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A180270 Integers of the form (k^12 - k^8 - k^4 + 1)/512. 1
0, 1025, 476073, 27022500, 551536100, 6129324225, 45502479225, 253405810448, 1137920432400, 4322847530025, 14366776735025, 42801847892100, 116415023802948, 293153032943225, 691043521403025, 1538402208782400 (list; graph; refs; listen; history; text; internal format)
OFFSET
1,2
LINKS
B. Berselli, Table of n, a(n) for n = 1..100000. [From Bruno Berselli, Sep 21 2010]
FORMULA
Integers of the form (k^4+1)*( (k-1)*(k+1)*(k^2+1) )^2/512.
a(n) = ((2*n-1)^4+1)*((n-1)*n*(n^2+(n-1)^2))^2/8 (for k=2n-1).
a(n) = A175110(n-1)*(A001844(n-1)*A000217(n-1))^2. - Bruno Berselli, Sep 21 2010
EXAMPLE
a(2) = 1025 is in the sequence because (3^12 - 3^8 - 3^4 + 1)/512 = 524800/512 = 1025.
MAPLE
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
MATHEMATICA
Select[Table[(k^12-k^8-k^4+1)/512, {k, 40}], IntegerQ] (* Harvey P. Dale, Jan 23 2011 *)
CROSSREFS
Sequence in context: A279643 A168119 A272672 * A103716 A291508 A031530
KEYWORD
nonn,easy
AUTHOR
Michel Lagneau, Aug 23 2010
EXTENSIONS
Comment converted to formula by R. J. Mathar, Aug 25 2010
Example corrected and general term of the sequence rewritten by Bruno Berselli, Sep 22 2010
STATUS
approved

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Last modified March 28 22:04 EDT 2024. Contains 371254 sequences. (Running on oeis4.)