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%I #30 Sep 02 2017 16:54:54
%S 1,4,9,24,35,138,133,528,855,2550,2057,12708,8203,45178,78645,182816,
%T 131087,933966,524305,3698220,4890627,13345794,8388629,67390440,
%U 60129575,225470518,279938133,1032462228,536870939,5018059170
%N a(n) = Sum_{k=1..n} binomial(n,k) * gcd(k,n).
%C Each term of the sum a(n) is divisible by n, so a(n) is a multiple of n for all positive integers n.
%H Seiichi Manyama, <a href="/A159068/b159068.txt">Table of n, a(n) for n = 1..1000</a>
%H Laszlo Toth, <a href="http://www.cs.uwaterloo.ca/journals/JIS/VOL14/Toth/toth9.html">Weighted gcd-sum functions</a>, J. Integer Sequences, 14 (2011), Article 11.7.7. [Notice that formula (26) contains error.]
%F a(n) = 2^n * Sum_{d|n} (phi(d)/d) Sum_{k=1..d} (-1)^(k*n/d)*cos(k*Pi/d)^n - n.
%e 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.
%p A159068 := proc(n) add(binomial(n,k)*gcd(k,n),k=1..n) ; end:
%p seq(A159068(n),n=1..80) ; # _R. J. Mathar_, Apr 06 2009
%t Table[Sum[Binomial[n, k] GCD[k, n], {k, n}], {n, 30}] (* _Michael De Vlieger_, Aug 29 2017 *)
%o (PARI) a(n) = sum(k=1, n, binomial(n,k) * gcd(k,n)); \\ _Michel Marcus_, Aug 30 2017
%Y Cf. A159069.
%K nonn
%O 1,2
%A _Leroy Quet_, Apr 04 2009
%E Formula corrected by _Max Alekseyev_, Jan 09 2015
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