A236385 -id:A236385 - cp's OEIS frontend

A114440 Numbers which divided by the sum of their digits (Harshad or Niven numbers) give integers which are also divisible by the sum of their digits (until a single-digit Harshad remains).

1, 2, 3, 4, 5, 6, 7, 8, 9, 12, 18, 21, 24, 27, 36, 42, 45, 48, 54, 63, 72, 81, 84, 108, 162, 216, 243, 324, 378, 405, 432, 486, 648, 756, 864, 972, 1296, 1458, 1944, 2916, 3402, 4374, 5832, 6804, 7290, 8748, 11664, 13122, 13608, 15552, 17496, 23328, 26244

Offset: 1

Piotr K. Olszewski (piotrkornelolszewski(AT)poczta.onet.pl), Feb 14 2006

The sequence is finite with a(15095), a 1434-digit number, being the final term. - Hans Havermann and Ray Chandler, Jan 21 2014

			The number 216 is a term of the sequence because it is divisible by the sum of its digits: 2+1+6=9; 216/9=24. Also, the successive quotients are divisible by the sum of their digits, until a single-digit Harshad remains: 24: 2+4=6; 24/6=4 and 4: 4/4=1.

Donovan Johnson, Table of n, a(n) for n = 1..235 (terms < 10^17)
Hans Havermann and Ray Chandler, Table of n, a(n) for n = 1..15095 (9.3 MB file)
Kornel, Ojciec i Syn (Polish) "Father and Son", mentions the term 216.
David W. Wilson, Ray Chandler, Alonso Del Arte, M. F. Hasler, Hans Havermann, Alex Meiburg, N. J. A. Sloane, Hugo Van Der Sanden, and Allan Wechsler, As much as I hate "base" sequences..., Copies of various posts to the Sequence Fans Mailing List, Circa January 2014. Assembled by _N. J. A. Sloane_, Dec 23 2024

Cf. A005349, A097569, A235600, A235601, A236295, A236362, A236363, A236385.

s=w={1}; Do[t={}; Do[v=s[[k]]; u={}; Do[If[Total[IntegerDigits[c*v]]==c, AppendTo[u,c*v]], {c,2,7000}]; t=Join[t,u], {k,Length[s]}]; s=Sort[t]; w=Join[w,s], {440}]; Union[w] (* Hans Havermann, Jan 21 2014 *)

v=vector(118); for(n=1, 9, v[n]=n; print1(n ", ")); c=9; for(n=10, 10^9, d=length(Str(n)); m=n; s=0; for(j=1, d, s=s+m%10; m=m\10); if(s==1, next); if(n%s==0, m=n/s, next); forstep(j=c, 1, -1, if(v[j]<=m, if(v[j]==m, c++; v[c]=n; print1(n ", ")); next(2)))) /* Donovan Johnson, Apr 09 2013 */

Offset corrected by Donovan Johnson, Apr 09 2013
a(54)-a(235) from Donovan Johnson, Apr 09 2013
a(236)-a(15095) from Hans Havermann and Ray Chandler, Jan 21 2014

A235601 Smallest number m such that repeated application of A235600 takes n steps to reach 1, where A235600(k) = k/A007953(k) if the digital sum A007953(k) divides k, A235600(k) = k otherwise.

Original entry on oeis.org

1, 2, 12, 108, 1944, 52488, 1102248, 44641044, 2008846980, 108477736920, 6508664215200, 421761441144960, 22142475660110400, 1793540528468942400, 160701231350817239040, 15909421903730906664960, 1874419162475932276162560

Offset: 0

N. J. A. Sloane and David W. Wilson, Jan 18 2014

Numbers m > 1 which never reach 1 are not candidates for a(n).
There is no analog in base 2 (cf. A235602).
Comment from David W. Wilson, Jan 20 2013: let S(0) = {1};  for each n >= 1, compute the set S(n) of possible predecessors of elements of S(n-1).  Then a(n) is the smallest element of S(n). Using this approach, I was able to compute up to a(100).
The sequence is finite with a(440), a 1434-digit number being the final term. - Hans Havermann and Ray Chandler, Jan 21 2014
Sequence A236338 gives the count of iterations of A235600 required to reach 1 when starting from any n. Otherwise said: This sequence is the RECORDS transform of A236338. - M. F. Hasler, Jan 22 2014
The terms are a proper subset of A114440. - Robert G. Wilson v, Jan 22 2014

			a(4) = 1944: 1944 ->1944/18 = 108 -> 108/9 = 12 -> 12/3 = 4 -> 4/4 = 1 in 4 steps.

Hans Havermann and Ray Chandler, Table of n, a(n) for n = 0..440 [First 100 terms were computed by David W. Wilson]
David W. Wilson, Ray Chandler, Alonso Del Arte, M. F. Hasler, Hans Havermann, Alex Meiburg, N. J. A. Sloane, Hugo Van Der Sanden, and Allan Wechsler, As much as I hate "base" sequences..., Copies of various posts to the Sequence Fans Mailing List, Circa January 2014. Assembled by _N. J. A. Sloane_, Dec 23 2024

Cf. A005349, A007953, A114440, A235600, A236385.

s={1}; Print[s[[1]]]; Do[t={}; Do[v=s[[k]]; u={}; Do[If[Total[IntegerDigits[c*v]]==c, AppendTo[u,c*v]], {c,2,7000}]; t=Join[t,u], {k,Length[s]}]; s=Sort[t]; Print[s[[1]]], {440}] (* Hans Havermann, Jan 21 2014 *)

a(8) from Hans Havermann, Jan 19 2014
a(9)-a(100) from David W. Wilson, Jan 21 2014
a(101)-a(440) from Hans Havermann and Ray Chandler, Jan 21 2014

cp's OEIS Frontend

A114440 Numbers which divided by the sum of their digits (Harshad or Niven numbers) give integers which are also divisible by the sum of their digits (until a single-digit Harshad remains).

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