A119397 -id:A119397 - cp's OEIS frontend

A094501 Smallest number that requires n iterations of the sum of digits of the divisors (A034690) to reach 15.

15, 8, 7, 4, 3, 2, 19, 12, 6, 5, 13, 9, 10, 16, 30, 18, 34, 36, 66, 162, 924, 71820, 127005777360

Offset: 0

Jason Earls, Jun 05 2004

			a(0)=15 trivially because 15 is reached in no steps (number of steps is 0);
a(1)=8 because divisors of 8 are 1,2,4,8 with sum of digits = 15 hence 15 is reached in 1 steps (number of steps is 1);
a(2)=7 because divisors of 7 are 1,7 with sum of digits =8 and we need another one step to reach 15 (number of steps is 2);
a(3)=4 because divisors of 4 are 1,2,4 with sum of digits =7 and we need another two steps to reach 15 (number of steps is 3);
a(20)=924 because starting with 924 we have the trajectory 924, 168, 102, 36, 46, 18, 30, 27, 22, 9, 13, 5, 6, 12, 19, 11, 3, 4, 7, 8, 15 reaching 15 in 20 steps.
a(21)=71820 because starting with 71820 we have the trajectory 71820, 1104, 168, 102, 36, 46, 18, 30, 27, 22, 9, 13, 5, 6, 12, 19, 11, 3, 4, 7, 8, 15 reaching 15 in 21 steps. - _Sean A. Irvine_, Oct 04 2009

Eric Angelini et al., List the dividers [sic], sum the digits, lost messages reconstructed by _N. J. A. Sloane_, Dec 21 2024

Cf. A086793, A095347, A119397.

See A260060 for another variant.

import Data.List (elemIndex); import Data.Maybe (fromJust)
a094501 = (+ 2) . fromJust . (`elemIndex` a086793_list)
-- Reinhard Zumkeller, Nov 08 2015

f[n_] := Block[{i = 0}, NestWhile[(i++; Plus @@ Flatten@ IntegerDigits@ Divisors@#) &, n, # != 15 &]; i]; t = Table[0, {100}]; Do[ a = f[n]; If[ t[[a]] < 101 && t[[a]] == 0, t[[a]] = n], {n, 2, 10^8}]; t (* Robert G. Wilson v, May 16 2006 *)

A094501(n)=for(k=2, 9e9, A086793(k)==n&&return(k)) \\ M. F. Hasler, Nov 08 2015

Examples provided by Zak Seidov, May 16 2006
Edited by N. J. A. Sloane at the suggestion of Andrew S. Plewe, May 10 2007
a(22) found by exhaustive search by Sean A. Irvine, Oct 04 2009
a(22) corrected by Donovan Johnson and Sean A. Irvine

A119396 Numbers n such that A086793(n)=20.

Original entry on oeis.org

924, 1104, 1134, 1540, 1650, 1760, 1820, 1908, 1992, 2016, 2288, 2556, 2632, 2744, 2860, 2940, 2970, 3000, 3192, 3204, 3220, 3248, 3400, 3630, 3738, 3784, 3840, 3852, 3880, 3968, 3990, 4134, 4260, 4410, 4464, 4674, 4736, 4860, 4875, 4930, 4992, 5016

Offset: 1

Zak Seidov, May 17 2006

Some trajectories are: 924,168,102,36,46,18,30,27,22,9,13,5,6,12,19,11,3,4,7,8,15 1104,168,102,... 1540,162,66,36,... 1650,162,66,36,... 2016,297,66,36,... 2940,297,66,36,... 3192,312,102,36,... All trajectories eventually join one of previous trajectories.

			924 is a term because it reaches 15 in 20 steps with this trajectory 924,168,102,36,46,18,30,27,22,9,13,5,6,12,19,11,3,4,7,8,15.

Robert Israel, Table of n, a(n) for n = 1..10000

Cf. A034690, A086793, A094501, A095347, A114527, A118358, A119397, A119398, A260059.

f:= proc(n) option remember; local t;
  if kernelopts(level) > 460 then return FAIL fi;
  t:= add(convert(convert(d,base,10),`+`),d=numtheory:-divisors(n));
  1+procname(t)
end proc:
f(15):= 0:
f(1):= FAIL:
Res:= NULL: count:= 0:
for n from 1 while count < 100 do
  if f(n) = 20 then
    count:= count+1;
    Res:= Res, n;
   fi
od:
Res; # Robert Israel, Apr 03 2018

Edited by Robert Israel, Apr 03 2018

cp's OEIS Frontend

A094501 Smallest number that requires n iterations of the sum of digits of the divisors (A034690) to reach 15.

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