|
|
A016850
|
|
a(n) = (5*n)^2.
|
|
11
|
|
|
0, 25, 100, 225, 400, 625, 900, 1225, 1600, 2025, 2500, 3025, 3600, 4225, 4900, 5625, 6400, 7225, 8100, 9025, 10000, 11025, 12100, 13225, 14400, 15625, 16900, 18225, 19600, 21025, 22500, 24025, 25600, 27225, 28900, 30625, 32400, 34225, 36100, 38025, 40000
(list;
graph;
refs;
listen;
history;
text;
internal format)
|
|
|
OFFSET
|
0,2
|
|
COMMENTS
|
If we define C(n) = (5*n)^2 (n > 0), the sequence is the first "square-sequence" such that for every n there exists p such that C(n) = C(p) + C(p+n). We observe in fact that p = 3n because 25 = 3^2 + 4^2. The sequence without 0 is linked with the first nontrivial solution (trivial: n^2 = 0^2 + n^2) of the equation X^2 = 2Y^2 + 2n^2 where X = 2*k and Y = 2*p + n which is equivalent to k^2 = p^2 + (p+n)^2 for n given. The second such "square-sequence" is (29*n)^2 (n > 0) because 29^2 = 20^2 + 21^2 and with this relation we obtain (29*n)^2 = (20*n)^2 + (20n+n)^2. - Richard Choulet, Dec 23 2007
|
|
LINKS
|
|
|
FORMULA
|
Sum_{n>=1} 1/a(n) = Pi^2/150.
Sum_{n>=1} (-1)^(n+1)/a(n) = Pi^2/300.
Product_{n>=1} (1 + 1/a(n)) = sinh(Pi/5)/(Pi/5).
Product_{n>=1} (1 - 1/a(n)) = sin(Pi/5)/(Pi/5) = 5*sqrt((5-sqrt(5))/2)/(2*Pi). (End)
|
|
MATHEMATICA
|
|
|
PROG
|
|
|
CROSSREFS
|
Similar sequences listed in A244630.
|
|
KEYWORD
|
nonn,easy
|
|
AUTHOR
|
|
|
STATUS
|
approved
|
|
|
|