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The first of two consecutive positive integers the sum of the squares of which is equal to the sum of the squares of ten consecutive positive integers.
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%I #10 Sep 07 2015 05:52:53

%S 26,48,68,126,468,866,1226,2268,8406,15548,22008,40706,150848,279006,

%T 394926,730448,2706866,5006568,7086668,13107366,48572748,89839226,

%U 127165106,235202148,871602606,1612099508,2281885248,4220531306,15640274168,28927951926

%N The first of two consecutive positive integers the sum of the squares of which is equal to the sum of the squares of ten consecutive positive integers.

%C For the first of the corresponding ten consecutive positive integers, see A261934.

%H Colin Barker, <a href="/A261932/b261932.txt">Table of n, a(n) for n = 1..1000</a>

%H <a href="/index/Rec#order_09">Index entries for linear recurrences with constant coefficients</a>, signature (1,0,0,18,-18,0,0,-1,1).

%F G.f.: -2*x*(4*x^8-x^7+x^5-63*x^4+29*x^3+10*x^2+11*x+13) / ((x-1)*(x^4-4*x^2-1)*(x^4+4*x^2-1)).

%F a(n) = a(n-1) + 18*a(n-4) - 18*a(n-5) - a(n-8) + a(n-9) for n>8. - _Vincenzo Librandi_, Sep 07 2015

%e 26 is in the sequence because 26^2 + 27^2 = 7^2 + 8^2 + ... + 16^2.

%t CoefficientList[Series[2 (4 x^8 - x^7 + x^5 - 63 x^4 + 29 x^3 + 10 x^2 + 11 x + 13)/((1 - x) (x^4 - 4 x^2 - 1) (x^4 + 4 x^2 - 1)), {x, 0, 45}], x] (* _Vincenzo Librandi_, Sep 07 2015 *)

%o (PARI) Vec(-2*x*(4*x^8-x^7+x^5-63*x^4+29*x^3+10*x^2+11*x+13)/((x-1)*(x^4-4*x^2-1)*(x^4+4*x^2-1)) + O(x^40))

%Y Cf. A001652, A031138, A261933, A261934, A261935.

%K nonn,easy

%O 1,1

%A _Colin Barker_, Sep 06 2015