cp's OEIS Frontend

This is a front-end for the Online Encyclopedia of Integer Sequences, made by Christian Perfect. The idea is to provide OEIS entries in non-ancient HTML, and then to think about how they're presented visually. The source code is on GitHub.

A118412 Denominator of sum of reciprocals of first n pentatope numbers A000332.

Original entry on oeis.org

1, 5, 15, 105, 42, 63, 90, 495, 55, 143, 91, 1365, 420, 510, 612, 2907, 855, 665, 385, 1771, 1518, 1725, 1950, 8775, 2457, 5481, 1015, 4495, 1240, 4092, 4488, 19635, 5355, 11655, 6327, 9139, 2470, 2665, 8610, 37023
Offset: 1

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Author

Jonathan Vos Post, Apr 27 2006

Keywords

Comments

Numerators are A118411. Fractions are: 1/1, 6/5, 19/15, 136/105, 83/63, 119/90, 656/495, 73/55, 190/143, 121/91, 1816/1365, 559/420, 679/510, 815/612, 3872/2907, 1139/855, 886/665, 513/385, 2360/1771, 2023/1518, 2299/1725, 2599/1950, 11696/8775, 3275/2457, 7306/5481, 1353/1015, 5992/4495, 1653/1240, 5455/4092, 5983/4488, 26176/19635, 7139/5355, 15538/11655, 8435/6327, 12184/9139, 3293/2470, 3553/2665, 11479/8610, 49360/37023. The denominator of sum of reciprocals of first n triangular numbers is A026741. The denominator of sum of reciprocals of first n tetrahedral numbers is A118392.

Examples

			a(1) = 1 = denominator of 1/1.
a(2) = 5 = denominator of 6/5 = 1/1 + 1/5.
a(3) = 15 = denominator of 19/15 = 1/1 + 1/5 + 1/15.
a(4) = 105 = denominator of 136/105 = 1/1 + 1/5 + 1/15 + 1/35.
a(5) = 42 = denominator of 55/42 = 1/1 + 1/5 + 1/15 + 1/35 + 1/70.
a(10) = 143 = denominator of 190/143 = 1/1 + 1/5 + 1/15 + 1/35 + 1/70 + 1/126 + 1/210 + 1/330 + 1/495 + 1/715.
a(20) = 1771 = denominator of 2360/1771 = 1/1 + 1/5 + 1/15 + 1/35 + 1/70 + 1/126 + 1/210 + 1/330 + 1/495 + 1/715 + 1/1001 + 1/1365 + 1/1820 + 1/2380 + 1/3060 + 1/3876 + 1/4845 + 1/5985 + 1/7315 + 1/8855.
		

Crossrefs

Programs

  • PARI
    s=0;for(i=4,50,s+=1/binomial(i,4);print(denominator(s))) /* Phil Carmody, Mar 27 2012 */

Formula

A118411(n)/A118412(n) = SUM[i=1..n] (1/A000332(n)). A118411(n)/A118412(n) = SUM[i=1..n] (1/C(n+2,4)). A118411(n)/A118412(n) = SUM[i=1..n] (1/(n*(n+1)*(n+2)*(n+3)/24)).