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A231619 E.g.f. A(x) satisfies: A'(x) = A(x*A'(x)) with A(0)=1.

%I A231619 #37 Jan 15 2018 08:53:06
%S A231619 1,1,1,3,18,174,2394,43488,993150,27560142,906516252,34681891932,
%T A231619 1520713138896,75519376934904,4206425439922920,260667523550924424,
%U A231619 17848073907441578484,1342326288499671643956,110319548590679184794880,9862994518551295719972264,955390360741496204485599576
%N A231619 E.g.f. A(x) satisfies: A'(x) = A(x*A'(x)) with A(0)=1.
%C A231619 Conjectures:
%C A231619 a(n) == 0 (mod 2) for n>=4.
%C A231619 a(n) == 0 (mod 3) for n>=3.
%C A231619 a(n) == 0 (mod 27) for n>=9.
%C A231619 a(n) == 0 (mod 2^2) for n>=10.
%C A231619 a(n) == 0 (mod 2^3) for n>=18.
%C A231619 a(n) == 0 (mod 2^k) for n>=(8*k-6), k>1.
%C A231619 a(8*n) == 2^n (mod 2^(n+1)) for n>=0.
%H A231619 Alois P. Heinz, <a href="/A231619/b231619.txt">Table of n, a(n) for n = 0..125</a>
%F A231619 E.g.f. satisfies: A(x) = A'(x/A(x)).
%F A231619 E.g.f. satisfies: A(x) = x / Series_Reversion( x*A'(x) ).
%F A231619 E.g.f. A(x) satisfies: A''(x) = A'(x) * A'(x*A'(x)) / (1 - x*A'(x*A'(x))).
%F A231619 E.g.f. A(x) satisfies: A''(x/A(x)) = A(x)^2 * A'(x) / (A(x) - x*A'(x)).
%F A231619 a(n) = [x^(n-1)/(n-1)!] A(x)^n/n for n>=1.
%e A231619 E.g.f.: A(x) = 1 + x + x^2/2! + 3*x^3/3! + 18*x^4/4! + 174*x^5/5! + 2394*x^6/6! +...
%e A231619 such that
%e A231619 A(x*A'(x)) = A'(x) = 1 + x + 3*x^2/2! + 18*x^3/3! + 174*x^4/4! + 2394*x^5/5! +...
%e A231619 To illustrate a(n) = [x^(n-1)/(n-1)!] A(x)^n/n, create a table of coefficients of x^k/k!, k>=0, in A(x)^n like so:
%e A231619 A^1: [1, 1,  1,   3,   18,   174,   2394,    43488,    993150, ...];
%e A231619 A^2: [1, 2,  4,  12,   66,   588,   7596,   131580,   2897316, ...];
%e A231619 A^3: [1, 3,  9,  33,  180,  1512,  18396,   303210,   6418602, ...];
%e A231619 A^4: [1, 4, 16,  72,  420,  3456,  40104,   630072,  12801888, ...];
%e A231619 A^5: [1, 5, 25, 135,  870,  7290,  82350,  1241820,  24234030, ...];
%e A231619 A^6: [1, 6, 36, 228, 1638, 14364, 161604,  2366388,  44519220, ...];
%e A231619 A^7: [1, 7, 49, 357, 2856, 26628, 304416,  4390470,  80167626, ...];
%e A231619 A^8: [1, 8, 64, 528, 4680, 46752, 551376,  7945200, 142078752, ...];
%e A231619 A^9: [1, 9, 81, 747, 7290, 78246, 961794, 14022072, 248041278, ...]; ...
%e A231619 then the diagonal in the above table generates this sequence shift left:
%e A231619 [1/1, 2/2, 9/3, 72/4, 870/5, 14364/6, 304416/7, 7945200/8, 248041278/9, ...].
%e A231619 SUMS OF TERM RESIDUES MODULO 2^n.
%e A231619 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 A231619 Let b(n) = Sum_{k>=0} a(k) (mod 2^n) for n>=1, then the sequence {b(n)} begins:
%e A231619 [4, 16, 44, 156, 428, 1068, 2476, 5804, 13484, 29868, 67756, 149676, 354476, 739500, 1558700, 3131564, 7129260, 14993580, 31246508, 68995244, 153929900, ...].
%t A231619 terms = 20; A[_] = 1; Do[A[x_] = 1 + Integrate[A[x*A'[x]] + O[x]^j, x] + O[x]^j // Normal, {j, 1, terms}]; CoefficientList[A[x] , x]*Range[0, terms-1]! (* _Jean-François Alcover_, Jan 15 2018 *)
%o A231619 (PARI) {a(n)=local(A=1+x);for(i=1,n,A=1+intformal(subst(A,x,x*A' +x*O(x^n))));n!*polcoeff(A,n)}
%o A231619 for(n=0,25,print1(a(n),", "))
%o A231619 (PARI) {a(n)=local(A=1+x);for(i=1,n,A=1+intformal(1/x*serreverse(x/A +x*O(x^n))));n!*polcoeff(A,n)}
%o A231619 for(n=0,25,print1(a(n),", "))
%Y A231619 Cf. A231866, A231899, A232694, A232695, A232696, A232686.
%K A231619 nonn
%O A231619 0,4
%A A231619 _Paul D. Hanna_, Nov 14 2013