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 154 results. Next

A002275 Repunits: (10^n - 1)/9. Often denoted by R_n.

Original entry on oeis.org

0, 1, 11, 111, 1111, 11111, 111111, 1111111, 11111111, 111111111, 1111111111, 11111111111, 111111111111, 1111111111111, 11111111111111, 111111111111111, 1111111111111111, 11111111111111111, 111111111111111111, 1111111111111111111, 11111111111111111111
Offset: 0

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Author

Keywords

Comments

R_n is a string of n 1's.
Base-4 representation of Jacobsthal bisection sequence A002450. E.g., a(4)= 1111 because A002450(4)= 85 (in base 10) = 64 + 16 + 4 + 1 = 1*(4^3) + 1*(4^2) + 1*(4^1) + 1. - Paul Barry, Mar 12 2004
Except for the first two terms, these numbers cannot be perfect squares, because x^2 != 11 (mod 100). - Zak Seidov, Dec 05 2008
For n >= 0: a(n) = (A000225(n) written in base 2). - Jaroslav Krizek, Jul 27 2009, edited by M. F. Hasler, Jul 03 2020
Let A be the Hessenberg matrix of order n, defined by: A[1,j]=1, A[i,i]:=10, (i>1), A[i,i-1]=-1, and A[i,j]=0 otherwise. Then, for n>=1, a(n)=det(A). - Milan Janjic, Feb 21 2010
Except 0, 1 and 11, all these integers are Brazilian numbers, A125134. - Bernard Schott, Dec 24 2012
Numbers n such that 11...111 = R_n = (10^n - 1)/9 is prime are in A004023. - Bernard Schott, Dec 24 2012
The terms 0 and 1 are the only squares in this sequence, as a(n) == 3 (mod 4) for n>=2. - Nehul Yadav, Sep 26 2013
For n>=2 the multiplicative order of 10 modulo the a(n) is n. - Robert G. Wilson v, Aug 20 2014
The above is a special case of the statement that the order of z modulo (z^n-1)/(z-1) is n, here for z=10. - Joerg Arndt, Aug 21 2014
From Peter Bala, Sep 20 2015: (Start)
Let d be a divisor of a(n). Let m*d be any multiple of d. Split the decimal expansion of m*d into 2 blocks of contiguous digits a and b, so we have m*d = 10^k*a + b for some k, where 0 <= k < number of decimal digits of m*d. Then d divides a^n - (-b)^n (see McGough). For example, 271 divides a(5) and we find 2^5 + 71^5 = 11*73*271*8291 and 27^5 + 1^5 = 2^2*7*31*61*271 are both divisible by 271. Similarly, 4*271 = 1084 and 10^5 + 84^5 = 2^5*31*47*271*331 while 108^5 + 4^5 = 2^12*7*31*61*271 are again both divisible by 271. (End)
Starting with the second term this sequence is the binary representation of the n-th iteration of the Rule 220 and 252 elementary cellular automaton starting with a single ON (black) cell. - Robert Price, Feb 21 2016
If p > 5 is a prime, then p divides a(p-1). - Thomas Ordowski, Apr 10 2016
0, 1 and 11 are only terms that are of the form x^2 + y^2 + z^2 where x, y, z are integers. In other words, a(n) is a member of A004215 for all n > 2. - Altug Alkan, May 08 2016
Except for the initial terms, the binary representation of the x-axis, from the left edge to the origin, of the n-th stage of growth of the two-dimensional cellular automaton defined by "Rule 737", based on the 5-celled von Neumann neighborhood, initialized with a single black (ON) cell at stage zero. - Robert Price, Mar 17 2017
The term "repunit" was coined by Albert H. Beiler in 1964. - Amiram Eldar, Nov 13 2020
q-integers for q = 10. - John Keith, Apr 12 2021
Binomial transform of A001019 with leading zero. - Jules Beauchamp, Jan 04 2022

References

  • Albert H. Beiler, Recreations in the Theory of Numbers: The Queen of Mathematics Entertains, New York: Dover Publications, 1964, chapter XI, p. 83.
  • Paulo Ribenboim, The Little Book of Bigger Primes, Springer-Verlag NY 2004. See pp. 235-237.
  • David Wells, The Penguin Dictionary of Curious and Interesting Numbers, Penguin Books, 1987, pp. 197-198.
  • Samuel Yates, Peculiar Properties of Repunits, J. Recr. Math. 2, 139-146, 1969.
  • Samuel Yates, Prime Divisors of Repunits, J. Recr. Math. 8, 33-38, 1975.

