A166536 A product of consecutive doubled Fibonacci numbers.
1, 3, 6, 16, 40, 105, 273, 715, 1870, 4896, 12816, 33553, 87841, 229971, 602070, 1576240, 4126648, 10803705, 28284465, 74049691, 193864606, 507544128, 1328767776, 3478759201, 9107509825, 23843770275, 62423800998, 163427632720
Offset: 0
Links
- G. C. Greubel, Table of n, a(n) for n = 0..1000
- Index entries for linear recurrences with constant coefficients, signature (3,0,-3,1).
Programs
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GAP
a:=[1,3,6,16];; for n in [5..30] do a[n]:=3*a[n-1]-3*a[n-3]+a[n-4]; od; a; # G. C. Greubel, Jan 09 2019
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Magma
/* From the sixth formula: */ F:=Fibonacci; [&+[F(i+1)*F(i-1): i in [0..n+1]]: n in [0..30]]; // Bruno Berselli, Feb 15 2017
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Mathematica
LinearRecurrence[{3, 0, -3, 1}, {1, 3, 6, 16}, 30] (* G. C. Greubel, May 16 2016 *) -
PARI
my(x='x+O('x^30)); Vec((1-3*x^2+x^3)/(1-3*x+3*x^3-x^4)) \\ G. C. Greubel, Jan 09 2019 -
Sage
((1-3*x^2+x^3)/(1-3*x+3*x^3-x^4)).series(x, 30).coefficients(x, sparse=False) # G. C. Greubel, Jan 09 2019
Formula
G.f.: (1 - 3*x^2 + x^3)/(1 - 3*x + 3*x^3 - x^4).
a(n) = F(n+1)*F(n+2) + (1 - (-1)^n)/2, where F = A000045.
a(n) = (F(n+2)*(1 + (-1)^n)/2 + F(n)*(1 - (-1)^n)/2)*(F(n+3)*(1 - (-1)^n)/2 + F(n+1)*(1 + (-1)^n)/2).
a(n)*a(n+2) - a(n+1)^2 = (-1)^n*(F(2*n+4) - 1).
a(n) = 3*a(n-1) - 3*a(n-3) + a(n-4). - G. C. Greubel, May 16 2016
a(n) = Sum_{i=0..n+1} F(i+1)*F(i-1), where F(-1) = 1. - Bruno Berselli, Feb 16 2017