OFFSET
1,1
COMMENTS
Note that the equation rad(k)=rad(k+n) has no solutions for odd n. Thus, parameterizing the difference as 2n is the natural formulation.
The existence of at least one solution for any n is guaranteed because k=2n always satisfies the condition. Thus, a(n)<=2n.
The prime factors of k must be a subset of the prime factors of 2n.
LINKS
Banri Ogawa, Table of n, a(n) for n = 1..10000
EXAMPLE
For n=10, rad(k)=rad(k+20). The minimum k satisfying this is a(10)=5 (rad(5)=rad(5+20)=5).
MATHEMATICA
a[n_]:=Module[{k=1}, Until[Times@@First/@FactorInteger@k==Times@@First/@FactorInteger@(k+2n), k++]; k]; Array[a, 73] (* James C. McMahon, Jun 01 2026 *)
PROG
(Python)
from math import prod
from sympy import primefactors
def a(n): return next(k for k in range(1, 2*n+1) if prod(primefactors(k)) == prod(primefactors(k+2*n)))
(PARI) rad(n) = factorback(factorint(n)[, 1]); \\ A007947
a(n) = my(k=1); while(rad(k+2*n) != rad(k), k++); k; \\ Michel Marcus, May 28 2026
CROSSREFS
KEYWORD
nonn
AUTHOR
Banri Ogawa, May 28 2026
STATUS
approved
