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A355579
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Numbers k such that A072079(k)/k sets a new record.
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2
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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
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OFFSET
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1,2
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COMMENTS
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All the terms are 3-smooth numbers (A003586).
Equivalently, 3-smooth numbers k such that A000203(k)/k sets a new record.
Analogous to superabundant numbers (A004394) with 3-smooth numbers only.
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LINKS
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FORMULA
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Limit_{n->oo} A072079(a(n))/a(n) = lim_{n->oo} A000203(a(n))/a(n) = 3.
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EXAMPLE
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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.
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MATHEMATICA
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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
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PROG
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(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<lim, (2**i for i in count(0))))
pows3 = list(takewhile(lambda x: x<lim, (3**i for i in count(0))))
return sorted(c*d for c in pows2 for d in pows3 if c*d <= lim)
def aupto(lim):
data, records, record = smooth3(lim), [], -1
for argv, v in zip(data, map(f, data)):
if v > record: record = v; records.append(argv)
return records
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CROSSREFS
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KEYWORD
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nonn
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AUTHOR
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STATUS
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approved
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