A235600 -id:A235600 - cp's OEIS frontend

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.

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

A236338 Number of iterations of A235600 to reach 1 when starting with n, or -1 if 1 is never reached. (A235600(x) = x/sum_of_digits(x) if this is an integer, otherwise A235600(x) = x.)

Original entry on oeis.org

0, 1, 1, 1, 1, 1, 1, 1, 1, -1, -1, 2, -1, -1, -1, -1, -1, 2, -1, -1, 2, -1, -1, 2, -1, -1, 2, -1, -1, -1, -1, -1, -1, -1, -1, 2, -1, -1, -1, -1, -1, 2, -1, -1, 2, -1, -1, 2, -1, -1, -1, -1, -1, 2, -1, -1, -1, -1, -1, -1, -1, -1, 2, -1, -1, -1, -1, -1

Offset: 1

M. F. Hasler, Jan 22 2014

Based on an idea from N. J. A. Sloane, cf. link.
Numbers n = 10^k and also numbers not divisible by their digital sum A007953, are fixed points of A235600, therefore a(n) = -1 for these, except for a(1) = 0, cf. Example.
A235601(k) is the smallest n for which a(n) = k.

			a(1) = 0 since no iteration of A235600 is needed to reach 1.
a(n) = 1 for 1 <= n <= 9, since these n are equal to (thus divisible by) their sum of digits A007953(n), and 1 is reached upon the first iteration of A235600 (which consists of dividing n by its sum of digits).
a(10) = -1 since A007953(10) = 1 and therefore application of A235600 yields a constant sequence that never reaches 1.
a(11) = -1 since 11 is not divisible by A007953(11) = 2 and therefore application of A235600 yields a constant sequence that never reaches 1.
a(12) = 2 since A235600(12) = 12/(1+2) = 4 and A235600(4) = 4/4 = 1, reached after 2 iterations.

N. J. A. Sloane, in reply to D. W. Wilson, Re: As much as I hate "base" sequences..., SeqFan list, Jan 18 2014
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, A235601.

A236338 = n -> for(i=0,999,n==1&&return(i);if(n%sumdigits(n)||n==n\=sumdigits(n),return(-1)))

A236385 Largest 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, 9, 84, 1458, 39366, 1062882, 47829690, 1721868840, 92980917360, 5020969537440, 361509806695680, 26028706082088960, 1874066837910405120, 58110713122393733760, 3643185932897827553280, 393464080752965375754240

Offset: 0

Ray Chandler, Jan 24 2014

Ray Chandler, Table of n, a(n) for n = 0..322 [terms <= 1000 digits]
Ray Chandler, Table of n, a(n) for n = 0..440 [complete sequence]

Cf. A005349, A007953, A114440, A235600, A235601.

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).

Original entry on oeis.org

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

A235602 a(n) = n/wt(n) if wt(n) divides n, otherwise a(n) = n, where wt(n) is the binary weight of n (A000120).

Original entry on oeis.org

1, 2, 3, 4, 5, 3, 7, 8, 9, 5, 11, 6, 13, 14, 15, 16, 17, 9, 19, 10, 7, 22, 23, 12, 25, 26, 27, 28, 29, 30, 31, 32, 33, 17, 35, 18, 37, 38, 39, 20, 41, 14, 43, 44, 45, 46, 47, 24, 49, 50, 51, 52, 53, 54, 11, 56, 57, 58, 59, 15, 61, 62, 63, 64, 65, 33, 67, 34, 23, 70, 71, 36, 73, 74, 75, 76, 77, 78, 79, 40, 27, 82

Offset: 1

N. J. A. Sloane, Jan 18 2014

Amiram Eldar, Table of n, a(n) for n = 1..10000

Cf. A000079, A000120, A049445, A065878, A235600, A235601.

bw[n_]:=Module[{w=DigitCount[n,2,1]},If[Divisible[n,w],n/w,n]]; Array[ bw,90] (* Harvey P. Dale, Nov 06 2016 *)

a(n) = my(s=hammingweight(n)); if (n % s, n, n/s); \\ Michel Marcus, Jul 15 2021

From Amiram Eldar, Aug 04 2025: (Start)
a(n) = n if and only if n is in A065878 or A000079.
a(n) < n if and only if n is in A049445 but not in A000079. (End)

A236362 Number of steps for A114440(n) to reach 1.

Original entry on oeis.org

0, 1, 1, 1, 1, 1, 1, 1, 1, 2, 2, 2, 2, 2, 2, 2, 2, 2, 2, 2, 2, 2, 2, 3, 3, 3, 3, 3, 3, 3, 3, 3, 3, 3, 3, 3, 3, 3, 4, 4, 4, 4, 4, 4, 4, 4, 4, 4, 4, 4, 4, 4, 4, 4, 4, 5, 5, 5, 5, 5, 5, 5, 5, 5, 5, 5, 5, 5, 5, 5, 5, 5, 5, 5, 5, 5, 6, 6, 6, 6, 6, 6, 6, 6, 6, 6, 6, 6, 6, 6, 6, 6, 6, 6, 6, 6, 6, 6, 7, 6, 7, 7

Offset: 1

Ray Chandler, Jan 23 2014

Sequence suggested by David W. Wilson on the SeqFan mailing list.
Because A114440 is finite, this sequence is necessarily also finite.
Sequence is mostly nondecreasing with the following three exceptions: a(99)>a(100), a(226)>a(227), a(258)>a(259).

Ray Chandler, Table of n, a(n) for n = 1..15095

Cf. A005349, A007953, A114440, A235600, A236338.

A236363 Sum of digits of A114440(n).

Original entry on oeis.org

1, 2, 3, 4, 5, 6, 7, 8, 9, 3, 9, 3, 6, 9, 9, 6, 9, 12, 9, 9, 9, 9, 12, 9, 9, 9, 9, 9, 18, 9, 9, 18, 18, 18, 18, 18, 18, 18, 18, 18, 9, 18, 18, 18, 18, 27, 18, 9, 18, 18, 27, 18, 18, 27, 27, 27, 18, 36, 27, 27, 27, 18, 27, 27, 27, 27, 27, 27, 27, 27, 27, 36, 36, 36, 36, 27

Offset: 1

Ray Chandler, Jan 23 2014

Because A114440 is finite, this sequence is necessarily also finite.

Note that after a(23), all terms are divisible by 9 since all the terms that take three steps to arrive at 1 are multiples of 9 so the sum of digits of their predecessors must also be a multiple of 9.

Ray Chandler, Table of n, a(n) for n = 1..15095

Cf. A005349, A007953, A114440, A235600.

a(n) = A007953(A114440(n)).

cp's OEIS Frontend

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.

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A236338 Number of iterations of A235600 to reach 1 when starting with n, or -1 if 1 is never reached. (A235600(x) = x/sum_of_digits(x) if this is an integer, otherwise A235600(x) = x.)

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A236385 Largest 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.

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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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A235602 a(n) = n/wt(n) if wt(n) divides n, otherwise a(n) = n, where wt(n) is the binary weight of n (A000120).

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A236362 Number of steps for A114440(n) to reach 1.

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A236363 Sum of digits of A114440(n).

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