[seqfan] Re: List the dividers, sum the digits

[Eric Angelini]

Hello SeqFans,
Jean-Marc Falcoz has computed the
22 first terms of the A(n) seq where
a(1) is the smallest integer looping in
1 stage, a(2) the smallest looping in
2 stages, a(3) in 3 stages, etc.
It seems that computing a(23) will
take a lot of time [the graph of A(n)
climbs exponentially after a(22)] --
not to say a(24)]. If someone wants
to give it a try and submit the seq to
the OEIS, please do...
Best,
É.

{1},
{8, 15},
{7, 8, 15},
{4, 7, 8, 15},
{3, 4, 7, 8, 15},
{2, 3, 4, 7, 8, 15},
{19, 11, 3, 4, 7, 8, 15},
{12, 19, 11, 3, 4, 7, 8, 15},
{6, 12, 19, 11, 3, 4, 7, 8, 15},
{5, 6, 12, 19, 11, 3, 4, 7, 8, 15},
{13, 5, 6, 12, 19, 11, 3, 4, 7, 8, 15},
{9, 13, 5, 6, 12, 19, 11, 3, 4, 7, 8, 15},
{10, 9, 13, 5, 6, 12, 19, 11, 3, 4, 7, 8, 15},
{16, 22, 9, 13, 5, 6, 12, 19, 11, 3, 4, 7, 8, 15},
{30, 27, 22, 9, 13, 5, 6, 12, 19, 11, 3, 4, 7, 8, 15},
{18, 30, 27, 22, 9, 13, 5, 6, 12, 19, 11, 3, 4, 7, 8, 15},
{34, 18, 30, 27, 22, 9, 13, 5, 6, 12, 19, 11, 3, 4, 7, 8, 15},
{36, 46, 18, 30, 27, 22, 9, 13, 5, 6, 12, 19, 11, 3, 4, 7, 8, 15},
{66, 36, 46, 18, 30, 27, 22, 9, 13, 5, 6, 12, 19, 11, 3, 4, 7, 8, 15},
{162, 66, 36, 46, 18, 30, 27, 22, 9, 13, 5, 6, 12, 19, 11, 3, 4, 7, 8, 15},
{924, 168, 102, 36, 46, 18, 30, 27, 22, 9, 13, 5, 6, 12, 19, 11, 3, 4, 7, 8, 15},
{71820, 1104, 168, 102, 36, 46, 18, 30, 27, 22, 9, 13, 5, 6, 12, 19, 11, 3, 4, 7, 8, 15}
...
Best,
É.

Le 8 nov. 2015 à 17:23, Neil Sloane <njasloane at gmail.com<mailto:njasloane at gmail.com>> a écrit :

Eric,
Bob is right, I think

You should have said 13 has divisors 1 and 13 so the digit sum is 5, not 14

Best regards
Neil


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