OFFSET
0,2
COMMENTS
Conjecture: if G(x) = exp( 1/G(x) * Integral G(x)^t dx ) for some fixed real t, then G(x) will consist entirely of nonnegative coefficients iff t >= 8/3.
FORMULA
E.g.f. A(x) satisfies: A(x)^(8/3) = A'(x) * (1 + log(A(x))).
EXAMPLE
E.g.f.: A(x) = Sum_{n>=0} a(n) * x^n / (n! * 3^n).
E.g.f.: A(x) = 1 + 3*x/3 + 15*x^2/(2!*3^2) + 132*x^3/(3!*3^3) + 1485*x^4/(4!*3^4) + 22248*x^5/(5!*3^5) + 389259*x^6/(6!*3^6) + 8252280*x^7/(7!*3^7) + 195922881*x^8/(8!*3^8) + 5374450440*x^9/(9!*3^9) + 161160732999*x^10/(10!*3^10) + 5426241074712*x^11/(11!*3^11) + 196597702744245*x^12/(12!*3^12) +...
A(x)^(8/3) = 1 + 8*x/3 + 80*x^2/(2!*3^2) + 1032*x^3/(3!*3^3) + 16320*x^4/(4!*3^4) + 307848*x^5/(5!*3^5) + 6738624*x^6/(6!*3^6) + 168294600*x^7/(7!*3^7) + 4718840256*x^8/(8!*3^8) + 146939658120*x^9/(9!*3^9) + 5028536908224*x^10/(10!*3^10) + 187710703997832*x^11/(11!*3^11) + 7587706672658880*x^12/(12!*3^12) +...
which equals A'(x) * (1 + log(A(x))).
log(A(x)) = 3*x/3 + 6*x^2/(2!*3^2) + 51*x^3/(3!*3^3) + 360*x^4/(4!*3^4) + 5715*x^5/(5!*3^5) + 70920*x^6/(6!*3^6) + 1634715*x^7/(7!*3^7) + 29206440*x^8/(8!*3^8) + 880341075*x^9/(9!*3^9) + 20578357800*x^10/(10!*3^10) + 765371614275*x^11/(11!*3^11) + 22128348997800*x^12/(12!*3^12) +...
which equals 1/A(x) * Integral A(x)^(8/3) dx.
The series expansion of the e.g.f. with reduced fractional coefficients begins:
A(x) = 1 + x + 5/6*x^2 + 22/27*x^3 + 55/72*x^4 + 103/135*x^5 + 14417/19440*x^6 + 283/378*x^7 + 115181/155520*x^8 + 184309/244944*x^9 + 94744699/125971200*x^10 + 2791276273/3637418400*x^11 + 14709891713/19046845440*x^12 + 1345444691309/1702311811200*x^13 + 127883500285/159993501696*x^14 + 2637791930244871/3217369323168000*x^15 +...
PROG
(PARI) {a(n) = my(A=1+x +x*O(x^n)); for(i=1, n, A = exp( 1/A * intformal( A^(8/3) +x*O(x^n)))); n!*3^n*polcoeff(A, n)}
for(n=0, 30, print1(a(n), ", "))
CROSSREFS
KEYWORD
nonn
AUTHOR
Paul D. Hanna, Jan 31 2017
STATUS
approved