Crossrefs

Programs

  • Haskell
    a002275 = (`div` 9) . subtract 1 . (10 ^)
    a002275_list = iterate ((+ 1) . (* 10)) 0
    -- Reinhard Zumkeller, Jul 05 2013, Feb 05 2012
    
  • Magma
    [(10^n-1)/9: n in [0..25]]; // Vincenzo Librandi, Nov 06 2014
    
  • Maple
    seq((10^k - 1)/9, k=0..30); # Wesley Ivan Hurt, Sep 28 2013
  • Mathematica
    Table[(10^n - 1)/9, {n, 0, 19}] (* Alonso del Arte, Nov 15 2011 *)
    Join[{0},Table[FromDigits[PadRight[{},n,1]],{n,20}]] (* Harvey P. Dale, Mar 04 2012 *)
  • Maxima
    a[0]:0$
    a[1]:1$
    a[n]:=11*a[n-1]-10*a[n-2]$
    A002275(n):=a[n]$
    makelist(A002275(n),n,0,30); /* Martin Ettl, Nov 05 2012 */
    
  • PARI
    a(n)=(10^n-1)/9; \\ Michael B. Porter, Oct 26 2009
    
  • PARI
    my(x='x+O('x^30)); concat(0, Vec(x/((1-10*x)*(1-x)))) \\ Altug Alkan, Apr 10 2016
    
  • Python
    print([(10**n-1)//9 for n in range(100)]) # Michael S. Branicky, Apr 30 2022
  • Sage
    [lucas_number1(n, 11, 10) for n in range(21)]  # Zerinvary Lajos, Apr 27 2009
    

Formula

a(n) = 10*a(n-1) + 1, a(0)=0.
a(n) = A000042(n) for n >= 1.
Second binomial transform of Jacobsthal trisection A001045(3n)/3 (A015565). - Paul Barry, Mar 24 2004
G.f.: x/((1-10*x)*(1-x)). Regarded as base b numbers, g.f. x/((1-b*x)*(1-x)). - Franklin T. Adams-Watters, Jun 15 2006
a(n) = 11*a(n-1) - 10*a(n-2), a(0)=0, a(1)=1. - Lekraj Beedassy, Jun 07 2006
a(n) = A125118(n,9) for n>8. - Reinhard Zumkeller, Nov 21 2006
a(n) = A075412(n)/A002283(n). - Reinhard Zumkeller, May 31 2010
a(n) = a(n-1) + 10^(n-1) with a(0)=0. - Vincenzo Librandi, Jul 22 2010
a(n) = A242614(n,A242622(n)). - Reinhard Zumkeller, Jul 17 2014
E.g.f.: (exp(9*x) - 1)*exp(x)/9. - Ilya Gutkovskiy, May 11 2016
a(n) = Sum_{k=0..n-1} 10^k. - Torlach Rush, Nov 03 2020
Sum_{n>=1} 1/a(n) = A065444. - Amiram Eldar, Nov 13 2020
From Elmo R. Oliveira, Aug 02 2025: (Start)
a(n) = A002283(n)/9 = A105279(n)/10.
a(n) = A010785(A017173(n-1)) for n >= 1. (End)

A004022 Primes of the form (10^k - 1)/9. Also called repunit primes or repdigit primes.

Original entry on oeis.org

11, 1111111111111111111, 11111111111111111111111
Offset: 1

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Author

Keywords

Comments

The next term corresponds to k = 317 and is too large to include: see A004023.
Also called repunit primes or prime repunits.
Also, primes with digital product = 1.
The number of 1's in these repunits must also be prime. Since the number of 1's in (10^k-1)/9 is k, if k = p*m then (10^(p*m)-1) = (10^p)^m-1 => (10^p-1)/9 = q and q divides (10^k-1). This follows from the identity a^k - b^k = (a-b)*(a^(k-1) + a^(k-2)*b + ... + b^(k-1)). - Cino Hilliard, Dec 23 2008
A subset of A020449, ..., A020457, A036953, ..., cf. link to OEIS index. - M. F. Hasler, Jul 27 2015
The terms in this sequence, except 11 which is not Brazilian, are prime repunits in base ten, so they are Brazilian primes belonging to A085104 and A285017. - Bernard Schott, Apr 08 2017

