OFFSET
0,4
COMMENTS
CONJECTURES.
a(n) == 0 (mod 3) for n>=3.
a(n) == 1 (mod 8) for n>=0.
FORMULA
E.g.f. satisfies: A(x) = A'(x/A(x)^4).
E.g.f. satisfies: A(x) = ( x / Series_Reversion( x*A'(x)^4 ) )^(1/4).
a(n) = [x^(n-1)/(n-1)!] A(x)^(4*n-3)/(4*n-3) for n>=1.
EXAMPLE
E.g.f.: A(x) = 1 + x + x^2/2! + 9*x^3/3! + 177*x^4/4! + 5601*x^5/5! +...
such that
A(x*A'(x)^4) = A'(x) = 1 + x + 9*x^2/2! + 177*x^3/3! + 5601*x^4/4! +...
To illustrate a(n) = [x^(n-1)/(n-1)!] A(x)^(4*n-3)/(4*n-3), create a table of coefficients of x^k/k!, k>=0, in A(x)^(4*n-3), n>=1, like so:
A^1 : [1, 1, 1, 9, 177, 5601, 249681, 14569545, ...];
A^5 : [1, 5, 25, 165, 2145, 55125, 2211225, 120873045, ...];
A^9 : [1, 9, 81, 801, 10449, 218889, 7501761, 373998465, ...];
A^13: [1, 13, 169, 2301, 35841, 731133, 21950409, 974182989, ...];
A^17: [1, 17, 289, 5049, 95217, 2102577, 60325809, 2417773881, ...];
A^21: [1, 21, 441, 9429, 211617, 5243301, 154446201, 5861076165, ...];
A^25: [1, 25, 625, 15825, 414225, 11585625, 364238625, 13752570225, ...];
A^29: [1, 29, 841, 24621, 738369, 23206989, 791747241, 30816074685, ...]; ...
then the diagonal in the above table generates this sequence shift left:
[1/1, 5/5, 81/9, 2301/13, 95217/17, 5243301/21, 364238625/25, 30816074685/29, ...].
PROG
(PARI) {a(n)=local(A=1+x); for(i=1, n, A=1+intformal(subst(A, x, x*A'^4 +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((1/x*serreverse(x/A^4 +x*O(x^n)))^(1/4))); n!*polcoeff(A, n)}
for(n=0, 25, print1(a(n), ", "))
CROSSREFS
KEYWORD
nonn
AUTHOR
Paul D. Hanna, Nov 27 2013
STATUS
approved