

A110096


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


1



1, 1, 3, 3, 15, 15, 1605, 1605, 19425, 2397347205, 153535525935, 29503289812425, 29503289812425
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OFFSET

1,3


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)
a(15) > 5*10^15.
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)


LINKS

Table of n, a(n) for n=1..13.


EXAMPLE

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.


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


PROG

(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))==(p1), s*=p)); forstep(k=s/2, oo, s, if(is(k, n), return(k))); } \\ Jinyuan Wang, Jul 30 2020


CROSSREFS

Cf. A193109.
Sequence in context: A055634 A133221 A232097 * A157526 A208229 A269956
Adjacent sequences: A110093 A110094 A110095 * A110097 A110098 A110099


KEYWORD

nonn,more


AUTHOR

Joseph L. Pe, Sep 05 2005


EXTENSIONS

a(10) from T. D. Noe, Sep 06 2005
a(11) from Don Reble, Sep 17 2005


STATUS

approved



