A232210 -id:A232210 - cp's OEIS frontend

A248919 "Stubborn primes" (see comments in A232210).

13, 131, 653, 883, 1279, 10739, 17669

Offset: 1

Vladimir Shevelev, Oct 16 2014

Terms a(2)-a(5) were obtained by Peter J. C. Moses.
Terms a(6)-a(7) were obtained by Hans Havermann (cf. b-file in A232210).
Hypothetically, a(8) = 26293 = A232210(2889).
However, there are two conjectures: 1) for every n, prime a(n) exists (Shevelev); 2) already prime a(8) does not exist (Havermann).
M. F. Hasler showed that, if a prime of the form 262933...3 > 26293 exists, then it has at least several thousand digits.
Note that, for a(n), n=1,...,7, the number of digits of the smallest prime of the form a(n)*10^k+3...3 (k 3's) respectively equals 16, 26, 53, 255, 4756, 6525, 9677. Judging from the ratio 4756/255 > 18.65, the smallest prime of the form 262933...3 could have more than 180000 digits.

Vladimir Shevelev, "Stubborn primes"

Cf. A000040, A232210.

A055782 Primes q of the form q = 10p + 3, where p is also prime.

Original entry on oeis.org

23, 53, 73, 113, 173, 193, 233, 293, 313, 373, 433, 593, 613, 673, 733, 1013, 1033, 1093, 1373, 1493, 1733, 1913, 1933, 1973, 1993, 2113, 2273, 2293, 2333, 2393, 2633, 2693, 2713, 2833, 3313, 3373, 3533, 3593, 3673, 3733, 3793, 3833, 4013, 4093, 4493

Offset: 1

Labos Elemer, Jul 13 2000

These primes correspond to resulting primes for A232210, when A232210(n)=1. - Vladimir Shevelev, Oct 16 2014

			5413 = 541*10 + 3, 3 appended to 541.

Michael De Vlieger, Table of n, a(n) for n = 1..10000

Cf. A005384, A005385, A023238. Apart from first term, same as A057667.

Select[Prime@ Range@ 630, PrimeQ[(# - 3)/10] &] (* Michael De Vlieger, Jul 14 2017 *)

lista(nn) = {forprime(p=2, nn, if (isprime(q=10*p+3), print1(q, ", ")););} \\ Michel Marcus, Oct 20 2014

a(n) = 10*A023238(n) + 3. - R. J. Mathar, Sep 21 2009

A242775 Let b_k=3...3 consist of k>=1 3's. Then a(n) is the smallest k such that the concatenation b_k and prime(n) is prime, or a(n)=0 if there is no such prime.

Original entry on oeis.org

0, 0, 0, 1, 1, 1, 1, 2, 2, 2, 1, 1, 4, 2, 1, 1, 1, 2, 1, 2, 1, 1, 1, 1, 1, 3, 3, 2, 1, 2, 7, 3, 1, 3, 2, 2, 8, 1, 1, 7, 2, 1, 1, 5, 3, 2, 2, 2, 3, 1, 3, 8, 5, 1, 1, 4, 3, 1, 4, 5, 3, 6, 1, 2, 1, 2, 1, 3, 1, 2, 2, 1, 3, 1, 6, 3, 1, 3, 4, 2, 3, 8, 4, 1, 3, 34, 1

Offset: 1

Vladimir Shevelev, Sep 13 2014

nonn

Conjecture: for n>=4, a(n)>0.

Records >=1: 1,2,4,7,8,34,... correspond to primes 7,19,41,127,157,443,...

			For n<=3, a(n) = 0, because 3..32, 3..33 and 3..35 can never be prime, whatever the number of 3's that are concatenated.
For n=4, prime(n)=7, 37 is prime. So a(4)=1.

Peter J. C. Moses, Table of n, a(n) for n = 1..2000

Cf. A232210, A247341, A247342.

a(n) = {if (n<=3, return (0)); p = prime(n); k = 1; while (! isprime(p = eval(concat("3", Str(p)))), k++); k; } \\ Michel Marcus, Sep 17 2014

More terms from Peter J. C. Moses, Sep 14 2014

A247341 Let b_k=3...3 consist of k>=1 3's. Then a(n) is the smallest k such that the concatenation 2^n b_k is prime, or a(n)=0 if there is no such prime.

