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Smallest exponent m of n such that A289280(n) | n^m.
1

%I #11 Jan 19 2023 22:44:17

%S 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,

%T 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,

%U 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

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

%C 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.

%C 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.

%C 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).

%C 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.

%H Michael De Vlieger, <a href="/A316990/b316990.txt">Table of n, a(n) for n = 2..10000</a>

%e For n = 2, A289280(n) = 4 = 2^2, the square of n = 2: thus a(2) = 2.

%e For n = 8, A289280(n) = 16 = 2^4; 2^4 | 8^2, thus a(8) = 2.

%e For n = 10, the least k > 10 that divides 10^e for e > 1 is 16. 16 | 10^4, thus a(n) = 4.

%t 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}]

%Y Cf. A162306, A289280.

%K nonn

%O 2,1

%A _Michael De Vlieger_, Jul 28 2018