References

  • T. M. Apostol, Introduction to Analytic Number Theory, Springer-Verlag, 1976, p. 11. Graham, Knuth and Patashnik, Concrete mathematics, Addison-Wesley, 1994; see p. 146, problem 22.
  • M. Barsanti, R. Dvornicich, M. Forti, T. Franzoni, M. Gobbino, S. Mortola, L. Pernazza and R. Romito, Il Fibonacci N. 8 (included in Il Fibonacci, Unione Matematica Italiana, 2011), 2004, Problem 8.10.
  • Clifford A. Pickover, A Passion for Mathematics, Wiley, 2005; see p. 60.
  • N. J. A. Sloane and Simon Plouffe, The Encyclopedia of Integer Sequences, Academic Press, 1995 (includes this sequence).
  • James J. Tattersall, Elementary Number Theory in Nine Chapters, Cambridge University Press, 1999, page 114.

Crossrefs

Subsequence of A020449.
A116692 is another version of repunit primes or repdigit primes. - N. J. A. Sloane, Jan 22 2023
See A004023 for the number of 1's.
Cf. A046413.

Programs

  • Magma
    [a: n in [0..300] | IsPrime(a) where a is (10^n - 1) div 9 ]; // Vincenzo Librandi, Nov 08 2014
    
  • Mathematica
    lst={}; Do[If[PrimeQ[p = (10^n - 1)/9], AppendTo[lst, p]], {n, 10^2}]; lst (* Vladimir Joseph Stephan Orlovsky, Aug 22 2008 *)
    Select[Table[(10^n - 1) / 9, {n, 500}], PrimeQ] (* Vincenzo Librandi, Nov 08 2014 *)
    Select[Table[FromDigits[PadRight[{},n,1]],{n,30}],PrimeQ] (* Harvey P. Dale, Apr 07 2018 *)
  • PARI
    forprime(x=2,20000,if(ispseudoprime((10^x-1)/9),print1((10^x-1)/9","))) \\ Cino Hilliard, Dec 23 2008
    
  • Python
    from sympy import isprime
    from itertools import count, islice
    def agen(): # generator of terms
        yield from (t for t in (int("1"*k) for k in count(1)) if isprime(t))
    print(list(islice(agen(), 4))) # Michael S. Branicky, Jun 09 2022

Formula

a(n) = A002275(A004023(n)).

Extensions

Edited by Max Alekseyev, Nov 15 2010
Name expanded by N. J. A. Sloane, Jan 22 2023

A003459 Absolute primes (or permutable primes): every permutation of the digits is a prime.

Original entry on oeis.org

2, 3, 5, 7, 11, 13, 17, 31, 37, 71, 73, 79, 97, 113, 131, 199, 311, 337, 373, 733, 919, 991, 1111111111111111111, 11111111111111111111111
Offset: 1

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Author

Keywords

Comments

From Bill Gosper, Jan 24 2003, in a posting to the Math Fun Mailing List: (Start)
Recall Sloane's old request for more terms of A003459 = (2 3 5 7 11 13 17 31 37 71 73 79 97 113 131 199 311 337 373 733 919 991 ...) and Richard C. Schroeppel's astonishing observation that the next term is 1111111111111111111. Absent Rich's analysis, trying to extend this sequence makes a great set of beginner's programming exercises. We may restrict the search to combinations of the four digits 1,3,7,9, only look at starting numbers with nondecreasing digits, generate only unique digit combinations, and only as needed. (We get the target sequence afterward by generating and merging the various permutations, and fudging the initial 2,3,5,7.)
To my amazement the (uncompiled, Macsyma) program printed 11,13,...,199,337, and after about a minute, 1111111111111111111!
And after a few more minutes, (10^23-1)/9! (End)
Boal and Bevis say that Johnson (1977) proves that if there is a term > 1000 with exactly two distinct digits then it must have more than nine billion digits. - N. J. A. Sloane, Jun 06 2015
Some authors require permutable or absolute primes to have at least two different digits. This produces the subsequence A129338. - M. F. Hasler, Mar 26 2008
See A039986 for a related problem with more sophisticated (PARI) code (iteration over only inequivalent digit permutations). - M. F. Hasler, Jul 10 2018

References

  • Richard C. Schroeppel, personal communication.
  • Wacław Sierpiński, Co wiemy, a czego nie wiemy o liczbach pierwszych. Warsaw: PZWS, 1961, pp. 20-21.
  • N. J. A. Sloane and Simon Plouffe, The Encyclopedia of Integer Sequences, Academic Press, 1995 (includes this sequence).
  • James J. Tattersall, Elementary Number Theory in Nine Chapters, Cambridge University Press, 1999, page 113.

