OFFSET
1,2
COMMENTS
As d^2 | N-d we have N = k*d^2 + d for some k >= 0 and d > 1. So gcd(k*d^2 + d - d, (N*d^2 + d)*d) = gcd(k*d^2, k*d^3 + d^2) = gcd(k*d^2, d^2) = d^2. So for any N such that d^2 | gcd(N - d, N*d) we have gcd(N - d, N*d) = d^2. - David A. Corneth, Apr 20 2022
Since gcd(N - d, N*d) is never larger than d^2 (if N = n*g, d = f*g with gcd(n,f) = 1, then gcd(N - d, N*d) = g*gcd(n-f,n*f*g) = g*gcd(n-f, f*f*g) <= g*g, since by assumption, no factor of f divides n), so one can also replace "=" by ">=" in the definition.
LINKS
Michael De Vlieger, Table of n, a(n) for n = 1..10000
FORMULA
EXAMPLE
N = 1 is in the sequence because d(N) = 1, gcd(1 - 1, 1*1) = 1 = d^2.
N = 2 is in the sequence because d(N) = 2, gcd(2 - 2, 2*2) = 4 = d^2.
N = 136 = 8*17 is in the sequence because d(N) = 4*2 = 8, gcd(8*17 - 8, 8*17*8) = gcd(8*16, 8*8*17) = 8*8 = d^2. Similarly for N = 8*p with any prime p = 8*k + 1.
N = 156 = 2^2*3*13 is in the sequence because d(n) = 3*2*2 = 12, gcd(12*13 - 12, 12*13*12) = gcd(12*12, 12*12*13) = 12*12 = d^2. Similarly for any N = 12*p with prime p = 12*k + 1.
More generally, when N = m*p^k with p^k == 1 (mod m) and m = (k+1)*d(m), then d(N) = d(m)*(k+1) = m and gcd(n - d, n*d) = gcd(m*p^k - m, m*p^k*m) = m*gcd(p^k - 1, p^k*m) = m^2. This holds for m = 8 and 12 with k = 1, for m = 9, 18 and 24 with k = 2, etc: see sequence A033950 for the m-values.
MATHEMATICA
Select[Range[4650], GCD[#1 - #2, #1 #2] == #2^2 & @@ {#, DivisorSigma[0, #]} &] (* Michael De Vlieger, Apr 21 2022 *)
PROG
(PARI) select( {is(n, d=numdiv(n))=gcd(n-d, d^2)==d^2}, [1..10^4])
CROSSREFS
KEYWORD
nonn
AUTHOR
M. F. Hasler, Apr 15 2022
STATUS
approved