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A060083
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Coefficients of even indexed Euler polynomials (rising powers without zeros).
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7
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1, -1, 1, 1, -2, 1, -3, 5, -3, 1, 17, -28, 14, -4, 1, -155, 255, -126, 30, -5, 1, 2073, -3410, 1683, -396, 55, -6, 1, -38227, 62881, -31031, 7293, -1001, 91, -7, 1, 929569, -1529080, 754572, -177320, 24310, -2184, 140, -8, 1, -28820619
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OFFSET
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0,5
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COMMENTS
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E(2*n,1/2)*(-4)^n = A000364(n) (signless Euler numbers without zeros).
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REFERENCES
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M. Abramowitz and I. A. Stegun, eds., Handbook of Mathematical Functions, National Bureau of Standards Applied Math. Series 55, 1964 (and various reprintings), p. 809.
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LINKS
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Table of n, a(n) for n=0..45.
M. Abramowitz and I. A. Stegun, eds., Handbook of Mathematical Functions, National Bureau of Standards, Applied Math. Series 55, Tenth Printing, 1972 [alternative scanned copy].
J. Cigler, q-Fibonacci polynomials and q-Genocchi numbers
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FORMULA
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E(2*n, x)= sum(a(n, m)*x^(2*m+1), m=0..n-1) + x^(2*n), n >= 1; E(0, x)=1.
T(n, k) = A102054(n, k+1) - A102054(n+1, k+1), where A102054 is matrix inverse. E.g.f.: A(x^2, y^2) = [cosh(xy)*(y-1) + exp(xy)/(exp(x)+1) + exp(-xy)/(exp(-x)+1)]/y. - Paul D. Hanna, Dec 28 2004
T(n,k) = 1/(2*k+1)*binomial(2*n,2*k)*A001469(n-k) for 0 <= k <= n-1.
Let F(n,x) = sum {k = 0..n-1} binomial(n-k-1,k)*x^k be a Fibonacci polynomial (see A011973 for coefficients). Then F(2*n,x) = - sum {k = 0..n-1} T(n,k)*F(2*k+1,x). For example, F(8,x) = -17*F(1,x) + 28*F(3,x) - 14*F(5,x) + 4*F(7,x). See Cigler, Corollary 1.3. - Peter Bala, Mar 14 2012
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PROG
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(PARI) {T(n, k)=local(X=x+x*O(x^(2*n)), Y=y+y*O(y^(2*k+1))); (2*n)!*polcoeff(polcoeff((cosh(X*Y)*(Y-1)+ exp(X*Y)/(exp(X)+1)+exp(-X*Y)/(exp(-X)+1))/Y, 2*n, x), 2*k, y)} (Hanna)
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CROSSREFS
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A060082 (falling powers).
Matrix inverse is A102054. Column 0 is A001469 (Genocchi numbers).
Cf. A102054, A001469.
Sequence in context: A019588 A193953 A201377 * A069931 A209152 A209158
Adjacent sequences: A060080 A060081 A060082 * A060084 A060085 A060086
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KEYWORD
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sign,easy,tabl
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AUTHOR
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Wolfdieter Lang, Mar 29 2001
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STATUS
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approved
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