%I #19 Dec 03 2013 20:46:22
%S 1,1,1,5,53,909,22149,711297,28687833,1405408841,81620841401,
%T 5516637014061,427699967681709,37595972586389109,3711295383595024221,
%U 408142117923542673737,49663409518409586541937,6647274714312311181770577,973638869018128380202018353
%N E.g.f. A(x) satisfies: A'(x) = A(x*A'(x)^2) with A(0)=1.
%F E.g.f. satisfies: A(x) = A'(x/A(x)^2).
%F E.g.f. satisfies: A(x) = sqrt( x / Series_Reversion( x*A'(x)^2 ) ).
%F a(n) = [x^(n-1)/(n-1)!] A(x)^(2*n-1)/(2*n-1) for n>=1.
%F a(n) == 1 (mod 4) for n>=0.
%e E.g.f.: A(x) = 1 + x + x^2/2! + 5*x^3/3! + 53*x^4/4! + 909*x^5/5! + 22149*x^6/6! +...
%e such that
%e A(x*A'(x)^2) = A'(x) = 1 + x + 5*x^2/2! + 53*x^3/3! + 909*x^4/4! + 22149*x^5/5! +...
%e The square of the e.g.f. begins:
%e A(x)^2 = 1 + 2*x + 4*x^2/2! + 16*x^3/3! + 152*x^4/4! + 2448*x^5/5! +...
%e To illustrate a(n) = [x^(n-1)/(n-1)!] A(x)^(2*n-1)/(2*n-1), create a table of coefficients of x^k/k!, k>=0, in A(x)^(2*n-1), n>=1, like so:
%e A^1 : [1, 1, 1, 5, 53, 909, 22149, 711297, ...];
%e A^3 : [1, 3, 9, 39, 333, 5007, 112101, 3395907, ...];
%e A^5 : [1, 5, 25, 145, 1205, 16065, 326525, 9235165, ...];
%e A^7 : [1, 7, 49, 371, 3437, 44163, 825741, 21682143, ...];
%e A^9 : [1, 9, 81, 765, 8181, 108981, 1952469, 47966553, ...];
%e A^11: [1, 11, 121, 1375, 16973, 243639, 4370069, 102669787, ...];
%e A^13: [1, 13, 169, 2249, 31733, 498537, 9246861, 213557877, ...];
%e A^15: [1, 15, 225, 3435, 54765, 945195, 18486525, 430317495, ...]; ...
%e then the diagonal in the above table generates this sequence shift left:
%e [1/1, 3/3, 25/5, 371/7, 8181/9, 243639/11, 9246861/13, 430317495/15, ...].
%o (PARI) {a(n)=local(A=1+x); for(i=1, n, A=1+intformal(subst(A, x, x*A'^2 +x*O(x^n)))); n!*polcoeff(A, n)}
%o for(n=0, 25, print1(a(n), ", "))
%o (PARI) {a(n)=local(A=1+x);for(i=1,n,A=1+intformal(sqrt(1/x*serreverse(x/A^2 +x*O(x^n)))));n!*polcoeff(A,n)}
%o for(n=0,25,print1(a(n),", "))
%Y Cf. A231619, A231899, A232694, A232695, A232696.
%K nonn
%O 0,4
%A _Paul D. Hanna_, Nov 14 2013