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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A093177 Primes of the form 90*R_k + 1, where R_k is the repunit (A002275) of length k.

Original entry on oeis.org

991, 99991, 9999991, 999999999999999999999999999999991, 999999999999999999999999999999999999999999991, 999999999999999999999999999999999999999999999999999999999999999999999999999999999999999999999999999999991
Offset: 1

Views

Author

Rick L. Shepherd, Mar 27 2004

Keywords

Comments

Primes of the form 10^k - 9 (k=3,5,7,...). - Vincenzo Librandi, Nov 16 2010

Crossrefs

Cf. A056696 (corresponding k), A095714.

Formula

a(n) = 10^(A056696(n)+1) - 9 = 10^A095714(n) - 9.

Extensions

Edited by Ray Chandler, Feb 26 2012

A055557 Numbers k such that 3*R_k - 2 is prime, where R_k = 11...1 is the repunit (A002275) of length k.

Original entry on oeis.org

2, 3, 4, 5, 6, 7, 8, 18, 40, 50, 60, 78, 101, 151, 319, 382, 784, 1732, 1918, 8855, 11245, 11960, 12130, 18533, 22718, 23365, 24253, 24549, 25324, 30178, 53718, 380976, 424861, 563535, 666903
Offset: 1

Views

Author

Labos Elemer, Jul 10 2000

Keywords

Comments

Also numbers k such that (10^k-7)/3 is prime.
Sierpiński attributes the primes for k = 2,...,8 to A. Makowski.
The history of the discovery of these numbers may be as follows: a(1)-a(7), Makowski; a(8)-a(18), Caldwell; a(19), Earls; a(20)-a(31), Kamada. (Corrections to this account will be welcomed.)
Concerning certifying primes, see the references by Goldwasser et al., Atkin et al. and Morain. - Labos
No more than 14 consecutive exponents can provide primes because for exponents 15m+2, 16m+9, 18m+12, 22m+21, terms are divisible by 31, 17, 19, 23 respectively. Here 7 of possible 14 is realized. - Labos Elemer, Jan 19 2005
(10^(15m+2)-7)/3 == 0 (mod 31). So 15m+2 isn't a term for m > 0. - Seiichi Manyama, Nov 05 2016

References

  • C. Caldwell, The near repdigit primes 333...331, J.Recreational Math. 21:4 (1989) 299-304.
  • S. Goldwasser and J. Kilian, Almost All Primes Can Be Quickly Certified. in Proc. 18th STOC, 1986, pp. 316-329.
  • W. Sierpiński, 200 Zadan z Elementarnej Teorii Liczb [200 Problems from the Elementary Theory of Numbers], Warszawa, 1964; Problem 88.

Crossrefs

Programs

  • Mathematica
    Do[ If[ PrimeQ[(10^n - 7)/3], Print[n]], {n, 50410}]
    One may run the prime certificate program as follows <True]}, {n, 1, 16}] (* Labos Elemer *)
  • PARI
    for(n=1,2000, if(isprime((10^n-7)/3),print(n)))

Formula

a(n) = A055520(n) + 1.

Extensions

Corrected and extended by Jason Earls, Sep 22 2001
a(20)-a(31) were found by Makoto Kamada (see links for details). At present they correspond only to probable primes.
a(32)-a(33) from Leonid Durman, Jan 09-10 2012
a(34)-a(35) from Kamada data by Tyler Busby, Apr 14 2024

A099814 Bisection of A002275.

Original entry on oeis.org

0, 11, 1111, 111111, 11111111, 1111111111, 111111111111, 11111111111111, 1111111111111111, 111111111111111111, 11111111111111111111, 1111111111111111111111, 111111111111111111111111, 11111111111111111111111111, 1111111111111111111111111111, 111111111111111111111111111111, 11111111111111111111111111111111
Offset: 0

Views

Author

N. J. A. Sloane, Nov 19 2004

Keywords

Comments

Except for the first term (replace 0 with 1) this is the binary representation of the n-th iteration of the elementary cellular automaton starting with a single ON (black) cell for Rule 189. - Robert Price, Feb 21 2016

Crossrefs

Cf. A002275, A100706 (other bisection).

Programs

Formula

For n>0, 1 repeated 2n times.
a(n) = a(n-1) + 11*100^(n-1) with a(0)=0. - Vincenzo Librandi, Nov 16 2010
a(n) = (100^n - 1)/9. - Bernard Schott, Apr 09 2023
From Stefano Spezia, Jul 27 2024: (Start)
G.f.: 11*x/((1 - x)*(1 - 100*x)).
E.g.f.: exp(x)*(exp(99*x) - 1)/9. (End)

Extensions

More terms from Stefan Steinerberger, Apr 29 2006

A092571 Primes of the form 60*R_k + 1, where R_k is the repunit (A002275) of length k.

