A159069 -id:A159069 - cp's OEIS frontend

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

Leroy Quet, Apr 04 2009

nonn

Each term of the sum a(n) is divisible by n, so a(n) is a multiple of n for all positive integers n.

			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.

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.]

Cf. A159069.

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

Table[Sum[Binomial[n, k] GCD[k, n], {k, n}], {n, 30}] (* Michael De Vlieger, Aug 29 2017 *)

a(n) = sum(k=1, n,  binomial(n,k) * gcd(k,n)); \\ Michel Marcus, Aug 30 2017

a(n) = 2^n * Sum_{d|n} (phi(d)/d) Sum_{k=1..d} (-1)^(k*n/d)*cos(k*Pi/d)^n - n.

Formula corrected by Max Alekseyev, Jan 09 2015

A159554 a(n) = A159553(n)/n.

Original entry on oeis.org

2, 3, 4, 7, 8, 24, 20, 67, 96, 256, 188, 1060, 632, 3228, 5244, 11427, 7712, 51888, 27596, 184912, 232888, 606628, 364724, 2807936, 2405184, 8671944, 10368080, 36873652, 18512792, 167268640, 69273668, 496472227, 551130064, 1856103040

Offset: 1

Leroy Quet, Apr 14 2009

nonn

a(n) = A159069(n) + 1.

Cf. A159069, A159553.

Extended by Ray Chandler, Jun 18 2009

A159458 Numbers m such that m^2 divides A159068(m), where A159068(m) = Sum_{k=1..m} binomial(m,k) * gcd(m,k).

Original entry on oeis.org

1, 2, 3, 11, 33, 309, 665, 1461, 2323, 6789

Offset: 1

Leroy Quet, Apr 12 2009

m divides A159068(m) for all positive integers m.

Cf. A159068, A159069.

is(m) = sum(k=1, m, binomial(m, k)*gcd(k, m))%m^2 == 0; \\ Jinyuan Wang, Aug 10 2021

4 more terms from R. J. Mathar, Apr 16 2009

a(10)=6789 from Ray Chandler, Jun 19 2009

A178623 Triangle T(n,m) read by rows: T(n,0)= prime(n); T(n,m)=1 if m>=1.

Original entry on oeis.org

1, 2, 1, 3, 1, 1, 5, 1, 1, 1, 1, 7, 1, 1, 1, 1, 1, 1, 11, 1, 1, 1, 1, 1, 1, 1, 1, 1, 1, 13, 1, 1, 1, 1, 1, 1, 1, 1, 1, 1, 1, 1, 17, 1, 1, 1, 1, 1, 1, 1, 1, 1, 1, 1, 1, 1, 1, 1, 1, 19, 1, 1, 1, 1, 1, 1, 1, 1, 1, 1, 1, 1, 1, 1, 1, 1, 1, 1, 23, 1, 1, 1, 1, 1, 1, 1, 1, 1, 1, 1, 1, 1, 1, 1, 1, 1, 1, 1, 1, 1, 1, 29

Offset: 0

Paul Curtz, May 31 2010

The sequence reflects a conjecture on the denominator of inverse Bernoulli polynomials in A178340: if the row index is one less than one of the primes in A008578, the row of denominators starts with that prime and contains 1's in the remaining entries.

[Row sums in A178252 are A159069(n+1), unless there is a common factor in numerator and denominator. The row sum over columns with index of the same parity as the row index in the table of fractions of the [x^m] B^{-1}(n,x) in A178252 are: 1, 1, 1/3+1=4/3, 1+1=2, 1/5+2+1=16/5, 1+10/3+1=16/3, 1/7+3+5+1=64/7, 16, 256/9, 256/5, 1024/11, 512/3, 496/13, ... =A084623(n+1)/A000265(n+1).]

			1;
2,1;
3,1,1;
5,1,1,1,1;
7,1,1,1,1,1,1;
11,1,1,1,1,1,1,1,1,1,1;
13,1,1,1,1,1,1,1,1,1,1,1,1;
17,1,1,1,1,1,1,1,1,1,1,1,1,1,1,1,1;
19,1,1,1,1,1,1,1,1,1,1,1,1,1,1,1,1,1,1;
23,1,1,1,1,1,1,1,1,1,1,1,1,1,1,1,1,1,1,1,1,1,1;
29,1,1,1,1,1,1,1,1,1,1,1,1,1,1,1,1,1,1,1,1,1,1,1,1,1,1,1,1;

Cf. A076274 (row sums).

T(n,0) = A008578(n+1). T(n,m) =1, 1<=m<=A008578(n+1)-1.

cp's OEIS Frontend

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

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A159554 a(n) = A159553(n)/n.

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A159458 Numbers m such that m^2 divides A159068(m), where A159068(m) = Sum_{k=1..m} binomial(m,k) * gcd(m,k).

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A178623 Triangle T(n,m) read by rows: T(n,0)= prime(n); T(n,m)=1 if m>=1.

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