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The first of seven consecutive positive integers the sum of the squares of which is equal to the sum of the squares of six consecutive positive integers.
2

%I #13 Sep 08 2022 08:46:13

%S 78,2100,54594,1417416,36798294,955338300,24801997578,643896598800,

%T 16716509571294,433985352254916,11266902649056594,292505483523216600,

%U 7593875668954575078,197148261909295735500,5118260933972734547994,132877636021381802512416

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

%C For the first of the corresponding six consecutive positive integers, see A262062.

%H Colin Barker, <a href="/A262063/b262063.txt">Table of n, a(n) for n = 1..706</a>

%H <a href="/index/Rec#order_03">Index entries for linear recurrences with constant coefficients</a>, signature (27,-27,1).

%F a(n) = 27*a(n-1)-27*a(n-2)+a(n-3) for n>3.

%F G.f.: 6*x*(x-13) / ((x-1)*(x^2-26*x+1)).

%F a(n) = (13+2*sqrt(42))^(-n)*(6-sqrt(42)+(6+sqrt(42))*(13+2*sqrt(42))^(2*n))/4-3. - _Colin Barker_, Mar 03 2016

%e 78 is in the sequence because 78^2 + ... + 84^2 = 45955 = 85^2 + ... + 90^2.

%t CoefficientList[Series[6 (x - 13)/((x - 1) (x^2 - 26 x + 1)), {x, 0, 20}], x] (* _Vincenzo Librandi_, Sep 10 2015 *)

%t LinearRecurrence[{27,-27,1},{78,2100,54594},30] (* _Harvey P. Dale_, May 17 2018 *)

%o (PARI) Vec(6*x*(x-13)/((x-1)*(x^2-26*x+1)) + O(x^20))

%o (Magma) I:=[78,2100,54594]; [n le 3 select I[n] else 27*Self(n-1)-27*Self(n-2)+Self(n-3): n in [1..20]]; // _Vincenzo Librandi_, Sep 10 2015

%Y Cf. A262062.

%K nonn,easy

%O 1,1

%A _Colin Barker_, Sep 09 2015