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)
Note that the other divisor does not necessarily have to divide the other, for example for n=60, we also have pairs d1=3, d2=5 and d1=4, d2=6 as possible solutions - Antti Karttunen, Nov 27 2024
LINKS
Antti Karttunen, Table of n, a(n) for n = 1..20000
FORMULA
a(n) = Sum_{d1|n, d2|n, d1<=d2, (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}]
PROG
(PARI) A337331(n) = { my(divs=divisors(n), d1, d2); sum(i=1, #divs, d1=divs[i]; sum(j=i, #divs, d2=divs[j]; !((2*n)%(d1+d2)) * !((d1+d2)%2))); }; \\ Antti Karttunen, Nov 27 2024
CROSSREFS
KEYWORD
nonn
AUTHOR
Wesley Ivan Hurt, Aug 23 2020
EXTENSIONS
Definition and formula clarified, and more terms added by Antti Karttunen, Nov 27 2024
STATUS
approved