|
%I #24 Mar 02 2023 15:12:46
%S 1,1,3,5,12,19,40,65,130,210,404,654,1227,1985,3653,5911,10720,17345,
%T 31090,50305,89316,144516,254568,411900,720757,1166209,2029095,
%U 3283145,5684340,9197455,15855964,25655489,44061862,71293590,122032508
%N a(n) = s(1)*s(n) + s(2)*s(n-1) + ... + s(k)*s(n+1-k), where k = floor((n+1)/2), s = (Fibonacci numbers).
%C From _Wolfdieter Lang_, Jan 02 2012: (Start)
%C chat(n):=a(n+1), n>=0, is the half-convolution of the sequence A000045(n+1), n>=0, with itself. For the definition of half-convolution see a comment on A201204, where also the rule to find the o.g.f. is given. Here the o.g.f. is obtained from (U(x)^2 + U2(x^2))/2 with U(x)=1/(1-x-x^2),the o.g.f. of A000045(n+1), n>=0, and U2(x):=(1-x)/((1+x)*(1-3*x+x^2) the o.g.f. of A007598(n+1), n>=0. This coincides with the o.g.f. given below in the formula section after x has been divided.
%C For the bisection of this half-convolution see A027991(n+1) and A001870(n), n>=0.
%C (End)
%H G. C. Greubel, <a href="/A024458/b024458.txt">Table of n, a(n) for n = 1..1000</a>
%H <a href="/index/Rec#order_08">Index entries for linear recurrences with constant coefficients</a>, signature (1,3,-2,0,-2,-3,1,1).
%F G.f.: x*(1-x^2+x^3)/((1+x^2)*(1+x-x^2)*(1-x-x^2)^2).
%F a(n) = ((13 - 5*(-1)^n + 10*n)*A000045(n) + (1 - (-1)^n + 2*n)*A000032(n) + 8*sin(Pi*n/2))/40. - _Vladimir Reshetnikov_, Oct 03 2016
%F From _G. C. Greubel_, Apr 06 2022: (Start)
%F a(2*n) = (1/5)*(n*Lucas(2*n+1) + Fibonacci(2*n)), n >= 1.
%F a(2*n+1) = (1/5)*((-1)^n + (n+1)*Lucas(2*n+2) + Fibonacci(2*n+1)), n >= 0.
%F a(n) = Sum_{j=0..floor((n-1)/2)} fibonacci(j+1)*Fibonacci(n-j). (End)
%t Table[((13-5(-1)^n +10n)Fibonacci[n] + (1-(-1)^n +2n)LucasL[n] +8Sin[Pi*n/2])/40, {n, 30}] (* _Vladimir Reshetnikov_, Oct 03 2016 *)
%t LinearRecurrence[{1,3,-2,0,-2,-3,1,1},{1,1,3,5,12,19,40,65},40] (* _Harvey P. Dale_, Mar 02 2023 *)
%o (Magma) [(&+[Fibonacci(j+1)*Fibonacci(n-j): j in [0..Floor((n-1)/2)]]): n in [1..50]]; // _G. C. Greubel_, Apr 06 2022
%o (SageMath)
%o def A024458(n): return sum(fibonacci(j+1)*fibonacci(n-j) for j in (0..((n-1)//2)) )
%o [A024458(n) for n in (1..50)] # _G. C. Greubel_, Apr 06 2022
%Y Cf. A000032, A000045, A058071.
%K nonn,easy
%O 1,3
%A _Clark Kimberling_
%E More terms from _James A. Sellers_, May 03 2000
|
