A171457 Irregular triangle T(n, k) = coefficients of p(n, x), where p(n, x) = p(n-1, x)*Sum_{j=0..n-1} x^i - x*Sum_{j=0..binomial(n,2)-2} x^i, read by rows.

1, 1, 1, 1, 1, 1, 1, 1, 1, 2, 3, 2, 1, 1, 1, 1, 3, 6, 8, 8, 8, 6, 3, 1, 1, 1, 1, 4, 10, 18, 26, 33, 38, 38, 33, 26, 18, 10, 4, 1, 1, 1, 1, 5, 15, 33, 59, 92, 129, 166, 195, 211, 211, 195, 166, 129, 92, 59, 33, 15, 5, 1, 1, 1, 1, 6, 21, 54, 113, 205, 334, 499, 693, 899, 1095, 1257, 1364, 1401, 1364, 1257, 1095, 899, 693, 499, 334, 205, 113, 54, 21, 6, 1, 1

Offset

1,10

References

J. Riordan, An Introduction to Combinatorial Analysis, Wiley, 1958, p. 146.

Links

G. C. Greubel, Rows n = 1..20 of the irregular triangle, flattened

Formula

T(n, k) = coefficients of p(n, x), where p(n, x) = p(n-1, x)*Sum_{j=0..n-1} x^i - x*Sum_{j=0..binomial(n,2)-2} x^i, p(1,x) = 1, p(2,x) = 1+x, p(3,x) = 1+x+x^2+x^3.

Example

The irregular triangle begins as:
  1;
  1, 1;
  1, 1, 1,  1;
  1, 1, 2,  3,  2,  1,  1;
  1, 1, 3,  6,  8,  8,  8,  6,  3,  1,  1;
  1, 1, 4, 10, 18, 26, 33, 38, 38, 33, 26, 18, 10, 4, 1, 1;

Mathematica

p[n_, x_]:= p[n, x]= If[n<4, (1-x^2^(n-1))/(1-x), ((1-x^n)*p[n-1, x] - x*(1 - x^(Binomial[n, 2] -1)))/(1-x)];
T[n_]:= CoefficientList[p[n, x], x];
Table[T[n], {n, 1, 10}] (* modified by G. C. Greubel, May 10 2021 *)

Prog

(Sage)
def p(n, x): return (1-x^2^(n-1))/(1-x) if (n<4) else ((1-x^n)*p(n-1, x) -x*(1-x^(binomial(n, 2)-1)))/(1-x)
def T(n): return ( p(n, x) ).full_simplify().coefficients(sparse=False)
[T(n) for n in (1..10)] # G. C. Greubel, May 10 2021

Crossrefs

Cf. A171456.

Sequence in context: A008406 A039735 A283761 * A376611 A129385 A217983

Adjacent sequences: A171454 A171455 A171456 * A171458 A171459 A171460

Keyword

nonn,tabf

Author

Roger L. Bagula, Dec 09 2009

Extensions

Edited by G. C. Greubel, May 10 2021

Status

approved