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A225076
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Triangle read by rows: absolute values of odd-numbered rows of A225356.
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5
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1, 1, 22, 1, 1, 236, 1446, 236, 1, 1, 2178, 58479, 201244, 58479, 2178, 1, 1, 19672, 1736668, 19971304, 49441990, 19971304, 1736668, 19672, 1, 1, 177134, 46525293, 1356555432, 9480267666, 19107752148, 9480267666, 1356555432, 46525293, 177134, 1
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OFFSET
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1,3
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COMMENTS
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An equivalent definition: take the polynomials corresponding to rows 2, 4, 6, 8, ... of A060187, divide by x+1, and extract the coefficients. [Corrected by Petros Hadjicostas, Apr 17 2020]
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LINKS
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FORMULA
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Triangle read by rows: row n gives coefficients in the expansion of the polynomial ((x - 1)^(2*n)/(x + 1)) * Sum_{k >= 0} (2*k + 1)^(2*n-1)*x^k. The infinite sum simplifies to a polynomial.
Sum_{m=0..2*n-2} T(n,m)*t^m = 2^(2*n-1) * (1-t)^(2*n) * LerchPhi(t, 1-2*n, 1/2)/(1 + t).
Sum_{k=1..n} T(n, k) = A002671(n-1).
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EXAMPLE
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Triangle T(n,m) (for n >= 1 and 0 <= m <= 2*n - 2) begins as follows:
1;
1, 22, 1;
1, 236, 1446, 236, 1;
1, 2178, 58479, 201244, 58479, 2178, 1;
1, 19672, 1736668, 19971304, 49441990, 19971304, 1736668, 19672, 1;
...
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MATHEMATICA
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(* Power series via an infinite sum *)
p[x_, n_] = (x-1)^(2*n)*Sum[(2*k+1)^(2*n-1)*x^k, {k, 0, Infinity}];
Table[CoefficientList[p[x, n]/(1+x), x], {n, 1, 10}]//Flatten
(* First alternative method: recurrence *)
t[n_, k_, m_]:= t[n, k, m]= If[k==1 || k==n, 1, (m*n-m*k+1)*t[n-1, k-1, m] + (m*k - (m-1))*t[n-1, k, m]];
T[n_, k_]:= T[n, k]= t[n+1, k+1, 2]; (* t(n, k, 2) = A060187 *)
Flatten[Table[CoefficientList[Sum[T[n, k]*x^k, {k, 0, n}]/(x+1), x], {n, 14, 2}]]
(* Second alternative method: polynomial expansion *)
p[t_] = Exp[t]*x/(-Exp[2*t] + x);
Flatten[Table[CoefficientList[(n!*(-1+x)^(n+1)/(x*(x+1)))*SeriesCoefficient[ Series[p[t], {t, 0, 30}], n], x], {n, 1, 13, 2}]]
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PROG
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(Sage)
def A060187(n, k): return sum( (-1)^(k-j)*(2*j-1)^(n-1)*binomial(n, k-j) for j in (1..k) )
def A225076(n, k): return sum( (-1)^(k-j-1)*A060187(2*n, j+1) for j in (0..k-1) )
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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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