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User: Alon Ran

Alon Ran has authored 3 sequences.

A336383 a(n) is the smallest number such that, with f(x) = x - (the product of the digits of x), f(a(n)) reaches a fixed point after n iterations.

0, 1, 21, 31, 42, 52, 73, 81, 319, 391, 463, 583, 2911, 3667, 6451, 8793, 9927, 237126, 254158, 278393, 2561363, 9398143, 9431623, 9951265, 83543869, 83896381, 83935261, 2843233127, 2847297383, 2853748583, 2885762663, 266998137657, 685718563667, 688373877587

Offset: 0

Alon Ran, Jul 19 2020

A fixed point occurs once the function returns a number that contains the digit 0. After that, the product of the digits will be 0, and so subtracting it from the number will be idempotent.

This sequence is conceptually similar to A003001, though unlike the latter, it is probably infinite.

			a(9) = 391 because:
   1: 391 - 3*9*1 = 364
   2: 364 - 3*6*4 = 292
   3: 292 - 2*9*2 = 256
   4: 256 - 2*5*6 = 196
   5: 196 - 1*9*6 = 142
   6: 142 - 1*4*2 = 134
   7: 134 - 1*3*4 = 122
   8: 122 - 1*2*2 = 118
   9: 118 - 1*1*8 = 110
After iteration 9, the function becomes idempotent:
  10: 110 - 1*1*0 = 110
  11: 110 - 1*1*0 = 110
  12: 110 - 1*1*0 = 110
  ...
Additionally, 391 is the smallest number with this property. Thus, it is a(9).

Cf. A003001, A070565.

nmax = 20; tab = ConstantArray[Null, nmax];
For[k = 0, k <= 1000000, k++,
 l=Length@  NestWhileList[#-Times @@ IntegerDigits[#] &,k,UnsameQ[##] &, 2]-2;
If[tab[[l+1]] == Null, tab[[l+1]] = k]]; tab  (* Robert Price, Sep 13 2020 *)

f(m) = m - vecprod(digits(m)) + (m==0);
lista(nn) = {my(c, m, t); for(k=0, nn, c=0; m=k; while(m!=(m=f(m)), c++); if(c==t, print1(k, ", "); t++)); } \\ Jinyuan Wang, Aug 14 2020

def f(x):
    prod = 1
    for digit in str(x):
        prod *= int(digit)
    return x - prod
def a(n):
    i = 0
    iteration = 0
    while iteration != n:
        i += 1
        j = i
        iteration = 0
        new_j = f(j)
        while j != new_j:
            iteration += 1
            j = new_j
            new_j = f(j)
    return i

a(27)-a(30) from Jinyuan Wang, Aug 14 2020

a(31)-a(33) added by Michael S. Branicky, Aug 29 2020

A331206 Numbers k such that A053985(k) divides k.

Original entry on oeis.org

1, 2, 3, 4, 5, 6, 8, 10, 12, 15, 16, 17, 20, 21, 24, 30, 32, 34, 40, 42, 48, 51, 60, 63, 64, 65, 68, 69, 80, 81, 84, 85, 96, 102, 120, 126, 128, 130, 136, 138, 160, 162, 168, 170, 192, 195, 204, 207, 240, 243, 252, 255, 256, 257, 260, 261, 272, 273, 276, 277

Offset: 1

Alon Ran, Jan 12 2020

It appears that A053985(a(n)) / a(n) is always either 1, -1, 3 or -3.

This sequence seems to contain A329000.

			15 is written 1111 base 2 and (-2)^3 + (-2)^2 + (-2)^1 +(-2)^0 = -8 + 4 - 2 + 1 = -5, 15 is divisible by -5.

is(k) = k%fromdigits(binary(k), -2) == 0; \\ Jinyuan Wang, Jan 15 2020

A329126 a(n) is the least positive number which yields a multiple of n when its decimal digits (which are necessarily 0's and 1's) are read in any base.

