This is a front-end for the Online Encyclopedia of Integer Sequences, made by Christian Perfect. The idea is to provide OEIS entries in non-ancient HTML, and then to think about how they're presented visually. The source code is on GitHub.
%I A106521 #52 Jul 09 2023 11:47:44
%S A106521 18,38,456,854,9192,17132,183474,341876,3660378,6820478,73024176,
%T A106521 136067774,1456823232,2714535092,29063440554,54154634156,579811987938,
%U A106521 1080378148118,11567176318296,21553408328294
%N A106521 Numbers m such that Sum_{k=0..10} (m+k)^2 is a square.
%C A106521 Equivalently, 11*a(n)^2 + 110*a(n) + 385 is a square.
%C A106521 11*((m+5)^2+10) is a square iff the second factor is divisible by 11 and the quotient is a square, i.e., iff m = 11*k - 4 or m = 11*k - 6 and 11*k^2 +- 2 k + 1 is a square. Thus a(n) == (7,5,5,7,7,5,5,7,...) (mod 11), repeating with period 4 and the values are obtained by solving these Pell-type equations (cf. link to Dario Alpern's quadratic solver). The corresponding recurrence equations (see PARI code) should make it possible to prove the conjectured g.f. - _M. F. Hasler_, Jan 27 2008
%C A106521 All sequences of this type (i.e., sequences with fixed offset k, and a discernible pattern: k=0...10 for this sequence, 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 A106521 Vincenzo Librandi, <a href="/A106521/b106521.txt">Table of n, a(n) for n = 1..500</a>
%H A106521 Dario Alpern, <a href="https://www.alpertron.com.ar/QUAD.HTM">Quadratic two integer variable equation solver</a>
%H A106521 <a href="/index/Rec#order_05">Index entries for linear recurrences with constant coefficients</a>, signature (1,20,-20,-1,1).
%F A106521 G.f.: 2*x*(9+10*x+29*x^2-x^3-2*x^4)/(1-x)/(1-20*x^2+x^4). - _Vladeta Jovovic_, May 31 2005; adapted to the offset by _Bruno Berselli_, May 16 2011
%F A106521 a(1)=18, a(2)=38, a(3)=456, a(4)=854, a(5)=9192; thereafter a(n)=a(n-1)+20*a(n-2)- 20*a(n-3)-a(n-4)+a(n-5). - _Harvey P. Dale_, May 07 2011
%F A106521 a(n) = A198949(n+1)-5. - _Bruno Berselli_, Feb 12 2012
%F A106521 a(1)=18, a(2)=38, a(3)=456, a(4)=854; thereafter a(n) = 20*a(n-2) - a(n-4) + 90. - _Daniel Mondot_, Aug 05 2016
%e A106521 Since 18^2 + 19^2 + ... + 28^2 = 5929 = 77^2, 18 is in the sequence. - _Michael B. Porter_, Aug 07 2016
%t A106521 LinearRecurrence[{1,20,-20,-1,1},{18,38,456,854,9192},30] (* _Harvey P. Dale_, May 07 2011 *)
%o A106521 (PARI) A106521(n)={local(xy=[ -4-2*(n%2);11],PQRS=[10,3;33,10],KL=[45;165]);until(0>=n-=2,xy=PQRS*xy+KL);xy[1]} \\ _M. F. Hasler_, Jan 27 2008
%Y A106521 Cf. A001032, A094196.
%K A106521 nonn,easy
%O A106521 1,1
%A A106521 _Ralf Stephan_, May 30 2005
%E A106521 Edited and extended by _M. F. Hasler_, Jan 27 2008