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"Stubborn primes" (see comments in A232210).
+20
1
13, 131, 653, 883, 1279, 10739, 17669
COMMENTS
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.
Primes q of the form q = 10p + 3, where p is also prime.
+10
6
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
EXAMPLE
5413 = 541*10 + 3, 3 appended to 541.
PROG
(PARI) lista(nn) = {forprime(p=2, nn, if (isprime(q=10*p+3), print1(q, ", ")); ); } \\ Michel Marcus, Oct 20 2014
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.
+10
5
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
COMMENTS
Conjecture: for n>=4, a(n)>0.
Records >=1: 1,2,4,7,8,34,... correspond to primes 7,19,41,127,157,443,...
EXAMPLE
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.
PROG
(PARI) 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
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.
+10
4
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
COMMENTS
Conjecture: for all n, a(n)>0.
PROG
(PARI) a(n) = {k = 0; while (! isprime(eval(concat(Str(2^n), Str((10^k-1)/3)))), k++); k; } \\ Michel Marcus, Sep 16 2014
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.
+10
4
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
COMMENTS
Conjecture: for all n, a(n)>0.
a(443) > 17000 if it is not 0.
EXAMPLE
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.
MAPLE
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:
PROG
(PARI) 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
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.
+10
3
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
COMMENTS
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.
FORMULA
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.
MATHEMATICA
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]
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.
+10
3
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
COMMENTS
The only unknown terms less than 10000, tested to 17500, are for n: 484, 1291, 2096, 2238, 3503, 3859, 6674, 7087, 7824, 8954.
FORMULA
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.
MATHEMATICA
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]
PROG
(PARI) 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
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.
+10
3
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
COMMENTS
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).
FORMULA
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.
MATHEMATICA
f[n_] := Block[{k = 1, p = Prime[n]}, While[ !PrimeQ[p*10^k + 10^k - 1], k++]; k]; f[2] = 0; Array[f, 86]
Primes p for which none of the concatenations p3, p9, 3p, 9p are primes.
+10
1
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
MATHEMATICA
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 *)
PROG
(PARI) 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 (Python)
import sympy
from sympy import isprime
from sympy import prime
for n in range(1, 10**3):
..p = str(prime(n))
..if not isprime(p+'3') and not isprime(p+'9') and not isprime('3'+p) and not isprime('9'+p):
....print(int(p), end=', ') # Derek Orr, Sep 16 2014
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