A344383 a(n) is the number of steps needed to reach 1 (or -1 if 1 is never reached) using the following rule: Begin with k = n. On each step, if k is divisible by 9, then divide it by 9; otherwise, if k has not already appeared during the iterative process, concatenate k with the digit d which makes the resulting number divisible by 9, and divide by 9; otherwise, concatenate d with k, and divide by 9.

0, 8, 7, 6, 5, 4, 3, 2, 1, 13, 16, 12, 15, 11, 14, 10, 13, 9, 12, 28, 10, 11, 27, 9, 10, 26, 8, 9, 25, 10, 37, 8, 24, 9, 36, 7, 23, 8, 44, 7, 70, 22, 7, 32, 6, 69, 21, 6, 47, 6, 6, 68, 20, 5, 43, 5, 5, 67, 19, 22, 25, 31, 4, 4, 66, 18, 21, 24, 92, 17, 22, 3, 65

Offset

1,2

Comments

It appears that all numbers will go to 1 eventually.

Count the iterations to reach 1, starting at k = n, using the following rule for each iteration: If k is divisible by 9, then divide it by 9; otherwise, concatenate the digit d = 9 - (k mod 9) to the end of k and divide by 9, unless k has already appeared during the iterative process; in this case, concatenate d with k and divide by 9.

Least n whose trajectory has no instance of repeated terms is 1; the trajectory of n=31 has 1 repeated term, and that of n=96 has 2 repeated terms. 2 repeated terms is the maximum for 1 <= n <= 2^18. Trajectories for all n in 1..2^20 eventually reach 1. - Michael De Vlieger, May 16 2021

Trajectories for all values of n between 1 and 10^8 eventually reach 1. - Kyle Peterson, May 19 2021

Links

Table of n, a(n) for n=1..73.

Michael De Vlieger, Scatterplot of a(n) for n = 1..2^16.

Michael De Vlieger, Scatterplot of a(n) for n = 1..10^4, showing n with 1 repeated step in red and 2 repeated steps in blue, otherwise, a(n) is plotted in black.

Michael De Vlieger, Log-log scatterplot of a(n) for n = 1..2^18.

Michael De Vlieger, Plot of the steps in a(n) for n = 1..10^4, showing condition A (divisible by 9) in gold, condition B (concatenation of n and d) in light blue, and condition C (repeated step, concatenation of d and n) in red. Magnification is 4X.

Example

n=7: start with k=7;
iteration 1: concatenate 7 with 9 - (7 mod 9) = 9 - 7 = 2 to get 72, then divide by 9, yielding k=8;
iteration 2: concatenate 8 with 9 - (8 mod 9) = 9 - 8 = 1 to get 81, then divide by 9, yielding k=9;
iteration 3: divide 9 by 9 to get k=1. k reaches 1 at iteration 3, so a(7)=3.
n=31: k=31 -> 315/9=35 -> 351/9=39 -> 396/9=44 -> 441/9=49 -> 495/9=55 -> 558/9=62 -> 621/9=69 -> 693/9=77 -> 774/9=86 -> 864/9=96 -> 963/9=107 -> 1071/9=119 -> 1197/9=133 -> 1332/9=148 -> 1485/9=165 -> 1656/9=184 -> 1845/9=205 -> 2052/9=228 -> 2286/9=254 -> 2547/9=283 -> 2835/9=315 -> 315/9=35 (*35 occurred at step 1*) -> 135/9=15 -> 153/9=17 -> 171/9=19 -> 198/9=22 -> 225/9=25 -> 252/9=28 -> 288/9=32 -> 324/9=36 -> 36/9=4 -> 45/9=5 -> 54/9=6 -> 63/9=7 -> 72/9=8 -> 81/9=9 -> 9/9=1, so a(31)=37.

Mathematica

Array[-1 + Length@ NestWhile[Append[#1, If[Mod[#2, 9] == 0, #2/9, If[FreeQ[Most@ #1, #2], Select[10*#2 + Range[9], Mod[#, 9] == 0 &][[1]]/9, Select[10^IntegerLength[#2]*Range[9] + #2, Mod[#, 9] == 0 &][[1]]/9]] ] & @@ {#, Last[#]} &, {#}, Last[#] != 1 &] &, 73] (* Michael De Vlieger, May 16 2021 *)

Prog

(Python)
def a(n):
    k, prevk, steps, seen = n, n, 0, set()
    while k != 1:
        q, r = divmod(k, 9)
        if r == 0: k = q
        elif  k not in seen: k = int(str(k) + str(9-r))//9
        else: k = int(str(9-r) + str(k))//9
        steps, seen, prevk = steps+1, seen|{prevk}, k
    return steps
print([a(n) for n in range(1, 74)]) # Michael S. Branicky, Jun 21 2021

Crossrefs

Sequence in context: A180598 A262703 A031311 * A055119 A265338 A090915

Adjacent sequences: A344380 A344381 A344382 * A344384 A344385 A344386

Keyword

nonn,easy,base

Author

Kyle Peterson, May 16 2021

Extensions

More terms from Michael De Vlieger, May 16 2021

Status

approved