|
%I #33 Mar 17 2022 07:04:15
%S 0,15,17,62,66,141,147,252,260,395,405,570,582,777,791,1016,1032,1287,
%T 1305,1590,1610,1925,1947,2292,2316,2691,2717,3122,3150,3585,3615,
%U 4080,4112,4607,4641,5166,5202,5757,5795,6380,6420,7035,7077,7722,7766,8441
%N One-eighth of triangular numbers (integers only).
%C From _Lamine Ngom_, Oct 27 2020: (Start)
%C Numbers of the form (4*k)^2-k (A157446) or (4*k)^2+k (A157474).
%C Also numbers k such that 1+64*k is a square. (End)
%H Robert Israel, <a href="/A157716/b157716.txt">Table of n, a(n) for n = 1..10000</a>
%H <a href="/index/Rec#order_05">Index entries for linear recurrences with constant coefficients</a>, signature (1,2,-2,-1,1).
%F G.f.: x^2*(15+2*x+15*x^2)/((1+x)^2*(1-x)^3 ). [Maksym Voznyy (voznyy(AT)mail.ru), Jul 26 2009; checked and corrected by _R. J. Mathar_, Sep 16 2009]
%F a(n) = (2*n-1 + 7/8*(-1)^n)^2 -1/64. - _Robert Israel_, Apr 20 2014
%F a(n) = a(n-1) + 2*a(n-2) - 2*a(n-3) - a(n-4) + a(n-5). - _Wesley Ivan Hurt_, Nov 10 2020
%F Sum_{n>=2} 1/a(n) = 16 - (sqrt(2*(2+sqrt(2))) + sqrt(2) + 1)*Pi. - _Amiram Eldar_, Mar 17 2022
%e The first three members of A000217 that are divisible by 8 are A000217(0), A000217(15) and A000217(16), so a(1) = A000217(0)/8 = 0, a(2) = A000217(15)/8 = 15, a(3) = A000217(16)/8 = 17.
%p seq((2*n-1 + 7/8*(-1)^n)^2 - 1/64, n = 1 .. 1000); # _Robert Israel_, Apr 20 2014
%t Array[(2 # - 1 + 7/8*(-1)^#)^2 - 1/64 &, 46] (* or *)
%t Rest@ CoefficientList[Series[x^2*(15 + 2 x + 15 x^2)/((1 + x)^2*(1 - x)^3), {x, 0, 46}], x] (* _Michael De Vlieger_, Nov 05 2020 *)
%Y Cf. A000217, A101877.
%K nonn,easy
%O 1,2
%A _Vladimir Joseph Stephan Orlovsky_, Mar 04 2009
%E Definition edited by _N. J. A. Sloane_, Mar 08 2009
|
