cp's OEIS Frontend

This is a front-end for the Online Encyclopedia of Integer Sequences, made by Christian Perfect. The idea is to provide OEIS entries in non-ancient HTML, and then to think about how they're presented visually. The source code is on GitHub.

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A334287 Smallest full reptend prime p such that there is a gap of exactly 2n between p and the next full reptend prime, or 0 if no such prime exists.

Original entry on oeis.org

17, 19, 23, 491, 7, 47, 419, 577, 29, 0, 1789, 233, 461, 433, 193, 509, 823, 61, 1979, 1327, 659, 269, 11503, 1381, 887, 14251, 3167, 8297, 3469, 0, 7247, 15073, 2473, 743, 19309, 4349, 21503, 12823, 14939, 3863, 5419, 6389, 24137, 27211, 10343, 13577, 18979
Offset: 1

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Author

Martin Raab, Apr 21 2020

Keywords

Comments

Gaps of length congruent to 20 mod 40 do not exist. All full reptend primes are either 7, 11, 17, 19, 21, 23, 29, or 33 mod 40, and no difference of 20 exists between any of these numbers.

Examples

			a(9) = 29 because there is a gap of 2*9 = 18 between 29 and the next full reptend prime 47.
a(10) = 0 because no gap of 2*10 = 20 exists between full reptend primes.
		

Crossrefs

Cf. A001913.

Programs

  • PARI
    is(p) = Mod(10, p)^(p\2)==-1 && znorder(Mod(10, p))+1==p;
    isok(p, n) = {if (! is(p), return (0)); if (isprime(p+n) && is(p+n), forprime(q=p+1, p+n-1, if (is(q), return (0));); return (1););}
    a(n) = {n *= 2; if ((n % 40) == 20, return (0)); my (p = 2); while (! isok(p, n), p = nextprime(p+1)); p;} \\ Michel Marcus, Apr 22 2020

A343833 Prime numbers of the form floor((j/7)*10^k) where 1 <= j <= 6 and k >= 1.

Original entry on oeis.org

2, 5, 7, 71, 571, 857, 2857, 28571, 1428571, 71428571, 7142857142857, 571428571428571, 1428571428571428571428571, 28571428571428571428571428571, 7142857142857142857142857142857, 2857142857142857142857142857142857, 42857142857142857142857142857142857142857
Offset: 1

Views

Author

Konstantin Kutsenko, May 01 2021

Keywords

Crossrefs

A021117 Decimal expansion of 1/113.

Original entry on oeis.org

0, 0, 8, 8, 4, 9, 5, 5, 7, 5, 2, 2, 1, 2, 3, 8, 9, 3, 8, 0, 5, 3, 0, 9, 7, 3, 4, 5, 1, 3, 2, 7, 4, 3, 3, 6, 2, 8, 3, 1, 8, 5, 8, 4, 0, 7, 0, 7, 9, 6, 4, 6, 0, 1, 7, 6, 9, 9, 1, 1, 5, 0, 4, 4, 2, 4, 7, 7, 8, 7, 6, 1, 0, 6, 1, 9, 4, 6, 9, 0, 2, 6, 5, 4, 8, 6, 7, 2, 5, 6, 6, 3, 7, 1, 6, 8, 1, 4, 1, 5, 9, 2, 9, 2, 0, 3, 5, 3, 9, 8, 2, 3
Offset: 0

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Author

Keywords

Comments

Periodic with period length 112. - Ray Chandler, Jan 23 2024

Examples

			0.0088495575221238938...
		

Crossrefs

Programs

Extensions

Extended to a full period by Sean A. Irvine, May 07 2019

A225488 Murai Chuzen numbers.

