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A269447 The first of 23 consecutive positive integers the sum of the squares of which is a square.

%I A269447 #29 May 21 2024 17:21:26
%S A269447 7,17,881,1351,42787,65337,2053401,3135331,98520967,150431057,
%T A269447 4726953521,7217555911,226795248547,346292253177,10881444977241,
%U A269447 16614810597091,522082563659527,797164616407697,25049081610680561,38247286776972871,1201833834749007907
%N A269447 The first of 23 consecutive positive integers the sum of the squares of which is a square.
%C A269447 Positive integers y in the solutions to 2*x^2-46*y^2-1012*y-7590 = 0.
%C A269447 All sequences of this type (i.e. sequences with fixed offset k, and a discernible pattern: k=0...22 for this sequence, k=0..1 for A001652, k=0...10 for A106521) 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 A269447 Colin Barker, <a href="/A269447/b269447.txt">Table of n, a(n) for n = 1..1000</a>
%H A269447 <a href="/index/Rec#order_05">Index entries for linear recurrences with constant coefficients</a>, signature (1,48,-48,-1,1).
%F A269447 a(n) = a(n-1)+48*a(n-2)-48*a(n-3)-a(n-4)+a(n-5) for n>5.
%F A269447 G.f.: x*(7+10*x+528*x^2-10*x^3-29*x^4) / ((1-x)*(1-48*x^2+x^4)).
%F A269447 a(1)=7, a(2)=17, a(3)=881, a(4)=1351, a(n) = 48*a(n-2)-a(n-4)+506. - _Daniel Mondot_, Aug 05 2016
%e A269447 7 is in the sequence because sum(k=7, 29, k^2) = 8464 = 92^2.
%t A269447 LinearRecurrence[{1,48,-48,-1,1},{7,17,881,1351,42787},30] (* _Harvey P. Dale_, May 21 2024 *)
%o A269447 (PARI) Vec(x*(7+10*x+528*x^2-10*x^3-29*x^4)/((1-x)*(1-48*x^2+x^4)) + O(x^30))
%Y A269447 Cf. A001032, A001652, A094196, A106521, A257761, A269448, A269449, A269451.
%K A269447 nonn,easy
%O A269447 1,1
%A A269447 _Colin Barker_, Feb 27 2016