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%I #65 Jul 08 2023 21:01:17
%S 0,25,100,225,400,625,900,1225,1600,2025,2500,3025,3600,4225,4900,
%T 5625,6400,7225,8100,9025,10000,11025,12100,13225,14400,15625,16900,
%U 18225,19600,21025,22500,24025,25600,27225,28900,30625,32400,34225,36100,38025,40000
%N a(n) = (5*n)^2.
%C 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
%H Vincenzo Librandi, <a href="/A016850/b016850.txt">Table of n, a(n) for n = 0..800</a>
%H <a href="/index/Rec#order_03">Index entries for linear recurrences with constant coefficients</a>, signature (3,-3,1).
%F a(n) = 25*n^2 = 25*A000290(n) = 5*A033429(n). - _Omar E. Pol_, Jul 03 2014
%F From _Amiram Eldar_, Jan 25 2021: (Start)
%F Sum_{n>=1} 1/a(n) = Pi^2/150.
%F Sum_{n>=1} (-1)^(n+1)/a(n) = Pi^2/300.
%F Product_{n>=1} (1 + 1/a(n)) = sinh(Pi/5)/(Pi/5).
%F Product_{n>=1} (1 - 1/a(n)) = sin(Pi/5)/(Pi/5) = 5*sqrt((5-sqrt(5))/2)/(2*Pi). (End)
%F a(n) = Sum_{i=0..n-1} A053742(i). - _John Elias_, Jun 30 2021
%F G.f.: 25*x*(1 + x)/(1 - x)^3. - _Stefano Spezia_, Jul 08 2023
%t (5Range[0, 31])^2 (* _Alonso del Arte_, Oct 08 2017 *)
%o (Magma) [(5*n)^2: n in [0..50]]; // _Vincenzo Librandi_, Apr 26 2011
%o (PARI) a(n)=(5*n)^2 \\ _Charles R Greathouse IV_, Jun 17 2017
%Y Cf. A000290, A033429, A053742 (first differences).
%Y Similar sequences listed in A244630.
%K nonn,easy
%O 0,2
%A _N. J. A. Sloane_
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