A110096 Least positive integer which, when added to each of 2^1, ..., 2^n, yields all primes; or 0 if none exists.

1, 1, 3, 3, 15, 15, 1605, 1605, 19425, 2397347205, 153535525935, 29503289812425, 29503289812425, 32467505340816975, 143924005810811655, 143924005810811655

Offset: 1

Author

Joseph L. Pe, Sep 05 2005

Keywords

• nonn
• more

Comments

Dickson's conjecture implies that a(n) > 0 for all n. - Charles R Greathouse IV, Oct 11 2011

From David A. Corneth, Jul 30 2020: (Start)

We can restrict the search using the Chinese Remainder Theorem as follows: there are 4 possible residues mod 7 for a term; 0, 1, 2 and 4. There are 4 possible residues mod 19 for a term; 0, 9, 14 and 18.

Combining this we can find all 4*4 = 16 possible residues mod (19*7) = mod 133. These residues are 0, 9, 14, 18, 28, 37, 56, 57, 71, 85, 95, 109, 113, 114, 123, 128.

I did this for the primes up to 41, except 31, giving 13837825 numbers to check per 9814524629910 such that on average 1 in about 7*10^5 numbers had to be checked. (End)

Examples

			a(5)=15 is the least positive integer which, when added to 2^1, 2^2, 2^3, 2^4, 2^5, yields all primes: 17, 19, 23, 31, 47.
		

Crossrefs

Cf. A193109.

Programs

Mathematica

p[n_] := Table[2^i, {i, 1, n}]; f[k_, n_] := MemberQ[PrimeQ[k + p[n]], False]; r = {}; For[n = 1, n <= 9, n++, k = 1; While[f[k, n], k = k + 1]; r = Append[r, k]]; r

PARI

is(k, n) = for(i=1, n, if(!isprime(k+2^i), return(0))); 1;
a(n) = {my(s=2); forprime(p=3, n, if(znorder(Mod(2, p))==(p-1), s*=p)); forstep(k=s/2, oo, s, if(is(k, n), return(k))); } \\ Jinyuan Wang, Jul 30 2020

Extensions

a(10) from T. D. Noe, Sep 06 2005

a(11) from Don Reble, Sep 17 2005

a(14)-a(16) from Bert Dobbelaere, Apr 24 2021