A165889
Irregular triangle T(n, k) = [x^k]( p(n, x) ), where p(n, x) = (1-x)^(2*n+4)*( Sum_{j >= 0} j^(n+1)*x^j )^2/x^2, read by rows.
Original entry on oeis.org
1, 1, 2, 1, 1, 8, 18, 8, 1, 1, 22, 143, 244, 143, 22, 1, 1, 52, 808, 3484, 5710, 3484, 808, 52, 1, 1, 114, 3853, 35032, 125746, 188908, 125746, 35032, 3853, 114, 1, 1, 240, 16782, 290672, 2000703, 6040992, 8702820, 6040992, 2000703, 290672, 16782, 240, 1
Offset: 0
Irregular triangle begins as:
1;
1, 2, 1;
1, 8, 18, 8, 1;
1, 22, 143, 244, 143, 22, 1;
1, 52, 808, 3484, 5710, 3484, 808, 52, 1;
1, 114, 3853, 35032, 125746, 188908, 125746, 35032, 3853, 114, 1;
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p[n_, x_]:= p[n, x]= (1/x^2)*(1-x)^(2*n+4)*Sum[k^(n+1)*x^k, {k, 0, Infinity}]^2;
Table[CoefficientList[p[n, x], x], {n,0,12}]//Flatten (* modified by G. C. Greubel, Mar 09 2022 *)
-
def p(n,x): return (1/x^2)*(1-x)^(2*n+4)*sum( j^(n+1)*x^j for j in (0..2*n+4) )^2
def T(n,k): return ( p(n,x) ).series(x, 2*n+1).list()[k]
flatten([[T(n,k) for k in (0..2*n)] for n in (0..12)])
A165883
Irregular triangle T(n, k) = [x^k]( p(n, x) ), where p(n, x) = (1-x)^(2*n+3)*( Sum_{j >= 0} (2*j+ 1)^n*x^j )*( Sum_{j >= 0} j^(n+1)*x^j ), read by rows.
Original entry on oeis.org
1, 1, 2, 1, 1, 10, 26, 10, 1, 1, 34, 287, 508, 287, 34, 1, 1, 102, 2272, 11098, 19134, 11098, 2272, 102, 1, 1, 294, 15493, 169432, 675706, 1042948, 675706, 169432, 15493, 294, 1, 1, 842, 98374, 2151026, 17138559, 55643460, 82178676, 55643460, 17138559, 2151026, 98374, 842, 1
Offset: 0
Irregular triangle begins as:
1;
1, 2, 1;
1, 10, 26, 10, 1;
1, 34, 287, 508, 287, 34, 1;
1, 102, 2272, 11098, 19134, 11098, 2272, 102, 1;
1, 294, 15493, 169432, 675706, 1042948, 675706, 169432, 15493, 294, 1;
-
p[n_, x_]:= p[n, x]= (1/x)*(1-x)^(2*n+3)*Sum[(2*k+1)^n*x^k, {k,0,Infinity}]*Sum[k^(n+1)*x^k, {k,0,Infinity}];
Table[CoefficientList[p[n, x], x], {n,0,10}]//Flatten (* modified by G. C. Greubel, Mar 08 2022 *)
-
def p(n,x): return (1/x)*(1-x)^(2*n+3)*sum( (2*j+1)^n*x^j for j in (0..2*n+3) )*sum( j^(n+1)*x^j for j in (0..2*n+3) )
def T(n,k): return ( p(n,x) ).series(x, 2*n+1).list()[k]
flatten([[T(n,k) for k in (0..2*n)] for n in (0..12)]) # G. C. Greubel, Mar 08 2022
A165890
Irregular triangle T(n, k) = [x^k]( p(n, x) ), where p(n, x) = ( (1-x)^(n+1) * Sum_{k >= 0} (2*k+1)^(n-1)*x^k )^2, read by rows.
Original entry on oeis.org
1, 1, -2, 1, 1, 0, -2, 0, 1, 1, 10, 15, -52, 15, 10, 1, 1, 44, 484, -44, -970, -44, 484, 44, 1, 1, 150, 5933, 22792, 466, -58684, 466, 22792, 5933, 150, 1, 1, 472, 58586, 682040, 2085135, -682512, -4287444, -682512, 2085135, 682040, 58586, 472, 1
Offset: 0
Irregular triangle begins as:
1;
1, -2, 1;
1, 0, -2, 0, 1;
1, 10, 15, -52, 15, 10, 1;
1, 44, 484, -44, -970, -44, 484, 44, 1;
1, 150, 5933, 22792, 466, -58684, 466, 22792, 5933, 150, 1;
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p[n_, x_]:= p[n, x]= If[n==0, 1, (2^(n-1)*(1-x)^(n+1)*LerchPhi[x, -n+1, 1/2])^2];
Table[CoefficientList[p[n, x], x], {n,0,12}]//Flatten (* modified by G. C. Greubel, Mar 09 2022 *)
-
def p(n,x): return (1-x)^(2*n+2)*sum( (2*j+1)^(n-1)*x^j for j in (0..2*n+2) )^2
def T(n,k): return ( p(n,x) ).series(x, 2*n+2).list()[k]
flatten([[T(n,k) for k in (0..2*n)] for n in (0..12)]) # G. C. Greubel, Mar 09 2022
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