Original entry on oeis.org

1, 1, 1, 1, 1, 4, 1, 1, 2, 3, 1, 1, 4, 4, 6, 30, 3, 1, 6, 1, 32, 3, 3, 2, 22, 1, 6, 1, 2, 14, 7, 1, 10, 1, 2, 6, 3, 4, 2, 5, 2, 6, 1, 1, 37, 53, 53, 13, 64, 1, 67, 1, 45, 29, 17, 12, 14, 1, 2, 5, 15, 36, 10, 7, 1, 1, 81, 4, 18, 5, 55, 8, 33, 19, 8, 6, 2, 11

Offset: 0

Vladimir Shevelev, Sep 14 2014

nonn

Conjecture: for all n, a(n)>0.

Cf. A000079, A232210, A242775.

a(n) = {k = 0; while (! isprime(eval(concat(Str(2^n), Str((10^k-1)/3)))), k++); k;} \\ Michel Marcus, Sep 16 2014

More terms from Michel Marcus, Sep 16 2014

A247342 Let b_k=3...3 consist of k>=1 3's. Then a(n) is the smallest k such that the odd part (A000265) of concatenation b_k 2^n is prime, or a(n)=0 if there is no such prime.

Original entry on oeis.org

1, 2, 1, 1, 1, 1, 4, 3, 2, 1, 3, 1, 1, 6, 1, 1, 1, 3, 1, 15, 29, 5, 1, 2, 3, 6, 1, 6, 20, 6, 3, 50, 3, 22, 8, 5, 5, 1, 84, 8, 7, 36, 3, 6, 7, 20, 6, 6, 8, 1, 6, 3, 2, 38, 1, 5, 3, 2, 5, 16, 1, 12, 13, 7, 1, 4, 16, 5, 32, 1, 6, 13, 4, 150, 7, 29, 17, 9, 12, 34

Offset: 0

Vladimir Shevelev, Sep 14 2014

Conjecture: for all n, a(n)>0.

a(443) > 17000 if it is not 0.

			2^0=1 and already 31 is prime. So a(0)=1;
2^1=2, but odd part of 32 is 1 (nonprime); then consider odd part of 332. It is 83 that is prime. So a(1)=2.

Robert Israel, Table of n, a(n) for n = 0..442

Cf. A000079, A232210, A242775, A247341.

f:= proc(n) local m,d,k,x;
    m:= 2^n;
    d:=ilog10(m);
    for k from 1 do
       x:= (10^k-1)/3*10^(d+1)+m;
       if isprime(x/2^padic:-ordp(x,2)) then return k fi
    od
end proc:
map(f, [$0..100]); # Robert Israel, Oct 30 2016

a(n) = {k = 0; while (! ((val = eval(concat(Str((10^k-1)/3), Str(2^n)))) && isprime(val/2^valuation(val, 2))), k++); k;} \\ Michel Marcus, Sep 15 2014

More terms from Michel Marcus, Sep 15 2014

A257459 Let b_k=1...1 consist of k>0 1's. Then a(n) is the smallest k such that the concatenation prime(n)b_k is prime, or a(n)=0 if there is no such prime.

Original entry on oeis.org

2, 1, 5, 1, 17, 1, 8, 1, 2, 6, 1, 0, 2, 1, 3, 9, 18, 4, 210, 6, 7, 3, 2, 6, 1, 2, 1, 2, 1, 2, 4, 3, 2, 24, 3, 1, 1, 6, 5, 11, 2, 1, 11, 1, 12, 6, 1, 7, 3, 39, 2, 2, 1, 2, 9, 3, 5, 1, 6, 2, 3, 2, 180, 3, 15, 17, 24, 1, 5, 1, 2, 2, 1, 64, 7, 6, 3, 24, 2, 1, 2, 1, 6, 16, 1, 9, 8, 6, 17, 4, 6, 2, 1, 9, 30, 2, 6, 44, 1, 6

Offset: 1

Vladimir Shevelev and Robert G. Wilson v, Apr 24 2015

The only unknown terms less than 10000, tested to 15000, are for n: 284, 714, 1257, 1618, 2248, 2450, 2779, 3886, 3891, 4007, 4359, 4784, 4912, 5364, 6108, 6356, 6371, 7570, 7668, 8446, 9606.

