%I M2153 N0858 #60 Jul 30 2024 02:14:38
%S 2,30,3522,1066590,604935042,551609685150,737740947722562,
%T 1360427147514751710,3308161927353377294082,
%U 10256718523496425979562270,39490468691102039103925777602,184856411587530526077816051412830,1033888847501229495999134528615701122
%N Generalized Euler numbers, c(5,n).
%D N. J. A. Sloane, A Handbook of Integer Sequences, Academic Press, 1973 (includes this sequence).
%D N. J. A. Sloane and Simon Plouffe, The Encyclopedia of Integer Sequences, Academic Press, 1995 (includes this sequence).
%H Sean A. Irvine, <a href="/A000187/b000187.txt">Table of n, a(n) for n = 0..250</a>
%H D. Shanks, <a href="http://dx.doi.org/10.1090/S0025-5718-1967-0223295-5">Generalized Euler and class numbers</a>. Math. Comp. 21 (1967) 689-694.
%H D. Shanks, <a href="http://dx.doi.org/10.1090/S0025-5718-1968-0227093-9">Corrigenda to: "Generalized Euler and class numbers"</a>, Math. Comp. 22 (1968), 699.
%H D. Shanks, <a href="/A000003/a000003.pdf">Generalized Euler and class numbers</a>, Math. Comp. 21 (1967), 689-694; 22 (1968), 699. [Annotated scanned copy]
%F From the Shanks paper: Consider the Dirichlet series L_a(s) = sum_{k>=0} (-a|2k+1) / (2k+1)^s, where (-a|2k+1) is the Jacobi symbol. Then the numbers c_(a,n) are defined by L_a(2n+1)= (Pi/(2a))^(2n+1)*sqrt(a)* c(a,n)/ (2n)! for a > 1 and n = 0,1,2,... - _Sean A. Irvine_, Mar 26 2012
%F From _Peter Bala_, Nov 18 2020: (Start)
%F a(n) = (-1)^n*10^(2*n)*( E(2*n,1/10) + E(2*n,3/10) ), where E(n,x) are the Euler polynomials - see A060096.
%F Row 5 of A235605.
%F G.f.: A(x) = 2*cos(x)*cos(3*x)/( 2*cos(x)*cos(4*x) - cos(3*x) ) = 2 + 30*x^2/2! + 3522*x^4/4! + ....
%F Alternative forms:
%F A(x) = (exp(i*x) + exp(3*i*x) + exp(7*i*x) + exp(9*i*x))/(1 + exp(10*i*x));
%F A(x) = (sqrt(5)/10)*( sec(x + Pi/5) + sec(x + 2*Pi/5) - sec(x + 3*Pi/5) - sec(x + 4*Pi/5) ). (End)
%F a(n) = (2*n)!*[x^(2*n)](sec(5*x)*(cos(2*x) + cos(4*x))). - _Peter Luschny_, Nov 21 2021
%F a(n) ~ 2^(4*n + 2) * 5^(2*n + 1/2) * n^(2*n + 1/2) / (Pi^(2*n + 1/2) * exp(2*n)). - _Vaclav Kotesovec_, Apr 15 2022
%e a(3) = 1066590: L_5(7) = Sum_{n >= 0} (-1)^n*( 1/(10*n+1)^7 + 1/(10*n+3)^7 + 1/(10*n+7)^7 + 1/(10*n+9)^7 ) = 1066590*( (1/6!)*sqrt(5)*(Pi/10)^7 ). - _Peter Bala_, Nov 18 2020
%p seq((-1)^n*(10)^(2*n)*(euler(2*n,1/10) + euler(2*n,3/10)), n = 0..11); - _Peter Bala_, Nov 18 2020
%p egf := sec(5*x)*(cos(2*x) + cos(4*x)): ser := series(egf, x, 26):
%p seq((2*n)!*coeff(ser, x, 2*n), n = 0..11); # _Peter Luschny_, Nov 21 2021
%t a0=5; nmax=20; km0 = nmax; Clear[cc]; L[a_, s_, km_] := Sum[JacobiSymbol[ -a, 2k+1]/(2k+1)^s, {k, 0, km}]; c[a_, n_, km_] := 2^(2n+1)*Pi^(-2n-1)*(2n)!*a^(2n+1/2)*L[a, 2n+1, km] // Round; cc[km_] := cc[km] = Table[ c[a0, n, km], {n, 0, nmax}]; cc[km0]; cc[km = 2km0]; While[cc[km] != cc[ km/2, km = 2km]]; A000187 = cc[km] (* _Jean-François Alcover_, Feb 05 2016 *)
%Y Cf. A000192, A000490, A060096, A235605, A000320, A349265, A349264.
%K nonn,easy
%O 0,1
%A _N. J. A. Sloane_
%E More terms from Kok Seng Chua (chuaks(AT)ihpc.nus.edu.sg), Jun 02 2000