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%I #12 Aug 16 2018 13:28:59
%S 1,10,130,2100,40950,943740,25269300,774635400,26836251750,
%T 1038607069500,44448725821500,2084869401615000,106355178306877500,
%U 5861473946222895000,346999395775257225000,21956626245257906202000,1478562610889805715023750,105561794005139231136877500,7963731010308915234880987500,632966979266333111428303275000,52862553418201438508049805852500
%N O.g.f. equals the square of the e.g.f. of A291561.
%C A291561 is a diagonal in triangle A291560: a(n) = -A291560(n+1, n) for n >= 1; the e.g.f. of triangle A291560 equals arcsin( k*sin(x) ).
%e O.g.f.: A(x) = x^2 + 10*x^3 + 130*x^4 + 2100*x^5 + 40950*x^6 + 943740*x^7 + 25269300*x^8 + 774635400*x^9 + 26836251750*x^10 + 1038607069500*x^11 + 44448725821500*x^12 + 2084869401615000*x^13 + 106355178306877500*x^14 + 5861473946222895000*x^15 + 346999395775257225000*x^16 + 21956626245257906202000*x^17 + 1478562610889805715023750*x^18 + ...
%e such that the square root of the g.f. equals the e.g.f. of A291561, which begins:
%e A(x)^(1/2) = x + 10*x^2/2! + 315*x^3/3! + 18900*x^4/4! + 1819125*x^5/5! + 255405150*x^6/6! + 49165491375*x^7/7! + 12417798393000*x^8/8! + 3981456609755625*x^9/9! + 1579311121869731250*x^10/10! + ... + A291561(n)*x^n/n! + ...
%o (PARI) {A291560(n, r) = (2*n-1)! * polcoeff( polcoeff( asin( k*sin(x + O(x^(2*n)))), 2*n-1, x), 2*r-1, k)}
%o {a(n) = polcoeff( sum(m=1,n,-A291560(m+1, m) * x^m / m! +x*O(x^n) )^2, n)}
%o for(n=2, 25, print1(a(n), ", "))
%Y Cf. A291561, A291560.
%K nonn
%O 2,2
%A _Paul D. Hanna_, Sep 18 2017