Original entry on oeis.org

9, 45, 3, 225, 18, 15, -1, 1125, 1, 99, 495, 33, 2475, 198, 165, -1, 12375, 11, 999, 4995, 333, 24975, 1998, 1665, -1, 124875, 111, 9999, 49995, 3333, 249975, 19998, 16665, -1, 1249875, 1111, 99999, 49995, 33333, 2499975, 199998, 166665, -1, 12499875, 11111, 999999, 4999995, 333333, 24999975, 1999998, 1666665, -1, 124999875, 111111
Offset: 1

Views

Author

Jonathan Sondow, May 10 2013

Keywords

Comments

"Murai Chuzen divides 9 by 1, 2, 3, 4, 5, 6, 7, 8, 9, getting the figures 9, 45, 3, 225, 18, 15, x (not divisible), 1125, 1, -- without reference to the decimal points. Similarly he divides 99 by 1, 2, 3, 4, 5, 6, 7, 8, 9, getting the figures 99, 495, 33, 2475, 198, 165, x, 12375, 11. Next he divides 999 by 1, 2, 3, 4, 5, 6, 7, 8, 9, getting the figures 999, 4995, 333, 24975, 1998, 1665, x, 124875, 111." (Smith and Mikami, expanded and corrected)
Smith and Mikami put "x" whenever a decimal does not terminate. In the data, I put -1 instead of "x".
Murai Chuzen concludes that if 1 is divided by 9, 45, 3, 225, 18, 15, 1125, and 1, the results will have one-digit repetends; if 1 is divided by 99, 495, 33, 2475, 198, 165, 12375, and 11, the results will have two-digit repetends; if 1 is divided by 999, 4995, 333, 24975, 1998, 1665, 124875, and 111, the results will have three-digit repetends; etc.

Examples

			9/1 = 9, so a(1) = 9; 9/2 = 4.5, so a(2) = 45; 9/7 does not terminate, so a(7) = -1; 9/8 = 1.125, so a(8) = 1125; 9/9 = 1, so a(9) = 1.
99/1 = 99, so a(10) = 99; 99/2 = 49.5, so a(11) = 495.
		

References

  • Murai Chuzen, Sampo Doshi-mon (Arithmetic for the Young), 1781.

Crossrefs

A229069 Group cyclic numbers.

Original entry on oeis.org

32258064516129, 354838709677419, 23255813953488372093, 46511627906976744186, 2830188679245, 5660377358490, 1698113207547, 2264150943396, 164179104477611940298507462686567, 283582089552238805970149253731343
Offset: 1

Views

Author

Y. Z. Chen, Sep 12 2013

Keywords

Examples

			Every group has the same digits, and they are "cyclic numbers" in its group, not only the number itself.
		

Crossrefs

Cf. A001913.

A307070 a(n) is the number of decimal places before the decimal expansion of 1/n terminates, or the period of the recurring portion of 1/n if it is recurring.

Original entry on oeis.org

0, 1, 1, 2, 1, 1, 6, 3, 1, 1, 2, 1, 6, 6, 1, 4, 16, 1, 18, 2, 6, 2, 22, 1, 2, 6, 3, 6, 28, 1, 15, 5, 2, 16, 6, 1, 3, 18, 6, 3, 5, 6, 21, 2, 1, 22, 46, 1, 42, 2, 16, 6, 13, 3, 2, 6, 18, 28, 58, 1, 60, 15, 6, 6, 6, 2, 33, 16, 22, 6, 35, 1, 8, 3, 1, 18, 6, 6, 13
Offset: 1

Views

Author

Luke W. Richards, Mar 22 2019

Keywords

Comments

If the decimal expansion of 1/n terminates, we will write it as ending with infinitely many 0's (rather than 9's). Then for any n > 1, the expansion of 1/n consists of a preamble whose length is given by A051628(n), followed by a periodic part with period length A007732(n). This sequence is defined as follows: If the only primes dividing n are 2 and 5 (see A003592), a(n) = A051628(n), otherwise a(n) = A007732(n) (and the preamble is ignored). - N. J. A. Sloane, Mar 22 2019
This sequence was discovered by a school class (aged 12-13) at Arden School, Solihull, UK.
Equally space the digits 0-9 on a circle. The digits of the decimal expansion of rational numbers can be connected on this circle to form data visualizations. This sequence is useful, cf. A007732 or A051626, for identifying the complexity of that visualization.

Examples

			1/1 is 1.0. There are no decimal digits, so a(1) = 0.
1/2 is 0.5. This is a terminating decimal. There is 1 digit, so a(2) = 1.
1/6 is 0.166666... This is a recurring decimal with a period of 1 (the initial '1' does not recur) so a(6) = 1.
1/7 is 0.142857142857... This is a recurring decimal, with a period of 6 ('142857') so a(7) = 6.
		

