OFFSET
0,4
FORMULA
E.g.f. satisfies: A(x) = A'(x/A(x)^2).
E.g.f. satisfies: A(x) = sqrt( x / Series_Reversion( x*A'(x)^2 ) ).
a(n) = [x^(n-1)/(n-1)!] A(x)^(2*n-1)/(2*n-1) for n>=1.
a(n) == 1 (mod 4) for n>=0.
EXAMPLE
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! +...
such that
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! +...
The square of the e.g.f. begins:
A(x)^2 = 1 + 2*x + 4*x^2/2! + 16*x^3/3! + 152*x^4/4! + 2448*x^5/5! +...
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:
A^1 : [1, 1, 1, 5, 53, 909, 22149, 711297, ...];
A^3 : [1, 3, 9, 39, 333, 5007, 112101, 3395907, ...];
A^5 : [1, 5, 25, 145, 1205, 16065, 326525, 9235165, ...];
A^7 : [1, 7, 49, 371, 3437, 44163, 825741, 21682143, ...];
A^9 : [1, 9, 81, 765, 8181, 108981, 1952469, 47966553, ...];
A^11: [1, 11, 121, 1375, 16973, 243639, 4370069, 102669787, ...];
A^13: [1, 13, 169, 2249, 31733, 498537, 9246861, 213557877, ...];
A^15: [1, 15, 225, 3435, 54765, 945195, 18486525, 430317495, ...]; ...
then the diagonal in the above table generates this sequence shift left:
[1/1, 3/3, 25/5, 371/7, 8181/9, 243639/11, 9246861/13, 430317495/15, ...].
PROG
(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)}
for(n=0, 25, print1(a(n), ", "))
(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)}
for(n=0, 25, print1(a(n), ", "))
CROSSREFS
KEYWORD
nonn
AUTHOR
Paul D. Hanna, Nov 14 2013
STATUS
approved