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.

Showing 1-10 of 26 results. Next

A004042 Periods of reciprocals of A006883, starting with first nonzero digit.

Original entry on oeis.org

0, 142857, 5882352941176470, 526315789473684210, 4347826086956521739130, 3448275862068965517241379310, 2127659574468085106382978723404255319148936170, 1694915254237288135593220338983050847457627118644067796610
Offset: 1

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Author

Keywords

Comments

A variant of A180340.

Crossrefs

Cf. A180340.

Programs

  • Mathematica
    f[n_]:=Block[{q},q=Last[First[RealDigits[1/n]]];If[IntegerQ[q],q={}];FromDigits[q]]; q=0;lst={};Do[If[StringLength[ToString[f[n]]]>q,AppendTo[lst,f[n]];q=StringLength[ToString[f[n]]]],{n,5!}];lst (* Vladimir Joseph Stephan Orlovsky, May 21 2009 *)

Extensions

Definition corrected by Max Alekseyev, Feb 12 2012

A002371 Period of decimal expansion of 1/(n-th prime) (0 by convention for the primes 2 and 5).

Original entry on oeis.org

0, 1, 0, 6, 2, 6, 16, 18, 22, 28, 15, 3, 5, 21, 46, 13, 58, 60, 33, 35, 8, 13, 41, 44, 96, 4, 34, 53, 108, 112, 42, 130, 8, 46, 148, 75, 78, 81, 166, 43, 178, 180, 95, 192, 98, 99, 30, 222, 113, 228, 232, 7, 30, 50, 256, 262, 268, 5, 69, 28, 141, 146, 153, 155, 312, 79, 110
Offset: 1

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Keywords

Comments

a(n) is the minimum solution x of modular equation 10^x == 1 (mod p), where p = prime(n). - Carmine Suriano, Oct 10 2012
a(n) = smallest m such that 111...11 (m 1's) is divisible by the n-th prime, or 0 if no such m exists (with the exception that a(2) = 3 instead of 1). E.g., the 5th prime, 11, divides 11, so a(5) = 2. - N. J. A. Sloane, Oct 03 2013 [Comment corrected by Derek Orr, Jun 14 2014]
Numbers n such that A071126(n) = A000040(n) - 1. - Hugo Pfoertner, Mar 18 2003
Except for n = 1 and 3, a(n) divides A006093(n). - Robert Israel, Jul 15 2016

Examples

			A002371(11) = 15 because the 11th prime is 31, and 1/31 = 0.03225806451612903225806451612903225806452... has period 15. - _Richard F. Lyon_, Mar 29 2022
		

References

  • Albert H. Beiler, Recreations in the Theory of Numbers, 2nd ed. New York: Dover, 1966, pages 65, 309. ISBN 0-486-21096-0.
  • John H. Conway and R. K. Guy, The Book of Numbers, Copernicus Press, 1996, p. 162. ISBN 978-0-387-97993-9.
  • D. H. Lehmer, Guide to Tables in the Theory of Numbers. Bulletin No. 105, National Research Council, Washington, DC, 1941, p. 15.
  • N. J. A. Sloane, A Handbook of Integer Sequences, Academic Press, 1973 (includes this sequence).
  • N. J. A. Sloane and Simon Plouffe, The Encyclopedia of Integer Sequences, Academic Press, 1995 (includes this sequence).

Crossrefs

See A048595 for another version. Cf. A006883, A007732, A051626, A071126, A000040, A002275, A097443.
Cf. A001913 (full repetend primes), A060257 (1/prime(n) has period prime(n) - 1).

Programs

  • Maple
    seq(subs(FAIL=0,numtheory:-order(10, ithprime(n))),n=1..100); # Robert Israel, Jul 15 2016
  • Mathematica
    Table[ Length[ RealDigits[1 / Prime[n]] [[1, 1]]], {n, 1, 70}]
    Table[If[IntegerQ[#], #, 0] &[MultiplicativeOrder[10, Prime[n]]], {n, 1, 70}] (* Jan Mangaldan, Jul 07 2020 *)
  • PARI
    a(n)=if(n<4,n==2,znorder(Mod(10, prime(n))))
    
  • Python
    from sympy import prime, n_order
    def A002371(n): return 0 if n == 1 or n == 3 else n_order(10,prime(n)) # Chai Wah Wu, Feb 07 2022

Formula

From Alexander Adamchuk, Jan 28 2007: (Start)
a(A000720(p)) = p - 1 for primes p in A001913.
a(A060257(n)) = prime(A060257(n)) - 1. (End)

Extensions

More terms from Arlin Anderson (starship1(AT)gmail.com)
Edited by Charles R Greathouse IV, Mar 24 2010

A001913 Full reptend primes: primes with primitive root 10.

