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A009843
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E.g.f. x/cos(x) (odd powers only).
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12
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1, 3, 25, 427, 12465, 555731, 35135945, 2990414715, 329655706465, 45692713833379, 7777794952988025, 1595024111042171723, 387863354088927172625, 110350957750914345093747
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OFFSET
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0,2
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COMMENTS
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Related to the formulae sum(k>0,sin(kx)/k^(2n+1))=(-1)^(n+1)/2*x^(2n+1)/(2n+1)!*sum(i=0,2n,(2Pi/x)^i*B(i)*C(2n+1,i)) and if x=Pi/2 sum(k>0,(-1)^(k+1)/k^(2n+1))=(-1)^n*E(2n)*Pi^(2n+1)/2^(2n+2)/(2n)! - Benoit Cloitre, May 01 2002
Expanding x/cosh(x) gives alt. signed values at odd positions.
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LINKS
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Table of n, a(n) for n=0..13.
Peter Luschny, The lost Bernoulli numbers.
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FORMULA
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a(n)=(2n+1)*A000364(n)=sum(i=0, 2n, B(i)*C(2n+1, i)*4^i)=(2n+1)*E(2n) where B(i) are the Bernoulli numbers, C(2n, i) the binomial numbers and E(2n) the Euler numbers. - Benoit Cloitre, May 01 2002
Recurrence: a(n) = -(-1)^n*Sum[i=0..n-1, (-1)^i*a(i)*C(2n+1, 2i+1) ]. - Ralf Stephan, Feb 24 2005
a(n) = 4^n |E_{2n}(1/2)+E_{2n}(1)| (2n+1) for n > 0; E_{n}(x) Euler polynomial. [Peter Luschny, Nov 25 2010]
a(n) = (2*n+1)! * [x^(2*n+1)] x/cos(x).
E.g.f.: x / cos(x) =x+x^3/Q(0); Q(k)=8k+2-x^2/(1+(2k+1)*(2k+2)/Q(k+1)); (continued fraction). - Sergei N. Gladkovskii, Nov 15 2011
E.g.f.: x + x^3/U(0) where U(k)= (2*k+1)*(2*k+2) - x^2 + x^2*(2*k+1)*(2*k+2)/U(k+1) ; (continued fraction, Euler's 1st kind, 1-step). - Sergei N. Gladkovskii, Oct 19 2012
G.f.: x/T(0) where T(k)= 1 + x^2 - x^2*(2*k+2)^2/(1 - x^2*(2*k+2)^2/T(k+1);(continued fraction, 2-step). - Sergei N. Gladkovskii, Nov 10 2012
G.f.: 1/G(0) where G(k) = 1 - x*(8*k^2+8*k+3)-16*x^2*(k+1)^4/G(k+1); (continued fraction). - Sergei N. Gladkovskii, Jan 14 2013.
E.g.f.: 2*x/(Q(0) + 1) where Q(k)= 1 - x/(2*k+1)/(2*k+2)/(1 - 1/(1 + 1/Q(k+1) )); (continued fraction ). - Sergei N. Gladkovskii, Apr 10 2013
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EXAMPLE
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x/cos(x) = x + 1/2*x^3 + 5/24*x^5 + 61/720*x^7 + 277/8064*x^9 + ...
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MAPLE
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seq((2*i+1)!*coeff(series(x/cos(x), x, 32), x, 2*i+1), i=0..13);
A009843 := n -> (-1)^n*(2*n+1)*euler(2*n): [Peter Luschny]
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PROG
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(PARI) for(n=0, 25, print1(sum(i=0, 2*n+1, binomial(2*n+1, i)*bernfrac(i)*4^i), ", "))
# (Python) The objective of this implementation is efficiency.
# n -> [a(0), a(1), ..., a(n)] for n > 0.
def A009843_list(n):
....S = [0 for i in range(0, n+1)]
....S[0] = 1
....for k in range(1, n+1):
........S[k] = k*S[k-1]
....for k in range(1, n+1):
........for j in range(k, n+1):
............S[j] = (j-k)*S[j-1]+(j-k+1)*S[j]
........S[k] = (2*k+1)*S[k]
....return S
print(A009843_list(10)) # - Peter Luschny, Aug 09 2011
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CROSSREFS
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Bisection of A009391, A009392, A065619, A083008.
Cf. A099028.
Sequence in context: A143925 A074708 A160143 * A182962 A223076 A136173
Adjacent sequences: A009840 A009841 A009842 * A009844 A009845 A009846
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KEYWORD
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nonn
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AUTHOR
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R. H. Hardin
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EXTENSIONS
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Extended and signs tested Mar 15 1997 by Olivier Gerard.
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STATUS
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approved
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