A217482 Quarter-square tetrahedrals: a(n) = k*(k - 1)*(k - 2)/6, k = A002620(n).

0, 0, 0, 0, 4, 20, 84, 220, 560, 1140, 2300, 4060, 7140, 11480, 18424, 27720, 41664, 59640, 85320, 117480, 161700, 215820, 287980, 374660, 487344, 620620, 790244, 988260, 1235780, 1521520, 1873200, 2275280, 2763520, 3317040, 3981264, 4728720, 5616324, 6608580

Offset

0,5

Comments

Observation: (3/2)*a(n) + 2 is a power of 2 up to n = 6 (giving {2, 2, 2, 2, 8, 32, 128}).

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).

Links

Table of n, a(n) for n=0..37.

Physics Forums, A Tetrahedral Counterpart to Ramanujan-Nagell Triangular Numbers?, Nov 2010.

Index entries for linear recurrences with constant coefficients, signature (2,4,-10,-5,20,0,-20,5,10,-4,-2,1).

Formula

a(n) = (1/6)*floor(n^2/4)*(floor(n^2/4)-1)*(floor(n^2/4)-2).

a(2n + 2) = A178208(n+1).

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

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

Maple

a:= n-> binomial(floor(n^2/4), 3):
seq(a(n), n=0..41);  # Alois P. Heinz, Feb 16 2024

Mathematica

(#*(#-1)*(#-2)/6)& /@ Table[Floor[n^2/4], {n, 0, 20}] (* Amiram Eldar, Feb 17 2024 *)

Prog

(PARI) a(n)=my(k=floor(n^2/4)); k*(k-1)*(k-2)/6 \\ Charles R Greathouse IV, Oct 05 2012

Crossrefs

Cf. A000292, A002620, A178208.

Sequence in context: A344063 A055296 A140532 * A099898 A003489 A167682

Adjacent sequences: A217479 A217480 A217481 * A217483 A217484 A217485

Keyword

nonn,easy

Author

Raphie Frank, Oct 04 2012

Extensions

a(24) corrected by Charles R Greathouse IV, Oct 05 2012

Status

approved