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 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. 2
 1, 1, 3, 16, 105, 856, 8433, 96916, 1272225, 18789136, 308335713, 5565837916, 109603592145, 2338198823416, 53718370204593, 1322292130204516, 34718481333932865, 968552056638097696, 28609403248435931073, 892022330159009036716, 29276492753074019702385 (list; graph; refs; listen; history; text; internal format)
 OFFSET 0,3 COMMENTS Inverse binomial transform of A370456. 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 LINKS Table of n, a(n) for n=0..20. FORMULA E.g.f.: 2*exp(x)/(1 + exp(x) + exp(2*x) - exp(3*x)). MATHEMATICA 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 *) PROG (SageMath) def a(m): if m==0: return 1 else: return (-1)^m+1/2*sum([(1-(-2)^j-(-1)^j)*binomial(m, j)*a(m-j) for j in [1, .., m]]) list(a(m) for m in [0, .., 20]) (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 CROSSREFS Cf. A370163, A370456. Sequence in context: A215931 A271777 A014304 * A063548 A157452 A369694 Adjacent sequences: A370089 A370090 A370091 * A370093 A370094 A370095 KEYWORD nonn AUTHOR Prabha Sivaramannair, Feb 09 2024 STATUS approved

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Last modified April 23 07:42 EDT 2024. Contains 371905 sequences. (Running on oeis4.)