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A105963
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Expansion of (1+4*x)/(1-x-3*x^2).
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3
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1, 5, 8, 23, 47, 116, 257, 605, 1376, 3191, 7319, 16892, 38849, 89525, 206072, 474647, 1092863, 2516804, 5795393, 13345805, 30731984, 70769399, 162965351, 375273548, 864169601, 1989990245, 4582499048, 10552469783, 24299966927
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OFFSET
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0,2
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COMMENTS
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Inversion of the periodic sequence with initial period (1,4,-1,-4). Sequence appears to have the property: for m > n, if s divides both a(n) and a(m) then s also divides a(2*m-n). E.g., 23 divides both a(3) = 23 and a(25) = 1989990245; 23 also divides a(2*25-3) = a(47) = 185518234185384428 = (2)^2*(23)*(131)*(15393149202239).
Floretion Algebra Multiplication Program, FAMP Code: 1jesforseq[.5'k + .5k' + 2'kk' + 2e]
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LINKS
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FORMULA
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a(n) = (2^(-1-n)*((1-sqrt(13))^n*(-9+sqrt(13)) + (1+sqrt(13))^n*(9+sqrt(13)))) / sqrt(13).
a(n) = a(n-1) + 3*a(n-2) for n > 1. (End)
a(n) = 3^((n-1)/2)*( sqrt(3)*Fibonacci(n+1, 1/sqrt(3)) + 4*Fibonacci(n, 1/sqrt(3)) ). - G. C. Greubel, Jan 15 2020
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MAPLE
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seq(coeff(series((1+4*x)/(1-x-3*x^2), x, n+1), x, n), n = 0 .. 40); # G. C. Greubel, Jan 15 2020
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MATHEMATICA
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CoefficientList[Series[(1+4x)/(1-x-3x^2), {x, 0, 40}], x] (* Vincenzo Librandi, Jul 20 2013 *)
Table[Round[3^((n-1)/2)*(Sqrt[3]*Fibonacci[n+1, 1/Sqrt[3]] + 4*Fibonacci[n, 1/Sqrt[3]] )], {n, 0, 40}] (* G. C. Greubel, Jan 15 2020 *)
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PROG
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(Magma) I:=[ 1, 5]; [n le 2 select I[n] else Self(n-1)+3*Self(n-2): n in [1..40]]; // Vincenzo Librandi, Jul 20 2013
(SageMath)
P.<x> = PowerSeriesRing(ZZ, prec)
return P( (1+4*x)/(1-x-3*x^2) ).list()
(GAP) a:=[1, 5];; for n in [3..40] do a[n]:=a[n-1]+3*a[n-2]; od; a; # G. C. Greubel, Jan 15 2020
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CROSSREFS
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KEYWORD
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nonn,easy
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AUTHOR
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STATUS
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approved
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