A233319 E.g.f. A(x) satisfies: A'(x) = A(x*A'(x))^3.

1, 1, 3, 33, 726, 25236, 1229328, 78167484, 6193726506, 592068123054, 66673324219176, 8685890001564984, 1290531658541292252, 216188985806157611520, 40449991773179254230432, 8386998677130790903212000, 1914263814914709029067344724, 478208364783447353623777136772

Offset: 0

Author

Paul D. Hanna, Dec 07 2013

Keywords

• nonn

Comments

CONJECTURES:

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

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

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

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

a(n) == 0 (mod 3) for n>=2.

a(n) == 0 (mod 3^2) for n>=5.

a(n) == 0 (mod 3^3) for n>=7.

a(n) == 0 (mod 3^4) for n>=10.

a(n) == 0 (mod 3^k) for n>=(3*k-2), k>=3.

a(n) == 0 (mod 13) for n>=13.

Examples

			E.g.f.: A(x) = 1 + x + 3*x^2/2! + 33*x^3/3! + 726*x^4/4! + 25236*x^5/5! +...
such that
A(x*A'(x))^3 = A'(x) = 1 + 3*x + 33*x^2/2! + 726*x^3/3! + 25236*x^4/4! +...
A(x*A'(x)) = (A'(x))^(1/3) = 1 + x + 9*x^2/2! + 186*x^3/3! + 6330*x^4/4! + 306846*x^5/5! + 19560006*x^6/6! + 1559472498*x^7/7! +...
To illustrate a(n) = [x^(n-1)/(n-1)!] A(x)^(3*n)/n, create a table of coefficients of x^k/k!, k>=0, in A(x)^(3*n) like so:
A^3: [1,  3,  15,   159,   3240,   106218,   4961250,   305900982, ...];
A^6: [1,  6,  48,   588,  11646,   357336,  15709968,   923153004, ...];
A^9: [1,  9,  99,  1449,  30078,   899964,  37750104,  2118453588, ...];
A^12:[1, 12, 168,  2904,  65340,  1977912,  80833248,  4365682056, ...];
A^15:[1, 15, 255,  5115, 126180,  3961350, 161145630,  8476536330, ...];
A^18:[1, 18, 360,  8244, 223290,  7375968, 304020000, 15786282132, ...];
A^21:[1, 21, 483, 12453, 369306, 12932136, 547172388, 28405637064, ...];
A^24:[1, 24, 624, 17904, 578808, 21554064, 944463744, 49549812048, ...]; ...
then the diagonal in the above table generates this sequence shift left:
[1/1, 6/2, 99/3, 2904/4, 126180/5, 7375968/6, 547172388/7, 49549812048/8, ...].
SUMS OF TERM RESIDUES MODULO 2^n.
Given a(k) == 0 (mod 2^n) for k>=(8*n-6) for n>=2, then it is interesting to consider the sums of the residues of all terms modulo 2^n for n>=1.
Let b(n) = Sum_{k>=0} a(k) (mod 2^n) for n>=1, then the sequence {b(n)} begins:
[4, 12, 40, 112, 336, 848, 2128, 5584, 13776, 29648, 64464, 136144, 316368, ...].
SUMS OF TERM RESIDUES MODULO 3^n.
Given a(k) == 0 (mod 3^n) for k>=(3*n-2) for n>=3, then it is interesting to consider the sums of the residues of all terms modulo 3^n for n>=1.
Let c(n) = Sum_{k>=0} a(k) (mod 3^n) for n>=1, then the sequence {c(n)} begins:
[2, 17, 71, 368, 1340, 4985, 13733, 59660, 217124, 689516, 2520035, 6594416, 18286118, 72493100, 206416232, 722976884, 2617032608, 8170059617, 25603981622, 93015146708, 256894013555, 832213439720, 2338504300952, 6292517811686, 24650437682951, 71251311202316, 249181919185346, 729594560739527, ...].
		

Links

• Paul D. Hanna, Table of n, a(n) for n = 0..100

Crossrefs

Cf. A232686, A231619, A231866, A231899, A232694, A232695, A232696.

Programs

PARI

{a(n)=local(A=1+x); for(i=1, n, A=1+intformal(subst(A^3, x, x*A' +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^3 +x*O(x^n)))); n!*polcoeff(A, n)}
for(n=0, 25, print1(a(n), ", "))

Formula

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

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

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