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%I #46 Sep 05 2022 09:11:06
%S 1,13,15,27,29,41,43,55,57,69,71,83,85,97,99,111,113,125,127,139,141,
%T 153,155,167,169,181,183,195,197,209,211,223,225,237,239,251,253,265,
%U 267,279,281,293,295,307,309,321,323,335,337,349,351,363,365,377,379
%N Numbers that are congruent to {1, 13} mod 14.
%C If 14k+1 is a perfect square..(0,12,16,52,60,120..) then the square root of 14k+1 = a(n) - _Gary Detlefs_, Feb 22 2010
%C More generally, these numbers are of the form (2*h*n+(h-4)*(-1)^n-h)/4 (h, n natural numbers), therefore ((2*h*n+(h-4)*(-1)^n-h)/4)^2-1==0 (mod h); in our case, a(n)^2-1==0 (mod 14). Also a(n)^2-1==0 (mod 28). - _Bruno Berselli_, Oct 26 2010 - Nov 17 2010
%H Reinhard Zumkeller, <a href="/A113801/b113801.txt">Table of n, a(n) for n = 1..10000</a>
%H <a href="/index/Rec#order_03">Index entries for linear recurrences with constant coefficients</a>, signature (1,1,-1).
%F a(n) = 14*(n-1)-a(n-1), n>1. - _R. J. Mathar_, Jan 30 2010
%F From _Bruno Berselli_, Oct 26 2010: (Start)
%F a(n) = -a(-n+1) = (14*n+5*(-1)^n-7)/2.
%F G.f.: x*(1+12*x+x^2)/((1+x)*(1-x)^2).
%F a(n) = a(n-2)+14 for n>2.
%F a(n) = 14*A000217(n-1)+1 - 2*sum[i=1..n-1] a(i) for n>1. (End)
%F a(0)=1, a(1)=13, a(2)=15, a(n)=a(n-1)+a(n-2)-a(n-3). - _Harvey P. Dale_, May 11 2011
%F Sum_{n>=1} (-1)^(n+1)/a(n) = (Pi/14)*cot(Pi/14). - _Amiram Eldar_, Dec 04 2021
%F E.g.f.: 1 + ((14*x - 7)*exp(x) + 5*exp(-x))/2. - _David Lovler_, Sep 04 2022
%t LinearRecurrence[{1,1,-1},{1,13,15},60] (* or *) Select[Range[500], MemberQ[{1,13},Mod[#,14]]&] (* _Harvey P. Dale_, May 11 2011 *)
%o (Haskell)
%o a113801 n = a113801_list !! (n-1)
%o a113801_list = 1 : 13 : map (+ 14) a113801_list
%o -- _Reinhard Zumkeller_, Jan 07 2012
%o (PARI) a(n)=n\2*14-(-1)^n \\ _Charles R Greathouse IV_, Sep 15 2015
%Y Cf. A000217, A113802, A113803, A113804, A113805, A113806, A113807, A008589, A045472 (primes), A195145 (partial sums), A005408, A047209, A007310, A047336, A047522, A056020, A090771, A175885, A091998, A175886, A175887.
%K nonn,easy
%O 1,2
%A _Giovanni Teofilatto_, Jan 22 2006
%E Corrected and extended by _Giovanni Teofilatto_, Nov 14 2008
%E Replaced the various formulas by a correct one - _R. J. Mathar_, Jan 30 2010
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