Prime(12)=37 and b_k for k == 2 (mod 3), the concatenation is divisible by 3; for k == 1 (mod 3), the concatenation is divisible by either 7 or 13; and finally for k == 0 (mod 3), the concatenation is divisible by 37.

Vladimir Shevelev and Robert G. Wilson v, Table of n, a(n) for n = 1..10000 with -1 for those entries where a(n) has not yet been found.

Cf. A232210, A257460, A257461.

f[n_] := Block[{k = 1, p = Prime[n]}, While[ !PrimeQ[p*10^k + (10^k - 1)/9], k++]; k]; f[12] = 0; Array[f, 100]

a(n)=k for the least k such that p(n)*10^k+(10^k-1)/9 is prime, where p(n) is the n_th prime.

A257460 Let b_k=7...7 consist of k>0 7's. Then a(n) is the smallest k such that the concatenation prime(n)b_k is prime, or a(n)=0 if there is no such prime.

Original entry on oeis.org

2, 1, 2, 0, 3, 1, 2, 1, 2, 48, 1, 10, 2, 3, 3, 3, 9, 1, 1, 2, 66, 1, 2, 8, 1, 2, 6, 3, 1, 3, 1, 2, 3, 6, 8, 9, 7, 1, 3, 2, 2, 3, 17, 4, 2, 1, 3, 1, 2, 1, 3, 2, 1, 5, 17, 5, 8, 16, 1, 3, 1, 8, 6, 2, 1, 3, 3, 2184, 6, 6, 3, 2, 1, 3, 1, 2, 2, 4, 2, 3, 3, 1, 2, 1, 1, 3, 6, 15, 5, 1, 48, 2, 1, 2, 7, 2, 47, 2, 1, 1

Offset: 1

Vladimir Shevelev and Robert G. Wilson v, Apr 24 2015

The only unknown terms less than 10000, tested to 17500, are for n: 484, 1291, 2096, 2238, 3503, 3859, 6674, 7087, 7824, 8954.

Vladimir Shevelev and Robert G. Wilson v, a(n) for n = 1..10000 with -1 for those entries where a(n) has not yet been found.

Cf. A232210, A257459, A257461.

f[n_] := Block[{k = 1, p = Prime[n]}, While[ !PrimeQ[p*10^k + 7(10^k - 1)/9], k++]; k]; f[4] = 0; Array[f, 100]

isok(k, dp) = ispseudoprime(fromdigits(concat(dp, vector(k, i, 7))));
a(n) = {if (prime(n) == 7, return(0)); my(k=1, p=prime(n)); while (!ispseudoprime(p*10^k+7*(10^k-1)/9), k++); k;} \\ Michel Marcus, Jan 20 2021

a(n)=k for the least k such that prime(n)*10^k+7*(10^k-1)/9 is prime, where prime(n) is the n-th prime.

A257461 Let b_k=9...9 consist of k>0 9's. Then a(n) is the smallest k such that the concatenation prime(n)b_k is prime, or a(n)=0 if there is no such prime.

Original entry on oeis.org

1, 0, 1, 1, 5, 1, 1, 1, 1, 2, 28, 1, 1, 1, 1, 2, 1, 1, 4, 1, 1, 3, 1, 2, 90, 1, 1, 2, 8, 2, 1, 1, 2, 1, 1, 2, 1, 4, 6, 8, 3, 2, 3, 4, 1, 1, 5, 1, 5, 60, 1, 1, 5, 6, 1, 2, 1, 1, 2, 1, 10, 1, 1, 1, 5, 2, 1, 3, 4, 1, 1, 2, 4, 1, 3, 4, 3, 2, 1, 1, 2, 1, 6, 1, 5, 3

Offset: 1

Vladimir Shevelev and Robert G. Wilson v, Apr 24 2015

The only unknown terms less than 10000, tested to 25000, are for n: 87, 5744, 8041, 9533.

