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A258377 O.g.f. satisfies A^2(z) = 1/(1 - z)*( BINOMIAL(BINOMIAL(A(z))) ).

%I A258377 #31 May 28 2025 11:55:24
%S A258377 1,3,13,79,649,6955,93813,1539991,29884881,669628819,17005862301,
%T A258377 482399018527,15108642099673,517599894435643,19247498583665029,
%U A258377 771922934908235751,33206411983713679009,1525025984109289947171,74466779211331635306029,3852255519421356879419631
%N A258377 O.g.f. satisfies A^2(z) = 1/(1 - z)*( BINOMIAL(BINOMIAL(A(z))) ).
%C A258377 For a fixed integer N, Hanna has considered the problem of finding an o.g.f. of the form A(z) = 1 + N*z + a(2)*z^2 + a(3)*z^3 + ..., with integer coefficients a(2), a(3), ... dependent on the parameter N, which is a solution to the functional equation A^(N+1) = ( BINOMIAL(A) )^N. Here BINOMIAL(F(z))= 1/(1 - z)*F(z/(1 - z)) denotes the binomial transform of the o.g.f. F(z).
%C A258377 The function A(z) is related to the triangle of ordered Stirling numbers A019538 via logarithmic differentiation. It can be shown that z*A'(z)/A(z) = Sum_{k >= 1} R(k,N)*z^k, where R(k,x) denotes the k-th row polynomial of A019538; equivalently, A(z) = exp( Sum_{k >= 1} R(k,N)*z^k/k ).
%C A258377 Cases include A084784 (N = 1), A090352 (N = 2), A090355 (N = 3), A090357 (N = 4), A090362 (N = 5) and a signed version of A084785 (N = -2).
%C A258377 It turns out that the o.g.f. B(z) := A(z)^(1/N) also has integer coefficients. It satisfies the functional equation B^(N+1) = BINOMIAL(B^N). For cases see A084784 (N = 1), A090351 (N = 2), A090353 (N = 3), A090356 (N = 4), A090358 (N = 5) and A084784 (N = -2).
%C A258377 There are similar results to the above associated with triangle A145901, which can be viewed as a type B analog of A019538.
%C A258377 For a fixed integer N, consider the problem of finding an o.g.f. with integer coefficients (depending on the parameter N) of the form A(z) = 1 + N*z + a(2)*z^2 + a(3)*z^3 + ..., which is a solution to the functional equation A^(N+1)(z) = 1/(1 - z) * ( BINOMIAL(BINOMIAL(A(z))) )^N; equivalently A^(N+1)(z) = 1/(1 - z)* 1/(1 - 2*z)^N*A^N(z/(1 - 2*z)). This is the type B analog of Hanna's type A functional equation above.
%C A258377 It can be shown that z*A'(z)/A(z) = Sum_{k >= 1} P(k,N)*z^k, where P(k,x) denotes the k-th row polynomial of A145901. However, unlike the type A situation, the type B function A(z)^(1/N) does not have all integer coefficients.
%C A258377 The present sequence is the case N = 1. For further examples of solutions to the type B functional equation see A258378 (N = 2), A258379 (N = 3), A258380 (N = 4) and A258381 (N = 5).
%C A258377 From _Peter Bala_, Dec 06 2017: (Start)
%C A258377 a(n) appears to be always odd. Calculation suggests that for k = 1,2,3,..., the sequence a(n) (mod 2^k) is purely periodic with period 2^(k-1). For example, a(n) (mod 4) = (1,3,1,3,...) seems to be purely periodic with period 2 and a(n) (mod 8) = (1,3,5,7,1,3,5,7,...) seems to be purely periodic with period 4 (both checked up to n = 1000).
%C A258377 (End)
%H A258377 Alois P. Heinz, <a href="/A258377/b258377.txt">Table of n, a(n) for n = 0..383</a>
%H A258377 N. J. A. Sloane, <a href="/transforms.txt">Transforms</a>.
%F A258377 a(0) = 1 and for n >= 1, a(n) = 1/n*Sum_{i = 0..n-1} R(i+1,1)*a(n-1-i), where R(n,x) denotes the n-th row polynomial of A145901.
%F A258377 O.g.f.: A(z) = 1 + 3*z + 13*z^2 + 79*z^3 + 649*z^4 + ... satisfies A^2(z) = 1/(1 - z)*1/(1 - 2*z)*A(z/(1 - 2*z)).
%F A258377 O.g.f.: A(z) = exp( Sum_{k >= 1} R(k,1)*z^k/k ).
%F A258377 1 + z*A'(z)/A(z) = 1 + 3*z + 17*z^2 + 147*z^3 + 1697*z^4 + ... is the o.g.f. for A080253.
%F A258377 a(n) = Sum_{j=0..n} binomial(n,j) * A084783(n,n-j). - _Alois P. Heinz_, Jun 09 2023
%F A258377 a(n) ~ (n-1)! * 2^(n - 1/2) / log(2)^(n+1). - _Vaclav Kotesovec_, May 28 2025
%p A258377 #A258377
%p A258377 with(combinat):
%p A258377 #recursively define row polynomials R(n,x) of A145901
%p A258377 R := proc (n, x) option remember; if n = 0 then 1 else 1 + x*add(binomial(n, i)*2^(n-i)*R(i,x), i = 0..n-1) end if; end proc:
%p A258377 #define a family of sequences depending on an integer parameter k
%p A258377 a := proc (n, k) option remember; if n = 0 then 1 else 1/n*add(R(i+1,k)*a(n-1-i,k), i = 0..n-1) end if; end proc:
%p A258377 # display the case k = 1
%p A258377 seq(a(n,1), n = 0..19);
%t A258377 R[n_, x_] := R[n, x] = If[n==0, 1, 1+x*Sum[Binomial[n, i]*2^(n-i)*R[i, x], {i, 0, n-1}]];
%t A258377 a[n_, k_] := a[n, k] = If[n==0, 1, 1/n*Sum[R[i+1, k]*a[n-1-i, k], {i, 0, n-1}]];
%t A258377 a[n_] := a[n, 1];
%t A258377 a /@ Range[0, 19] (* _Jean-François Alcover_, Oct 02 2019 *)
%Y A258377 Cf. A019538, A080253, A084784, A084785, A090351, A090352, A090353, A090355, A090356, A090357, A090358, A090362, A145901, A258378 (N = 2), A258379 (N = 3), A258380 (N = 4), A258381 (N = 5).
%Y A258377 Cf. A084783.
%K A258377 nonn,easy
%O A258377 0,2
%A A258377 _Peter Bala_, May 28 2015