Original entry on oeis.org

1, 110, 101010, 111100, 100010001000100010, 1111110, 10000010000010000010000010000010000010, 11111111000, 1010101010101010100, 1100110011001100110, 100000000010000000001000000000100000000010000000001000000000100000000010000000001000000000100000000010, 11111111111100

Offset: 1

Alon Ran, Nov 05 2019

a(n) might be called the "smallest trans-basic multiple of n."
In order to be a valid binary number, the terms may contain only 0's and 1's.
The number of 1's in a(n) is conjectured to be n; the number of 0's separating each one digit is usually A268336(n)-1 for small n. The number of trailing 0's is A051903(n).
The first 21 terms have been found and verified to be minimal via an advanced search; the 13th term (see b-file) contains 146 digits, and in general for every prime n the corresponding a(n) is conjectured to contain 2+(n-1)^2 0's and 1's.
A lower bound for a(n) is given by a(A032742(n)). Proof: If a(n) were smaller than a(A032742(n)), then a(A032742(n)) would not be the smallest trans-basic multiple of A032742(n); a(n) would be. By definition a(n) is the smallest trans-basic multiple of n, so we have a contradiction; QED.
To verify a trans-basic multiple of n for n > 2, one must only: A) make sure the string has some multiple of n of '1' digits; B) make sure the string ends with at least one '0' digit; and C) check that, for all prime bases below n, the resulting number is divisible by n. If these three conditions are met, the string is a trans-basic multiple of n.
While the formula given below is guaranteed to provide a trans-basic multiple of n, it does not always yield a(n) which by definition is the smallest such number. [Corrected by M. F. Hasler, Nov 14 2019]
From N. J. A. Sloane, Nov 12 2019: (Start)
For each n, the values of (string a(n) read in base b)/n for b = 1,2,3,... give a sequence of integers.
For n=1 this is the all-1's sequence A000012.
For n=2, a(2) = 110 which in base b is b+b^2. Divided by 2 we get (b+b^2)/2, which evaluated at b = 1,2,3,4,... is 1,3,6,10,..., the triangular numbers A000217.
For n=3, we get (b+b^3+b^5)/3, which is A220892.
For n=4, we get A328994. (End)
See A329000 = (1, 6, 42, 60, 139810, 126, ...) for a(n) converted from base 2 to base 10, i.e., the numbers which yield the terms here when written in base 2. - M. F. Hasler, Nov 09 2021

			a(3) = 101010:
  101010_2  =     42 =    14*3;
  101010_3  =    273 =    91*3;
  101010_4  =   1092 =   364*3;
  101010_5  =   3255 =  1085*3;
  101010_6  =   7998 =  2666*3;
  101010_7  =  17157 =  5719*3;
  101010_8  =  33288 = 11096*3;
  101010_9  =  59787 = 19929*3;
  101010_10 = 101010 = 33670*3;
  101010_11 = 162393 = 54131*3;
  101010_12 = 250572 = 83524*3;
and so on. All the resulting values are multiples of 3.

Alon Ran, Table of n, a(n) for n = 1..21
Alon Ran, Comments on this sequence

Cf. A000012, A000217, A051903, A220892, A268336, A329338 (an upper bound), A328994.
A329000 gives a(n) read in base 2 and converted to base 10.
See also A329443.

\\ See A329338 for an upper bound which equals a(n) in many cases, e.g., all n < 14 except for n = 10. - M. F. Hasler, Nov 10 2021

To generate an upper bound on a(n), start with n 1's (this is required to ensure that it is divisible by n in bases n+1, 2n+1, etc.)
Next, place A268336(n)-1 0's in between the 1's (this ensures that the powers that are added will always sum to 0 (mod n)).
Finally, add A051903(n) 0's on the right (this is to ensure that the number will be divisible by n in bases that are roots of factors of n).
Note that this formula does not always yield the minimal solution a(n). For instance, a(10) is obtained from the above result by grouping the 1's in pairs and separating the pairs by two 0's.
a(n) <= A329338(n), with equality except for n = 10, 14, 15, ... - M. F. Hasler, Nov 14 2019

I have weakened some of the assertions in the Comments section, since they seemed to be unproved. See Alon Ran's comments (see Links). - N. J. A. Sloane, Dec 02 2019
Definition corrected, following a remark by Don Reble, by M. F. Hasler, Nov 09 2021
The present definition has been reworded by Peter Munn, Nov 17 2021, and by N. J. A. Sloane, Nov 29 2021

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User: Alon Ran

A336383 a(n) is the smallest number such that, with f(x) = x - (the product of the digits of x), f(a(n)) reaches a fixed point after n iterations.

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A331206 Numbers k such that A053985(k) divides k.

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A329126 a(n) is the least positive number which yields a multiple of n when its decimal digits (which are necessarily 0's and 1's) are read in any base.

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