A290883 - OEIS

%I #23 Oct 01 2022 19:46:46

%S 1,2,-20,920,-95600,17588000,-5034785600,2068322672000,

%T -1153339941728000,838147215114560000,-769492266756037760000,

%U 870869784123573927680000,-1191080747725445120960000000,1936606018449416970940544000000,-3692030834904045806243452160000000,8156631422332715861303860160000000000,-20671774666617006397027638099614720000000

%N E.g.f. A(x) = sqrt(C(x)^2 + S(x)^2) such that C(x)^2 - S(x)^2 = 1 and C'(x)^2 + S'(x)^2 = 1, where C(x) is the e.g.f. of A290880 and S(x) is the e.g.f. of A290881.

%C Conjecture: If p is a prime congruent to 1 mod 4 and k is a positive integer, then p^k divides a(n) for n >= (k/2)*p-1. Furthermore, if p^k divides (2n+2)!, then p^k divides a(n). - _Andrew Slattery_, Aug 21 2022

%H Paul D. Hanna, <a href="/A290883/b290883.txt">Table of n, a(n) for n = 0..100</a>

%F E.g.f.: A(x) = sqrt( cosh( 2*Series_Reversion( Integral sqrt( cosh(2*x) ) dx ) ) ).

%e E.g.f.: A(x) = 1 + 2*x^2/2! - 20*x^4/4! + 920*x^6/6! - 95600*x^8/8! + 17588000*x^10/10! - 5034785600*x^12/12! + 2068322672000*x^14/14! - 1153339941728000*x^16/16! + 838147215114560000*x^18/18! +...

%e such that A(x) = sqrt(C(x)^2 + S(x)^2) where series C(x) and S(x) begin:

%e S(x) = x - x^3/3! + 25*x^5/5! - 1705*x^7/7! + 227665*x^9/9! - 50333425*x^11/11! + 16655398825*x^13/13! - 7711225809625*x^15/15! + 4760499335502625*x^17/17! - 3779764853639958625*x^19/19! + 3752942823715824285625*x^21/21! +...

%e C(x) = 1 + x^2/2! - 7*x^4/4! + 265*x^6/6! - 24175*x^8/8! + 4037425*x^10/10! - 1070526775*x^12/12! + 412826556025*x^14/14! - 218150106913375*x^16/16! + 151297155973926625*x^18/18! - 133288452772763494375*x^20/20! +...

%e These series satisfy: C(x)^2 - S(x)^2 = 1 and C'(x)^2 + S'(x)^2 = 1.

%o (PARI) {a(n) = my(C=1,S=x); for(i=1,n, C = 1 + intformal( S/sqrt(C^2 + S^2 + O(x^(n+2))) ); S = intformal( C/sqrt(C^2 + S^2)) ); n!*polcoeff(C + S,n)}

%o for(n=0,30, print1(a(n),", "))

%o (PARI) {a(n) = my(E=1); A = sqrt( cosh( 2*serreverse( intformal( sqrt(cosh(2*x + O(x^(2*n+2)))) ) ))); (2*n)!*polcoeff(A,2*n)}

%o for(n=0,30, print1(a(n),", "))

%Y Cf. A290879, A290880, A290881, A290882, A153302.

%K sign

%O 0,2

%A _Paul D. Hanna_, Aug 13 2017