Original entry on oeis.org

61, 661, 6661, 6666666661, 666666666666666661, 666666666666666666661, 6666666666666666666661, 6666666666666666666666666661, 6666666666666666666666666666666666666666661
Offset: 1

Views

Author

Rick L. Shepherd, Feb 28 2004

Keywords

Comments

The number of 6's in each term is given by the corresponding term of A056658.
Primes of the form (6*10^k - 51)/9. - Vincenzo Librandi, Nov 17 2010

Crossrefs

Cf. A002275, A056658 (corresponding k).

Programs

  • Mathematica
    Select[Table[FromDigits[PadLeft[{1},n,6]],{n,2,50}],PrimeQ] (* Harvey P. Dale, Sep 21 2023 *)

A093170 Primes of the form 60*R_k + 7, where R_k is the repunit (A002275) of length k.

Original entry on oeis.org

7, 67, 666667, 66666667, 666666667, 66666666667, 66666666666666666667, 66666666666666666666667, 66666666666666666666666666666666666666667, 666666666666666666666666666666666666666666666666666666666666667
Offset: 1

Views

Author

Rick L. Shepherd, Mar 26 2004

Keywords

Comments

Primes of the form (2*10^k + 1)/3. - Vincenzo Librandi, Nov 16 2010
Occur in the factorization of some of the numbers of the form 13...3 not in A093671, cf. second Kamada link. - M. F. Hasler, Sep 14 2014

Crossrefs

Cf. A002275, A056657 (corresponding k), A093671, A096507.

Programs

  • Maple
    A093170:=n->`if`(isprime((2*10^n+1)/3),(2*10^n+1)/3,NULL): seq(A093170(n), n=1..70); # Wesley Ivan Hurt, Sep 14 2014
  • Mathematica
    Select[Table[FromDigits[PadLeft[{7},n,6]],{n,70}],PrimeQ] (* Harvey P. Dale, Jan 26 2013 *)

Formula

a(n) = (20*10^A056657(n)+1)/3 = (2*10^A096507(n)+1)/3.

Extensions

Edited by Ray Chandler, Feb 23 2012

A093176 Primes of the form 70*R_k + 1, where R_k is the repunit (A002275) of length k.

Original entry on oeis.org

71, 7777777777771, 77777777777777777771, 77777777777777777777771, 7777777777777777777777777777771, 7777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777771
Offset: 1

Views

Author

Rick L. Shepherd, Mar 27 2004

Keywords

Comments

Primes of the form (7*10^k - 61)/9. - Vincenzo Librandi, Nov 16 2010
The next term (a(7)) has 241 digits. - Harvey P. Dale, Jul 01 2022

Crossrefs

Cf. A002275, A056688 (corresponding k), A099419.

Programs

  • Mathematica
    Select[Table[FromDigits[PadLeft[{1},n,7]],{n,100}],PrimeQ] (* Harvey P. Dale, Jul 01 2022 *)

Formula

a(n) = (70*10^A056688(n) - 61)/9 = (7*10^A099419(n) - 61)/9.

Extensions

Edited by Ray Chandler, Mar 06 2012

A096846 Numbers n for which 8*R_n - 1 is prime, where R_n = 11...1 is the repunit (A002275) of length n.

Original entry on oeis.org

1, 3, 4, 6, 9, 12, 72, 118, 124, 190, 244, 304, 357, 1422, 2691, 5538, 7581, 21906, 32176, 44358, 120552, 137073, 152260
Offset: 1

Views

Author

Labos Elemer, Jul 15 2004

Keywords

Comments

Also numbers n such that (8*10^n-17)/9 is prime.
The numbers corresponding to a(1)-a(15) are certified prime, the numbers corresponding to a(16)-a(20) are probable primes. a(21) > 10^5. - Robert Price, May 20 2014

Examples

			n=6: a(4)=888887 which is prime.
		

Crossrefs

Programs

  • Mathematica
    Do[ If[ PrimeQ[ 8(10^n - 1)/9 - 1], Print[n]], {n, 0, 5000}] (* Robert G. Wilson v, Oct 15 2004; corrected by Derek Orr, Sep 06 2014 *)
  • PARI
    for(n=1,10^4,if(ispseudoprime(8*(10^n-1)/9-1),print1(n,", "))) \\ Derek Orr, Sep 06 2014

Formula

a(n) = A056695(n) + 1. - Herman Jamke (hermanjamke(AT)fastmail.fm), Jan 01 2008

Extensions

More terms from Herman Jamke (hermanjamke(AT)fastmail.fm), Jan 01 2008
a(18)-a(20) discovered and reported to Makoto Kamada by Erik Branger; added to OEIS by Robert Price, May 20 2014
a(21)-a(23) from Kamada data by Tyler Busby, Apr 23 2024

A331860 Numbers k such that R(k) + 10^floor(k/2-1) is prime, where R(k) = (10^k-1)/9 (repunit: A002275).

