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A171456
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Triangle T(n, k) = coefficients of p(n, x) where p(n, x) = ((1-x^n)/(1-x))*p(n-1, x), read by rows.
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2
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1, 1, 1, 1, 1, 1, 1, 1, 2, 3, 4, 3, 2, 1, 1, 3, 6, 10, 13, 14, 13, 10, 6, 3, 1, 1, 4, 10, 20, 33, 47, 59, 66, 66, 59, 47, 33, 20, 10, 4, 1, 1, 5, 15, 35, 68, 115, 174, 239, 301, 350, 377, 377, 350, 301, 239, 174, 115, 68, 35, 15, 5, 1, 1, 6, 21, 56, 124, 239, 413, 652, 952, 1297, 1659, 2001, 2283, 2469, 2534, 2469, 2283, 2001, 1659, 1297, 952, 652, 413, 239, 124, 56, 21, 6, 1
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OFFSET
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1,9
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REFERENCES
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J. Riordan, An Introduction to Combinatorial Analysis, Wiley, 1958, p. 146.
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LINKS
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FORMULA
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T(n, k) = coefficients of p(n, x) where p(n, x) = ((1-x^n)/(1-x))*p(n-1, x), p(1,x) = 1, p(2,x) = 1+x, p(3,x) = 1+x+x^2+x^3.
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EXAMPLE
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The irregular triangle begins as:
1;
1, 1;
1, 1, 1, 1;
1, 2, 3, 4, 3, 2, 1;
1, 3, 6, 10, 13, 14, 13, 10, 6, 3, 1;
1, 4, 10, 20, 33, 47, 59, 66, 66, 59, 47, 33, 20, 10, 4, 1;
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MATHEMATICA
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p[x_, n_]:= p[x, n]= If[n<4, (1-x^2^(n-1))/(1-x), ((1-x^4)/(1-x))*Product[(1 - x^j)/(1-x), {j, 4, n}]];
T[n_]:= CoefficientList[p[x, n], x];
Table[T[n], {n, 1, 10}]//Flatten (* modified by G. C. Greubel, May 10 2021 *)
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PROG
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(Sage)
def p(n, x): return (1-x^2^(n-1))/(1-x) if (n<4) else ((1-x^4)/(1-x))*product( (1-x^j)/(1-x) for j in (4..n) )
def T(n): return ( p(n, x) ).full_simplify().coefficients(sparse=False)
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CROSSREFS
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KEYWORD
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nonn,tabf
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AUTHOR
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EXTENSIONS
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STATUS
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approved
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