cp's OEIS Frontend

A370092 a(0) = 1, a(n) = (-1)^n + (1/2) * Sum_{j=1..n} (1-(-1)^j-(-2)^j) * binomial(n,j) * a(n-j) for n > 0.

%I A370092 #37 Feb 28 2024 20:40:05
%S A370092 1,1,3,16,105,856,8433,96916,1272225,18789136,308335713,5565837916,
%T A370092 109603592145,2338198823416,53718370204593,1322292130204516,
%U A370092 34718481333932865,968552056638097696,28609403248435931073,892022330159009036716,29276492753074019702385
%N A370092 a(0) = 1, a(n) = (-1)^n + (1/2) * Sum_{j=1..n} (1-(-1)^j-(-2)^j) * binomial(n,j) * a(n-j) for n > 0.
%C A370092 Inverse binomial transform of A370456.
%C A370092 Conjecture: Let k > 2 be a positive integer. The sequence obtained by reducing a(n) modulo k is eventually periodic with the period dividing phi(k) = A000010(k). For example, modulo 10 we obtain the sequence [1, 1, 3, 6, 5, 6, 3, 6, 5, 6, 3, 6, 5, 6, 3, 6, 5, 6, ...] with an apparent period of 4 beginning at a(2). See A000670 for a more general conjecture. - _Peter Bala_, Feb 16 2024
%F A370092 E.g.f.: 2*exp(x)/(1 + exp(x) + exp(2*x) - exp(3*x)).
%t A370092 a[0]=1;Table[(-1)^n+Sum[ (1-(-1)^j-  (-2) ^j) *Binomial[n,j]*a[n-j]/2,{j,1,n} ],{n,0,20}] (* _James C. McMahon_, Feb 10 2024 *)
%o A370092 (SageMath)
%o A370092 def a(m):
%o A370092     if m==0:
%o A370092         return 1
%o A370092     else:
%o A370092         return (-1)^m+1/2*sum([(1-(-2)^j-(-1)^j)*binomial(m,j)*a(m-j) for j in [1,..,m]])
%o A370092 list(a(m) for m in [0,..,20])
%o A370092 (PARI) seq(n)={my(p=exp(x + O(x*x^n))); Vec(serlaplace(2*p/(1 + p + p^2 - p^3)))} \\ _Andrew Howroyd_, Feb 10 2024
%Y A370092 Cf. A370163, A370456.
%K A370092 nonn
%O A370092 0,3
%A A370092 _Prabha Sivaramannair_, Feb 09 2024