A159069 a(n) = A159068(n)/n.
1, 2, 3, 6, 7, 23, 19, 66, 95, 255, 187, 1059, 631, 3227, 5243, 11426, 7711, 51887, 27595, 184911, 232887, 606627, 364723, 2807935, 2405183, 8671943, 10368079, 36873651, 18512791, 167268639, 69273667, 496472226, 551130063, 1856103039
Offset: 1
Keywords
Examples
Row 6 of Pascal's triangle is 1,6,15,20,15,6,1. The greatest common divisors of n and each integer from 1 to 6 are gcd(1,6)=1, gcd(2,6)=2, gcd(3,6)=3, gcd(4,6)=2, gcd(5,6)=1, and gcd(6,6)=6. So a(6) = (1/6)*( 6*1 + 15*2 + 20*3 + 15*2 + 6*1 + 1*6) = 138/6 = 23. Note that each term of the sum in parentheses is a multiple of 6, so 138 is a multiple of 6.
Links
- Seiichi Manyama, Table of n, a(n) for n = 1..1000
Crossrefs
Cf. A159068.
Programs
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Maple
A159068 := proc(n) add(binomial(n,k)*gcd(k,n),k=1..n) ; end: A159069 := proc(n) A159068(n)/n ; end: seq(A159069(n),n=1..80) ; # R. J. Mathar, Apr 06 2009
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Mathematica
Table[Sum[Binomial[n, k] GCD[k, n], {k, n}]/n, {n, 34}] (* Michael De Vlieger, Aug 29 2017 *)
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PARI
a(n) = sum(k=1, n, binomial(n,k) * gcd(k,n))/n; \\ Michel Marcus, Aug 30 2017
Extensions
Extended by R. J. Mathar, Apr 06 2009