%I #19 Dec 03 2013 20:48:12

%S 1,1,1,7,106,2590,89500,4061830,230971450,15931539370,1301850919000,

%T 123784442479000,13503177242345500,1670758163868804100,

%U 232275860883184444000,35996619826480459259200,6176482540368771725898700,1166516670494844414929980300,241253958437270751756621102400

%N E.g.f. A(x) satisfies: A'(x) = A(x*A'(x)^3) with A(0)=1.

%C CONJECTURES.

%C a(n) == 0 (mod 2) for n>=4.

%C a(n) == 1 (mod 3) for n>=0.

%C a(n) == 0 (mod 5) for n>=5.

%C a(n) == 0 (mod 100) for n>=10.

%C a(n) == 0 (mod 2^2) for n>=10.

%C a(n) == 0 (mod 2^3) for n>=18.

%C a(n) == 0 (mod 2^k) for n>=(8*k-6) for k>1.

%H Paul D. Hanna, <a href="/A231899/b231899.txt">Table of n, a(n) for n = 0..160</a>

%F E.g.f. satisfies: A(x) = A'(x/A(x)^3).

%F E.g.f. satisfies: A(x) = ( x / Series_Reversion( x*A'(x)^3 ) )^(1/3).

%F a(n) = [x^(n-1)/(n-1)!] A(x)^(3*n-2)/(3*n-2) for n>=1.

%e E.g.f.: A(x) = 1 + x + x^2/2! + 7*x^3/3! + 106*x^4/4! + 2590*x^5/5! + 89500*x^6/6! +...

%e such that

%e A(x*A'(x)^3) = A'(x) = 1 + x + 7*x^2/2! + 106*x^3/3! + 2590*x^4/4! + 89500*x^5/5! +...

%e To illustrate a(n) = [x^(n-1)/(n-1)!] A(x)^(3*n-2)/(3*n-2), create a table of coefficients of x^k/k!, k>=0, in A(x)^(3*n-2), n>=1, like so:

%e A^1 : [1, 1, 1, 7, 106, 2590, 89500, 4061830, ...];

%e A^4 : [1, 4, 16, 88, 964, 19840, 622480, 26528920, ...];

%e A^7 : [1, 7, 49, 385, 4144, 72100, 1990030, 78187060, ...];

%e A^10: [1, 10, 100, 1060, 13210, 221740, 5476000, 195409300, ...];

%e A^13: [1, 13, 169, 2275, 33670, 599170, 14113840, 461992570, ...];

%e A^16: [1, 16, 256, 4192, 72976, 1432000, 34173760, 1061820640, ...];

%e A^19: [1, 19, 361, 6973, 140524, 3074200, 77174770, 2368268680, ...];

%e A^22: [1, 22, 484, 10780, 247654, 6035260, 162421600, 5081371900, ...]; ...

%e then the diagonal in the above table generates this sequence shift left:

%e [1/1, 4/4, 49/7, 1060/10, 33670/13, 1432000/16, 77174770/19, 5081371900/22, ...].

%e SUMS OF TERM RESIDUES MODULO 2^n.

%e Given a(k) == 0 (mod 2^n) for k>=(8*n-6) for n>1, then it is interesting to consider the sums of the residues of all terms modulo 2^n for n>=1.

%e Let b(n) = Sum_{k>=0} a(k) (mod 2^n) for n>=1, then the sequence {b(n)} begins:

%e [4, 16, 48, 152, 424, 936, 2472, 5672, 14376, 32296, 72232, 145960, 350760, 809512, 1677864, 3512872, 7182888, 16489000, 34314792, 72587816, 147036712, ...].

%o (PARI) {a(n)=local(A=1+x); for(i=1, n, A=1+intformal(subst(A, x, x*A'^3 +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^3 +x*O(x^n)))^(1/3)));n!*polcoeff(A,n)}

%o for(n=0,25,print1(a(n),", "))

%Y Cf. A231619, A231866, A232694, A232695.

%K nonn

%O 0,4

%A _Paul D. Hanna_, Nov 14 2013