Smallest pandigital reptend of 1/n in base b

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Joshua Searle

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Mar 27, 2025, 5:48:16 PMMar 27
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In base 10, 1/17 = 0.(0588235294117647) with the bracketed section repeating. It does contain every digit 0-9 at least once and is the smallest unit fraction to do so.

Starting with base 2, the sequence appears to begin:
3,5,13,7,11,11,11,43,17,13,17...

When the base is a perfect square there do not exist maximal length reptends so that explains why the terms base 4 and 9 are unusually large.

Some questions:
1. Will this always be a prime?
- It might be possible if the base is a square and would need to be at least 4 times the base.

2. Does every number appear at least once?
- If n isn't in the sequence by base n-1 then it won't be in the sequence.

3a. Can a number appear an arbitrary number of times?
3b. Can a number appear consecutively an arbitrary number of times?
- I notice that 11 appears three times - and consecutively too. You can then create a derived sequence for the smallest number that appears k times in the sequence: 3, 13, 11... 

If this seems interesting enough to make an OEIS entry/-ies for I'll do so.

Joshua.

Joshua Searle

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Mar 28, 2025, 1:34:28 PMMar 28
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> 2. Does every number appear at least once?
I meant to say 'prime number' here.

> ...k times in the sequence: 3, 13, 11... 
This should be 3, 11, 11

Managed to cobble together some code to generate the results up to base 36.
It's not quite raw code output, I've edited it so the fractional part is the repeating string.

[base, 1/k, expansion]:

[2, 3, '0.01']
[3, 5, '0.012']
[4, 13, '0.010323']
[5, 7, '0.032412']
[6, 11, '0.0313452421']
[7, 11, '0.0431162355']
[8, 11, '0.0564272135']
[9, 43, '0.017852074728331226435']
[10, 17, '0.0588235294117647']
[11, 13, '0.093425a17685']
[12, 17, '0.08579214b36429a7']
[13, 19, '0.08b82976ac414a3562']
[14, 17, '0.0b75a9c4d2683419']
[15, 19, '0.0bc9718a3e3257d64b']
[16, 79, '0.033d91d2a2067b23a5440cf6474a8819ec8e951']
[17, 23, '0.0c9a5f8ed52g476b1823be']
[18, 29, '0.0b31f95a9gdbe4h6eg28c781463d']
[19, 23, '0.0fd4291c784i35eg9h6bae']
[20, 23, '0.0h7ga8di546j2c39b61efd']
[21, 23, '0.0j3decg92fak1h7684bi5a']
[22, 31, '0.0fdae45ejj3c194l68b7hg722i9kch']
[23, 47, '0.0b5k1ahe496jd4kcgeff3l0mbh2lc58idg39i2a6877j1m']
[24, 31, '0.0idmak327hj8c96n5a1d3klg64fbeh']
[25, 73, '0.08e10h3219642ic85bogano7lmnfikm6cgjd']
[26, 29, '0.0n81kg3f674ce8p2ho59majildbh']
[27, 29, '0.0p3jeo5fm98a6dq1n7c2lb4higkd']
[28, 41, '0.0j3bh21a6n642kdic85cr8ogapqhl4lnp7e9fjmf']
[29, 41, '0.0keolqpdcl6ahjn9q4rgs8e4723fg7mib95j2o1c']
[30, 41, '0.0lsg2ro4bl6hgoqa79fat81dr25pi8ncd53jmkej']
[31, 47, '0.0kdqbr19rlnn2jocgf58hp1uah4j3tl3977sb6iefpmd5t']
[32, 37, '0.0rljsh9gdqpu8ko6tcv4ac3emfi561nb7p2j']
[33, 43, '0.0paoidqsd1hlg3rknq329w7m8ej64jvfbgt5c96tun']
[34, 41, '0.0s6lj2gjun7fpo1md944x5rcevhe3aqi89wbkott']
[35, 37, '0.0x3rf4piw5nmoksd8hy1v7ju9g2tbcae6lqh']
[36, 137, '0.09gjyy5s47cvj6khv9q0ix3xwbk8epr2d4zqjg11u7vsn4gtfi4q9zh2w23ofrla8xmv']

There is also the derived sequence for the first number to appear k times, which is now: 3, 11, 11, 23.

