OFFSET
1,2
COMMENTS
Each term of the sum a(n) is divisible by n, so a(n) is a multiple of n for all positive integers n.
LINKS
Seiichi Manyama, Table of n, a(n) for n = 1..1000
Laszlo Toth, Weighted gcd-sum functions, J. Integer Sequences, 14 (2011), Article 11.7.7. [Notice that formula (26) contains error.]
FORMULA
a(n) = 2^n * Sum_{d|n} (phi(d)/d) Sum_{k=1..d} (-1)^(k*n/d)*cos(k*Pi/d)^n - n.
EXAMPLE
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) = 6*1 + 15*2 + 20*3 + 15*2 + 6*1 + 1*6 = 138. Note that each term of the sum is a multiple of 6, so 138 is a multiple of 6.
MAPLE
A159068 := proc(n) add(binomial(n, k)*gcd(k, n), k=1..n) ; end:
seq(A159068(n), n=1..80) ; # R. J. Mathar, Apr 06 2009
MATHEMATICA
Table[Sum[Binomial[n, k] GCD[k, n], {k, n}], {n, 30}] (* Michael De Vlieger, Aug 29 2017 *)
PROG
(PARI) a(n) = sum(k=1, n, binomial(n, k) * gcd(k, n)); \\ Michel Marcus, Aug 30 2017
CROSSREFS
KEYWORD
nonn
AUTHOR
Leroy Quet, Apr 04 2009
EXTENSIONS
Formula corrected by Max Alekseyev, Jan 09 2015
STATUS
approved