A288994 a(n) = n*(n+3) when n is congruent to 0 or 3 (mod 4), and n*(n+3)/2 otherwise.
0, 2, 5, 18, 28, 20, 27, 70, 88, 54, 65, 154, 180, 104, 119, 270, 304, 170, 189, 418, 460, 252, 275, 598, 648, 350, 377, 810, 868, 464, 495, 1054, 1120, 594, 629, 1330, 1404, 740, 779, 1638, 1720, 902, 945, 1978, 2068, 1080, 1127, 2350, 2448, 1274, 1325
Offset: 0
Links
- Colin Barker, Table of n, a(n) for n = 0..1000
- Index entries for linear recurrences with constant coefficients, signature (3,-6,10,-12,12,-10,6,-3,1).
Programs
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Mathematica
a[n_] := n (n+3) Switch[Mod[n, 4], 0|3, 1, _, 1/2]; Table[a[n], {n, 0, 50}] Table[If[MemberQ[{0,3},Mod[n,4]],n(n+3),(n(n+3))/2],{n,0,50}] (* or *) LinearRecurrence[{3,-6,10,-12,12,-10,6,-3,1},{0,2,5,18,28,20,27,70,88},60] (* Harvey P. Dale, Jun 05 2021 *)
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PARI
concat(0, Vec(x*(2 - x + 15*x^2 - 16*x^3 + 18*x^4 - 9*x^5 + 5*x^6 - 2*x^7) / ((1 - x)^3*(1 + x^2)^3) + O(x^60))) \\ Colin Barker, Jun 21 2017
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PARI
i=I; a(n) = (1/8 + i/8)*(((3 - 3*i) - i*(-i)^n + i^n)*n*(3 + n)) \\ Colin Barker, Jun 21 2017
Formula
a(n) = n*(n+3)/2 * (2 - floor((n+1)/2) mod 2), where n*(n+3)/2 is A000096(n).
a(n) = (2*n*(n+3))/(GCD(4, n+2)*GCD(4, n+3)).
a(n) = A227316(n+1) - (period 4 repeat 2,1,1,2).
From Colin Barker, Jun 21 2017: (Start)
G.f.: x*(2 - x + 15*x^2 - 16*x^3 + 18*x^4 - 9*x^5 + 5*x^6 - 2*x^7) / ((1 - x)^3*(1 + x^2)^3).
a(n) = (1/8 + i/8)*(((3 - 3*i) - i*(-i)^n + i^n)*n*(3 + n)), where i=sqrt(-1). (End)
Sum_{n>=1} 1/a(n) = 17/18 + log(2)/6. - Amiram Eldar, Aug 12 2022