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A337686
a(n) is the least multiplier k such that n*k has twice as many divisors as n.
4
2, 3, 2, 3, 2, 4, 2, 3, 2, 3, 2, 5, 2, 3, 2, 3, 2, 4, 2, 3, 2, 3, 2, 5, 2, 3, 2, 3, 2, 4, 2, 3, 2, 3, 2, 5, 2, 3, 2, 3, 2, 4, 2, 3, 2, 3, 2, 5, 2, 3, 2, 3, 2, 4, 2, 3, 2, 3, 2, 6, 2, 3, 2, 3, 2, 4, 2, 3, 2, 3, 2, 5, 2, 3, 2, 3, 2, 4, 2, 3, 2, 3, 2, 5, 2, 3, 2, 3, 2, 4
OFFSET
1,1
COMMENTS
The zeros in A139315 are the missing values in this sequence (see A337709).
There are no 1's in this sequence. a(n) = 2 for all odd n and a(n) >= 3 for all even n. - J. Lowell, Sep 15 2020
Empirical observation: A007978(n) - a(n) = 1 for n = 60*A206547(n), = 2 for n = 420*A007310(n), else = 0. - Hugo Pfoertner, Sep 30 2020
LINKS
Hugo Pfoertner, Illustration of ratio A007978(n) / a(n), using Plot 2.
FORMULA
a(n) = A129902(n)/n.
EXAMPLE
a(1) = 2 because 1 has 1 divisor, 1*2 has 2 divisors, so 2 is the least multiplier to apply to 1 to get twice as many divisors.
MATHEMATICA
nn = 105; Do[d[i] = DivisorSigma[0, i], {i, 12 nn}]; Reap[Do[m = 2; While[d[m i] != 2 d[i], m++]; Sow[m ], {i, nn}]][[-1, -1]] (* Michael De Vlieger, Jan 10 2022 *)
PROG
(PARI) a(n) = {my(k=1); while (numdiv(n*k) != 2*numdiv(n), k++); k; }
CROSSREFS
Cf. A000005, A129902, A139315, A337709 (missing values).
Sequence in context: A259940 A228829 A341982 * A007978 A245575 A333600
KEYWORD
nonn
AUTHOR
Michel Marcus, Sep 15 2020
STATUS
approved