Which should be enough for an entry. What should I do with the digit strings... add them as a file I suppose?

Joshua

Ed Pegg

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Mar 30, 2025, 12:06:12 AMMar 30
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Note the large jumps at square bases.  I bet there's an explanation.

On Fri, Mar 28, 2025 at 12:34 PM Joshua Searle <jprs...@gmail.com> wrote:
[4, 13, '0.010323']
[9, 43, '0.017852074728331226435']
[16, 79, '0.033d91d2a2067b23a5440cf6474a8819ec8e951']
[25, 73, '0.08e10h3219642ic85bogano7lmnfikm6cgjd']
[36, 137, '0.09gjyy5s47cvj6khv9q0ix3xwbk8epr2d4zqjg11u7vsn4gtfi4q9zh2w23ofrla8xmv']

Joshua Searle

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Mar 30, 2025, 5:32:32 AMMar 30
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In my initial post I wrote:
> When the base is a perfect square there do not exist maximal length reptends so that explains why the terms base 4 and 9 are unusually large.

For squarefree n, a(n) > n but for square n, a(n) > 2n

A draft entry now exists at A382498

Joshua.

Martin Fuller

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Mar 30, 2025, 8:37:35 AMMar 30
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It is not clear whether "pandigital" includes the integer part, or only the fractional part, or only the repetend.  I have assumed fractional part.

a(2)-a(100): 3, 5, 13, 7, 11, 11, 11, 43, 17, 13, 17, 19, 17, 19, 79, 23, 29, 23, 23, 23, 31, 47, 31, 73, 29, 29, 41, 41, 41, 47, 37, 43, 41, 37, 137, 59, 47, 47, 47, 47, 59, 47, 47, 47, 67, 59, 53, 241, 53, 53, 59, 71, 59, 59, 59, 67, 73, 61, 73, 67, 71, 67, 383, 71, 79, 71, 71, 71, 79, 83, 83, 83, 79, 79, 83, 79, 103, 83, 83, 343, 89, 89, 97, 103, 89, 97, 103, 101, 97, 107, 97, 101, 101, 107, 103, 107, 101, 101, 577

a(n) is composite for n=81, 121, 144, 169, 196, 225, 256, 289, 324, 361, 400, 484, 529, 576, 625, 729, 784, 841, 961, 1024, 1089, 1156, 1225, 1296, 1444, 1521, 1600, 1681, 1849, 1936, 2025, 2116, 2209, 2304, 2401, 2500...
a(81)=7^3, a(121)=a(144)=a(169)=5^4, a(196)=3^6

Missing primes: 2, 109, 193, 1033, 1489, 1789, 2161, 2341, 2689, 2713, 3121, 3361, 4801, 5113, 5281, 5641, 5881, 6361, 6553, 6841, 6961, 7561, 8681, 8761, 9241, 9769...

(PARI)
ispandigitalfractionalpart(n,d,b=10)={
  my(residueset=0,digitset=0); n %= d;
  while(n && !bittest(residueset,n),
    residueset += 1<<n;
    digitset = bitor(digitset, 1<<(n*b\d));
    n = (n*b)%d;
  );
  digitset==(1<<b)-1
};
a(n,limit=oo)=for(k=n+1,limit,if(ispandigitalfractionalpart(1,k,n),return(k)));

Joshua Searle

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Mar 30, 2025, 9:10:13 AMMar 30
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Thanks for the additional terms, I'm updating the entry.
Is it worth adding those two sequences to the OEIS as well? (composite a(n) and missing primes)
The first number to appear n times in the sequence: 3, 11, 11, 23, 47, 47, 47, 47, 47... and the first to appear at least n times consecutively 3, 11, 11, 47... probably isn't worthwhile though unless you think otherwise?

Josh.

Dmitriy Alexandrovich

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Apr 4, 2025, 12:22:54 AMApr 4
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bnb1wc0shltd98jhyh6n90chxqx2qxe8f8qrw4af0e
 
Please help me to find the way to earn some money. 
 
Cuz had some trubles with total understanding.
 
Also my enother email d011...@pm.me
30 марта 2025 г., 16:10 +0300, Joshua Searle <jprs...@gmail.com>, писал:
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L. Edson Jeffery

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Apr 4, 2025, 1:17:12 AMApr 4
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Um, get a job and leave us alone!

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