A090624 If n = Product(pj^ej), i.e., written in its prime factorization, then a(n) = max_j{(pj-1)*ej}.

1, 2, 2, 4, 2, 6, 3, 4, 4, 10, 2, 12, 6, 4, 4, 16, 4, 18, 4, 6, 10, 22, 3, 8, 12, 6, 6, 28, 4, 30, 5, 10, 16, 6, 4, 36, 18, 12, 4, 40, 6, 42, 10, 4, 22, 46, 4, 12, 8, 16, 12, 52, 6, 10, 6, 18, 28, 58, 4, 60, 30, 6, 6, 12, 10, 66, 16, 22, 6, 70, 4, 72, 36, 8, 18, 10, 12, 78, 4, 8, 40, 82, 6

Offset: 2

Author

Henry Bottomley, Dec 06 2003

Keywords

• nonn

Comments

The highest power of k dividing n! (A090622) is close to, but below, n/a(k).

Examples

			72 = 2^3*3^2 so a(72) = max((2-1)*3, (3-1)*2) = max(3,4) = 4.
		

Links

• David W. Wilson, Table of n, a(n) for n = 2..10000
• A. M. Oller-Marcen and J. Maria Grau, On the Base-b Expansion of the Number of Trailing Zeros of b^k!, J. Int. Seq. 14 (2011) 11.6.8, propos. 3.

Programs

Maple

seq(max(map(t ->(t[1]-1)*t[2], ifactors(n)[2])),n=2..100); # Robert Israel, Sep 06 2016

Mathematica

a[n_] := Module[{aux = FactorInteger[n]},Max[Table[aux[[i, 2]]*(aux[[i, 1]] - 1), {i, 1, Length[aux]}]]] (* José María Grau Ribas, Feb 15 2010 *)

PARI

a(n)=my(f=factor(n)); vecmax(vector(#f~,i,(f[i,1]-1)*f[i,2])) \\ Charles R Greathouse IV, Sep 07 2016

Python

from sympy import factorint
def A090624(n): return max((p-1)*e for p, e in factorint(n).items()) # Chai Wah Wu, Apr 28 2023

Formula

a(p) = p-1; a(p^m) = (p-1)*m.

a(b*c) = max(a(b), a(c)) for b and c coprime.

a(n) = lim_{k->inf} k/A090622(k, n) = lim_{k->inf} (k/highest power of k dividing n!). - David W. Wilson, Sep 05 2016