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%I #7 Aug 05 2021 08:48:31
%S 1,1,1,5,10,17,26,37,50,65,82,101,122,145,170,197,226,257,290,325,362,
%T 401,442,485,530,577,626,677,730,785,842,901,962,1025,1090,1157,1226,
%U 1297,1370,1445,1522,1601,1682,1765,1850,1937,2026,2117,2210,2305,2402
%N Let f[n_]=((n^4-n^3-1)/ (n^3-n-1))^2; then a(n) = Floor[f[n]].
%F It is easy to show that Floor[((n^4-n^3-1)/ (n^3-n-1))^2] = (n-1)^2 + 1 for n >= 3. So this is essentially the same sequence as A002522. - Juan Jose Alba Gonzalez, Nov 09 2006.
%t (* polynomial sequence with Theta1 to Theta0 pattern*) digits=200 f[n_]=((n^4-n^3-1)/ (n^3-n-1))^2 a=Table[Floor[f[n]], {n, 0, digits}]
%Y Cf. A002522.
%K nonn,easy,less
%O 0,4
%A _Roger L. Bagula_, Oct 01 2004