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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, 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.
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
KEYWORD
nonn,tabf
AUTHOR
Roger L. Bagula, Dec 09 2009
EXTENSIONS
Edited by G. C. Greubel, May 10 2021
STATUS
approved