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A231619
E.g.f. A(x) satisfies: A'(x) = A(x*A'(x)) with A(0)=1.
9
1, 1, 1, 3, 18, 174, 2394, 43488, 993150, 27560142, 906516252, 34681891932, 1520713138896, 75519376934904, 4206425439922920, 260667523550924424, 17848073907441578484, 1342326288499671643956, 110319548590679184794880, 9862994518551295719972264, 955390360741496204485599576
OFFSET
0,4
COMMENTS
Conjectures:
a(n) == 0 (mod 2) for n>=4.
a(n) == 0 (mod 3) for n>=3.
a(n) == 0 (mod 27) for n>=9.
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>1.
a(8*n) == 2^n (mod 2^(n+1)) for n>=0.
LINKS
FORMULA
E.g.f. satisfies: A(x) = A'(x/A(x)).
E.g.f. satisfies: A(x) = x / Series_Reversion( x*A'(x) ).
E.g.f. A(x) satisfies: A''(x) = A'(x) * A'(x*A'(x)) / (1 - x*A'(x*A'(x))).
E.g.f. A(x) satisfies: A''(x/A(x)) = A(x)^2 * A'(x) / (A(x) - x*A'(x)).
a(n) = [x^(n-1)/(n-1)!] A(x)^n/n for n>=1.
EXAMPLE
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! +...
such that
A(x*A'(x)) = A'(x) = 1 + x + 3*x^2/2! + 18*x^3/3! + 174*x^4/4! + 2394*x^5/5! +...
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:
A^1: [1, 1, 1, 3, 18, 174, 2394, 43488, 993150, ...];
A^2: [1, 2, 4, 12, 66, 588, 7596, 131580, 2897316, ...];
A^3: [1, 3, 9, 33, 180, 1512, 18396, 303210, 6418602, ...];
A^4: [1, 4, 16, 72, 420, 3456, 40104, 630072, 12801888, ...];
A^5: [1, 5, 25, 135, 870, 7290, 82350, 1241820, 24234030, ...];
A^6: [1, 6, 36, 228, 1638, 14364, 161604, 2366388, 44519220, ...];
A^7: [1, 7, 49, 357, 2856, 26628, 304416, 4390470, 80167626, ...];
A^8: [1, 8, 64, 528, 4680, 46752, 551376, 7945200, 142078752, ...];
A^9: [1, 9, 81, 747, 7290, 78246, 961794, 14022072, 248041278, ...]; ...
then the diagonal in the above table generates this sequence shift left:
[1/1, 2/2, 9/3, 72/4, 870/5, 14364/6, 304416/7, 7945200/8, 248041278/9, ...].
SUMS OF TERM RESIDUES MODULO 2^n.
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.
Let b(n) = Sum_{k>=0} a(k) (mod 2^n) for n>=1, then the sequence {b(n)} begins:
[4, 16, 44, 156, 428, 1068, 2476, 5804, 13484, 29868, 67756, 149676, 354476, 739500, 1558700, 3131564, 7129260, 14993580, 31246508, 68995244, 153929900, ...].
MATHEMATICA
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 *)
PROG
(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)}
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 +x*O(x^n)))); n!*polcoeff(A, n)}
for(n=0, 25, print1(a(n), ", "))
CROSSREFS
KEYWORD
nonn
AUTHOR
Paul D. Hanna, Nov 14 2013
STATUS
approved