Idea for a sequence

[Dave Consiglio]

Hello all,

The sequence is: numbers n such that all of the digits of n are not encountered in the Collatz pathway from n to 1.

Example: 5 is a member of this sequence because 5 --> 16 --> 8 --> 4 --> 2 --> 1 and the digit 5 is not encountered in this pathway (after the initial 5)

Similarly, 29 is in this sequence because 29 88 44 22 11 34 17 52 26 13 40 20 10 5 16 8 4 2 1 covers the 2 in 29, but the 9 never appears.

Finally, 77 is a member because 77 232 116 58 29 88 44 22 11 34 17 52 26 13 40 20 10 5 16 8 4 2 1 has one 7 in it, but both 7s need to appear.

30 is not a member because 30 15 46 23 70 has both a 3 and a 0.

The terms grow very sparse very quickly.

[1, 2, 3, 4, 5, 8, 9, 10, 13, 16, 17, 29, 32, 49, 53, 69, 70, 77, 88, 90, 92, 93, 96, 149, 170, 227, 256, 276, 277, 296, 320, 336, 369, 469, 672, 739, 753, 768, 896, 904, 906, 908, 909, 936, 953, 960, 964, 966, 1024, 1109, 1492, 1493, 1696, 1704, 1706, 2275, 2728, 2730, 2896, 2900, 2901, 2954, 3392, 3413, 4017, 4096, 4277, 4292, 4293, 4437, 4549, 5376, 5909, 6784, 7272, 7281, 7509, 8872, 8960, 9024, 9096, 9098, 9100, 9101, 9600, 9664, 9668, 9669, 9670, 10752, 10912, 13653, 16070, 16077, 16177, 17172, 17744, 17749, 21427, 25770, 26692, 26693, 26694, 27296, 27304, 27306, 29008, 29012, 29013, 29096, 30946, 32768, 35496, 38797, 40960, 47776, 53389, 64276, 64277, 68692, 68693, 69632, 70912, 70992, 70997, 72192, 72320, 72784, 72817, 77056, 77120, 77312, 77669, 81349, 87376, 90112, 91249, 91296, 91349, 91520, 91589, 92840, 94549, 94913, 97077, 102912, 106496, 109056, 109216, 162696, 212992, 257128, 257130, 273749, 274773, 309474, 310677, 327680, 427092, 427093, 488096, 549546, 567978, 569429, 569484, 569485, 569494, 618949, 696320, 728177, 771072, 771328, 771392, 771712, 776644, 776645, 776716, 776717, 854229, 873728, 873792, 873808, 928424, 961024, 976128, 976192, 990292, 990293, 1094912, 1094992, 1099092, 1099093, 1138915, 1649664, 1703936, 1747616, 2989169, 3297280, 3299328, 3407872, 3416912, 3416917, 3494912, 3727872, 4694469, 5124096, 6607872, 6833492, 6833493, 6941909, 8098949, 9110869, 9111779, 9388941, 9389169, 13667669]

From 10,000,000 to 20,000,000 there is only one additional term: 13667669

The terms don't seem to have a lot in common. though repeated digits are well represented (for obvious reasons). Evens and odds are both common. Is there rhyme or reason here? My computer ran all night and didn't get to 250 terms...is the sequence finite? And, of course, is it worthy of OEIS?

I'd love to hear your thoughts.

Best,

Dave Consiglio

[Geoffrey Caveney]

To me the most interesting phenomenon in your sequence is the Collatz pathway of 13667669. One might naturally expect that four 6's fail to occur as digits among all the integers in its pathway, but this is not the case. Rather, it is remarkable that the digit 9 fails to occur even a single time in its entire Collatz pathway!! I would consider the sequence of integers in the Collatz pathway of 13667669 to be interesting in its own right for this incredibly improbable property alone.

Although you observe that evens and odds are both common, the ten decimal digits do not occur with nearly equal frequency at all among the integers in your sequence:

digit 9: 179 occurrences

digit 7: 141 occurrences

digit 2: 123 occurrences

digit 6: 119 occurrences

digit 1: 102 occurrences

digit 0: 90 occurrences

digit 4: 87 occurrences

digit 3: 76 occurrences

digit 8: 61 occurrences

digit 5: 35 occurrences

I would expect that it is very likely that your sequence is finite.