Crossrefs

See A114205 and A051628 for the preamble, A036275 and A051626 for the periodic part.

Programs

  • PARI
    a(n) = my (t=valuation(n,2), f=valuation(n,5), r=n/(2^t*5^f)); if (r==1, max(t,f), znorder(Mod(10, r))) \\ Rémy Sigrist, May 08 2019
  • Python
    def sequence(n):
      count = 0
      dividend = 1
      remainder = dividend % n
      remainders = [remainder]
      no_recurrence = True
      while remainder != 0:
        count += 1
        dividend = remainder * 10
        remainder = dividend % n
        if remainder in remainders:
          if no_recurrence:
            no_recurrence = False
            remainders = [remainder]
          else:
            return len(remainders)
        else:
          remainders.append(remainder)
      else:
        return count
    

Extensions

More terms from Rémy Sigrist, May 08 2019

A347225 Lesser of twin primes (A001359) being both half-period primes (A097443).

Original entry on oeis.org

197, 599, 881, 1277, 1997, 2081, 2237, 2801, 2999, 3359, 4721, 5279, 5879, 6197, 6959, 7877, 8837, 9239, 9719, 12161, 12239, 13721, 17921, 17957, 18521, 21839, 22637, 24917, 28277, 30557, 31319, 31721, 32117, 32441, 32717, 34757, 35081, 35279, 35837, 38921, 39239, 39839
Offset: 1

Views

Author

Lamine Ngom, Aug 24 2021

Keywords

Comments

A proper subset of both A001359 and A097443.
Number of terms < 10^k: 0, 0, 3, 19, 86, 516, 3686, 27834, 216758, 1739358, …
A243096 provides lesser of twin primes being both full reptend primes (A001913, A006883): in other words, lesser of twin primes whose periods difference is 2.
This sequence lists lesser of twin primes whose periods difference is 1. Equivalently, these twin primes are both half-period primes (A097443).
The twin primes conjecture being true should imply that these two sequences are infinite.
Surprisingly, apart from 1 and 2, for any other value of k integer, it appears that the sequence "lesser of twin primes whose periods difference is k" is empty or contains at most two terms (no counterexample found for twin primes below 10^9).

Examples

			The decimal expansion 1/p for the prime p = 1277 has a periodic part length equal to (p-1)/2. 1277 is thus a half-period prime. The same applies for p + 2 = 1279 (prime). Hence 1277 is in sequence.
		

Crossrefs

Programs

  • Maple
    select(t -> isprime(t) and isprime(t + 2) and numtheory:-order(10, t) = (t - 1)/2 and numtheory:-order(10, t + 2) = (t + 1)/2, [seq(t, t = 3 .. 40000, 2)]);

Formula

a(n) is congruent to {11, 17, 29} mod 30.

A347226 Safe primes (A005385) that are half-period primes (A097443).

Original entry on oeis.org

83, 107, 227, 347, 359, 467, 479, 563, 587, 719, 839, 1187, 1283, 1307, 1319, 1439, 1523, 1907, 2027, 2039, 2879, 2963, 2999, 3119, 3203, 3467, 3803, 3947, 4079, 4283, 4547, 4679, 4787, 4799, 4919, 5387, 5399, 5483, 5507, 5639, 5879, 6599, 6719, 6827, 7079, 7187, 7523
Offset: 1

Views

Author

Lamine Ngom, Aug 24 2021

Keywords

Comments

Apart from 5 and 11, a safe prime p is necessarily either a full reptend prime (A001913) or a half-period prime (A097443) since (p-1) is semiprime: 2*A005384(n) (Sophie Germain primes).
Safe primes being full reptend primes are listed in A000353.
a(n) is of the form 100*k + 10*{0, 2, 4, 6, 8} + {3, 7} or 100*k + 10*{1, 3, 5, 7, 9} + 9.
Number of terms < 10^k: 0, 1, 11, 56, 343, 2138, 15267, 114847, 886907, 7079602, ...

Examples

			(107-1)/2 = 53 is a prime, and the periodic part of the decimal expansion of 1/107 is of length 53.
Hence the safe prime 107 is in the sequence.
		

