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%I #37 Feb 05 2021 09:42:39
%S 1,11,41,91,161,251,361,491,641,811,1001,1211,1441,1691,1961,2251,
%T 2561,2891,3241,3611,4001,4411,4841,5291,5761,6251,6761,7291,7841,
%U 8411,9001,9611,10241,10891,11561,12251,12961,13691,14441,15211,16001,16811,17641
%N a(n) = 10*n^2 + 1.
%C Sequence found by reading the segment (1, 11) together with the line from 11, in the direction 11, 41, ..., in the square spiral whose vertices are the generalized heptagonal numbers A085787. - _Omar E. Pol_, Sep 10 2011
%C The identity (10n^2 + 1)^2 - (25n^2 + 5)*(2n)^2 = 1 can be written as a(n)^2 - A158445(n)*A005843(n)^2 = 1. - _Vincenzo Librandi_, Jan 03 2012
%H Vincenzo Librandi, <a href="/A158187/b158187.txt">Table of n, a(n) for n = 0..10000</a>
%H <a href="/index/Rec#order_03">Index entries for linear recurrences with constant coefficients</a>, signature (3,-3,1).
%F a(n) = A033583(n) + 1.
%F For n > 0: a(n) = A010010(n)/2.
%F From _Vincenzo Librandi_, Jan 03 2012: (Start)
%F G.f: x*(11 + 8*x + x^2)/(1-x)^3.
%F a(n) = a(-n) = 3*a(n-1) - 3*a(n-2) + a(n-3). (End)
%F From _Amiram Eldar_, Jul 15 2020: (Start)
%F Sum_{n>=0} 1/a(n) = (1 + (Pi/sqrt(10))*coth(Pi/sqrt(10)))/2.
%F Sum_{n>=0} (-1)^n/a(n) = (1 + (Pi/sqrt(10))*csch(Pi/sqrt(10)))/2. (End)
%F From _Amiram Eldar_, Feb 05 2021: (Start)
%F Product_{n>=0} (1 + 1/a(n)) = sqrt(2)*csch(Pi/sqrt(10))*sinh(Pi/sqrt(5)).
%F Product_{n>=1} (1 - 1/a(n)) = (Pi/sqrt(10))*csch(Pi/sqrt(10)). (End)
%F E.g.f.: exp(x)*(1 + 10*x + 10*x^2). - _Stefano Spezia_, Feb 05 2021
%t Table[10*n^2+1,{n,0,50}] (* _Vincenzo Librandi_, Jan 03 2012 *)
%o (PARI) a(n)=10*n^2+1 \\ _Charles R Greathouse IV_, Oct 16 2015
%Y Cf. A158445, A005843. - _Vincenzo Librandi_, Mar 19 2009
%K nonn,easy
%O 0,2
%A _Reinhard Zumkeller_, Mar 13 2009
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