OFFSET
1,5
COMMENTS
There is a formula, but no closed form, for computing the entries of the sequence.
REFERENCES
N. R. Baeth, L. Luther and R. McKee, Variations on a Putnam Problem, preprint, 2016.
FORMULA
a(n) = (2*floor(n*(sqrt(2)-1)/2)+1)^2+4*Sum_{i=ceiling(-n*(sqrt(2)-1)/2)..floor(n*(sqrt(2)-1)/2)} ceiling(n/4-i^2/n)-1-floor(n*(sqrt(2)-1)/2).
MAPLE
A275485:=n->(2*floor(n*(sqrt(2)-1)/2)+1)^2+4*add(ceil(n/4-i^2/n)-1-floor(n*(sqrt(2)-1)/2), i=ceil(-n*(sqrt(2)-1)/2)..floor(n*(sqrt(2)-1)/2)): seq(A275485(n), n=1..100); # Wesley Ivan Hurt, Sep 27 2016
PROG
(PARI) a(n)=(2*floor(n*(sqrt(2)-1)/2)+1)^2+4*sum(i=ceil(-n*(sqrt(2)-1)/2), floor(n*(sqrt(2)-1)/2), ceil(n/4-i^2/n)-1-floor(n*(sqrt(2)-1)/2)); \\ Joerg Arndt, Sep 27 2016
CROSSREFS
KEYWORD
nonn
AUTHOR
Nicholas Baeth, Sep 26 2016
STATUS
approved