OFFSET
0,2
COMMENTS
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 = 4.
From Peter Bala, Dec 06 2017: (Start)
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).
The sequences a(n) (mod k), for other values of k, appear to have interesting but more complicated patterns. An example is given below.
(End)
LINKS
N. J. A. Sloane, Transforms.
FORMULA
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.
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)).
O.g.f.: A(z) = exp( Sum_{k >= 1} R(k,4)*z^k/k ).
EXAMPLE
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
MAPLE
with(combinat):
#recursively define the row polynomials R(n, x) of A145901
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:
#define a family of sequences depending on an integer parameter k
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:
# display the case k = 4
seq(a(n, 4), n = 0..16);
MATHEMATICA
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}]];
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}]];
a[n_] := a[n, 4];
a /@ Range[0, 16] (* Jean-François Alcover, Oct 02 2019 *)
CROSSREFS
KEYWORD
nonn,easy
AUTHOR
Peter Bala, May 28 2015
STATUS
approved