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a(3*n) = 1 + 4*n^2, a(1+3*n) = 2 + 4*n*(n+1), a(2+3*n) = 5 + 4*n*(n+1).
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%I #25 Jan 19 2022 13:51:47

%S 1,2,5,5,10,13,17,26,29,37,50,53,65,82,85,101,122,125,145,170,173,197,

%T 226,229,257,290,293,325,362,365,401,442,445,485,530,533,577,626,629,

%U 677,730,733,785,842,845,901,962

%N a(3*n) = 1 + 4*n^2, a(1+3*n) = 2 + 4*n*(n+1), a(2+3*n) = 5 + 4*n*(n+1).

%C A261327 sorted in nondecreasing order.

%H <a href="/index/Rec#order_07">Index entries for linear recurrences with constant coefficients</a>, signature (1,0,2,-2,0,-1,1).

%F a(-n) = a(n) - A099838(n+2).

%F a(n) = a(n-3) + 4*A004523(n-1) for n >= 3

%F = a(n-6) + 8*A004396(n-3) for n >= 6

%F = a(n-9) + 12*A004523(n-4) for n >= 9

%F = a(n-12) + 16*A004396(n-6) for n >= 12

%F ...

%t nterms=100;LinearRecurrence[{1,0,2,-2,0,-1,1},{1,2,5,5,10,13,17},nterms] (* _Paolo Xausa_, Dec 01 2021 *)

%Y Cf. A261327.

%Y Trisections: A053755, A069894, A078370.

%Y Cf. A004396, A004523, A099838.

%K nonn,easy

%O 0,2

%A _Paul Curtz_, Dec 01 2021