The same concept should certainly be investigated for the digits of binary and ternary representations of integers, since the Collatz function itself is based on division by 2 and multiplication by 3. I expect that the integers satisfying your condition in binary and ternary will be even much sparser than your sequence based on decimal digits. For example, binary 100 (4 in decimal) satisfies your condition, since there are two 0's in 100 and only one digit 0 in binary 10 (decimal 2). But already binary 1000 (decimal 8) fails your condition, since its three 0's are matched by the three 0's in 100 and 10 together.

Geoffrey

--
You received this message because you are subscribed to the Google Groups "SeqFan" group.
To unsubscribe from this group and stop receiving emails from it, send an email to seqfan+un...@googlegroups.com.
To view this discussion visit https://groups.google.com/d/msgid/seqfan/CAE7Bzue7byGZrWo1b4dVpjCLNYFVcXgiAN9MevAZba48MU1Yew%40mail.gmail.com.

[Jeff Sponaugle]

I did a quick search up to 10^12 and only found 231 terms.  Is that the same as what you found?  The last term I found was 15,078,777,741.   -jeff

[Dave Consiglio]

Hi Jeff,

I'm realizing that there are some optimizations that could make it much faster. I just threw the original version together. But I submitted the sequence - if it gets approved please add more terms. I'm going to try and see if I can improve on my code but I don't anticipate getting many more terms. 

Thanks for letting me know!

Dave

On Sat, Sep 5, 2026 at 7:47 AM Jeff Sponaugle <je...@sponaugle.com> wrote:

I did a quick search up to 10^12 and only found 231 terms.  Is that the same as what you found?  The last term I found was 15,078,777,741.   -jeff

The terms don't seem to have a lot in common. though repeated digits are well represented (for obvious reasons). Evens and odds are both common. Is there rhyme or reason here? My computer ran all night and didn't get to 250 terms...is the sequence finite? And, of course, is it worthy of OEIS?

--
You received this message because you are subscribed to the Google Groups "SeqFan" group.
To unsubscribe from this group and stop receiving emails from it, send an email to seqfan+un...@googlegroups.com.

To view this discussion visit https://groups.google.com/d/msgid/seqfan/3c3cdf4d-a346-4d82-bb1f-5d27a3777af0n%40googlegroups.com.

[Geoffrey Caveney]

My instinct is that a complete proof could be found in the binary case, that only 1, 10, and 100 (decimal 1, 2, 4) satisfy the condition. The ternary case may be attackable as well, but more difficult to prove that the sequence is finite.

To view this discussion visit https://groups.google.com/d/msgid/seqfan/CAE7BzufL0YCzVa_TDyvqsFJdfBp1pGCGMfzi4qgsfYq58gtrMA%40mail.gmail.com.

[Dave Consiglio]

Thanks Jeff and Geoff :) Seems like storing the digits covered by previous numbers and using that to expedite this process might allow for faster calculation. The memory overhead becomes problematic at some point, I'm sure. But still, fun stuff. 

To view this discussion visit https://groups.google.com/d/msgid/seqfan/CAFhKorVtBnxNTo7C3WkUpRdXj_u2N8y92Cb-J%2Bw4jMYLxvJM7A%40mail.gmail.com.

[Jeff Sponaugle]

Yea, there certainly are a few other optimizations possible.   Here is my current version:
https://github.com/jeffsponaugle/newcol

Right now it is running at about 850-900 M tests/sec, so 10T tests will take 3.5 hours.

Jeffs-Mac-Studio newcol % ./collatzdig 1 10000B > terms_10000B.txt
 11.16%  at 1.116T    found 231      845.8 M/s  elapsed 0:22:59  ETA 3:02:56

[David Corneth]

How are you guys searching? From the code I see one is looking from the trajectory from the number itself to 1 (or abort if the number is found). 

I imagine if you look for terms of at most 12 digits one can treat 8 as a root, saying you saw a digit 1, a digit 2 and a digit 4.

At every number t you may double and test 2t. And also, if the number == 1 (mod 3) you can go to (t - 1)/3.

Keep track of digits seen. If each digits is seen 12 times, stop. That may save a bit of searching but I haven't implemented this (yet).

Could this help?

Best,

David  

[David Corneth]

Okay I implemented the idea above. My prog is over at A398712. I got to 10^25 and no new terms. Search took some 11 minutes so I guess you guys with more patience and or better machinery can push this limit up more if you care. 

Best,

David