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
Vincenzo Librandi, Table of n, a(n) for n = 0..800
Index entries for linear recurrences with constant coefficients, signature (3,-3,1).
FORMULA
From Amiram Eldar, Jan 25 2021: (Start)
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)
a(n) = Sum_{i=0..n-1} A053742(i). - John Elias, Jun 30 2021
G.f.: 25*x*(1 + x)/(1 - x)^3. - Stefano Spezia, Jul 08 2023
MATHEMATICA
(5Range[0, 31])^2 (* Alonso del Arte, Oct 08 2017 *)
PROG
(Magma) [(5*n)^2: n in [0..50]]; // Vincenzo Librandi, Apr 26 2011
(PARI) a(n)=(5*n)^2 \\ Charles R Greathouse IV, Jun 17 2017
CROSSREFS
KEYWORD
nonn,easy
AUTHOR
STATUS
approved