%I #7 Dec 03 2013 20:49:42
%S 1,1,1,9,177,5601,249681,14569545,1062623265,93853717761,
%T 9810385567329,1192614883442889,166310354311947345,
%U 26308546859152889697,4677436610087462937393,927353710845763536487305,203648424149429271943770945,49245501579619466882211194625,13045520297945193508654786790337
%N E.g.f. A(x) satisfies: A'(x) = A(x*A'(x)^4) with A(0)=1.
%C CONJECTURES.
%C a(n) == 0 (mod 3) for n>=3.
%C a(n) == 1 (mod 8) for n>=0.
%F E.g.f. satisfies: A(x) = A'(x/A(x)^4).
%F E.g.f. satisfies: A(x) = ( x / Series_Reversion( x*A'(x)^4 ) )^(1/4).
%F a(n) = [x^(n-1)/(n-1)!] A(x)^(4*n-3)/(4*n-3) for n>=1.
%e E.g.f.: A(x) = 1 + x + x^2/2! + 9*x^3/3! + 177*x^4/4! + 5601*x^5/5! +...
%e such that
%e A(x*A'(x)^4) = A'(x) = 1 + x + 9*x^2/2! + 177*x^3/3! + 5601*x^4/4! +...
%e 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:
%e A^1 : [1, 1, 1, 9, 177, 5601, 249681, 14569545, ...];
%e A^5 : [1, 5, 25, 165, 2145, 55125, 2211225, 120873045, ...];
%e A^9 : [1, 9, 81, 801, 10449, 218889, 7501761, 373998465, ...];
%e A^13: [1, 13, 169, 2301, 35841, 731133, 21950409, 974182989, ...];
%e A^17: [1, 17, 289, 5049, 95217, 2102577, 60325809, 2417773881, ...];
%e A^21: [1, 21, 441, 9429, 211617, 5243301, 154446201, 5861076165, ...];
%e A^25: [1, 25, 625, 15825, 414225, 11585625, 364238625, 13752570225, ...];
%e A^29: [1, 29, 841, 24621, 738369, 23206989, 791747241, 30816074685, ...]; ...
%e then the diagonal in the above table generates this sequence shift left:
%e [1/1, 5/5, 81/9, 2301/13, 95217/17, 5243301/21, 364238625/25, 30816074685/29, ...].
%o (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)}
%o for(n=0, 25, print1(a(n), ", "))
%o (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)}
%o for(n=0, 25, print1(a(n), ", "))
%Y Cf. A231619, A231866, A231899, A232695, A232696.
%K nonn
%O 0,4
%A _Paul D. Hanna_, Nov 27 2013