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A258380 O.g.f. satisfies A^5(z) = 1/(1 - z)*( BINOMIAL(BINOMIAL(A(z))) )^4.

%I A258380 #20 May 28 2025 12:19:43
%S A258380 1,9,121,2289,58561,1954281,82055449,4190913201,252934661569,
%T A258380 17620643974921,1390978843729657,122629436549879473,
%U A258380 11935272648323364097,1270531043409588667753,146799401794935250517017,18292108113357605085295345,2444763748582590165449000065
%N A258380 O.g.f. satisfies A^5(z) = 1/(1 - z)*( BINOMIAL(BINOMIAL(A(z))) )^4.
%C A258380 The binomial transform of an o.g.f. A(z) is given by BINOMIAL(A(z)) = 1/(1 - z)*A(z/(1 - z)).
%C A258380 For general remarks on a solution to the functional equation A^(N+1)(z) = 1/(1 - z)*( BINOMIAL(BINOMIAL(A(z))) )^N for integer N, and the connection with triangle A145901 see A258377 (case N = 1). This is the case N = 4.
%C A258380 From _Peter Bala_, Dec 06 2017: (Start)
%C A258380 a(n) appears to be of the form 8*m + 1. Calculation suggests that for k = 1,2,3,..., the sequence a(n) (mod 2^k) is purely periodic with period length a divisor of 2^(k-1). For example, a(n) (mod 16) = (1,9,9,1,1,9,9,1,...) seems to be purely periodic with period length 4 and a(n) (mod 32) = (1,9,25,17,1,9,25,17,...) seems to be purely periodic with period length 4 (both checked up to n = 1000).
%C A258380 The sequences a(n) (mod k), for other values of k, appear to have interesting but more complicated patterns. An example is given below.
%C A258380 (End)
%H A258380 N. J. A. Sloane, <a href="/transforms.txt">Transforms</a>.
%F A258380 a(0) = 1 and for n >= 1, a(n) = 1/n*Sum_{i = 0..n-1} R(i+1,4)*a(n-1-i), where R(n,x) denotes the n-th row polynomial of A145901.
%F A258380 O.g.f.: A(z) = 1 + 9*z + 121*z^2 + 2289*z^3 + 58561*z^4 + ... satisfies A^5(z) = 1/(1 - z)*1/(1 - 2*z)^4*A^4(z/(1 - 2*z)).
%F A258380 O.g.f.: A(z) = exp( Sum_{k >= 1} R(k,4)*z^k/k ).
%F A258380 a(n) ~ (n-1)! * 2^(n-1) / (sqrt(5) * log(5/4)^(n+1)). - _Vaclav Kotesovec_, May 28 2025
%e A258380 a(n) (mod 5) = [1, 4, 1, 4, 1, 1, 4, 1, 4, 1, 2, 3, 2, 3, 2, 0, 0, 0, 0, 0, 3, 2, 3, 2, 3, 4, 1, 4, 1, 4, 4, 1, 4, 1, 4, 3, 2, 3, 2, 3, 0, 0, 0, 0, 0, 2, 3, 2, 3, 2, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 1, 4, 1, 4, 1, 1, 4, 1, 4, 1, 2, 3, 2, 3, 2, 0, 0, 0, 0, 0, 3, 2, 3, 2, 3, 3, 2, 3, 2, 3, 3, 2, 3, 2, 3, 1, 4, 1, 4, 1, 0, 0, 0, 0, 0, 4, 1, 4, 1, 4, 3, 2, 3, 2, 3, 3, 2, 3, 2, 3, 1, 4, 1, 4, 1, 0, 0, 0, 0, 0, 4, 1, 4, 1, 4, 1, 4, 1, 4, 1, 1, 4, 1, 4, 1, 2, 3, 2, 3, 2, 0, 0, 0, 0, 0, 3, 2, 3, 2, 3, ...]. - _Peter Bala_, Dec 06 2017
%p A258380 #A258380
%p A258380 with(combinat):
%p A258380 #recursively define the row polynomials R(n,x) of A145901
%p A258380 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 A258380 #define a family of sequences depending on an integer parameter k
%p A258380 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 A258380 # display the case k = 4
%p A258380 seq(a(n,4), n = 0..16);
%t A258380 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 A258380 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 A258380 a[n_] := a[n, 4];
%t A258380 a /@ Range[0, 16] (* _Jean-François Alcover_, Oct 02 2019 *)
%Y A258380 Cf. A019538, A145901, A258377 (N = 1), A258378 (N = 2), A258379 (N = 3), A258381 (N = 5).
%K A258380 nonn,easy
%O A258380 0,2
%A A258380 _Peter Bala_, May 28 2015