%I #48 Sep 08 2022 08:45:09
%S 0,1,4,13,37,101,269,710,1865,4890,12810,33546,87834,229963,602062,
%T 1576231,4126639,10803695,28284455,74049680,193864595,507544116,
%U 1328767764,3478759188,9107509812,23843770261,62423800984,163427632705
%N a(n) = Sum_{m=1..n} Sum_{i=1..m} F(i)*F(i+1) where F(n)=Fibonacci numbers A000045.
%C This is the 2-fold convolution of A001654 with the sequence 1,1,1,....
%C Equivalently, partial sums of A064831 which is the partial sums of A001654. - _Joerg Arndt_, Oct 01 2021
%C 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
%H Vincenzo Librandi, <a href="/A080145/b080145.txt">Table of n, a(n) for n = 0..1000</a>
%H Kenneth Edwards and Michael A. Allen, <a href="https://cs.uwaterloo.ca/journals/JIS/VOL24/Allen/edwards2.html">New combinatorial interpretations of the Fibonacci numbers squared, golden rectangle numbers, and Jacobsthal numbers using two types of tile</a>, J. Int. Seq. 24 (2021) Article 21.3.8.
%H Justin M. Troyka and Yan Zhuang, <a href="https://arxiv.org/abs/2109.14774">Fibonacci numbers, consecutive patterns, and inverse peaks</a>, arXiv:2109.14774 [math.CO], 2021.
%H Yan Zhuang, <a href="https://arxiv.org/abs/2108.10309">A lifting of the Goulden-Jackson cluster method to the Malvenuto-Reutenauer algebra</a>, arXiv:2108.10309 [math.CO], 2021.
%H <a href="/index/Rec#order_05">Index entries for linear recurrences with constant coefficients</a>, signature (4,-3,-3,4,-1).
%F a(n) = F(n+1)*F(n+2) - floor((n+2)/2).
%F G.f.: x/((1 - 2*x - 2*x^2 + x^3)*(1-x)^2).
%F a(n) = (4*Lucas(2*n + 3) + (-1)^(n+1) - 10*n - 15)/20. - _Ehren Metcalfe_, Aug 21 2017
%F a(n) = (4*Fibonacci(n+1)*Fibonacci(n+2) - 2*n - 3 - (-1)^n)/4. - _G. C. Greubel_, Jul 23 2019
%F a(n) = Sum_{j=1..n} j*F(n+1-j)*F(n+2-j). - _Michael A. Allen_, Jan 07 2022
%t CoefficientList[Series[x/((1-2x-2x^2+x^3)(1-x)^2), {x, 0, 30}], x] (* _Vladimir Joseph Stephan Orlovsky_, Jul 21 2009 *)
%t 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 *)
%o (Magma) [(4*Lucas(2*n+3)+(-1)^(n+1)-10*n-15)/20: n in [0..30]]; // _Vincenzo Librandi_, Aug 22 2017
%o (PARI) L(n)=fibonacci(n-1)+fibonacci(n+1)
%o a(n)=(4*L(2*n+3)-(-1)^n-10*n-15)/20 \\ _Charles R Greathouse IV_, Aug 26 2017
%o (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
%o (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
%Y Cf. A000032, A000045, A001654, A064831.
%K easy,nonn
%O 0,3
%A Mario Catalani (mario.catalani(AT)unito.it), Jan 31 2003