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%I #6 Dec 07 2013 12:43:14
%S 1,1,3,33,726,25236,1229328,78167484,6193726506,592068123054,
%T 66673324219176,8685890001564984,1290531658541292252,
%U 216188985806157611520,40449991773179254230432,8386998677130790903212000,1914263814914709029067344724,478208364783447353623777136772
%N E.g.f. A(x) satisfies: A'(x) = A(x*A'(x))^3.
%C CONJECTURES:
%C a(n) == 0 (mod 2) for n>=4.
%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), k>=2.
%C a(n) == 0 (mod 3) for n>=2.
%C a(n) == 0 (mod 3^2) for n>=5.
%C a(n) == 0 (mod 3^3) for n>=7.
%C a(n) == 0 (mod 3^4) for n>=10.
%C a(n) == 0 (mod 3^k) for n>=(3*k-2), k>=3.
%C a(n) == 0 (mod 13) for n>=13.
%H Paul D. Hanna, <a href="/A233319/b233319.txt">Table of n, a(n) for n = 0..100</a>
%F E.g.f. A(x) satisfies: A(x)^3 = A'(x/A(x)^3).
%F E.g.f. A(x) satisfies: A(x) = ( x / Series_Reversion( x*A'(x) ) )^(1/3).
%F a(n) = [x^(n-1)/(n-1)!] A(x)^(3*n)/n for n>=1.
%e E.g.f.: A(x) = 1 + x + 3*x^2/2! + 33*x^3/3! + 726*x^4/4! + 25236*x^5/5! +...
%e such that
%e A(x*A'(x))^3 = A'(x) = 1 + 3*x + 33*x^2/2! + 726*x^3/3! + 25236*x^4/4! +...
%e 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! +...
%e 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:
%e A^3: [1, 3, 15, 159, 3240, 106218, 4961250, 305900982, ...];
%e A^6: [1, 6, 48, 588, 11646, 357336, 15709968, 923153004, ...];
%e A^9: [1, 9, 99, 1449, 30078, 899964, 37750104, 2118453588, ...];
%e A^12:[1, 12, 168, 2904, 65340, 1977912, 80833248, 4365682056, ...];
%e A^15:[1, 15, 255, 5115, 126180, 3961350, 161145630, 8476536330, ...];
%e A^18:[1, 18, 360, 8244, 223290, 7375968, 304020000, 15786282132, ...];
%e A^21:[1, 21, 483, 12453, 369306, 12932136, 547172388, 28405637064, ...];
%e A^24:[1, 24, 624, 17904, 578808, 21554064, 944463744, 49549812048, ...]; ...
%e then the diagonal in the above table generates this sequence shift left:
%e [1/1, 6/2, 99/3, 2904/4, 126180/5, 7375968/6, 547172388/7, 49549812048/8, ...].
%e SUMS OF TERM RESIDUES MODULO 2^n.
%e 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.
%e Let b(n) = Sum_{k>=0} a(k) (mod 2^n) for n>=1, then the sequence {b(n)} begins:
%e [4, 12, 40, 112, 336, 848, 2128, 5584, 13776, 29648, 64464, 136144, 316368, ...].
%e SUMS OF TERM RESIDUES MODULO 3^n.
%e 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.
%e Let c(n) = Sum_{k>=0} a(k) (mod 3^n) for n>=1, then the sequence {c(n)} begins:
%e [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, ...].
%o (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)}
%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)))); n!*polcoeff(A, n)}
%o for(n=0, 25, print1(a(n), ", "))
%Y Cf. A232686, A231619, A231866, A231899, A232694, A232695, A232696.
%K nonn
%O 0,3
%A _Paul D. Hanna_, Dec 07 2013
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