Crossrefs

Includes all of A004022 = A002275(A004023).
A258706 gives minimal representatives of the permutation classes.
Cf. A039986.

Programs

  • Haskell
    import Data.List (permutations)
    a003459 n = a003459_list !! (n-1)
    a003459_list = filter isAbsPrime a000040_list where
       isAbsPrime = all (== 1) . map (a010051 . read) . permutations . show
    -- Reinhard Zumkeller, Sep 15 2011
    
  • Mathematica
    f[n_]:=Module[{b=Permutations[IntegerDigits[n]],q=1},Do[If[!PrimeQ[c=FromDigits[b[[m]]]],q=0;Break[]],{m,Length[b]}];q];Select[Range[1000],f[#]>0&] (* Vladimir Joseph Stephan Orlovsky, Feb 03 2011 *)
    (* Linear complexity: can't reach R(19). See A258706. - Bill Gosper, Jan 06 2017 *)
  • PARI
    for(n=1, oo, my(S=[],r=10^n\9); for(a=1, 9^(n>1), for(b=if(n>2, 1-a), 9-a, for(j=0, if(b, n-1), ispseudoprime(a*r+b*10^j)||next(2)); S=concat(S,vector(if(b,n,1),k,a*r+10^(k-1)*b))));apply(t->printf(t","),Set(S))) \\ M. F. Hasler, Jun 26 2018

Formula

Conjecture: for n >= 23, a(n) = A004022(n-21). - Max Alekseyev, Oct 08 2018

Extensions

The next terms are a(25)=A002275(317), a(26)=A002275(1031), a(27)=A002275(49081).

A077775 Odd numbers k such that (10^k - 1)/3 - 2*10^floor(k/2) is a palindromic wing prime (a.k.a. near-repdigit palindromic prime) of the form 3...313...3.

Original entry on oeis.org

3, 7, 15, 123, 181, 185, 539, 597, 643, 743, 1553, 3135, 4769, 5133, 6177, 11733, 16103, 18997, 25271, 49025, 65043, 87965
Offset: 1

Views

Author

Patrick De Geest, Nov 16 2002

Keywords

Comments

Prime versus probable prime status and proofs are given in the author's table.
a(23) > 2*10^5. - Robert Price, Jan 29 2016
Primes of the form (10^k-1)/3 - 2*10^floor(k/2) are obtained for k in (2, 3, 6, 7, 8, 10, 15, 22, 34, 123, 126, 144, 181, 185, 198, 534, 539, 597, 606, ...). For example (10^2 - 1)/3 - 2*10^1 = 13 is also prime. However, for even k the result is not palindromic. See A077775-A077798, A107123-A107127 for PWP's with digits other than 3 and 1. - M. F. Hasler, Mar 03 2019

Examples

			a(3) = 15 corresponds to the prime (10^15 - 1)/3 - 2*10^7 = 333333313333333.
		

References

  • C. Caldwell and H. Dubner, "Journal of Recreational Mathematics", Volume 28, No. 1, 1996-97, pp. 1-9.

Crossrefs

Programs

  • Mathematica
    Do[ If[ PrimeQ[(10^n - 6*10^Floor[n/2] - 1)/3], Print[n]], {n, 3, 49100, 2}] (* Robert G. Wilson v, Dec 16 2005 *)
  • PARI
    is(n)=bittest(n,0)&&ispseudoprime(10^n\3-2*10^(n\2)) \\ M. F. Hasler, Mar 03 2019

Formula

a(n) = 2*A183174(n) + 1.

Extensions

a(21)-a(22) from Robert Price, Jan 29 2016
a(21) corrected by Robert Price, Feb 03 2016
Name corrected by Jon E. Schoenfield, Oct 31 2018
Name edited by M. F. Hasler, Mar 03 2019

A183187 Numbers k such that 10^(2k+1)-10^k-1 is prime.

Original entry on oeis.org

26, 378, 1246, 1798, 2917, 23034, 47509, 52140, 67404
Offset: 1

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Author

Ray Chandler, Dec 28 2010

Keywords

Comments

944264 is a term but its position is not known. - Jeppe Stig Nielsen, Jan 12 2024
a(10) > 300000. - Serge Batalov, Jan 17 2024

References

  • C. Caldwell and H. Dubner, "Journal of Recreational Mathematics", Volume 28, No. 1, 1996-97, pp. 1-9.

