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

[M. F. Hasler]

One can also partition the positive numbers in an infinite table where
row n holds the numbers k for which A034690(k)=n :
n=0: (the only finite row, I think)
1,15
n=1: (this is your sequence without the fixed point 15)
8,14,20,26,59,62,122,123,143,149,167,206,239,257,293,302,341,347,383,419,422,491,509,563,617,653,743,761,941,1049,1133,1193,1202,...
n=2:
7,21,39,43,52,57,61,67,84,93,112,124,139,151,157,189,193,199,201,219,223,229,241,283,309,313,327,331,337,373,378,379,381,397,409,...
n=3:
4,35,44,48,49,50,65,99,104,105,116,121,125,132,140,141,147,150,152,155,161,169,195,215,225,242,256,260,266,273,275,305,332,344,348,...
n=4:
3,54,56,128,171,182,188,216,248,252,261,264,268,270,333,387,408,434,462,468,476,484,489,495,500,507,522,528,540,548,549,558,597,...
n=5:
2,11,101,136,138,160,184,190,208,232,238,255,282,290,318,328,345,351,356,357,370,374,388,406,418,470,477,539,561,663,670,705,715,...
n=6:
19,37,73,109,127,163,181,271,307,396,433,523,541,613,631,784,798,811,876,...
-- 
Maximilian

On Mon, Nov 9, 2015 at 1:00 PM, Veikko Pohjola <veikko at nordem.fi> wrote:
> Hi seqfans,
>
> The following sequence is composed of numbers n such that the sum of digits of all divisors of n equals 15:
> 8, 14, 15, 20, 26, 59, 62, …
> It actually depicts the positions of number 15 in A034690.
> Cute but maybe without much interest.
>
> Veikko
>
>
> M. F. Hasler kirjoitti 9.11.2015 kello 2.07:
>
>> Eric,
>>
>> I found that this is  A094501.
>> I did not find at once this and
>> A086793 = Iterate the map n -> sum of digits of all divisors of n (cf.
>> A034690); sequence gives number of steps to reach 15.
>> because I considered the version "...to reach a fixed point (1 or 15)"
>> which starts with 1.
>> I hope it will be accepted to change A086793 as to prefix a(1)=0. [CORRECTED]
>>
>> Maximilian
>>
>>
>> On Sun, Nov 8, 2015 at 1:11 PM, Eric Angelini <Eric.Angelini at kntv.be> wrote:
>>>
>>> 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},