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A080833
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E.g.f.: exp( x/(1 - x - x^2) ).
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2
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1, 1, 3, 19, 145, 1401, 16051, 213403, 3223809, 54514225, 1019601091, 20890209891, 465156779473, 11181638663209, 288536019179955, 7953590111627371, 233211718410856321, 7246720953253750113, 237849724555558441219, 8221578401608012672435, 298505383888840158941841
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OFFSET
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0,3
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COMMENTS
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The sequence terms are odd. 3 divides a(3*n-1), 9 divides a(9*n-1) and 27 divides a(27*n-1); 5 divides a(5*n+4), 25 divides a(25*n+9) and 125 divides a(125*n+34); 7 divides a(7*n+6), 49 divides a(49*n+34) and 343 divides a(343*n + 83); 15 divides a(15*n+14) and 17 divides a(17*n+13).
More generally, the congruence a(n+k) == a(n) (mod k) holds for all n and k. It follows that the sequence obtained by taking a(n) modulo a fixed positive integer k is periodic with exact period dividing k. For example, taken modulo 7 the sequence becomes [1, 1, 3, 5, 5, 1, 0, 1, 1, 3, 3, 5, 5, 1, 0, ...], a purely periodic sequence with period 7. (End)
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LINKS
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FORMULA
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E.g.f.: exp( x/(1 - x - x^2) ).
a(n) = n!*sum{i=0..n, sum{j=0..n, C(i+j-1, j)*C(j, n-i-j)/i!}}. - Paul Barry, Aug 29 2005
E.g.f.: 1 + x*(E(0)-1)/(x+1) where E(k) = 1 + 1/(k+1)/(1-x-x^2)/(1-x/(x+1/E(k+1) )); (recursively defined continued fraction). - Sergei N. Gladkovskii, Jan 27 2013
Recurrence: a(n) = (2*n-1)*a(n-1) + (n-2)*(n-1)*a(n-2) - (n-2)*(n-1)*(2*n-7)*a(n-3) - (n-4)*(n-3)*(n-2)*(n-1)*a(n-4). - Vaclav Kotesovec, Jun 27 2013
a(n) ~ ((1+sqrt(5))/2)^n*exp(2*sqrt(n)/5^(1/4)-n-1/10)*n^(n-1/4)/(sqrt(2)*5^(1/8)). - Vaclav Kotesovec, Jun 27 2013
a(0) = 1; a(n) = Sum_{k=1..n} binomial(n-1,k-1) * k! * Fibonacci(k) * a(n-k). - Ilya Gutkovskiy, Jun 07 2021
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MATHEMATICA
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CoefficientList[Series[E^(x/(1-x-x^2)), {x, 0, 20}], x]* Range[0, 20]! (* Vaclav Kotesovec, Jun 27 2013 *)
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PROG
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(PARI) my(x='x+O('x^30)); Vec(serlaplace(exp(x/(1-x-x^2)))) \\ Michel Marcus, Jun 07 2021
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CROSSREFS
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KEYWORD
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easy,nonn
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AUTHOR
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STATUS
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approved
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