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A258379 O.g.f. satisfies A^4(z) = 1/(1 - z)*( BINOMIAL(BINOMIAL(A(z))) )^3.

%I A258379 #18 May 28 2025 12:13:43
%S A258379 1,7,73,1071,21249,549927,17907177,709326255,33202983873,
%T A258379 1794040660359,109844961440841,7511188035994479,567027585432472641,
%U A258379 46818521577433459239,4195842793686119552361,405529683304196611790703,42039822952112350680798849,4652599937163116610404900871
%N A258379 O.g.f. satisfies A^4(z) = 1/(1 - z)*( BINOMIAL(BINOMIAL(A(z))) )^3.
%C A258379 The binomial transform of an o.g.f. A(z) is given by BINOMIAL(A(z)) = 1/(1 - z)*A(z/(1 - z)). 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 = 3.
%H A258379 N. J. A. Sloane, <a href="/transforms.txt">Transforms</a>.
%F A258379 a(0) = 1 and for n >= 1, a(n) = 1/n*Sum_{i = 0..n-1} R(i+1,3)*a(n-1-i), where R(n,x) denotes the n-th row polynomial of A145901.
%F A258379 O.g.f.: A(z) = 1 + 7*z + 73*z^2 + 1071*z^3 + 21249*z^4 + ... satisfies A^4(z) = 1/(1 - z)*1/(1 - 2*z)^3*A^3(z/(1 - 2*z)).
%F A258379 O.g.f.: A(z) = exp( Sum_{k >= 1} R(k,3)*z^k/k ).
%F A258379 From _Peter Bala_, Dec 06 2017: (Start)
%F A258379 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 length a divisor of 2^(k-1). For example, a(n) (mod 4) = (1,3,1,3,...) seems to be purely periodic with period 2, a(n) (mod 8) = (1,7,1,7,...) seems to be purely periodic also with period 2 while a(n) (mod 16) = (1,7,9,15,1,7,9,15,...) seems to be purely periodic with period 4 (all three checked up to n = 1000).
%F A258379 The sequences a(n) (mod k), for other values of k, appear to have interesting but more complicated patterns. An example is given below.
%F A258379 (End)
%F A258379 a(n) ~ (n-1)! * 2^(n-1) / (sqrt(3) * log(4/3)^(n+1)). - _Vaclav Kotesovec_, May 28 2025
%e A258379 a(n) (mod 3) = (1,1,1,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,1,1,1, 0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,1,1,1,0,0,0,0,0,0,0,0,0,0, 0,0,0,0,0,1,1,1,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,1,1,1,0,0, 0,0,0,0,1,1,1,0,0,0,0,0,0,1,1,1,0,0,0,0,0,0,1,1,1,0,0,0, 0,0,0,1,1,1,0,0,0,0,0,0,1,1,1,0,0,0,0,0,0,1,1,1,0,0,0,0, 0,0,1,1,1,0,0,0,0,0,0,1,1,1,0,0,0,0,0,0,1,1,1,...). - _Peter Bala_, Dec 06 2017
%p A258379 #A258379
%p A258379 with(combinat):
%p A258379 #recursively define the row polynomials R(n,x) of A145901
%p A258379 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 A258379 #define a family of sequences depending on an integer parameter k
%p A258379 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 A258379 # display the case k = 3
%p A258379 seq(a(n,3), n = 0..17);
%t A258379 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 A258379 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 A258379 a[n_] := a[n, 3];
%t A258379 a /@ Range[0, 17] (* _Jean-François Alcover_, Oct 02 2019 *)
%Y A258379 Cf. A019538, A145901, A258377 (N = 1), A258378 (N = 2), A258380 (N = 4), A258381 (N = 5).
%K A258379 nonn,easy
%O A258379 0,2
%A A258379 _Peter Bala_, May 28 2015