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A390582
a(n) = (2*n - sigma(n))/gcd(n, sigma(n)).
3
1, 1, 2, 1, 4, 0, 6, 1, 5, 1, 10, -1, 12, 2, 2, 1, 16, -1, 18, -1, 10, 4, 22, -1, 19, 5, 14, 0, 28, -2, 30, 1, 6, 7, 22, -19, 36, 8, 22, -1, 40, -2, 42, 1, 4, 10, 46, -7, 41, 7, 10, 3, 52, -2, 38, -1, 34, 13, 58, -4, 60, 14, 22, 1, 46, -2, 66, 5, 14, -2, 70, -17
OFFSET
1,3
COMMENTS
Empirical observation: mumbers in this sequence that are positive and close to 1 are more likely to produce regular amicable pairs under the Wiethaus's Rule (see Garcia et al).
The positions of -1's in the sequence are given by A153501.
The positions of 0's in the sequence are given by A000396.
The positions of 1's in the sequence are given by A271816.
If a(n) is negative, n is abundant (A005101).
If a(n) is positive, n is deficient (A005100).
If instead a(n) is 0, n is perfect (A000396).
LINKS
M. GarcĂ­a, J.M. Pedersen and H.J.J. te Riele, Amicable pairs : a survey, Fields Institute Communications, 41 (2004), 1-19. See p. 193.
FORMULA
a(n) = A033879(n)/A009194(n).
a(prime(n)) = prime(n) - 1, and those are also the records. - Peter Luschny, Nov 13 2025
EXAMPLE
For n = 1, (2*n - sigma(n)) = 1 and gcd(n,sigma(n)) = 1 so (2*n - sigma(n))/gcd(n, sigma(n)) = 1.
For n = 9, a(prime(9)) = 22 = prime(9) - 1;
MAPLE
a := proc(n) NumberTheory:-SumOfDivisors(n); (2*n - %)/igcd(n, %) end: seq(a(n), n = 1..72); # Peter Luschny, Nov 11 2025
MATHEMATICA
a[n_] := (2*n - #)/GCD[n, #]& @ DivisorSigma[1, n]; Array[a, 100] (* Amiram Eldar, Nov 11 2025 *)
PROG
(Python)
from sympy import divisor_sigma, gcd
lists = []
for A1 in range(1, 1000):
A2 = divisor_sigma(A1)
A3 = gcd(A1, A2)
lists.append((A1*2-A2)/A3)
print(lists[:100])
(PARI) a(n) = my(s=sigma(n)); (2*n-s)/gcd(n, s); \\ Michel Marcus, Nov 11 2025
CROSSREFS
KEYWORD
sign,easy
AUTHOR
Hayden Chesnut, Nov 11 2025
STATUS
approved