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A290168
If n is even then a(n) = n^2*(n+2)/8, otherwise a(n) = (n-1)*n*(n+1)/8.
1
0, 0, 2, 3, 12, 15, 36, 42, 80, 90, 150, 165, 252, 273, 392, 420, 576, 612, 810, 855, 1100, 1155, 1452, 1518, 1872, 1950, 2366, 2457, 2940, 3045, 3600, 3720, 4352, 4488, 5202, 5355, 6156, 6327, 7220, 7410, 8400
OFFSET
0,3
COMMENTS
Bisection of a(n) [0, 2, 12, 36, 80, 150, 252, ...] is A011379.
Bisection [0, 3, 15, 42, 90, 165, 273, ...] is A059270.
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.
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).
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).
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.
FORMULA
G.f.: x^2*(2 + x + 3*x^2)/((x-1)^4*(x+1)^3).
a(n) = (1/16)*(-1)^n*n*(1 + (-1)^(n+1) + 2*(1 + (-1)^n)*n + 2*(-1)^n*n^2).
Sum_{n>=2} 1/a(n) = 5 + Pi^2/6 - 8*log(2). - Amiram Eldar, Sep 17 2022
MATHEMATICA
a[n_] := If[EvenQ[n], n^2*(n + 2)/8, (n - 1)*n*(n + 1)/8]; Table[a[n], {n, 0, 40}]
PROG
(PARI) a(n) = if(n%2==0, n^2*(n+2)/8, (n-1)*n*(n+1)/8) \\ Felix Fröhlich, Jul 23 2017
KEYWORD
nonn
AUTHOR
STATUS
approved