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%I #25 Sep 08 2022 08:45:54
%S 3,9,41,59,123,153,249,291,419,473,633,699,891,969,1193,1283,1539,
%T 1641,1929,2043,2363,2489,2841,2979,3363,3513,3929,4091,4539,4713,
%U 5193,5379,5891,6089,6633,6843,7419,7641,8249,8483,9123,9369,10041,10299,11003,11273
%N Positive integers of the form (2*m^2+1)/11.
%C Here m = (11*(2*n - 1) - 5*(-1)^n)/4 for n>0.
%C All terms of A113338 are in the sequence.
%H Bruno Berselli, <a href="/A179088/b179088.txt">Table of n, a(n) for n = 1..1000</a>
%H <a href="/index/Rec#order_05">Index entries for linear recurrences with constant coefficients</a>, signature (1,2,-2,-1,1).
%F a(n) = (22*n*(n - 1) - 5*(2*n - 1)*(-1)^n + 7)/4.
%F G.f.: x*(3 + 6*x + 26*x^2 + 6*x^3 + 3*x^4)/((1 + x)^2*(1 - x)^3).
%F a(n) = a(-n+1) = a(n-1) + 2*a(n-2) - 2*a(n-3) - a(n-4) + a(n-5).
%F a(2*n)+a(2*n+1)= 11*A000290(2*n)+6; a(2*n-1)+a(2*n) = 11*A000290(2*n-1)+1.
%p t1:=[]; for n from 0 to 1000 do i:=2*n^2+1; if i mod 11 = 0 then t1:=[op(t1),i/11]; fi; od: # _N. J. A. Sloane_, Jul 05 2010
%t Select[Table[(2m^2 + 1)/11, {m, 0, 300}], IntegerQ] (* _Harvey P. Dale_, Dec 17 2010 *)
%o (Magma) [i/11: m in [1..250] | i mod 11 eq 0 where i is 2*m^2+1]; // _Bruno Berselli_, Jun 26 2011
%o (PARI) for(m=1, 250, i=2*m^2+1; if(Mod(i,11)==0, print1(i/11", "))); \\ _Bruno Berselli_, Jun 26 2011
%Y Cf. A113338, A179337-A179370.
%K nonn,easy
%O 1,1
%A _Bruno Berselli_, Jun 29 2010 - Jul 09 2010
%E Edited by _N. J. A. Sloane_, Jul 05 2010: offset changed to 1 (since this is a list), and terms verified.
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