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A376140
The number of divisors of n whose prime factorization has maximum exponent that is smaller than the maximum exponent in the prime factorization of n.
1
0, 1, 1, 2, 1, 1, 1, 3, 2, 1, 1, 4, 1, 1, 1, 4, 1, 4, 1, 4, 1, 1, 1, 6, 2, 1, 3, 4, 1, 1, 1, 5, 1, 1, 1, 4, 1, 1, 1, 6, 1, 1, 1, 4, 4, 1, 1, 8, 2, 4, 1, 4, 1, 6, 1, 6, 1, 1, 1, 8, 1, 1, 4, 6, 1, 1, 1, 4, 1, 1, 1, 9, 1, 1, 4, 4, 1, 1, 1, 8, 4, 1, 1, 8, 1, 1, 1
OFFSET
1,4
COMMENTS
The maximum exponent in the prime factorization of 1 is considered to be A051903(1) = 0.
FORMULA
a(n) = Sum_{d|n} [m(d) = m(n)], where m(n) = A051903(n) and [] is the Iverson bracket.
If n = Product_{i} p_i^e_i (where p_i are distinct primes), then a(n) = Product_{i} (e_i + [e_i < Max_{i}(e_i)]).
a(n) <= 1 if and only if n is squarefree (A005117), and a(n) = 0 only for n = 1.
a(n) < A000005(n).
a(n) = A000005(A375932(n)) * A005361(A375931(n)) for n >= 2.
a(n) = A000005(n) * (A051903(n)/(A051903(n)+1))^A362611(n) for n >= 2.
MATHEMATICA
a[n_] := Module[{e = FactorInteger[n][[;; , 2]], m}, m = Max[e]; Times@@ ((Min[#, m-1] & /@ e) + 1)]; a[1] = 0; Array[a, 100]
PROG
(PARI) a(n) = if(n == 1, 0, my(e = factor(n)[, 2], m = vecmax(e)); vecprod(apply(x -> 1 + min(x, m-1), e)));
KEYWORD
nonn,easy
AUTHOR
Amiram Eldar, Sep 11 2024
STATUS
approved