Original entry on oeis.org

7, 17, 19, 23, 29, 47, 59, 61, 97, 109, 113, 131, 149, 167, 179, 181, 193, 223, 229, 233, 257, 263, 269, 313, 337, 367, 379, 383, 389, 419, 433, 461, 487, 491, 499, 503, 509, 541, 571, 577, 593, 619, 647, 659, 701, 709, 727, 743, 811, 821, 823, 857, 863, 887, 937, 941, 953, 971, 977, 983
Offset: 1

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Comments

Primes p such that the decimal expansion of 1/p has period p-1, which is the greatest period possible for any integer.
Primes p such that the corresponding entry in A002371 is p-1.
Pieter Moree writes (Oct 20 2004): Assuming the Generalized Riemann Hypothesis it can be shown that the density of primes p such that a prescribed integer g has order (p-1)/t, with t fixed exists and, moreover, it can be computed. This density will be a rational number times the so-called Artin constant. For 2 and 10 the density of primitive roots is A, the Artin constant itself.
R. K. Guy writes (Oct 20 2004): MR 2004j:11141 speaks of the unearthing by Lenstra & Stevenhagen of correspondence concerning the density of this sequence between the Lehmers & Artin.
Also called long period primes, long primes or maximal period primes.
The base-10 cyclic numbers A180340, (b^(p-1) - 1) / p, with b = 10, are obtained from the full reptend primes p. - Daniel Forgues, Dec 17 2012
The number of terms < 10^n: A086018(n). - Robert G. Wilson v, Aug 18 2014

Examples

			7 is in the sequence because 1/7 = 0.142857142857... and the period = 7-1 = 6.
		

References

  • M. Abramowitz and I. A. Stegun, eds., Handbook of Mathematical Functions, National Bureau of Standards Applied Math. Series 55, 1964 (and various reprintings), p. 864.
  • Albert H. Beiler, Recreations in the Theory of Numbers, 2nd ed. New York: Dover, 1966, pages 65, 309.
  • John H. Conway and R. K. Guy, The Book of Numbers, Copernicus Press, p. 161.
  • C. F. Gauss, Disquisitiones Arithmeticae, Yale, 1965; see p. 380.
  • G. H. Hardy and E. M. Wright, An Introduction to the Theory of Numbers. 3rd ed., Oxford Univ. Press, 1954, p. 115.
  • M. Kraitchik, Recherches sur la Théorie des Nombres. Gauthiers-Villars, Paris, Vol. 1, 1924, Vol. 2, 1929, see Vol. 1, p. 61.
  • H. Rademacher and O. Toeplitz, Von Zahlen und Figuren (Springer 1930, reprinted 1968), Ch. 19, 'Die periodischen Dezimalbrüche'.
  • N. J. A. Sloane, A Handbook of Integer Sequences, Academic Press, 1973 (includes this sequence).
  • N. J. A. Sloane and Simon Plouffe, The Encyclopedia of Integer Sequences, Academic Press, 1995 (includes this sequence).

Crossrefs

Apart from initial term, identical to A006883.
Other definitions of cyclic numbers: A003277, A001914, A180340.

Programs

  • Maple
    A001913 := proc(n) local st, period:
    st := ithprime(n):
    period := numtheory[order](10,st):
    if (st-1 = period) then
       RETURN(st):
    fi: end:  seq(A001913(n), n=1..200); # Jani Melik, Feb 25 2011
  • Mathematica
    pr=10; Select[Prime[Range[200]], MultiplicativeOrder[pr, # ] == #-1 &]
    (* Second program: *)
    Join[{7},Select[Prime[Range[300]],PrimitiveRoot[#,10]==10&]] (* Harvey P. Dale, Feb 01 2018 *)
  • PARI
    forprime(p=7,1e3,if(znorder(Mod(10,p))+1==p,print1(p", "))) \\ Charles R Greathouse IV, Feb 27 2011
    
  • PARI
    is(n)=Mod(10,n)^(n\2)==-1 && isprime(n) && znorder(Mod(10,n))+1==n \\ Charles R Greathouse IV, Oct 24 2013
    
  • Python
    from itertools import count, islice
    from sympy import nextprime, n_order
    def A001913_gen(startvalue=1): # generator of terms >= startvalue
        p = max(startvalue-1,1)
        while (p:=nextprime(p)):
            if p!=2 and p!=5 and n_order(10,p)==p-1:
                yield p
    A001913_list = list(islice(A001913_gen(),20)) # Chai Wah Wu, Mar 03 2025

A051626 Period of decimal representation of 1/n, or 0 if 1/n terminates.

