|
|
A113467
|
|
Least k such that k, k+n and k+2n have the same number of divisors.
|
|
3
|
|
|
33, 3, 119, 3, 77, 5, 8, 3, 77, 3, 35, 5, 8, 3, 187, 6, 21, 5, 8, 3, 145, 33, 39, 5, 8, 39, 8, 3, 33, 7, 15, 12, 189, 3, 28, 7, 21, 3, 55, 3, 33, 5, 8, 66, 209, 69, 35, 5, 8, 3, 115, 39, 141, 5, 51, 6, 8, 27, 15, 7, 21, 66, 95, 3, 40, 5, 27, 3, 8, 15, 35, 7, 69, 55, 287, 6, 65, 11, 8, 3, 24
(list;
graph;
refs;
listen;
history;
text;
internal format)
|
|
|
OFFSET
|
1,1
|
|
COMMENTS
|
|
|
LINKS
|
|
|
EXAMPLE
|
a(7) = 8 because 8, 15 and 22 each have 4 divisors.
|
|
MATHEMATICA
|
snd[n_]:=Module[{k=1}, While[Length[Union[DivisorSigma[0, {k, k+n, k+2n}]]]>1, k++]; k]; Array[snd, 90] (* Harvey P. Dale, Aug 20 2017 *)
|
|
PROG
|
(PARI) a(n) = {k = 1; until ((numdiv(k) == numdiv(k+n)) && (numdiv(k) == numdiv(k+2*n)), k++); return (k); } \\ Michel Marcus, Jun 16 2013
|
|
CROSSREFS
|
|
|
KEYWORD
|
easy,nonn
|
|
AUTHOR
|
|
|
STATUS
|
approved
|
|
|
|