A316990 Smallest exponent m of n such that A289280(n) | n^m.

2, 2, 2, 2, 3, 2, 2, 2, 4, 2, 2, 2, 4, 2, 2, 2, 3, 2, 2, 3, 5, 2, 3, 2, 5, 2, 3, 2, 5, 2, 2, 4, 6, 2, 2, 2, 6, 4, 2, 2, 4, 2, 3, 2, 6, 2, 3, 2, 6, 4, 3, 2, 6, 2, 2, 4, 6, 2, 3, 2, 6, 2, 2, 3, 3, 2, 4, 4, 4, 2, 2, 2, 7, 4, 4, 2, 4, 2, 2, 2, 7, 2, 3, 3, 7, 5, 2, 2, 5, 2, 4, 5, 7, 3, 3, 2, 4, 2, 2, 2, 3, 2, 3, 3, 7

Offset: 2

Author

Michael De Vlieger, Jul 28 2018

Keywords

• nonn

Comments

Consider the least k > n such that k | n^m for m > 1. (We note that k cannot divide n if k exceeds n.) Values of k appear in A289280, while this sequence lists values of m.

If row n of A162306 were extended to include terms greater than n, A289280(n) would be the first term to follow those already in the row.

a(n) = 2 for n with omega(n) = 1. In other words, A289280(n) | n^2 for n = p^e with one distinct prime divisor, since A289280(p^e) = p^(e+1).

First indices of {2, 3, 4, 5, ..., m} are {2, 6, 10, 22, 34, 74, 134, 262, 514, 1042, 2062, 4106, 8198, 16418, 32822, 65542, ...}, i.e., the least even squarefree semiprime s > 2^(m - 1) for m > 2. This is because 2 is the smallest prime, and minimal multiplicity of 2 increases a(n) most efficiently. Let n = Product(p^e) and A289280(n) = Product(p^d), knowing there may be different values of p. a(n) = max(ceiling(d/e)) for d and e that pertain to the same prime p. Examples: for n = 10 = 2*5, A289280(10) = 16 = 2^4. Thus we are concerned with the ratio 4/1, and a(10) = 4. For n = 12 = 2^2*3 we have A289280(12) = 16; here we have the ratio 4/2 = 2. The greater multiplicity of 2 reduces a(n) for n = 12.

Examples

			For n = 2, A289280(n) = 4 = 2^2, the square of n = 2: thus a(2) = 2.
For n = 8, A289280(n) = 16 = 2^4; 2^4 | 8^2, thus a(8) = 2.
For n = 10, the least k > 10 that divides 10^e for e > 1 is 16. 16 | 10^4, thus a(n) = 4.
		

Links

• Michael De Vlieger, Table of n, a(n) for n = 2..10000

Crossrefs

Cf. A162306, A289280.

Programs

Mathematica

Table[If[PrimePowerQ@ n, 2, Block[{k = n + 1, m = 1}, While[PowerMod[n, k, k] != 0, k++]; While[PowerMod[n, m, k] != 0, m++]; m]], {n, 2, 106}]