A355579 Numbers k such that A072079(k)/k sets a new record.
1, 2, 4, 6, 12, 24, 36, 48, 72, 144, 288, 432, 864, 1728, 2592, 3456, 5184, 10368, 20736, 31104, 41472, 62208, 124416, 248832, 373248, 746496, 1492992, 2239488, 2985984, 4478976, 8957952, 17915904, 26873856, 53747712, 107495424, 161243136, 214990848, 322486272
Offset: 1
Keywords
Examples
The numbers of 3-smooth divisors of the first 6 positive integers are 1, 3, 4, 7, 1 and 12. The corresponding values of A072079(k)/k are 1, 3/2, 4/3, 7/4, 1/5 and 2. The record values, 1, 3/2, 7/4 and 2, occur at 1, 2, 4 and 6, the first 4 terms of this sequence.
Links
- Michael S. Branicky, Table of n, a(n) for n = 1..4370
Programs
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Mathematica
s[n_] := Module[{e = IntegerExponent[n, {2, 3}], p}, p = {2, 3}^e; If[Times @@ p == n, (2^(e[[1]] + 1) - 1)*(3^(e[[2]] + 1) - 1)/(2*n), 0]]; sm = 0; seq = {}; Do[sn = s[n]; If[sn > sm, sm = sn; AppendTo[seq, n]], {n, 1, 10^6}]; seq -
PARI
lista(nmax) = {my(list = List(), rmax = 0, e2, e3, r); for(n = 1, nmax, e2 = valuation(n, 2); e3 = valuation(n, 3); r = if(2^e2 * 3^e3 == n, (2^(e2 + 1) - 1)*(3^(e3 + 1) - 1)/(2*n), 0); if(r > rmax, rmax = r; listput(list, n))); Vec(list)}; -
Python
from fractions import Fraction from sympy import multiplicity as v from itertools import count, takewhile def f(n): return Fraction((2**(v(2, n)+1)-1) * (3**(v(3, n)+1)-1)//2, n) def smooth3(lim): pows2 = list(takewhile(lambda x: x
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