Original entry on oeis.org

6, 7, 12, 31, 58, 127, 454, 556, 558, 604, 2944, 8118, 12078, 16942, 26268, 45198
Offset: 1

Views

Author

M. F. Hasler, Jan 30 2020

Keywords

Comments

The corresponding primes are near-repunit primes, cf. A105992.
In base 10, R(k) + 10^floor(k/2-1) has ceiling(k/2) digits 1, one digit 2 and again floor(k/2-1) digits 1: for even as well as odd k, there is a digit 2 just left of the middle of the repunit of length k.
No term can be congruent to 2 (mod 3). - Chai Wah Wu, Feb 07 2020

Examples

			For n = 6,  R(6)  + 10^(3-1) = 111211 is prime.
For n = 7,  R(7)  + 10^(3-1) = 1111211 is prime.
For n = 12, R(12) + 10^(6-1) = 111111211111 is prime.
		

Crossrefs

Cf. A105992 (near-repunit primes), A002275 (repunits), A011557 (powers of 10).
Cf. A331861 (variant with floor(n/2) instead of floor(n/2-1)), A331863 (variant with - (digit 0) instead of + (digit 2)).

Programs

  • PARI
    for(n=2,999,isprime(p=10^n\9+10^(n\2-1))&&print1(n","))

Extensions

a(8)-a(14) from Giovanni Resta, Jan 31 2020
a(15)-a(16) from Michael S. Branicky, Jul 23 2024

A378761 Irregular triangle read by rows: T(n,k) for k <= n/2 is the number of partitions of the repunit A002275(n) into k nonzero complementary binary vectors having a common divisor > 1 in base 10.

Original entry on oeis.org

1, 1, 1, 3, 1, 0, 1, 19, 6, 1, 0, 0, 1, 47, 98, 29, 1, 84, 280, 0, 1, 141, 650, 600, 120, 1, 0, 0, 0, 0, 1, 1135, 16734, 28063, 5922, 756, 1, 130, 130, 13, 0, 0, 1, 1779, 43757, 161700, 161700, 52920, 5040, 1, 6183, 263386, 1401900, 1401400, 0, 0, 1, 9919, 438582, 2634549, 4381246, 2587326, 577612, 40913, 1, 0, 0, 0, 0, 0, 0, 0, 1, 75433
Offset: 2

Views

Author

Dmytro Inosov, Dec 06 2024

Keywords

Comments

We call a k-tuple of binary vectors of length n complementary if for every position m (1 <= m <= n) the digit "1" occurs on that position in exactly one of the vectors. For example, {1010, 0100, 0001} is a triple (k=3) of complementary binary vectors of length n=4. The sum of complementary binary vectors of length n is always a repunit of the same length, A002275(n).
T(n,k) gives the number of distinct unordered k-tuples of complementary binary vectors of length n that have a common divisor > 1 as integers in base 10.
For k > n/2, at least one of the binary vectors must contain just a single "1" (with all other digits zero) and is, therefore, a power of 10 (A011557). Hence it cannot have nontrivial common divisors with the repunit A002275(n), which implies T(n,k) = 0. The requirement k <= n/2 acts to skip the corresponding trivial zero terms.
The partitions for k = 1 are trivial and consist of one element -- the repunit itself, which is its own greatest common divisor. Therefore, T(n,1) = 1 for n >= 2.
If T(n,k)=0 for some n and k, then T(n,m)=0 also for any m >= k. Indeed, if some m-tuple of binary vectors existed that is counted toward T(n,m), then an (m-1)-tuple obtained by summing any two of its vectors while leaving others unchanged would be counted toward T(n,m-1). By induction, this leads to T(n,k)>0, which is a contradiction.
Consequently, T(n,k) = 0 for all k > 1 whenever A378511(n) = 1. This holds, in particular, for all n in A004023 (indices of prime repunits).

