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A088874
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T(n, k) = [x^k] (2*n)! [z^(2*n)] 1/cos(z)^x, triangle read by rows, for 0 <= k <= n.
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4
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1, 0, 1, 0, 2, 3, 0, 16, 30, 15, 0, 272, 588, 420, 105, 0, 7936, 18960, 16380, 6300, 945, 0, 353792, 911328, 893640, 429660, 103950, 10395, 0, 22368256, 61152000, 65825760, 36636600, 11351340, 1891890, 135135, 0, 1903757312
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OFFSET
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0,5
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COMMENTS
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Previous name was: Triangle read by rows, given by [0, 2, 6, 12, 20, 30, 42, 56, ...] DELTA [1, 2, 3, 4, 5, 6, 7, 8, ...] where Delta is the operator defined in A084938.
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LINKS
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FORMULA
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T(n, k) = A085734(n-1, k-1) for n>0 and k>0.
T(n, k) = [x^k] (2*n)! [z^(2*n)] sec(z)^x. - Peter Luschny, Jul 01 2019
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EXAMPLE
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Triangle starts:
[0] 1
[1] 0, 1
[2] 0, 2, 3
[3] 0, 16, 30, 15
[4] 0, 272, 588, 420, 105
[5] 0, 7936, 18960, 16380, 6300, 945
[6] 0, 353792, 911328, 893640, 429660, 103950, 10395
[7] 0, 22368256, 61152000, 65825760, 36636600, 11351340, 1891890, 135135
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MAPLE
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ser := series(sec(z)^x, z, 24): row := n -> n!*coeff(ser, z, n):
seq(seq(coeff(row(2*n), x, k), k=0..n), n=0..8); # Peter Luschny, Jul 01 2019
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MATHEMATICA
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T[1, 1] = 1; T[n_, k_] := Sum[(1/2^(j-1))*StirlingS1[j, k-1]*Sum[(-1)^(i + k + n)*(i-j)^(2(n-1)) Binomial[2j, i], {i, 0, j-1}]/j!, {j, 1, n-1}];
a[n_] := (2n)! SeriesCoefficient[Sec[z]^x, {z, 0, 2n}] // CoefficientList[#, x] &;
Table[a[n], {n, 0, 8}] // Flatten (* Peter Luschny, Jul 01 2019 *)
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PROG
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def fr2_row(n):
if n == 0: return [1]
S = sum(A241171(n, k)*(x-1)^(n-k) for k in (0..n))
L = expand(S).list()
return sum(L[k]*binomial(x+k, n) for k in (0..n-1)).list()
A088874_row = lambda n: [(-1)^(n-k)*m for k, m in enumerate(fr2_row(n))]
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CROSSREFS
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KEYWORD
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AUTHOR
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EXTENSIONS
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STATUS
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approved
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