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A290883
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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.
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5
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1, 2, -20, 920, -95600, 17588000, -5034785600, 2068322672000, -1153339941728000, 838147215114560000, -769492266756037760000, 870869784123573927680000, -1191080747725445120960000000, 1936606018449416970940544000000, -3692030834904045806243452160000000, 8156631422332715861303860160000000000, -20671774666617006397027638099614720000000
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OFFSET
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0,2
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COMMENTS
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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
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LINKS
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FORMULA
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E.g.f.: A(x) = sqrt( cosh( 2*Series_Reversion( Integral sqrt( cosh(2*x) ) dx ) ) ).
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EXAMPLE
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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! +...
such that A(x) = sqrt(C(x)^2 + S(x)^2) where series C(x) and S(x) begin:
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! +...
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! +...
These series satisfy: C(x)^2 - S(x)^2 = 1 and C'(x)^2 + S'(x)^2 = 1.
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PROG
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(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)}
for(n=0, 30, print1(a(n), ", "))
(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)}
for(n=0, 30, print1(a(n), ", "))
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CROSSREFS
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KEYWORD
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sign
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AUTHOR
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STATUS
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approved
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