A110096 - OEIS

%I #25 Apr 25 2021 03:21:10

%S 1,1,3,3,15,15,1605,1605,19425,2397347205,153535525935,29503289812425,

%T 29503289812425,32467505340816975,143924005810811655,

%U 143924005810811655

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

%C Dickson's conjecture implies that a(n) > 0 for all n. - _Charles R Greathouse IV_, Oct 11 2011

%C From _David A. Corneth_, Jul 30 2020: (Start)

%C 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.

%C 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.

%C 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)

%e 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.

%t 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

%o (PARI) is(k, n) = for(i=1, n, if(!isprime(k+2^i), return(0))); 1;

%o 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

%Y Cf. A193109.

%K nonn,more

%O 1,3

%A _Joseph L. Pe_, Sep 05 2005

%E a(10) from _T. D. Noe_, Sep 06 2005

%E a(11) from _Don Reble_, Sep 17 2005

%E a(14)-a(16) from _Bert Dobbelaere_, Apr 24 2021