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A158782
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Irregular triangle of coefficients of p(n, x) = (1 - x^2)^(n+1)*Sum_{j >= 0} (4*j+ 1)^n*x^(2*j), read by rows.
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4
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1, 1, 0, 3, 1, 0, 22, 0, 9, 1, 0, 121, 0, 235, 0, 27, 1, 0, 620, 0, 3446, 0, 1996, 0, 81, 1, 0, 3119, 0, 40314, 0, 63854, 0, 15349, 0, 243, 1, 0, 15618, 0, 422087, 0, 1434812, 0, 963327, 0, 112546, 0, 729, 1, 0, 78117, 0, 4157997, 0, 26672209, 0, 37898739, 0, 12960063, 0, 806047, 0, 2187
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OFFSET
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0,4
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COMMENTS
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Define the series q(x, n) = (1 - x^2)^(n+1)*Sum_{j >= 1} (4*k+1)^n*x^(2*k) then the sum r(x, n) = p(x, n) + q(x, n) is symmetrical and gives r(x, n) =(x+1)^(2*n+1)*A060187(x, n).
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LINKS
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FORMULA
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T(n, k) = [x^k]( p(n, x) ), where p(n, x) = (1 - x^2)^(n+1)*Sum_{j >= 0} (4*j+ 1)^n*x^(2*j).
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EXAMPLE
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The irregular triangle begins as:
1;
1, 0, 3;
1, 0, 22, 0, 9;
1, 0, 121, 0, 235, 0, 27;
1, 0, 620, 0, 3446, 0, 1996, 0, 81;
1, 0, 3119, 0, 40314, 0, 63854, 0, 15349, 0, 243;
1, 0, 15618, 0, 422087, 0, 1434812, 0, 963327, 0, 112546, 0, 729;
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MATHEMATICA
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p[n_, x_]= (1-x^2)^(n+1)*Sum[(4*k+1)^n*x^(2*k), {k, 0, Infinity}];
Table[FullSimplify[p[n, x]], {n, 0, 12}];
Table[CoefficientList[p[n, x], x], {n, 0, 12}]//Flatten (* modified by G. C. Greubel, Mar 08 2022 *)
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PROG
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(Sage)
def p(n, x): return (1-x^2)^(n+1)*sum( (4*j+1)^n*x^(2*j) for j in (0..n+1) )
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
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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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