Original entry on oeis.org

0, 0, 1, 0, 0, 1, 6, 0, 1, 0, 2, 1, 6, 6, 1, 0, 16, 1, 18, 0, 6, 2, 22, 1, 0, 6, 3, 6, 28, 1, 15, 0, 2, 16, 6, 1, 3, 18, 6, 0, 5, 6, 21, 2, 1, 22, 46, 1, 42, 0, 16, 6, 13, 3, 2, 6, 18, 28, 58, 1, 60, 15, 6, 0, 6, 2, 33, 16, 22, 6, 35, 1, 8, 3, 1, 18, 6, 6, 13, 0, 9, 5, 41, 6, 16, 21, 28, 2, 44, 1
Offset: 1

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Keywords

Comments

Essentially same as A007732.
For any prime number p: if a(p) > 0, a(p) divides p-1. - David Spitzer, Jan 09 2017

Examples

			From _M. F. Hasler_, Dec 14 2015: (Start)
a(1) = a(2) = 0 because 1/1 = 1 and 1/2 = 0.5 have a finite decimal expansion.
a(3) = a(6) = a(9) = a(12) = 1 because 1/3 = 0.{3}*, 1/6 = 0.1{6}*, 1/9 = 0.{1}*, 1/12 = 0.08{3}* where the sequence of digits {...}* which repeats indefinitely is of length 1.
a(7) = 6 because 1/7 = 0.{142857}* with a period of 6.
a(17) = 16 because 1/17 = 0.{0588235294117647}* with a period of 16.
a(19) = 18 because 1/19 = 0.{052631578947368421}* with a period of 18. (End)
		

Crossrefs

Essentially same as A007732. Cf. A002371, A048595, A006883, A036275, A114205, A114206, A001913.

Programs

  • Maple
    A051626 := proc(n) local lpow,mpow ;
        if isA003592(n) then
           RETURN(0) ;
        else
           lpow:=1 ;
           while true do
              for mpow from lpow-1 to 0 by -1 do
                  if (10^lpow-10^mpow) mod n =0 then
                     RETURN(lpow-mpow) ;
                  fi ;
              od ;
              lpow := lpow+1 ;
           od ;
        fi ;
    end: # R. J. Mathar, Oct 19 2006
  • Mathematica
    r[x_]:=RealDigits[1/x]; w[x_]:=First[r[x]]; f[x_]:=First[w[x]]; l[x_]:=Last[w[x]]; z[x_]:=Last[r[x]];
    d[x_] := Which[IntegerQ[l[x]], 0, IntegerQ[f[x]]==False, Length[f[x]], True, Length[l[x]]]; Table[d[i], {i,1,90}] (* Hans Havermann, Oct 19 2006 *)
    fd[n_] := Block[{q},q = Last[First[RealDigits[1/n]]];If[IntegerQ[q], q = {}]; Length[q]];Table[fd[n], {n, 100}] (* Ray Chandler, Dec 06 2006 *)
    Table[Length[RealDigits[1/n][[1,-1]]],{n,90}] (* Harvey P. Dale, Jul 03 2011 *)
    a[n_] := If[ PowerMod[10, n, n] == 0, 0, MultiplicativeOrder[10, n/2^IntegerExponent[n, 2]/5^IntegerExponent[n, 5]]]; Array[a, 90] (* myself in A003592 and T. D. Noe in A007732 *) (* Robert G. Wilson v, Feb 20 2025 *)
  • PARI
    A051626(n)=if(1M. F. Hasler, Dec 14 2015
    
  • Python
    def A051626(n):
        if isA003592(n):
            return 0
        else:
            lpow=1
            while True:
                for mpow in range(lpow-1,-1,-1):
                    if (10**lpow-10**mpow) % n == 0:
                        return lpow-mpow
                lpow += 1 # Kenneth Myers, May 06 2016
    
  • Python
    from sympy import multiplicity, n_order
    def A051626(n): return 0 if (m:=(n>>(~n & n-1).bit_length())//5**multiplicity(5,n)) == 1 else n_order(10,m) # Chai Wah Wu, Aug 11 2022

Formula

a(n)=A132726(n,1); a(n)=a(A132740(n)); a(A132741(n))=a(A003592(n))=0. - Reinhard Zumkeller, Aug 27 2007

Extensions

More terms from James Sellers

A036275 The periodic part of the decimal expansion of 1/n. Any initial 0's are to be placed at end of cycle.

