A386376 a(n) is the smallest integer k such that b(n,k) is squarefree, where b(n,1) = n and b(n, k+1) = b(n,k) + rad(b(n, k)) for k >= 1.

1, 1, 1, 2, 1, 1, 1, 2, 5, 1, 1, 4, 1, 1, 1, 4, 1, 3, 1, 2, 1, 1, 1, 2, 2, 1, 2, 2, 1, 1, 1, 2, 1, 1, 1, 2, 1, 1, 1, 8, 1, 1, 1, 2, 7, 1, 1, 8, 3, 7, 1, 2, 1, 7, 1, 2, 1, 1, 1, 6, 1, 1, 5, 2, 1, 1, 1, 2, 1, 1, 1, 2, 1, 1, 6, 2, 1, 1, 1, 6, 5, 1, 1, 4, 1, 1, 1

Offset: 1

Author

Yifan Xie, Aug 20 2025

Keywords

• nonn
• easy
• new

Comments

rad(k) = A007947(k) is the largest squarefree integer dividing n.

Proof of a(n)'s existence: (Start)

Let c(n,k) = b(n,k) / rad(b(n,k)). Suppose that c(n,k) > 1 for all positive integers k. Pick k such that c(n,k) = m is the smallest. Let p be the smallest prime not dividing b(n,k). There exists an integer 1 <= i' <= p-1 such that p divides m + i'. So there exists a smallest integer 1 <= i <= i' such that m + i has a prime divisor not dividing b(n,k). From the minimality of i, for 1 <= j <= i-1, c(n,k+j) = m + j does not produce prime divisors not dividing b(n,k).

From the choice of i, there exists a prime q >= p such that q does not divide b(n,k) but q divides m + i. Then c(n,k+i) = (rad(b(n,k+i-1) * (c(n,k+i-1) + 1)) / rad(rad(b(n,k+i-1) * (c(n,k+i-1) + 1)) <= (c(n,k+i-1) + 1) / q <= (m + i) / p <= (m + p - 1) / p < m, because m > 1, a contradiction from the minimality of m. (End)

Examples

			For n = 9, b(9,1) = 9, b(9,2) = 9 + rad(9) = 12, b(9,3) = 12 + rad(12) = 18, b(9,4) = 18 + rad(18) = 24, b(9,5) = 24 + rad(24) = 30 is squarefree, so a(9) = 5.
		

Crossrefs

Cf. A005117, A007947.

Programs

Mathematica

rad[m_]:=Times@@(First@# & /@ FactorInteger@ m);a[n_]:=Module[{k=1,b=n},While[!SquareFreeQ[b],k++;b=b+rad[b]];k];Array[a,87] (* James C. McMahon, Aug 25 2025 *)

PARI

A007947(n) = factorback(factorint(n)[, 1]);
a(n) = {my(cnt = 1, x = n); while(x != A007947(x), x += A007947(x); cnt++); cnt};