|
%I #22 Sep 17 2022 03:45:22
%S 0,0,2,3,12,15,36,42,80,90,150,165,252,273,392,420,576,612,810,855,
%T 1100,1155,1452,1518,1872,1950,2366,2457,2940,3045,3600,3720,4352,
%U 4488,5202,5355,6156,6327,7220,7410,8400
%N If n is even then a(n) = n^2*(n+2)/8, otherwise a(n) = (n-1)*n*(n+1)/8.
%C Bisection of a(n) [0, 2, 12, 36, 80, 150, 252, ...] is A011379.
%C Bisection [0, 3, 15, 42, 90, 165, 273, ...] is A059270.
%C Considering s(n) = [0, 0, 0, 0, 1, 1, 3, 3, 6, 6, 10, 10, 15, 15, ...] (triangular numbers repeated - see A008805), a(n) = n*s(n+2) holds.
%C Considering the first differences of a(n), b(n) = [0, 2, 1 , 9, 3, 21, 6, 38, 10, 60, 15, 87, ...], b(n) shows bisections A000217 and A005476. In addition, b(n) begins like A249264 up to 12th term, and is an alternation of 4 multiples of 3 and 2 not multiples; b(n) is also such that b(2n) + b(2n+1) = A049450(n).
%C Considering the second differences c(n), c(n) shows bisections A001105(n+1) and -A000384(n+1), c(n) has 3 consecutive terms multiples of 3 alternating with 3 not multiples; in addition, c(2n) + c(2n+1) = A000027(n).
%C Considering a(n)/c(n) = [0, 0, 1/4, -1/2, 2/3, -1, 9/8, -3/2, 8/5, -2, 25/12, -5/2, ...], it appears that it is A129194(n)/A022998(n+1) and -A026741(n)/A000034(n) alternating.
%H <a href="/index/Rec#order_07">Index entries for linear recurrences with constant coefficients</a>, signature (1,3,-3,-3,3,1,-1).
%F G.f.: x^2*(2 + x + 3*x^2)/((x-1)^4*(x+1)^3).
%F a(n) = (1/16)*(-1)^n*n*(1 + (-1)^(n+1) + 2*(1 + (-1)^n)*n + 2*(-1)^n*n^2).
%F Sum_{n>=2} 1/a(n) = 5 + Pi^2/6 - 8*log(2). - _Amiram Eldar_, Sep 17 2022
%t a[n_] := If[EvenQ[n], n^2*(n + 2)/8, (n - 1)*n*(n + 1)/8]; Table[a[n], {n, 0, 40}]
%o (PARI) a(n) = if(n%2==0, n^2*(n+2)/8, (n-1)*n*(n+1)/8) \\ _Felix Fröhlich_, Jul 23 2017
%Y Cf. A000027, A000034, A000217, A000384, A001105, A005476, A008805, A011379, A022998, A026741, A049450, A059270, A129194, A135713, A161680, A249264.
%K nonn
%O 0,3
%A _Jean-François Alcover_ and _Paul Curtz_, Jul 23 2017
|