Crossrefs

Programs

  • Mathematica
    Do[If[PrimeQ[10^(2n + 1) - 10^n - 1], Print[n]], {n, 3000}]
  • PARI
    for(n=1,1e3,if(ispseudoprime(10^(2*n+1)-10^n-1),print1(n", "))) \\ Charles R Greathouse IV, Jul 15 2011

Formula

a(n) = (A077794(n)-1)/2.

A115073 Numbers k such that 10^(2*k+1)-7*10^k-1 is prime.

Original entry on oeis.org

1, 8, 9, 352, 530, 697, 1315, 1918, 2874, 5876, 6768, 62938, 134739
Offset: 1

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Author

N. J. A. Sloane, Mar 03 2006

Keywords

References

  • C. Caldwell and H. Dubner, "Journal of Recreational Mathematics", Volume 28, No. 1, 1996-97, pp. 1-9.

Crossrefs

Programs

  • Mathematica
    Do[If[PrimeQ[10^(2n + 1) - 7*10^n - 1], Print[n]], {n, 3000}]
  • PARI
    for(n=0,1e4,if(ispseudoprime(t=10^(2*n+1)-7*10^n-1),print1(t", "))) \\ Charles R Greathouse IV, Jul 15 2011

Formula

a(n) = (A077778(n)-1)/2.

Extensions

Edited by Ray Chandler, Dec 28 2010
Added two more terms from PWP table, by Patrick De Geest, Nov 05 2014

A107123 Numbers k such that (10^(2*k+1)+18*10^k-1)/9 is prime.

Original entry on oeis.org

0, 1, 2, 19, 97, 9818
Offset: 1

Views

Author

Farideh Firoozbakht, May 19 2005

Keywords

Comments

A number k is in the sequence iff the palindromic number 1(k).3.1(k) is prime (1(k) means k copies of 1; dot between numbers means concatenation). If k is a positive term of the sequence then k is not of the form 3m, 6m+4, 12m+10, 28m+5, 28m+8, etc. (the proof is easy).
The palindromic number 1(k).2.1(k) is never prime for k > 0 because it is (1.0(k-1).1)*(1(k+1)). - Robert Israel, Jun 11 2015
a(7) > 10^5. - Robert Price, Apr 02 2016

Examples

			19 is in the sequence because the palindromic number (10^(2*19+1)+18*10^19-1)/9 = 1(19).3.1(19) = 111111111111111111131111111111111111111 is prime.
		

References

  • C. Caldwell and H. Dubner, "Journal of Recreational Mathematics", Volume 28, No. 1, 1996-97, pp. 1-9.

Crossrefs

Programs

  • Maple
    select(n -> isprime((10^(2*n+1)+18*10^n-1)/9), [$0..100]); # Robert Israel, Jun 11 2015
  • Mathematica
    Do[If[PrimeQ[(10^(2n + 1) + 18*10^n - 1)/9], Print[n]], {n, 2500}]
  • PARI
    for(n=0,1e4,if(ispseudoprime(t=(10^(2*n+1)+18*10^n)\9),print1(t", "))) \\ Charles R Greathouse IV, Jul 15 2011

Formula

a(n) = (A077779(n-1)-1)/2, for n > 1. [Corrected by M. F. Hasler, Feb 06 2020]

Extensions

Edited by Ray Chandler, Dec 28 2010

A107127 Numbers n such that (10^(2n+1)+54*10^n-1)/9 is prime.

Original entry on oeis.org

0, 3, 33, 311, 2933, 22235, 39165, 41585
Offset: 1

Views

Author

Farideh Firoozbakht, May 19 2005

Keywords

Comments

n is in the sequence iff the palindromic number 1(n).7.1(n) is prime (dot between numbers means concatenation). If n is in the sequence then n is not of the forms 3m+1, 6m, 6m+2, 7m+2, 16m+9, 16m+14, 18m+1, 18m+7, 22m+13, 22m+19, etc. (the proof is easy).
a(9) > 10^5. - Robert Price, Apr 30 2017

Examples

			3 is in the sequence because (10^(2*3+1)+54*10^3-1)/9=1(3).7.1(3)=1117111 is prime.
2933 is in the sequence because (10^(2*2933+1)+54*10^2933-1)/9=1(2933).7.1(2933) is prime.
		

References

  • C. Caldwell and H. Dubner, "Journal of Recreational Mathematics", Volume 28, No. 1, 1996-97, pp. 1-9.

