A284167 a(n) = Sum_{i=1..A000005(n)} d(n+k(i)), where d(t) is the number of divisors of t and k(i) is the i-th divisor of n.

2, 5, 7, 10, 8, 15, 8, 18, 16, 18, 10, 29, 8, 19, 25, 28, 10, 33, 10, 35, 26, 20, 12, 50, 18, 20, 31, 36, 12, 51, 10, 42, 27, 23, 33, 62, 8, 22, 30, 60, 12, 53, 10, 40, 52, 22, 14, 78, 20, 41, 28, 38, 12, 63, 36, 63, 30, 24, 16, 95, 8, 23, 59, 60, 32, 54, 10

Offset: 1

Author

Ctibor O. Zizka, Mar 21 2017

Keywords

• nonn

Comments

Let S(n,n) be the number of solutions of the equation n/x + n/y = c where n, c, x, and y are positive integers. Then S(n,n) = Sum_{i=1..A000005(n)} d(n+k(i)), where d(t) is the number of divisors of t and k(i) is the i-th divisor of n.

For c = 1 , S(n,n) = A000005(n).

Let S(n,m) be the number of solutions of the equation n/x + m/y = c where n, m, c, x, and y are positive integers, n not equal to m. Let k(i) be the i-th divisor of n, and k(j) the j-th divisor of m. Let d(t) be the number of divisors of t. Let R = d(k(i) + k(j)). Then S(n,m) = Sum_{i=1..A000005(n)} Sum_{j=1..A000005(m)} [R*1 if gcd(k(i),k(j)) = 1 , R*0 else].

For c = 1 , S(n,m) = A000005(n) * A000005(m) - P, where P is the number of divisor pairs such that gcd(k(i),k(j)) >= 2.

Examples

			For n = 4, divisors of 4 are 1, 2, 4; thus a(4) = d(4+1) + d(4+2) + d(4+4) = d(5) + d(6) + d(8) = 2 + 4 + 4 = 10.
		

Links

• R. L. Graham, Paul Erdos and Egyptian Fractions, Bolyai Society Mathematical Studies 25, pp 289-309, 2013.

Crossrefs

Cf. A000005, A018892, A063647, A089233, A275387.

Programs

Mathematica

a[n_] := Sum[DivisorSigma[0, d + n], {d, Divisors@n}]; Array[a, 67] (* Giovanni Resta, Mar 21 2017 *)

PARI

for(n=1, 101, print1(sumdiv(n, d, numdiv(d + n)),", ")) \\ Indranil Ghosh, Mar 22 2017

Python

from sympy import divisor_count, divisors
def a(n):
    return sum(divisor_count(n + d) for d in divisors(n)) # Indranil Ghosh, Mar 22 2017

Extensions

a(21)-a(67) from Giovanni Resta, Mar 21 2017