A023552 Convolution of natural numbers >= 3 and Fibonacci numbers.
3, 7, 15, 28, 50, 86, 145, 241, 397, 650, 1060, 1724, 2799, 4539, 7355, 11912, 19286, 31218, 50525, 81765, 132313, 214102, 346440, 560568, 907035, 1467631, 2374695, 3842356, 6217082, 10059470, 16276585, 26336089, 42612709, 68948834, 111561580, 180510452
Offset: 1
Links
- Colin Barker, Table of n, a(n) for n = 1..1000
- Index entries for linear recurrences with constant coefficients, signature (3,-2,-1,1).
Programs
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GAP
F:=Fibonacci; List([1..40], n-> F(n+4)+2*F(n+2)-n-5); # G. C. Greubel, Jul 08 2019
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Magma
F:=Fibonacci; [F(n+4)+2*F(n+2)-n-5: n in [1..40]]; // G. C. Greubel, Jul 08 2019
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Mathematica
LinearRecurrence[{3,-2,-1,1},{3,7,15,28},40] (* or *) Rest[ CoefficientList[Series[(x(3-2x))/((1-x-x^2)(1-x)^2),{x,0,40}],x]] (* Harvey P. Dale, Apr 24 2011 *) With[{F=Fibonacci}, Table[F[n+4]+2*F[n+2]-n-5, {n,40}]] (* G. C. Greubel, Jul 08 2019 *) -
PARI
Vec(x*(3-2*x)/((1-x-x^2)*(1-x)^2) + O(x^40)) \\ Colin Barker, Mar 11 2017
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PARI
vector(40, n, f=fibonacci; f(n+4)+2*f(n+2)-n-5) \\ G. C. Greubel, Jul 08 2019
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Sage
f=fibonacci; [f(n+4)+2*f(n+2)-n-5 for n in (1..40)] # G. C. Greubel, Jul 08 2019
Formula
G.f.: x*(3-2*x)/((1-x-x^2)*(1-x)^2). - Ralf Stephan, Apr 28 2004
From Colin Barker, Mar 11 2017: (Start)
a(n) = 3*a(n-1) - 2*a(n-2) - a(n-3) + a(n-4) for n>4.
a(n) = -5 + (2^(-1-n)*((1-sqrt(5))^n*(-13+5*sqrt(5)) + (1+sqrt(5))^n*(13+5*sqrt(5)))) / sqrt(5) - n. (End)
a(n) = Fibonacci(n+4) + 2*Fibonacci(n+2) - (n+5). - G. C. Greubel, Jul 08 2019