A080145 a(n) = Sum_{m=1..n} Sum_{i=1..m} F(i)*F(i+1) where F(n)=Fibonacci numbers A000045.

0, 1, 4, 13, 37, 101, 269, 710, 1865, 4890, 12810, 33546, 87834, 229963, 602062, 1576231, 4126639, 10803695, 28284455, 74049680, 193864595, 507544116, 1328767764, 3478759188, 9107509812, 23843770261, 62423800984, 163427632705

Offset: 0

Author

Mario Catalani (mario.catalani(AT)unito.it), Jan 31 2003

Keywords

• easy
• nonn

Comments

This is the 2-fold convolution of A001654 with the sequence 1,1,1,....

Equivalently, partial sums of A064831 which is the partial sums of A001654. - Joerg Arndt, Oct 01 2021

a(n) is the number of permutations p in Sn(321) such that p^(-1) has exactly one left peak. See Troyka and Zhuang. - Michel Marcus, Oct 01 2021

Links

• Vincenzo Librandi, Table of n, a(n) for n = 0..1000
• Kenneth Edwards and Michael A. Allen, New combinatorial interpretations of the Fibonacci numbers squared, golden rectangle numbers, and Jacobsthal numbers using two types of tile, J. Int. Seq. 24 (2021) Article 21.3.8.
• Justin M. Troyka and Yan Zhuang, Fibonacci numbers, consecutive patterns, and inverse peaks, arXiv:2109.14774 [math.CO], 2021.
• Yan Zhuang, A lifting of the Goulden-Jackson cluster method to the Malvenuto-Reutenauer algebra, arXiv:2108.10309 [math.CO], 2021.
• Index entries for linear recurrences with constant coefficients, signature (4,-3,-3,4,-1).

Crossrefs

Cf. A000032, A000045, A001654, A064831.

Programs

GAP

F:=Fibonacci;; List([0..30], n-> (4*F(n+1)*F(n+2)-2*n-3-(-1)^n)/4); # G. C. Greubel, Jul 23 2019

Magma

[(4*Lucas(2*n+3)+(-1)^(n+1)-10*n-15)/20: n in [0..30]]; // Vincenzo Librandi, Aug 22 2017

Mathematica

CoefficientList[Series[x/((1-2x-2x^2+x^3)(1-x)^2), {x, 0, 30}], x] (* Vladimir Joseph Stephan Orlovsky, Jul 21 2009 *)
With[{F=Fibonacci}, Table[(4*F[n+1]*F[n+2]-2*n-3-(-1)^n)/4, {n,0,30}]] (* G. C. Greubel, Jul 23 2019 *)

PARI

L(n)=fibonacci(n-1)+fibonacci(n+1)
a(n)=(4*L(2*n+3)-(-1)^n-10*n-15)/20 \\ Charles R Greathouse IV, Aug 26 2017

Sage

f=fibonacci; [(4*f(n+1)*f(n+2)-2*n-3-(-1)^n)/4 for n in (0..30)] # G. C. Greubel, Jul 23 2019

Formula

a(n) = F(n+1)*F(n+2) - floor((n+2)/2).

G.f.: x/((1 - 2*x - 2*x^2 + x^3)*(1-x)^2).

a(n) = (4*Lucas(2*n + 3) + (-1)^(n+1) - 10*n - 15)/20. - Ehren Metcalfe, Aug 21 2017

a(n) = (4*Fibonacci(n+1)*Fibonacci(n+2) - 2*n - 3 - (-1)^n)/4. - G. C. Greubel, Jul 23 2019

a(n) = Sum_{j=1..n} j*F(n+1-j)*F(n+2-j). - Michael A. Allen, Jan 07 2022