

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, 32467505340816975, 143924005810811655, 143924005810811655
(list;
graph;
refs;
listen;
history;
text;
internal format)



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)
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..16.


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


STATUS

approved



