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A232695 E.g.f. A(x) satisfies: A'(x) = A(x*A'(x)^5) with A(0)=1.

%I A232695 #22 Dec 03 2013 20:44:17
%S A232695 1,1,1,11,266,10326,562926,40058076,3554828286,381374161166,
%T A232695 48366170807276,7128626213386476,1204840675597360776,
%U A232695 230986547885416953936,49777541426984300127816,11964954349177005321013976,3186498480002528225295506276,934756070179948684556233837476
%N A232695 E.g.f. A(x) satisfies: A'(x) = A(x*A'(x)^5) with A(0)=1.
%C A232695  CONJECTURES.
%C A232695 a(n) == 1 (mod 5) for n>=0.
%C A232695 a(n) == 0 (mod 2) for n>=4.
%C A232695 a(n) == 0 (mod 2^2) for n>=10.
%C A232695 a(n) == 0 (mod 2^3) for n>=18.
%C A232695 a(n) == 0 (mod 2^k) for n>=(8*k-6) for k>1.
%H A232695 Paul D. Hanna, <a href="/A232695/b232695.txt">Table of n, a(n) for n = 0..100</a>
%F A232695 E.g.f. satisfies: A(x) = A'(x/A(x)^5).
%F A232695 E.g.f. satisfies: A(x) = ( x / Series_Reversion( x*A'(x)^5 ) )^(1/5).
%F A232695 a(n) = [x^(n-1)/(n-1)!] A(x)^(5*n-4)/(5*n-4) for n>=1.
%e A232695 E.g.f.: A(x) = 1 + x + x^2/2! + 11*x^3/3! + 266*x^4/4! + 10326*x^5/5! +...
%e A232695 such that
%e A232695 A(x*A'(x)^5) = A'(x) = 1 + x + 11*x^2/2! + 266*x^3/3! + 10326*x^4/4! +...
%e A232695 To illustrate a(n) = [x^(n-1)/(n-1)!] A(x)^(5*n-4)/(5*n-4), create a table of coefficients of x^k/k!, k>=0, in A(x)^(5*n-4), n>=1, like so:
%e A232695 A^1 : [1,  1,   1,    11,    266,    10326,     562926, ...];
%e A232695 A^6 : [1,  6,  36,   276,   4086,   124476,    6058956, ...];
%e A232695 A^11: [1, 11, 121,  1441,  21956,   530376,   21460736, ...];
%e A232695 A^16: [1, 16, 256,  4256,  79376,  1891776,   66002016, ...];
%e A232695 A^21: [1, 21, 441,  9471, 216846,  5697426,  191016546, ...];
%e A232695 A^26: [1, 26, 676, 17836, 489866, 14636076,  510313076, ...];
%e A232695 A^31: [1, 31, 961, 30101, 968936, 32971476, 1241800356, ...]; ...
%e A232695 then the diagonal in the above table generates this sequence shift left:
%e A232695 [1/1, 6/6, 121/11, 4256/16, 216846/21, 14636076/26, 1241800356/31, ...].
%e A232695 SUMS OF TERM RESIDUES MODULO 2^n.
%e A232695 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 A232695 Let b(n) = Sum_{k>=0} a(k) (mod 2^n) for n>=1, then the sequence {b(n)} begins:
%e A232695 [4, 16, 52, 180, 388, 868, 2532, 5860, 13028, 27364, 63204, 157412, 370404, 780004, 1730276, 3630820, 7431908, 14509796, 32597732, 72967908, ...].
%o A232695 (PARI) {a(n)=local(A=1+x); for(i=1, n, A=1+intformal(subst(A, x, x*A'^5 +x*O(x^n)))); n!*polcoeff(A, n)}
%o A232695 for(n=0, 25, print1(a(n), ", "))
%o A232695 (PARI) {a(n)=local(A=1+x); for(i=1, n, A=1+intformal((1/x*serreverse(x/A^5 +x*O(x^n)))^(1/5))); n!*polcoeff(A, n)}
%o A232695 for(n=0, 25, print1(a(n), ", "))
%Y A232695 Cf. A231619, A231866, A231899, A232694, A232696.
%K A232695 nonn
%O A232695 0,4
%A A232695 _Paul D. Hanna_, Nov 27 2013