A114440 -id:A114440 - cp's OEIS frontend

A236295 Number of terms of A114440 that require n steps to arrive at 1.

1, 8, 14, 15, 17, 21, 23, 21, 22, 27, 27, 30, 32, 26, 20, 23, 25, 22, 17, 20, 22, 32, 41, 34, 35, 38, 43, 45, 49, 51, 52, 43, 44, 41, 51, 40, 38, 37, 30, 30, 35, 26, 24, 24, 30, 22, 21, 25, 21, 18, 24, 19, 25, 23, 23, 22, 31, 22, 27, 28, 25, 21, 21, 25

Offset: 0

Hans Havermann, Jan 21 2014

Because A114440 is finite, this sequence is necessarily also finite. The sum of all 441 terms of this sequence, a(0) to a(440), is 15095, the number of terms in A114440.

			There is 1 terms of A114440 that requires no steps to reach 1: {1}. So a(0) = 1.
There are 8 terms of A114440 that require one step to reach 1: {2,3,4,5,6,7,8,9}. So a(1) = 8.
There are 14 terms of A114440 that require two steps to reach 1: {12,18,21,24,27,36,42,45,48,54,63,72,81,84}. So a(2) = 14.
...
There are 4 terms of A114440 that require 440 steps to reach 1, so a(440) = 4. There are no terms that require more than 440 steps to reach 1.

Hans Havermann, Table of n, a(n) for n = 0..440
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. A114440, A235601.

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)).

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

A334416 Numbers m such that (m / sum of digits of m) is a palindrome.

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, 110, 132, 198, 220, 264, 330, 396, 440, 550, 594, 605, 660, 715, 770, 792, 825, 880, 935, 990, 1010, 1056, 1188, 1212, 1310, 1386, 1452, 1584, 1782, 1810, 1812, 1815, 1818, 1848

Offset: 1

Bernard Schott, Apr 28 2020

Not to be confused with A114440 whose first 23 terms are identical to the terms of this sequence, while A114440(24) = 108 and a(24) = 110.

			The number 264 is a term of the sequence because it is divisible by the sum of its digits: 2+6+4=12; 264/12=22 and 22 is a palindrome.

Giovanni Resta, Table of n, a(n) for n = 1..10000
Index entries for sequences related to palindromes

Cf. A001101 (similar for primes).
Cf. A334417 (resulting palindromic quotients).
Subsequence of A005349 (Niven (or Harshad) numbers).
Subsequence: A276142 (palindromic terms).

Select[Range[2000], PalindromeQ[# / Plus @@ IntegerDigits[#]] &] (* Amiram Eldar, Apr 28 2020 *)

isok(m) = iferr(my(d=digits(m/sumdigits(m))); d==Vecrev(d), E, 0); \\ Michel Marcus, Apr 29 2020

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.

A343680 Niven (or Harshad) numbers which when divided by sum of their digits, give a quotient which is a Zuckerman number.

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, 135, 198, 216, 324, 648, 1008, 1035, 1050, 1152, 1215, 1344, 1380, 1680, 1725, 2016, 2376, 2592, 2625, 2688, 2997, 3675, 3816, 3888, 5616, 5670, 6912, 10008, 10017, 10035, 10044

Offset: 1

Bernard Schott, Apr 26 2021

The first 24 terms of A114440 are the first 24 terms of this sequence, then A114440(25) = 162, while a(25) = 135.

			84 is a Niven number as 84/(8+4) = 7, 7/7 = 1 so 7 is a Zuckerman number, and 84 is a term.
108 is a Niven number as 108/(1+0+8) = 12, 12/(1*2) = 6 so 12 is a Zuckerman number, and 108 is a term.

Giovanni Resta, Zuckerman Numbers, Numbers Aplenty.

Cf. A005349, A007602, A114440, A235507.

zuckQ[n_] := IntegerQ[n] && (prod = Times @@ IntegerDigits[n]) > 0 && Divisible[n, prod]; Select[Range[10^4], zuckQ[#/Plus @@ IntegerDigits[#]] &] (* Amiram Eldar, Apr 26 2021 *)

isz(n) = my(p=vecprod(digits(n))); p && !(n % p); \\ A007602
isok(n) = my(s=sumdigits(n)); !(n%s) && isz(n/s); \\ Michel Marcus, Apr 26 2021

More terms from Michel Marcus, Apr 26 2021

A343744 Zuckerman numbers which divided by the product of their digits give integers which are also divisible by the product of their digits, and so on, until result is 1.

Original entry on oeis.org

1, 2, 3, 4, 5, 6, 7, 8, 9, 12, 15, 24, 36, 128, 135, 144, 175, 384, 672, 735, 1296, 1575, 82944, 139968, 1492992, 27869184

Offset: 1

Bernard Schott, Apr 27 2021

Repunits >= 11 (A002275) are not in the sequence because, as they are fixed points of this map, they don't fit the definition.
Question: is this sequence finite as the similar sequence with Niven numbers (A114440) that has 15095 terms?
No other terms up to 2*10^9. - Michel Marcus, Apr 27 2021
From David A. Corneth, Apr 27 2021: (Start)
Terms are 7-smooth. Any prime factor > 7 will not be divided away by dividing by product of digits.
Any number k > a(26)*10^163 with product of digits vp > 0 has k/vp > a(26) so it suffices to check all candidates <= a(26)*10^163. Doing so gives no more terms so this sequence is finite and full. (End)
The number of steps needed to reach 1, has a maximum of 3, which occurs for n = 21, 23..26. - A.H.M. Smeets, Apr 29 2021

			The integer 1296 is divisible by the product of its digits as 1296/(1*2*9*6) = 12, then 12/(1*2) = 6 and 6/6 = 1; hence, 1296 is a term of this sequence.

Giovanni Resta, Zuckerman numbers, Numbers Aplenty.

Cf. A007602, A288069.
Cf. A114440 (similar for Harshad numbers).
Subsequence of A002473 and of A343681.

f[n_] := If[(prod = Times @@ IntegerDigits[n]) > 0 && Divisible[n, prod], n/prod, 0]; Select[Range[10^5], FixedPointList[f, #][[-1]] == 1 &] (* Amiram Eldar, Apr 27 2021 *)

isz(n) = my(p=vecprod(digits(n))); p && !(n % p); \\ A007602
isok(n) = if (n==1, return(1)); my(m=n); until(m==1, if (isz(m), my(nm = m/vecprod(digits(m))); if (nm==m, return (0), m = nm), return(0))); return(1); \\ Michel Marcus, Apr 27 2021

def proddigit(n):
    p = 1
    while n > 0:
        n, p = n//10, p*(n%10)
    return p
n, a = 1, 1
while n > 0:
    aa, pa = a, proddigit(a)
    while pa > 1 and aa%pa == 0 and aa > 1:
        aa = aa//pa
        pa = proddigit(aa)
    if aa == 1:
        print(n,a)
        n = n+1
    a = a+1 # A.H.M. Smeets, Apr 29 2021

a(26) from Michel Marcus, Apr 27 2021

cp's OEIS Frontend

A236295 Number of terms of A114440 that require n steps to arrive at 1.

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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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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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A334416 Numbers m such that (m / sum of digits of m) is a palindrome.

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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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A343680 Niven (or Harshad) numbers which when divided by sum of their digits, give a quotient which is a Zuckerman number.

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A343744 Zuckerman numbers which divided by the product of their digits give integers which are also divisible by the product of their digits, and so on, until result is 1.

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