%I
%S 18,38,456,854,9192,17132,183474,341876,3660378,6820478,73024176,
%T 136067774,1456823232,2714535092,29063440554,54154634156,579811987938,
%U 1080378148118,11567176318296,21553408328294
%N Numbers a(n) such that Sum[k=0..10, (a(n)+k)^2 ] is square.
%C Equivalently, 11*a(n)^2 + 110*a(n) + 385 is square.
%C 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 Pelltype equations, e.g. using http://www.alpertron.com.ar/QUAD.HTM. The corresponding recurrence equations (see PARI code) should make it possible to prove the conjectured g.f.  _M. F. Hasler_, Jan 27 2008
%C 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(np)  x(n2p) + 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 Vincenzo Librandi, <a href="/A106521/b106521.txt">Table of n, a(n) for n = 1..500</a>
%H <a href="/index/Rec#order_05">Index entries for linear recurrences with constant coefficients</a>, signature (1,20,20,1,1).
%F G.f.: 2*x*(9+10*x+29*x^2x^32*x^4)/(1x)/(120*x^2+x^4).  _Vladeta Jovovic_, May 31 2005
%F a(1)=18, a(2)=38, a(3)=456, a(4)=854, a(5)=9192; thereafter a(n)=a(n1)+20*a(n2) 20*a(n3)a(n4)+a(n5).  _Harvey P. Dale_, May 07 2011
%F a(n) = A198949(n+1)5.  _Bruno Berselli_, Feb 12 2012
%F a(1)=18, a(2)=38, a(3)=456, a(4)=854; thereafter a(n) = 20*a(n2)  a(n4) + 90.  _Daniel Mondot_, Aug 05 2016
%e Since 18^2 + 19^2 + ... + 28^2 = 5929 = 77^2, 18 is in the sequence.  _Michael B. Porter_, Aug 07 2016
%t LinearRecurrence[{1,20,20,1,1},{18,38,456,854,9192},30] (* _Harvey P. Dale_, May 07 2011 *)
%o (PARI) A106521(n)={local(xy=[ 42*(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 Cf. A001032, A094196.
%K nonn,easy
%O 1,1
%A _Ralf Stephan_, May 30 2005
%E Edited and extended by _M. F. Hasler_, Jan 27 2008
%E G.f. adapted to the offset by _Bruno Berselli_, May 16 2011
