|
| |
| |
|
|
|
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
(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
|
Vincenzo Librandi, Table of n, a(n) for n = 0..800
Index entries for linear recurrences with constant coefficients, signature (3,-3,1).
|
|
|
FORMULA
|
a(n) = 25*n^2 = 25*A000290(n) = 5*A033429(n). - Omar E. Pol, Jul 03 2014
|
|
|
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
|
Similar sequences listed in A244630.
Sequence in context: A237202 A198385 A134422 * A309779 A221274 A042220
Adjacent sequences: A016847 A016848 A016849 * A016851 A016852 A016853
|
|
|
KEYWORD
|
nonn,easy
|
|
|
AUTHOR
|
N. J. A. Sloane
|
|
|
STATUS
|
approved
|
| |
|
|