For p(87)=449, the concatenation is divisible by 11 if k is odd and is divisible by 7 if k == 4 (mod 6).

Vladimir Shevelev and Robert G. Wilson v, Table of n, a(n) for n = 1..10000 with -1 for those entries where a(n) has not yet been found.

Cf. A257459, A232210, A257460.

f[n_] := Block[{k = 1, p = Prime[n]}, While[ !PrimeQ[p*10^k + 10^k - 1], k++]; k]; f[2] = 0; Array[f, 86]

a(n)=k for the least k such that p(n)*10^k+10^k-1 is prime, where p(n) is the n_th prime.

a(87) from Eric Chen, Apr 24 2015

A245657 Primes p for which none of the concatenations p3, p9, 3p, 9p are primes.

Original entry on oeis.org

3, 107, 113, 179, 317, 443, 487, 599, 641, 653, 751, 773, 937, 977, 991, 1021, 1087, 1103, 1187, 1201, 1213, 1217, 1301, 1409, 1427, 1439, 1483, 1553, 1559, 1579, 1609, 1637, 1693, 1747, 1777, 1787, 1789, 1861, 1949, 1987, 1993, 2081, 2129, 2239, 2281, 2287, 2293, 2351, 2393, 2477

Offset: 1

Vladimir Shevelev, Sep 13 2014

Harvey P. Dale, Table of n, a(n) for n = 1..10000

Cf. A232210, A242775, A247341, A247342.

Select[Prime[Range[400]],NoneTrue[{10#+3,10#+9,3*10^IntegerLength[#]+#, 9*10^IntegerLength[ #]+#},PrimeQ]&] (* Requires Mathematica version 10 or later *) (* Harvey P. Dale, Sep 06 2020 *)

lista(nn) = {forprime(p=2, nn, if (!isprime(eval(concat(Str(p), Str(3)))) && ! isprime(eval(concat(Str(p), Str(9)))) && ! isprime(eval(concat(Str(3), Str(p)))) && ! isprime(eval(concat(Str(9), Str(p)))), print1(p, ", ")););} \\ Michel Marcus, Sep 14 2014

import sympy
from sympy import isprime
from sympy import prime
for n in range(1,10**3):
  p = str(prime(n))
  if not isprime(int(p+'3')) and not isprime(int(p+'9')) and not isprime(int('3'+p)) and not isprime(int('9'+p)):
    print(int(p),end=', ') # Derek Orr, Sep 16 2014

More terms from Derek Orr, Sep 16 2014

cp's OEIS Frontend

A248919 "Stubborn primes" (see comments in A232210).

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A055782 Primes q of the form q = 10p + 3, where p is also prime.

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A242775 Let b_k=3...3 consist of k>=1 3's. Then a(n) is the smallest k such that the concatenation b_k and prime(n) is prime, or a(n)=0 if there is no such prime.

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A247341 Let b_k=3...3 consist of k>=1 3's. Then a(n) is the smallest k such that the concatenation 2^n b_k is prime, or a(n)=0 if there is no such prime.

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A247342 Let b_k=3...3 consist of k>=1 3's. Then a(n) is the smallest k such that the odd part (A000265) of concatenation b_k 2^n is prime, or a(n)=0 if there is no such prime.

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Maple

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A257459 Let b_k=1...1 consist of k>0 1's. Then a(n) is the smallest k such that the concatenation prime(n)b_k is prime, or a(n)=0 if there is no such prime.

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A257460 Let b_k=7...7 consist of k>0 7's. Then a(n) is the smallest k such that the concatenation prime(n)b_k is prime, or a(n)=0 if there is no such prime.

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A257461 Let b_k=9...9 consist of k>0 9's. Then a(n) is the smallest k such that the concatenation prime(n)b_k is prime, or a(n)=0 if there is no such prime.

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