|
%I #53 Feb 17 2024 04:04:09
%S 0,0,0,0,4,20,84,220,560,1140,2300,4060,7140,11480,18424,27720,41664,
%T 59640,85320,117480,161700,215820,287980,374660,487344,620620,790244,
%U 988260,1235780,1521520,1873200,2275280,2763520,3317040,3981264,4728720,5616324,6608580
%N Quarter-square tetrahedrals: a(n) = k*(k - 1)*(k - 2)/6, k = A002620(n).
%C Observation: (3/2)*a(n) + 2 is a power of 2 up to n = 6 (giving {2, 2, 2, 2, 8, 32, 128}).
%C Conjecture: There are no other tetrahedral numbers (Tetra_n = A000292) > 84 such that (3/2)*Tetra_n + 2 is a power of 2. This is true to at least 1.41*10^1505 per computer check by _Charles R Greathouse IV_ on Physics Forums (Nov 2010).
%H Physics Forums, <a href="http://www.physicsforums.com/showthread.php?t=443958">A Tetrahedral Counterpart to Ramanujan-Nagell Triangular Numbers?</a>, Nov 2010.
%H <a href="/index/Rec#order_12">Index entries for linear recurrences with constant coefficients</a>, signature (2,4,-10,-5,20,0,-20,5,10,-4,-2,1).
%F a(n) = (1/6)*floor(n^2/4)*(floor(n^2/4)-1)*(floor(n^2/4)-2).
%F a(2n + 2) = A178208(n+1).
%F G.f.: -4*x^4*(x^4+3*x^3+7*x^2+3*x+1)/((x-1)^7*(x+1)^5). - _Colin Barker_, Oct 11 2012
%F Sum_{n>=4} 1/a(n) = Pi^2/2 - 5/12 - 3*Pi*cot(sqrt(2)*Pi)/(2*sqrt(2)) - 6*Pi*tan(sqrt(5)*Pi/2)/sqrt(5). - _Amiram Eldar_, Feb 17 2024
%p a:= n-> binomial(floor(n^2/4), 3):
%p seq(a(n), n=0..41); # _Alois P. Heinz_, Feb 16 2024
%t (#*(#-1)*(#-2)/6)& /@ Table[Floor[n^2/4], {n, 0, 20}] (* _Amiram Eldar_, Feb 17 2024 *)
%o (PARI) a(n)=my(k=floor(n^2/4));k*(k-1)*(k-2)/6 \\ _Charles R Greathouse IV_, Oct 05 2012
%Y Cf. A000292, A002620, A178208.
%K nonn,easy
%O 0,5
%A _Raphie Frank_, Oct 04 2012
%E a(24) corrected by _Charles R Greathouse IV_, Oct 05 2012
|
