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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 (list; graph; refs; listen; history; text; internal format)
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

Alois P. Heinz, Table of n, a(n) for n = 0..125

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-Fran├žois 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

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

Sequence in context: A177447 A005192 A080687 * A223895 A111465 A247029

Adjacent sequences:  A231616 A231617 A231618 * A231620 A231621 A231622

KEYWORD

nonn

AUTHOR

Paul D. Hanna, Nov 14 2013

STATUS

approved

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Last modified June 17 14:09 EDT 2019. Contains 324185 sequences. (Running on oeis4.)