Prime Powers Sequence

[Ali Sada]

Hi everyone,

Hope all is well.

a(n) is obtained by inserting exactly one decimal digit into the decimal expansion of n. Among all resulting integers m, maximize k such that p^k ∣ m. If more than one choice attains the same maximum exponent k, choose the one with the largest prime p.

A few examples:
a(1) = 81, since 81 = 3^4 and 16 = 2^4; both have exponent 4, but 3 > 2.
a(12) = 512, since 512 = 2^9.
a(23) = 243, since 243 = 3^5.
a(29) = 729, since 729 = 3^6.

81,32,32,64,54,64,27,81,96,160,112,512,135,144,125,160,176,128,192,320,216,224,243,243,256,256,272,128,729,320,351,320,336,384,352,736,375,384,392,640,416,432,243,448,405,486,472,448,496,560.

The second sequence is the associated primes: 3,2,2,2,3,2,3,3,2,2,2,2,3,2,5,2,2,2,2,2,3,2,3,3,2,2,2,2,3,2,3,2,2,2,2,2,5,2,7,2,2,3,3,2,3,3,2,2,2,2.

Conjecture: All prime numbers appear in this sequence.

I would really appreciate it if you could tell me whether these two sequences are appropriate for the OEIS. Also, please check the notes below and tell me if they were suitable as comments.

1) The power k should be at least 3 because we can always add a digit to n and make it a multiple of 8.

2) Powers of odd primes are more likely to appear when n is odd. 

3) For n = 2^i, a(n) = 10*n, j >4.
4) For n = 2^i*10^j, j > 5,  is there a possibility of getting a power of 3 or 7 instead of 10*n?


Best,

Ali

[Geoffrey Caveney]

Ali,

You state that a(39) = 392, and you give the associated prime as 7. But 392 = 2^3 * 7^2, so k=3 and the prime associated with k is 2, not 7.

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/CACOfRNpk_3LxLTLwAoLov_HUwPD%3Dg9UTHgTuAeQ8vYOsNq2hsQ%40mail.gmail.com.

[Allan Wechsler]

The conjecture that every prime appears in the second sequence seems heuristically reasonable. If we want a number N for which the highest prime-power dividing any one-digit insertion of N is, say, 17, we can calculate 17^7 = 410338673, get rid of a digit, to get, say, 10338673, and be reasonably confident that none of the approximately 90 possible insertions with "scoop" 17^7 as the winning prime-power factor. And if we do have bad luck, we can delete a different digit, and if we have bad luck with every such deletion, we can look at 2*17^7 instead ... I hope this is convincing, though it is certainly only a heuristic argument for why to suspect the conjecture is true. It's likely impossible to prove.

I'm usually skeptical of "digit-twiddling" sequences, and my usual question is, "Is this sequence likely to be more interesting, for some reason, than the corresponding one for base 9 or base 11?" I feel like the binary case should always be considered. But OEIS already contains thousands of digit-twiddling sequences anyway, so it feels like that particular ship has sailed. I don't see why these sequences are less interesting than some others that are already archived.

-- Allan

On Sat, Jul 4, 2026 at 11:15 AM Ali Sada <ali....@gmail.com> wrote:

[Ali Sada]

Thank you, Geoffry and Allan for your responses. I really appreciate them. And yes, a2(39) = 2, not 7. I apologize for the mistake. 

I thought of a similar proof for the conjecture, but I thought of 17^8 instead of 2*17*7 in Allan's example.  

While I agree with Allan regarding the binary case, I respectfully disagree when it comes to base 10. In my opinion, one of the main reasons base 10 is so important is that it is familiar to the human eye. For people with less mathematical training, like myself, it is much easier to come up with sequences in base 10 without requiring deep mathematical knowledge. Such sequences can still lead to interesting questions, beautiful graphs, and perhaps even unexpected discoveries. For example, in the two sequences I posted, I used only two concepts that we all learn in elementary school: powers and primes. Knowledge of binary and other bases comes much later.

To make my point more convincing mathematically, I will borrow the wise words of Mark Wahlberg in The Other Guys: "I'm a peacock! You gotta let me fly!"

Best,

Ali 

To view this discussion visit https://groups.google.com/d/msgid/seqfan/CADy-sGGirTqL53ydsw1FuPDionZpeNjnWak3T54XcW47Z4gU5g%40mail.gmail.com.

[Geoffrey Caveney]

Ali and Allan,

The binary case appears to have interesting properties in its own right. For example, unlike the base 10 case, in binary when n is odd, the associated prime of a(n) must also be odd. Proof: An odd binary number has final digit 1, and adding one digit can produce only one digit 0 at the end of the number a(n). Thus, a(n) must end in either ...1 or ...10. The former is odd and the latter is even but not divisible by 4 (100 in binary). Thus, in the prime factorization of a(n), the exponent k of the prime 2 (10 in binary) cannot be greater than 1. There must be an odd prime factor of a(n) for which either k>1 or k=1, which by the definition of the sequence must be chosen because any odd prime is greater than the prime 2.

However, when n is even, the associated prime of a(n) may be either even or odd. For example, a(1010) = 10010 [that is, a(10) = 18 = 3^2 * 2].

Interesting small values in the binary case include 

a(11) = 111  [that is, a(3)=7^1]

a(101) = 1001  [a(5) = 3^2]

a(111) = 1101  [a(7) = 13^1]

a(1001) = 11001  [a(9) = 5^2]

a(10001) = 110001  [a(17) = 7^2]

a(111001) = 1111001  [a(57) = 11^2]

It is even possible that the square of every prime appears in this sequence. With multiples of higher powers of 2 ruled out as candidates to be a(n) for odd n, the only competition for any p^2 would be multiples of cubes of odd primes and larger squares of primes, both of which have very low density among the integers.

Geoffrey

To view this discussion visit https://groups.google.com/d/msgid/seqfan/CACOfRNoc9b6Jb_bKCcHSXaVFb6gqi9%3DTKVStzcG7GNMbL90jKQ%40mail.gmail.com.

[Ali Sada]

Thank you, Geoffrey. This is fascinating. I have submitted the two sequences  A397698  and  A397699. I will wait for the editors' approval and then co-submit the binary version if you like. 

Best,

Ali 

To view this discussion visit https://groups.google.com/d/msgid/seqfan/CAFhKorVYeivrPZZYEX6zX4%2BtLDRvG1h4p66i-wr0P8%3DPsxt2kQ%40mail.gmail.com.