Original entry on oeis.org

0, 0, 3, 0, 0, 6, 142857, 0, 1, 0, 90, 3, 769230, 714285, 6, 0, 5882352941176470, 5, 526315789473684210, 0, 476190, 45, 4347826086956521739130, 6, 0, 384615, 370, 571428, 3448275862068965517241379310, 3, 322580645161290, 0, 30, 2941176470588235, 285714, 7
Offset: 1

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Comments

a(n) = 0 iff n = 2^i*5^j (A003592). - Jon Perry, Nov 19 2014
a(n) = n iff n = 3 or 6 (see De Koninck & Mercier reference). - Bernard Schott, Dec 02 2020

Examples

			1/28 = .03571428571428571428571428571428571428571... and digit-cycle is 571428, so a(28)=571428.
		

References

  • Jean-Marie De Koninck & Armel Mercier, 1001 Problèmes en Théorie Classique des Nombres, Problème 347 pp. 50 and 205, Ellipses, Paris, 2004.

Crossrefs

Programs

  • Maple
    isCycl := proc(n) local ifa,i ; if n <= 2 then RETURN(false) ; fi ; ifa := ifactors(n)[2] ; for i from 1 to nops(ifa) do if op(1,op(i,ifa)) <> 2 and op(1,op(i,ifa)) <> 5 then RETURN(true) ; fi ; od ; RETURN(false) ; end: A036275 := proc(n) local ifa,sh,lpow,mpow,r ; if not isCycl(n) then RETURN(0) ; else lpow:=1 ; while true do for mpow from lpow-1 to 0 by -1 do if (10^lpow-10^mpow) mod n =0 then r := (10^lpow-10^mpow)/n ; r := r mod (10^(lpow-mpow)-1) ; while r*10 < 10^(lpow-mpow) do r := 10*r ; od ; RETURN(r) ; fi ; od ; lpow := lpow+1 ; od ; fi ; end: for n from 1 to 60 do printf("%d %d ",n,A036275(n)) ; od ; # R. J. Mathar, Oct 19 2006
  • Mathematica
    fc[n_]:=Block[{q=RealDigits[1/n][[1,-1]]},If[IntegerQ[q],0,While[First[q]==0,q=RotateLeft[q]];FromDigits[q]]];
    Table[fc[n],{n,36}] (* Ray Chandler, Nov 19 2014, corrected Jun 27 2017 *)
    Table[FromDigits[FindTransientRepeat[RealDigits[1/n,10,120][[1]],3] [[2]]],{n,40}] (* Requires Mathematica version 10 or later *) (* Harvey P. Dale, Mar 12 2019 *)

Extensions

Corrected and extended by N. J. A. Sloane
Corrected a(92), a(208), a(248), a(328), a(352) and a(488) which missed a trailing zero (see the table). - Philippe Guglielmetti, Jun 20 2017

A097443 Half-period primes, i.e., primes p for which the decimal expansion of 1/p has period (p-1)/2.

Original entry on oeis.org

3, 13, 31, 43, 67, 71, 83, 89, 107, 151, 157, 163, 191, 197, 199, 227, 283, 293, 307, 311, 347, 359, 373, 401, 409, 431, 439, 443, 467, 479, 523, 557, 563, 569, 587, 599, 601, 631, 653, 677, 683, 719, 761, 787, 827, 839, 877, 881, 883, 911, 919, 929, 947, 991
Offset: 1

Views

Author

Julien Peter Benney (jpbenney(AT)ftml.net), Aug 23 2004

Keywords

Comments

Primes p for which 10 has multiplicative order (p-1)/2. - Robert Israel, Jul 15 2016

Examples

			13 is a half-period prime because 1/13 = 0.076923076923076923076923..., which has period 6, or (13-1)/2.
		

