OFFSET
0,3
COMMENTS
Compare to the trivial identity:
G(x) = -1/(1+x) + G(x) + 1/G(x) is satisfied when G(x) = 1+x.
LINKS
Vaclav Kotesovec, Table of n, a(n) for n = 0..400
FORMULA
G.f. satisfies: 0 = 1 - A(x)^2/(1+x) - A(x)^3 + A(x)^4.
Recurrence: 23*(n-2)*(n-1)*n*(65968*n^3 - 732009*n^2 + 2405345*n - 1988196)*a(n) = 6*(n-2)*(n-1)*(2045008*n^4 - 25759791*n^3 + 108645019*n^2 - 176119856*n + 103642224)*a(n-1) + (n-2)*(82789840*n^5 - 1249830655*n^4 + 7012352348*n^3 - 17983400513*n^2 + 20705525148*n - 8104455072)*a(n-2) + 2*(86813888*n^6 - 1614428004*n^5 + 11971039469*n^4 - 44825710320*n^3 + 87744280463*n^2 - 81702244032*n + 25071660432)*a(n-3) + (n-4)*(170659216*n^5 - 2747003363*n^4 + 16636540700*n^3 - 45902853403*n^2 + 53537638698*n - 15743101752)*a(n-4) + 6*(n-5)*(n-4)*(13523440*n^4 - 170347005*n^3 + 722770326*n^2 - 1098529717*n + 288859368)*a(n-5) + 229*(n-6)*(n-5)*(n-4)*(65968*n^3 - 534105*n^2 + 1139231*n - 248892)*a(n-6). - Vaclav Kotesovec, Nov 18 2017
a(n) ~ 1/((1+r) * sqrt(2*Pi*(2 + 6/s^4)) * n^(3/2) * r^(n - 1/2)), where r = 0.07686347945553067708794965527390972514673847319498... and s = 1.152776580718308026971178013130640009569368009247... are roots of the system of equations 1/s^2 + s^2 = 1/(1 + r) + s, 2*s = 1 + 2/s^3. - Vaclav Kotesovec, Nov 18 2017
EXAMPLE
G.f.: A(x) = 1 + x + 3*x^2 + 21*x^3 + 172*x^4 + 1575*x^5 + 15414*x^6 +...
where
A(x)^2 = 1 + 2*x + 7*x^2 + 48*x^3 + 395*x^4 + 3620*x^5 + 35451*x^6 +...
1/A(x)^2 = 1 - 2*x - 3*x^2 - 28*x^3 - 222*x^4 - 2046*x^5 - 20036*x^6 -...
so that A(x) = -1/(1+x) + A(x)^2 + 1/A(x)^2.
PROG
(PARI) {a(n)=local(A=1+x+x*O(x^n)); for(i=1, n, A=-1/(1+x) + A^2 + 1/(A^2 +x*O(x^n))); polcoeff(A, n)}
for(n=0, 30, print1(a(n), ", "))
CROSSREFS
KEYWORD
nonn
AUTHOR
Paul D. Hanna, Sep 08 2013
STATUS
approved