%I #60 Jul 26 2024 08:58:20
%S 0,3,8,11,12,15,20,23,24,27,32,35,36,39,44,47,48,51,56,59,60,63,68,71,
%T 72,75,80,83,84,87,92,95,96,99,104,107,108,111,116,119,120,123,128,
%U 131,132,135,140,143,144,147,152,155,156,159,164,167,168,171,176,179,180
%N Numbers k such that 12 divides k*(k+1).
%C First differences are 3, 5, 3, 1, 3, 5, 3, 1, 3, 5, 3, 1, 3, 5, 3, 1, ..., . - _Robert G. Wilson v_, May 31 2017
%C Numbers that are congruent to {0, 3, 8, 11} mod 12. - _Amiram Eldar_, Jul 26 2024
%H Vincenzo Librandi, <a href="/A108752/b108752.txt">Table of n, a(n) for n = 1..5000</a>
%H Robert Phillips, <a href="https://web.archive.org/web/20060919055917/http://www.usca.edu/math/~mathdept/bobp/pdf/tnumbers.pdf">Triangular numbers which are sums of two triangular numbers</a>, 2006.
%H <a href="/index/Rec#order_04">Index entries for linear recurrences with constant coefficients</a>, signature (2,-2,2,-1).
%F From _R. J. Mathar_, Jan 07 2009: (Start)
%F a(n) = 2*a(n-1) - 2*a(n-2) + 2*a(n-3) - a(n-4) = A016777(n) - A057077(n).
%F G.f.: x*(3 + 2*x + x^2)/((1 + x^2)*(1 - x)^2). (End)
%F a(n) = 3*n - 2 - (-1)^((2*n-3-(-1)^n)/4). - _Luce ETIENNE_, Apr 04 2015
%F Sum_{n>=2} 1/a(n) = log(2)/2 + arccoth(sqrt(3))/(2*sqrt(3)) - Pi*(3+2*sqrt(3))/72. - _Amiram Eldar_, Jul 26 2024
%p a:= proc(n) if is(n*(n+1)/12, integer) then n fi end: seq(a(n), n=0..200); # _Emeric Deutsch_, Jun 25 2005
%t Select[ Range[0, 182], Mod[ #(# + 1), 12] == 0 &] (* _Robert G. Wilson v_, Jun 25 2005 *)
%t LinearRecurrence[{2, -2, 2, -1}, {0, 3, 8, 11}, 200] (* _Vincenzo Librandi_, Jun 04 2017 *)
%o (Magma) [3*n-2-(-1)^((2*n-3-(-1)^n) div 4): n in [1..80]]; // _Vincenzo Librandi_, May 04 2017
%Y Equals A112652-1, A218155-3, A174398-5, A072833+3.
%Y Cf. A016777, A057077, A287765.
%K nonn,easy
%O 1,2
%A Robert Phillips (bobp(AT)usca.edu), Jun 23 2005
%E More terms from _Robert G. Wilson v_ and _Emeric Deutsch_, Jun 25 2005