Crossrefs

Programs

  • Mathematica
    Do[If[PrimeQ[(10^(2n + 1) + 54*10^n - 1)/9], Print[n]], {n, 3250}]
  • PARI
    for(n=0,1e4,if(ispseudoprime(t=(10^(2*n+1)+54*10^n)\9),print1(t", "))) \\ Charles R Greathouse IV, Jul 15 2011

Formula

a(n) = (A077789(n)-1)/2.

Extensions

Edited by Ray Chandler, Dec 28 2010
a(6)-a(8) from Robert Price, Apr 30 2017

A107648 Numbers n such that (10^(2n+1)+63*10^n-1)/9 is prime.

Original entry on oeis.org

1, 4, 6, 7, 384, 666, 675, 3165, 131020
Offset: 1

Views

Author

Farideh Firoozbakht, May 19 2005

Keywords

Comments

n is in the sequence iff the palindromic number 1(n).8.1(n) is prime (dot between numbers means concatenation). Let f(n)=(10^(2n+1)+63*10^n-1)/9 then for all nonnegative integers m we have: I. 3 divides f(3m+2) II. 19 divides f(18m+13) III. 29 divides f(28*m+16) & 29 divides f(28*m+25) IV. 31 divides f(30*m+2) & 31 divides f(30*m+17) V. 41 divides f(5m+3), etc. So if n is in the sequence then n is not of the forms 3m+2, 18m+13, 28m+16 28m+25, 30m+2, 30m+17, 5m+3, etc.
a(9) > 10^5. - Robert Price, Oct 30 2017

Examples

			7 is in the sequence because (10^15+63*10^7-1)/9=1(7).8.1(7)=111111181111111 is prime.
666 is in the sequence because (10^(2*666+1)+63*10^666-1)/9=1(666).8.1(666) is prime.
		

References

  • C. Caldwell and H. Dubner, "Journal of Recreational Mathematics", Volume 28, No. 1, 1996-97, pp. 1-9.
  • Ivan Niven, Herbert S. Zuckerman and Hugh L. Montgomery, An Introduction to the Theory Of Numbers, Fifth Edition, John Wiley and Sons, Inc., NY 1991, p. 141.

Crossrefs

Programs

  • Mathematica
    Do[If[PrimeQ[(10^(2n + 1) + 63*10^n - 1)/9], Print[n]], {n, 4000}]
  • PARI
    for(n=0,1e4,if(ispseudoprime(t=(10^(2*n+1)+63*10^n)\9),print1(t", "))) \\ Charles R Greathouse IV, Jul 15 2011

Formula

a(n) = (A077791(n)-1)/2.

Extensions

Edited by Ray Chandler, Dec 28 2010
a(9) from Robert Price, Aug 03 2024

A107649 Numbers n such that (10^(2n+1)+72*10^n-1)/9 is prime.

Original entry on oeis.org

1, 4, 26, 187, 226, 874, 13309, 34016, 42589
Offset: 1

Views

Author

Farideh Firoozbakht, May 19 2005

Keywords

Comments

n is in the sequence iff the palindromic number 1(n).9.1(n) is prime (dot between numbers means concatenation). If n is in the sequence then n is not of the forms 3m, 6m+5, 22m+3, 22m+7, etc. (the proof is easy).
a(10) > 200000. - _Robert Price, Jan 30 2025

Examples

			26 is in the sequence because (10^(2*26+1)+72*10^26-1)/9=1(26).9.1(26)
= 11111111111111111111111111911111111111111111111111111 is prime.
		

References

  • C. Caldwell and H. Dubner, "Journal of Recreational Mathematics", Volume 28, No. 1, 1996-97, pp. 1-9.

Crossrefs

Programs

  • Mathematica
    Do[If[PrimeQ[(10^(2n + 1) + 72*10^n - 1)/9], Print[n]], {n, 3000}]
    prQ[n_]:=Module[{c=PadRight[{},n,1]},PrimeQ[FromDigits[Join[c,{9},c]]]]; Select[Range[13500],prQ] (* Harvey P. Dale, Jan 19 2014 *)
  • PARI
    for(n=0,1e4,if(ispseudoprime(t=(10^(2*n+1)+72*10^n)\9),print1(t", "))) \\ Charles R Greathouse IV, Jul 15 2011

Formula

a(n) = (A077795(n)-1)/2.

Extensions

Edited by Ray Chandler, Dec 28 2010
a(8)-a(9) from Robert Price, Sep 28 2017
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