;
     5,       0,       -5;
    21,      87,      -87,      -21;
    85,    1330,        0,    -1330,       -85;
   341,   15045,    28160,   -28160,    -15045,      -341;
  1365,  152040,   816825,        0,   -816825,   -152040,    -1365;
  5461, 1457323, 16786931, 21064365, -21064365, -16786931, -1457323, -5461;

Mathematica

T[n_, k_, p_, q_, r_, t_]:= SeriesCoefficient[(1/p)*(1-x)^(n+1)*Sum[((q*j+r)^n - (q*j+t)^n)*x^j, {j, 0, n}], {x, 0, k}];
Table[T[n, k, 3, 5, 4, 1, 3], {n, 0, 12}, {k, 0, n}]//Flatten (* modified by G. C. Greubel, Mar 11 2021 *)

Prog

(Sage)
def f(n, p, q, r, t, x) : return (1/p)*(1-x)^(n+1)*sum( ((q*j+r)^n - (q*j+t)^n )*x^j for j in (0..n))
[[( f(n, 3, 5, 4, 1, x) ).series(x, n+1).list()[k] for k in (0..n)] for n in (0..12)] # G. C. Greubel, Mar 11 2021

Crossrefs

Cf. A154852, A154853, A154854.

Sequence in context: A341482 A055953 A165051 * A137562 A021668 A004552

Adjacent sequences: A154852 A154853 A154854 * A154856 A154857 A154858

Keyword

sign,tabl

Author

Roger L. Bagula, Jan 16 2009

Extensions

Edited by G. C. Greubel, Mar 11 2021

Status

approved