Examples

			The triangle T(n,k) starts (omitting terms with k > n/2, which are zero):
-----------------------------------------------------------------------------------------
n\k: 1,       2,         3,         4,         5,         6,        7,       8,   9,  ...
-----------------------------------------------------------------------------------------
 2 | 1;
 3 | 1;
 4 | 1,       3;
 5 | 1,       0;
 6 | 1,      19,         6;
 7 | 1,       0,         0;
 8 | 1,      47,        98,        29;
 9 | 1,      84,       280,         0;
10 | 1,     141,       650,       600,       120;
11 | 1,       0,         0,         0,         0;
12 | 1,    1135,     16734,     28063,      5922,       756;
13 | 1,     130,       130,        13,         0,         0;
14 | 1,    1779,     43757,    161700,    161700,     52920,     5040;
15 | 1,    6183,    263386,   1401900,   1401400,         0,        0;
16 | 1,    9919,    438582,   2634549,   4381246,   2587326,   577612,   40913;
17 | 1,       0,         0,         0,         0,         0,        0,       0;
18 | 1,   75433,  10808037, 140403209, 391178517, 290493433, 39663279, 6540609, 362880;
19 | 1,       0,         0,         0,         0,         0,        0,       0,      0;
20 | 1,  124467,  26825456, 514583021, ...
... (for more terms, see the A-file).
T(6,3) = 6 because among the {n,k} = 90 possible triples of nonzero binary vectors of length 6 there are exactly 6 with a common divisor > 1:
  {100001, 010010, 001100}: GCD(100001, 10010, 1100) = 11;
  {100001, 011000, 000110}: GCD(100001, 11000, 110) = 11;
  {100100, 010010, 001001}: GCD(100100, 10010, 1001) = 1001;
  {100100, 011000, 000011}: GCD(100100, 11000, 11) = 11;
  {110000, 001001, 000110}: GCD(110000, 1001, 110) = 11;
  {110000, 001100, 000011}: GCD(110000, 1100, 11) = 11.
The quadruple of binary vectors {1100000001000, 0010001100000, 0001100000001, 0000010010110} counts toward T(13,4) because in base 10, GCD(1100000001000, 10001100000, 1100000001, 10010110) = 53. In total, there are 13 such quadruples of length 13. This exemplifies the smallest prime n with nontrivial T(n,k).
T(17,k) = 0 for k >= 2 since A378511(17) = 1 (though 17 isn't a term in A004023).
T(317,k) = 0 for k >= 2 since 317 is a term in A004023.
		

Crossrefs

Programs

  • Mathematica
    Clear[SubListNonCoprimes];
    SubListNonCoprimes[bnum_, m_] := SubListNonCoprimes[bnum, m] =
      (If[m == 1, Return[If[bnum == Repunit, Nothing, {Repunit - bnum}]]];
      ListOfParts2 = Select[Total[10^(ResourceFunction["KSetPartitions"][(#)[[Range[Length[#]]]], 2] &[Position[IntegerDigits[bnum] // Reverse, 0] // Flatten] - 1), {3}] /. 0 -> {}, GCD @@ Prepend[#, bnum] > 1 &];
      If[m == 2, ListOfParts2, Select[Flatten[MapApply[Append]@*Thread@*
    Comap[{SubListNonCoprimes[# + bnum, m-1] &, Identity}] @* Max /@ ListOfParts2, 1], GCD @@ Prepend[#, bnum] > 1 &]]);
    SubCountNonCoprimes10[n_, m_, k_, totk_] := (Result = 0; Do[If[!CoprimeQ[#, Repunit-#],
    Result += Length[SubListNonCoprimes[#, m-1]]] &[FromDigits[IntegerDigits[i, 2]]], {i, #[[k]], #[[k+1]]-1}] &[Round[Subdivide[2^(n-1), 2^n, totk]]];
      Result);
    CountNonCoprimes10[n_, m_] := (If[m > n/2, Return[0], If[m == 1, Return[1]]];
      Repunit = (10^n - 1)/9; ParallelSum[SubCountNonCoprimes10[n, m, k, #], {k, #}, Method -> "FinestGrained", ProgressReporting -> (n >= 15)] &[If[n >= 15, 100, 1] $KernelCount]);
    Table[CountNonCoprimes10[n, k], {n, 2, 16}, {k, 1, 8}] // TableForm

Formula

T(n,k) = 0 for k > n/2 (such terms are skipped as trivial zeros).
T(n,1) = 1 for n >= 2.
T(n,1) + T(n,2) = A378511(n).
Sum_{k} T(n,k) = A385539(n) (row sums).
T(n,k) = Stirling2(n,k) - A378154(n,k) for 2 <= k <= 9.
T(A004023(n),k) = 0 for k >= 2.

A093676 Primes of the form 8*10^k + 3*R_k, where R_k is the repunit (A002275) of length k.

Original entry on oeis.org

83, 83333333, 833333333333333333333333, 833333333333333333333333333333
Offset: 1

Views

Author

Rick L. Shepherd, Apr 08 2004

Keywords

Crossrefs

Cf. A002275, A056723 (corresponding k).
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