login
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.
0
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
1,1
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.
LINKS
R. L. Graham, Paul Erdos and Egyptian Fractions, Bolyai Society Mathematical Studies 25, pp 289-309, 2013.
EXAMPLE
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.
MATHEMATICA
a[n_] := Sum[DivisorSigma[0, d + n], {d, Divisors@n}]; Array[a, 67] (* Giovanni Resta, Mar 21 2017 *)
PROG
(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
CROSSREFS
KEYWORD
nonn
AUTHOR
Ctibor O. Zizka, Mar 21 2017
EXTENSIONS
a(21)-a(67) from Giovanni Resta, Mar 21 2017
STATUS
approved