A088874 T(n, k) = [x^k] (2*n)! [z^(2*n)] 1/cos(z)^x, triangle read by rows, for 0 <= k <= n.

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

Offset: 0

Author

Philippe Deléham, Nov 26 2003

Keywords

• nonn
• tabl
• easy

Comments

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.

Examples

			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
		

Links

• Paul Barry, Continued fractions and transformations of integer sequences, JIS 12 (2009) #09.7.6.
• Ghislain R. Franssens, On a Number Pyramid Related to the Binomial, Deleham, Eulerian, MacMahon and Stirling number triangles , JIS 9 (2006) #06.4.1.
• Alan D. Sokal, The Euler and Springer numbers as moment sequences, arXiv:1804.04498 [math.CO], 2018.

Crossrefs

Another version of the triangle A085734. A signed version is A318146.

Diagonals give: A000007 A000182 A001147, row sums A000364.

Programs

Maple

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

Mathematica

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}];
Table[T[n, k], {n, 1, 10}, {k, 1, n}] // Flatten (* Jean-François Alcover, Jul 14 2018, after Vladimir Kruchinin *)
a[n_] := (2n)! SeriesCoefficient[Sec[z]^x, {z, 0, 2n}] // CoefficientList[#, x] &;
Table[a[n], {n, 0, 8}] // Flatten (* Peter Luschny, Jul 01 2019 *)

Sage

# uses [A241171]
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))]
for n in (0..7): print(A088874_row(n)) # Peter Luschny, Sep 19 2017

Formula

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

Extensions

New name by Peter Luschny, Jul 01 2019