

A337331


Number of pairs of divisors of n, (d1,d2), with d1 <= d2, whose average divides n.


1



1, 2, 2, 3, 2, 5, 2, 4, 3, 4, 2, 9, 2, 4, 5, 5, 2, 8, 2, 6, 4, 4, 2, 13, 3, 4, 4, 7, 2, 12, 2, 6, 4, 4, 4, 15, 2, 4, 4, 9, 2, 10, 2, 6, 9, 4, 2, 17, 3, 6, 4, 6, 2, 11, 4, 10, 4, 4, 2, 23, 2, 4, 6, 7, 4, 11, 2, 6, 4, 8, 2, 22, 2, 4, 8, 6, 4, 10, 2, 12, 5, 4, 2, 21, 4, 4, 4, 8, 2, 22
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OFFSET

1,2


COMMENTS

a(n) >= d(n), where d is the number of divisors of n (A000005).
From Bernard Schott, Aug 24 2020: (Start)
a(n) = 2 if and only if n is prime.
Also, a(2^k) = k+1 for k >= 0. (End)


LINKS

Table of n, a(n) for n=1..90.


FORMULA

a(n) = Sum_{d1n, d2n, (d1+d2)(2*n), 2(d1+d2)} 1.


EXAMPLE

a(6) = 5; The divisors of 6 are {1,2,3,6}. The pairs of divisors, (d1,d2), with d1 <= d2, whose average divides 6 are: (1,1), (1,3), (2,2), (3,3) and (6,6). So a(6) = 5.
a(7) = 2; The divisors of 7 are {1,7}. The pairs of divisors, (d1,d2), with d1 <= d2, whose average divides 7 are: (1,1) and (7,7). So a(7) = 2.
a(8) = 4; The divisors of 8 are {1,2,4,8}. The pairs of divisors, (d1,d2), with d1 <= d2, whose average divides 8 are: (1,1), (2,2), (4,4), and (8,8). So a(8) = 4.
a(9) = 3; The divisors of 9 are {1,3,9}. The pairs of divisors, (d1,d2), with d1 <= d2, whose average divides 9 are: (1,1), (3,3) and (9,9). So a(9) = 3.


MATHEMATICA

Table[Sum[Sum[(1  Ceiling[(i + k)/2] + Floor[(i + k)/2]) (1  Ceiling[2 n/(i + k)] + Floor[2 n/(i + k)]) (1  Ceiling[n/k] + Floor[n/k]) (1  Ceiling[n/i] + Floor[n/i]), {i, k}], {k, n}], {n, 100}]


CROSSREFS

Cf. A000005.
Sequence in context: A141822 A033099 A330833 * A018892 A100565 A244098
Adjacent sequences: A337328 A337329 A337330 * A337332 A337333 A337334


KEYWORD

nonn


AUTHOR

Wesley Ivan Hurt, Aug 23 2020


STATUS

approved



