A159068 a(n) = Sum_{k=1..n} binomial(n,k) * gcd(k,n).

1, 4, 9, 24, 35, 138, 133, 528, 855, 2550, 2057, 12708, 8203, 45178, 78645, 182816, 131087, 933966, 524305, 3698220, 4890627, 13345794, 8388629, 67390440, 60129575, 225470518, 279938133, 1032462228, 536870939, 5018059170

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

Cf. A159069.

Sequence in context: A288099 A288103 A286729 * A255876 A158141 A056575

Adjacent sequences: A159065 A159066 A159067 * A159069 A159070 A159071

Keyword

nonn

Author

Leroy Quet, Apr 04 2009

Extensions

Formula corrected by Max Alekseyev, Jan 09 2015

Status

approved