Crossrefs

Programs

  • Maple
    select(t -> isprime(t) and numtheory:-order(10, t) = (t-1)/2,
    [seq(t,t = 3..1000,2)]); # Robert Israel, Jul 15 2016
  • Mathematica
    f[n_Integer] := Block[{ds = Divisors[n - 1]}, (n - 1)/Take[ ds, Position[ PowerMod[ 10, ds, n], 1] [[1, 1]]] [[ -1]]]; Select[ Prime[ Range[4, 200]], f[ # ] == 2 &] (* Robert G. Wilson v, Sep 14 2004 *)
  • PARI
    is(n)= gcd(10,n)==1 && isprime(n) && znorder(Mod(10,n))==(n-1)/2 \\ Dana Jacobsen, Jul 19 2016
    
  • Perl
    use ntheory ":all"; forprimes { say if znorder(10,$) == ($-1)/2; } 1,1000; # Dana Jacobsen, Jul 19 2016

Extensions

Edited (including prepending 3), at the suggestion of Georg Fischer, by N. J. A. Sloane, Oct 19 2018

A040017 Prime 3 followed by unique period primes (the period r of 1/p is not shared with any other prime) of the form A019328(r)/gcd(A019328(r),r) in order (periods r are given in A051627).

Original entry on oeis.org

3, 11, 37, 101, 9091, 9901, 333667, 909091, 99990001, 999999000001, 9999999900000001, 909090909090909091, 1111111111111111111, 11111111111111111111111, 900900900900990990990991, 909090909090909090909090909091
Offset: 1

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Comments

Prime p=3 is the only known example of a unique period prime such that A019328(r)/gcd(A019328(r),r) = p^k with k > 1 (cf. A323748). It is plausible to assume that no other such prime exists. Under this (unproved) assumption, the current sequence lists all unique period primes in order and represents a sorted version of A007615. - Max Alekseyev, Oct 14 2022

Examples

			The decimal expansion of 1/101 is 0.00990099..., having a period of 4 and it is the only prime with that period.
		

References

  • J.-P. Delahaye, Merveilleux nombres premiers ("Amazing primes"), p. 324, Pour la Science Paris 2000.

Crossrefs

Programs

  • Mathematica
    lst = {}; Do[c = Cyclotomic[n, 10]; q = c/GCD[c, n]; If[PrimeQ[q], AppendTo[lst, q]], {n, 62}]; Prepend[Sort[lst], 3] (* Arkadiusz Wesolowski, May 13 2012 *)

Formula

For n >= 2, a(n) = A019328(r) / gcd(A019328(r), r), where r = A051627(n). - Max Alekseyev, Oct 14 2022

Extensions

Missing term a(45) inserted in b-file at the suggestion of Eric Chen by Max Alekseyev, Oct 13 2022
Edited by Max Alekseyev, Oct 14 2022

A007498 Unique period lengths of primes mentioned in A007615.

Original entry on oeis.org

1, 2, 3, 4, 9, 10, 12, 14, 19, 23, 24, 36, 38, 39, 48, 62, 93, 106, 120, 134, 150, 196, 294, 317, 320, 385, 586, 597, 654, 738, 945, 1031, 1172, 1282, 1404, 1426, 1452, 1521, 1752, 1812, 1836, 1844, 1862, 2134, 2232, 2264, 2667, 3750, 3903, 3927, 4274, 4354
Offset: 1

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Comments

Let {Zs(m, 10, 1)} be the Zsigmondy numbers for a = 10, b = 1: Zs(m, 10, 1) is the greatest divisor of 10^m - 1^m that is coprime to 10^r - 1^r for all positive integers r < m. Then this sequence gives m such that Zs(m, 10, 1) is a prime power (e.g., Zs(1, 10, 1) = 9 = 3^2, Zs(2, 10, 1) = 11, Zs(3, 10, 1) = 37, Zs(4, 10, 1) = 101). It is very likely that Zs(m, 10, 1) is prime if m > 1 is in this sequence (note that the Mathematica and PARI programs below are based on this assumption). - Jianing Song, Aug 12 2020

References

  • N. J. A. Sloane and Simon Plouffe, The Encyclopedia of Integer Sequences, Academic Press, 1995 (includes this sequence).
  • Samuel Yates, Period Lengths of Exactly One or Two Prime Numbers, J. Rec. Math., 18 (1985), 22-24.

Crossrefs

Cf. A161508 (unique period lengths in base 2).

Programs

  • Mathematica
    lst={1}; Do[p=Cyclotomic[n, 10]/GCD[n, Cyclotomic[n, 10]]; If[PrimeQ[p], AppendTo[lst, n]], {n, 3000}]; lst (* T. D. Noe, Sep 08 2005 *)
  • PARI
    isok(n) = if (n==1, 1, my(p = polcyclo(n, 10)); isprime(p/gcd(p, n))); \\ Michel Marcus, Jun 20 2018

Extensions

More terms from T. D. Noe, Sep 08 2005
a(48)-a(52) from Ray Chandler, Jul 09 2008

A054471 Smallest prime p having n different cycles in decimal expansions of k/p, k=1..p-1.

