A118391 Numerator of sum of reciprocals of first n tetrahedral numbers A000292.

1, 5, 27, 7, 10, 81, 35, 22, 81, 65, 77, 135, 52, 119, 405, 76, 85, 567, 209, 115, 378, 275, 299, 486, 175, 377, 1215, 217, 232, 1485, 527, 280, 891, 629, 665, 1053, 370, 779, 2457, 430, 451, 2835, 989, 517, 1620, 1127, 1175, 1836, 637, 1325, 4131, 715, 742

Offset: 1

Author

Jonathan Vos Post, Apr 27 2006

Keywords

• easy
• frac
• nonn

Comments

Denominators are A118392. Fractions are: 1/1, 5/4, 27/20, 7/5, 10/7, 81/56, 35/24, 22/15, 81/55, 65/44, 77/52, 135/91, 52/35, 119/80, 405/272, 76/51, 85/57, 567/380, 209/140, 115/77, 378/253, 275/184, 299/200, 486/325, 175/117, 377/252, 1215/812, 217/145, 232/155, 1485/992.

2n+3 divides a(2n). 2n-1 divides a(2n-1). p divides a(p) for prime p>2. The only primes in a(n) are a(2) = 5 and a(4) = 7. - Alexander Adamchuk, May 08 2007

Examples

			a(1) = 1 = numerator of 1/1.
a(2) = 5 = numerator of 5/4 = 1/1 + 1/4.
a(3) = 27 = numerator of 27/20 = 1/1 + 1/4 + 1/10.
a(4) = 7 = numerator of 7/5 = 1/1 + 1/4 + 1/10 + 1/20.
a(5) = 10 = numerator of 10/7 = 1/1 + 1/4 + 1/10 + 1/20 + 1/35.
a(20) = 115 = numerator of 115/77 = 1/1 + 1/4 + 1/10 + 1/20 + 1/35 + 1/56 + 1/84 + 1/120 + 1/165 + 1/220 + 1/286 + 1/364 + 1/455 + 1/560 + 1/680 + 1/816 + 1/969 + 1/1140 + 1/1330 + 1/1540.
		

Links

• G. C. Greubel, Table of n, a(n) for n = 1..1000

Crossrefs

Cf. A000292, A022998, A118392.

Programs

Magma

[Numerator(3*n*(n+3)/(2*(n+1)*(n+2))): n in [1..60]]; // G. C. Greubel, Feb 18 2021

Maple

A118391:= n-> numer(3*n*(n+3)/(2*(n+1)*(n+2))); seq(A118391(n), n=1..60) # G. C. Greubel, Feb 18 2021

Mathematica

Table[ Numerator[3n(n+3)/(2(n+1)(n+2))], {n,1,100} ] (* Alexander Adamchuk, May 08 2007 *)
Accumulate[1/Binomial[Range[60]+2,3]]//Numerator (* Harvey P. Dale, Aug 31 2023 *)

PARI

s=0;for(i=3,50,s+=1/binomial(i,3);print(numerator(s))) /* Phil Carmody, Mar 27 2012 */

Sage

[numerator(3*n*(n+3)/(2*(n+1)*(n+2))) for n in (1..60)] # G. C. Greubel, Feb 18 2021

Formula

A118391(n)/A118392(n) = Sum_{i=1..n} 1/A000292(n).

A118391(n)/A118392(n) = Sum_{i=1..n} 1/C(n+2,3).

A118391(n)/A118392(n) = Sum_{i=1..n} 6/(n*(n+1)*(n+2)).

a(n) = Numerator( 3*n*(n+3)/(2*(n+1)*(n+2)) ). - Alexander Adamchuk, May 08 2007

Extensions

More terms from Alexander Adamchuk, May 08 2007