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A119879
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Exponential Riordan array (sech(x),x).
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18
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1, 0, 1, -1, 0, 1, 0, -3, 0, 1, 5, 0, -6, 0, 1, 0, 25, 0, -10, 0, 1, -61, 0, 75, 0, -15, 0, 1, 0, -427, 0, 175, 0, -21, 0, 1, 1385, 0, -1708, 0, 350, 0, -28, 0, 1, 0, 12465, 0, -5124, 0, 630, 0, -36, 0, 1, -50521, 0, 62325, 0, -12810, 0, 1050, 0, -45, 0, 1
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OFFSET
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0,8
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COMMENTS
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Row sums have e.g.f. exp(x)*sech(x) (signed version of A009006). Inverse of masked Pascal triangle A119467. Transforms the sequence with e.g.f. g(x) to the sequence with e.g.f. g(x)*sech(x).
Coefficients of the Swiss-Knife polynomials for the computation of Euler, tangent and Bernoulli number (triangle read by rows). Another version in A153641. - Philippe Deléham, Oct 26 2013
Relations to Green functions and raising/creation and lowering/annihilation/destruction operators are presented in Hodges and Sukumar and in Copeland's discussion of this sequence and 2020 pdf. - Tom Copeland, Jul 24 2020
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LINKS
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FORMULA
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Number triangle whose k-th column has e.g.f. sech(x)*x^k/k!.
T(n,k) = C(n,k)*2^(n-k)*E_{n-k}(1/2) where C(n,k) is the binomial coefficient and E_{m}(x) are the Euler polynomials. - Peter Luschny, Jan 25 2009
The coefficients in ascending order of x^i of the polynomials p{0}(x) = 1 and p{n}(x) = Sum_{k=0..n-1; k even} binomial(n,k)*p{k}(0)*((n mod 2) - 1 + x^(n-k)). - Peter Luschny, Jul 16 2012
With all offsets 0, let A_n(x;y) = (y + E.(x))^n, an Appell sequence in y where E.(x)^k = E_k(x) are the Eulerian polynomials of A123125. Then the row polynomials of A046802 (the h-polynomials of the stellahedra) are given by h_n(x) = A_n(x;1); the row polynomials of A248727 (the face polynomials of the stellahedra), by f_n(x) = A_n(1 + x;1); the Swiss-knife polynomials of this entry, A119879, by Sw_n(x) = A_n(-1;1 + x); and the row polynomials of the Worpitsky triangle (A130850), by w_n(x) = A(1 + x;0). Other specializations of A_n(x;y) give A090582 (the f-polynomials of the permutohedra, cf. also A019538) and A028246 (another version of the Worpitsky triangle). - Tom Copeland, Jan 24 2020
Triangle equals P*((I + P^2)/2)^(-1), where P denotes Pascal's triangle A007318. - Peter Bala, Mar 07 2024
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EXAMPLE
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Triangle begins:
1;
0, 1;
-1, 0, 1;
0, -3, 0, 1;
5, 0, -6, 0, 1;
0, 25, 0, -10, 0, 1;
-61, 0, 75, 0, -15, 0, 1;
0, -427, 0, 175, 0, -21, 0, 1;
1385, 0, -1708, 0, 350, 0, -28, 0, 1;
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MAPLE
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T := (n, k) -> binomial(n, k)*2^(n-k)*euler(n-k, 1/2): # Peter Luschny, Jan 25 2009
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MATHEMATICA
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T[n_, k_] := Binomial[n, k] 2^(n-k) EulerE[n-k, 1/2];
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PROG
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(Sage)
@CachedFunction
return 1 if n == 0 else add(A119879_poly(k, 0)*binomial(n, k)*(x^(n-k)-1+n%2) for k in range(n)[::2])
R = PolynomialRing(ZZ, 'x')
# Alternatively:
(Sage) # uses[riordan_array from A256893]
riordan_array(sech(x), x, 9, exp=true) # Peter Luschny, Apr 19 2015
(PARI)
{T(n, k) = binomial(n, k)*2^(n-k)*(2/(n-k+1))*(subst(bernpol(n-k+1, x), x, 1/2) - 2^(n-k+1)*subst(bernpol(n-k+1, x), x, 1/4))};
for(n=0, 5, for(k=0, n, print1(T(n, k), ", "))) \\ G. C. Greubel, Feb 25 2019
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CROSSREFS
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KEYWORD
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AUTHOR
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STATUS
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approved
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