Original entry on oeis.org

7, 3, 103, 53, 11, 79, 211, 41, 73, 281, 353, 37, 2393, 449, 3061, 1889, 137, 2467, 16189, 641, 3109, 4973, 11087, 1321, 101, 7151, 7669, 757, 38629, 1231, 49663, 12289, 859, 239, 27581, 9613, 18131, 13757, 33931, 9161, 118901, 6763, 18233
Offset: 1

Views

Author

Robert G. Wilson v, 1994; Antreas P. Hatzipolakis (xpolakis(AT)otenet.gr), May 22 2000

Keywords

Comments

First cyclic number of n-th degree (or n-th order): the reciprocals of these numbers belong to one of n different cycles. Each cycle has (a(n) - 1)/n digits.
From Robert G. Wilson v, Aug 21 2014: (Start)
recursive by indices:
1, 7, 211, 79337, 634776923741, ...
2, 3, 103, 2368589, 785245568161181, ...
4, 53, 135257, 2332901103899, ...
5, 11, 353, 3795457, 693814982285339, ...
6, 79, 26861, 23947548497, ...
8, 41, 118901, 1015118238709, ...
9, 73, 142789, 267291583927, ...
10, 281, 3097183, 66880786504811, ...
12, 37, 18131, 105385168331, ...
13, 2393, 11160953, 7140939250711817, ...
14, 4999, 2148340247, > 10^19,
... .
(End)

References

  • John H. Conway and R. K. Guy, The Book of Numbers, Copernicus Press, p. 162.
  • M. Gardner, Mathematical Circus, Cambridge University Press (1996).

Crossrefs

First time n appears in A006556.
Cf. A006883, A097443, A055628, A056157, A056210, A056211, A056212, A056213, A056214, A056215, A056216, A056217, A098680, which are sequences of primes p where the period of the reciprocal is (p-1)/n for n=1, 2, 3, 4, 5, 6, 7, 8, 9, 10, 11, 12, 13.
Cf. A101208, A101209 (similar sequences for base 2 and base 3).

Programs

  • Mathematica
    a[n_Integer] := Block[{m = If[ OddQ@ n, 2n, n]}, p = m +1; While[ !PrimeQ@ p || p != 1 + n*MultiplicativeOrder[10, p], p = p += m]; p]; a[1] = 7; a[4] = 53; Array[f, 50] (* Robert G. Wilson v, Apr 19 2005; revised Aug 20 2014 and Feb 14 2025 *)

Extensions

More terms from David W. Wilson, May 22 2000

A007615 Primes with unique period length (the periods are given in A007498).

Original entry on oeis.org

3, 11, 37, 101, 333667, 9091, 9901, 909091, 1111111111111111111, 11111111111111111111111, 99990001, 999999000001, 909090909090909091, 900900900900990990990991, 9999999900000001, 909090909090909090909090909091, 900900900900900900900900900900990990990990990990990990990991
Offset: 1

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Author

Keywords

Comments

Additional terms are Phi(n,10)/gcd(n,Phi(n,10)) for the n in A007498, where Phi(n,10) is the n-th cyclotomic polynomial evaluated at 10.

Examples

			3 is the only prime p such that decimal expansion of 1/p has (nontrivial) period exactly 1.
		

References

  • N. J. A. Sloane and Simon Plouffe, The Encyclopedia of Integer Sequences, Academic Press, 1995 (includes this sequence).
  • Samuel Yates, Period Lengths of Exactly One or Two Prime Numbers, J. Rec. Math., 18 (1985), 22-24.

Crossrefs

Programs

  • Mathematica
    nmax = 50; periods = Reap[ Do[ p = Cyclotomic[n, 10] / GCD[n, Cyclotomic[n, 10]]; If[ PrimeQ[p], Sow[n]], {n, 1, nmax}]][[2, 1]]; Cyclotomic[#, 10] / GCD[#, Cyclotomic[#, 10]]& /@ periods // Prepend[#, 3]& (* Jean-François Alcover, Mar 28 2013 *)

Formula

a(n) = A061075(A007498(n)). - Max Alekseyev, Oct 16 2010
a(n) = A006530(A019328(A007498(n))). - Ray Chandler, May 10 2017
Showing 1-10 of 26 results. Next