Crossrefs

Programs

  • Maple
    select(t -> isprime(t) and isprime((t - 1)/2) and numtheory:-order(10, t) = (t - 1)/2, [seq(t, t = 3 .. 10000, 2)]);
  • Mathematica
    Select[Prime@Range@1000,PrimeQ[(#-1)/2]&&Length[First@@RealDigits[1/#]]==(#-1)/2&] (* Giorgos Kalogeropoulos, Sep 14 2021 *)

Formula

A005385 INTERSECTION A097443.
a(n) == {17, 23, 29} mod 30.
a(n) == 11 (mod 12). - Hugo Pfoertner, Aug 24 2021

A373407 Smallest positive integer k such that no more than n numbers (formed by multiplying k by a digit) are anagrams of k, or -1 if no such number exists.

Original entry on oeis.org

1, 1035, 123876, 1402857, 1037520684, 142857
Offset: 1

Views

Author

Jean-Marc Rebert, Jun 04 2024

Keywords

Comments

For n = 2..6 all terms are divisible by 9.
For n >= 4, a(n) must be divisible by 9, or a(n) = -1, because all anagrams d*k of k for d = 2, 3, 5, 6, 8 and 9 are divisible by 9. Thus there are only 3 values of d, i.e., 1, 4 and 7, for which k*d must not be divisible by 9.
If a(n) exists for n > 1 then 9|a(n). Holds for n = 2 and n = 3 by inspection. Proof for n >= 4: if k*d is an anagram of k where 2 <= d <= 9 then k*d - k = k*(d-1) is a multiple of 9. For this to be true, k must be a multiple of 9 as d is not of the form 1 (mod 3) for all d. - David A. Corneth, Jun 04 2024
From Michael S. Branicky, Jun 07 2024: (Start)
The following were constructed from multiples of cyclic numbers (cf. A180340, Wikipedia):
a(6) = 142857 = (10^6 - 1) / 7;
a(7) <= 1304347826086956521739 = 3*(10^22 - 1) / 23;
a(8) <= 1176470588235294 = 2*(10^16 - 1) / 17;
a(9) <= 105263157894736842 = 2*(10^18 - 1) / 19. (End)

Examples

			a(2) = 1035, because 1035 * 1 = 1035 and 1035 * 3 = 3105 are anagrams of 1035, and no other number 1035 * i with digit i is an anagram of 1035, and no lesser number verifies this property.
Table n, k, set of multipliers.
  1   1          [1]
  2   1035       [1, 3]
  3   123876     [1, 3, 7]
  4   1402857    [1, 2, 3, 5]
  5   1037520684 [1, 2, 4, 5, 8]
  6   142857     [1, 2, 3, 4, 5, 6]
		

Crossrefs

Programs

  • PARI
    isok(k, n) = my(d=vecsort(digits(k))); sum(i=1, 9, vecsort(digits(k*i)) == d) == n; \\ Michel Marcus, Jun 04 2024

A381590 Primes with primitive root -100.

Original entry on oeis.org

3, 7, 19, 23, 31, 43, 47, 59, 67, 71, 83, 107, 131, 151, 163, 167, 179, 191, 199, 223, 227, 263, 283, 307, 311, 347, 359, 367, 379, 383, 419, 431, 439, 443, 467, 479, 487, 491, 499, 503, 523, 563, 571, 587, 599, 619, 631, 647, 659, 683, 719, 727, 743, 787, 811
Offset: 1

Views

Author

Davide Rotondo, Feb 28 2025

Keywords

Comments

Union of long period primes (A006883) of the form 4k-1 and half period primes (A097443) of the form 4k-1.
Complement of A007349 in the union of A007348 and A001913. - Davide Rotondo, May 23 2025

Crossrefs

Programs

  • Mathematica
    Select[Prime[Range[150]], MultiplicativeOrder[-100, #] == # - 1 &]  (* Amiram Eldar, Mar 02 2025 *)
  • PARI
    is(n)=gcd(n,10)==1 && znorder(Mod(-100, n))==n-1 \\ Charles R Greathouse IV, Mar 01 2025
    
  • PARI
    list(lim)=my(v=List([3])); forprime(p=7,lim, if(znorder(Mod(-100, p))==p-1, listput(v,p))); Vec(v) \\ Charles R Greathouse IV, Mar 01 2025
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