%I #26 Aug 30 2016 05:23:13

%S 25,301,454,3850,31966,47569,393925,3261481,4852834,40177750,

%T 332640346,494942749,4097737825,33926055061,50479308814,417929081650,

%U 3460124977126,5148394557529,42624668591725,352898821613041,525085765560394,4347298267275550

%N The first of 26 consecutive positive integers the sum of the squares of which is a square.

%C Positive integers y in the solutions to 2*x^2-52*y^2-1300*y-11050 = 0.

%C All sequences of this type (i.e. sequences with fixed offset k, and a discernible pattern: k=0...25 for this sequence, k=0...22 for A269447, k=0..1 for A001652) can be continued using a formula such as x(n) = a*x(n-p) - x(n-2p) + b, where a and b are various constants, and p is the period of the series. Alternatively 'p' can be considered the number of concurrent series. - _Daniel Mondot_, Aug 05 2016

%H Colin Barker, <a href="/A269448/b269448.txt">Table of n, a(n) for n = 1..1000</a>

%H <a href="/index/Rec#order_07">Index entries for linear recurrences with constant coefficients</a>, signature (1,0,102,-102,0,-1,1).

%F G.f.: x*(25+276*x+153*x^2+846*x^3-36*x^4-3*x^5-11*x^6) / ((1-x)*(1-102*x^3+x^6)).

%F a(1)=25, a(2)=301, a(3)=454, a(4)=3850, a(5)=31966, a(6)=47569, a(n)=102*a(n-3) - a(n-6) + 1250. - _Daniel Mondot_, Aug 05 2016

%e 25 is in the sequence because sum(k=25, 50, k^2) = 38025 = 195^2.

%t Rest@ CoefficientList[Series[x (25 + 276 x + 153 x^2 + 846 x^3 - 36 x^4 - 3 x^5 - 11 x^6)/((1 - x) (1 - 102 x^3 + x^6)), {x, 0, 22}], x] (* _Michael De Vlieger_, Aug 07 2016 *)

%o (PARI) Vec(x*(25+276*x+153*x^2+846*x^3-36*x^4-3*x^5-11*x^6)/((1-x)*(1-102*x^3+x^6)) + O(x^30))

%Y Cf. A001032, A001652, A094196, A106521, A257765, A269447, A269449, A269451.

%K nonn,easy

%O 1,1

%A _Colin Barker_, Feb 27 2016