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A368740
a(n) = Sum_{k = 1..n} gcd(5*k, n).
0
1, 3, 5, 8, 25, 15, 13, 20, 21, 75, 21, 40, 25, 39, 125, 48, 33, 63, 37, 200, 65, 63, 45, 100, 225, 75, 81, 104, 57, 375, 61, 112, 105, 99, 325, 168, 73, 111, 125, 500, 81, 195, 85, 168, 525, 135, 93, 240, 133, 675, 165, 200, 105, 243, 525, 260, 185, 171, 117, 1000, 121, 183, 273, 256, 625
OFFSET
1,2
COMMENTS
a(n) equals the number of solutions to the congruence 5*x*y == 0 (mod n) for 1 <= x, y <= n.
FORMULA
a(5*n) = 25*A018804(n); a(5*n+r) = A018804(5*n+r) for 1 <= r <= 4.
a(n) = Sum_{d divides n} gcd(5, d)*phi(d)*n/d, where phi(n) = A000010(n).
Multiplicative: a(5^k) = (4*k + 1)*5^k and for prime p not equal to 5, a(p^k) = (k + 1)*p^k - k*p^(k-1).
Define D(n) = Sum_{d divides n} a(d). Then
D(5*n+r) = (5*n + r)*tau(5*n+r) for 1 <= r <= 4, where tau(n) = A000005(n), the number of divisors of n.
The sequence {(1/25)*( D(5*n) - D(n) ) : n >= 1} begins {1, 4, 6, 12, 9, 24, 14, 32, 27, 36, 22, 72, 26, 56, 54, 80, 34, 108, 38, 108, 84, 88, ...} and appears to be multiplicative.
Dirichlet g.f.: (1 + 15/5^s)/(1 - 1/5^s) * zeta(s-1)^2/zeta(s).
Sum_{k=1..n} a(k) ~ n^2 * (5*log(n) - 5/2 + 10*gamma - 25*log(5)/12 - 30*zeta'(2)/Pi^2) / Pi^2, where gamma is the Euler-Mascheroni constant A001620. - Vaclav Kotesovec, Jan 12 2024
EXAMPLE
a(10) = 75: each of the 100 pairs (x, y), 1 <= x, y <= 10, is a solution to the congruence 5*x*y == 0 (mod 10) except for the 25 pairs (x, y) with both x and y odd.
MAPLE
seq(add(gcd(5*k, n), k = 1..n), n = 1..70);
# alternative faster program for large n
with(numtheory): seq(add(gcd(5, d)*phi(d)*n/d, d in divisors(n)), n = 1..70);
MATHEMATICA
Table[Sum[GCD[5*k, n], {k, 1, n}], {n, 1, 100}] (* Vaclav Kotesovec, Jan 12 2024 *)
CROSSREFS
KEYWORD
nonn,mult,easy
AUTHOR
Peter Bala, Jan 07 2024
STATUS
approved