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%I #17 May 20 2019 07:08:25
%S 0,1025,476073,27022500,551536100,6129324225,45502479225,253405810448,
%T 1137920432400,4322847530025,14366776735025,42801847892100,
%U 116415023802948,293153032943225,691043521403025,1538402208782400
%N Integers of the form (k^12 - k^8 - k^4 + 1)/512.
%H B. Berselli, <a href="/A180270/b180270.txt">Table of n, a(n) for n = 1..100000</a>. [From _Bruno Berselli_, Sep 21 2010]
%F Integers of the form (k^4+1)*( (k-1)*(k+1)*(k^2+1) )^2/512.
%F a(n) = ((2*n-1)^4+1)*((n-1)*n*(n^2+(n-1)^2))^2/8 (for k=2n-1).
%F a(n) = A175110(n-1)*(A001844(n-1)*A000217(n-1))^2. - _Bruno Berselli_, Sep 21 2010
%e a(2) = 1025 is in the sequence because (3^12 - 3^8 - 3^4 + 1)/512 = 524800/512 = 1025.
%p for n from 1 by 2 to 60 do: x:= (n^12-n^8 -n^4+1)/512: printf(`%d, `, x):od: # incomplete program which also prints rationals, _R. J. Mathar_
%t Select[Table[(k^12-k^8-k^4+1)/512,{k,40}],IntegerQ] (* _Harvey P. Dale_, Jan 23 2011 *)
%K nonn,easy
%O 1,2
%A _Michel Lagneau_, Aug 23 2010
%E Comment converted to formula by _R. J. Mathar_, Aug 25 2010
%E Example corrected and general term of the sequence rewritten by _Bruno Berselli_, Sep 22 2010
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