A167604 A variant of Euclid-Mullin (A000945): a(1)=2, a(n+1) is the least prime dividing [Product_{i in I} a(i) + Product_{i not in I} a(i)], minimized over all subsets I of {1..n}.
2, 3, 5, 11, 37, 13, 7, 29, 17, 19, 43, 23, 47, 41, 53, 31, 61, 59, 67, 79, 83, 73, 97, 71, 101, 89, 103, 127, 107, 113, 137, 131, 139, 109, 149, 151, 163, 157, 167, 173, 193, 211, 179, 191, 181, 223, 199, 197, 233, 227, 229, 239, 241, 251, 257, 307, 281, 269, 271, 293
Offset: 1
Keywords
Examples
a(4)=11 which is the smallest prime dividing the 4 partitions 2+3*5=17, 3+2*5=13, 5+2*3=11, 1+2*3*5=31.
Links
- Max Alekseyev, Table of n, a(n) for n = 1..2500
- Andrew R. Booker, A variant of the Euclid-Mullin sequence containing every prime, arXiv preprint arXiv:1605.08929 [math.NT], 2016.
Crossrefs
Programs
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Maple
with(numtheory):p:=proc(N) local S, d : S:=NULL:for d in divisors(N) while d^2<=N do S:=S,divisors(d+N/d)[2] od : return(min(S)) end: a :=n->if n = 1 then 2 else p(mul(a(i),i = 1 .. n-1)) fi : seq(a(n), n=1..15); # Robert FERREOL, Oct 01 2019
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Mathematica
p[N_Integer] := Module[{S = {}, d, divisorsList}, For[d = 1, d^2 <= N, d++, If[Divisible[N, d], divisorsList = Divisors[d + N/d]; If[Length[divisorsList] >= 2, AppendTo[S, divisorsList[[2]]]];]]; Min[S]]; a[n_Integer] := If[n == 1, 2, p[Times @@ Table[a[i], {i, 1, n - 1}]]]; Table[a[n], {n, 1, 14}] (* Hilko Koning, Oct 30 2024 *) -
PARI
{ A167604_list() = my(a,A,p,b,q,z,m); a = []; A=1; while(1, p=2; while( kronecker(-A,p)!=1, p=nextprime(p+1) ); b=lift(sqrt(-A+O(p))); z=znprimroot(p); m=nextprime(random(10^6)); q=lift(prod(i=1,#a, Mod(1+x^znlog(Mod(a[i],p),z,p-1),(1-x^(p-1))*Mod(1,m)) )); if( polcoeff(q,znlog(Mod(b,p),z,p-1),x)==0, error("conjecture failed mod",m)); a=concat(a,[p]); A*=p; print1(p,", ") ) } /* Max Alekseyev, May 20 2015 */
Formula
For any n, we have Legendre symbol (-a(1)*a(2)*...*a(n-1) / a(n)) = 1. If p is the smallest prime such that (-a(1)*a(2)*...*a(n-1) / p) = 1, then a(n) >= p. Conjecture: For all n, a(n) = p. Note that if b is such that b^2 == -a(1)*a(2)*...*a(n-1) (mod p) and for some I, b == prod_{i in I} a(i) (mod p), then a(n) = p. Heuristically, I must exist for large enough n, since the number of possible subsets I is much larger than p. - Max Alekseyev, Nov 11 2009, May 20 2015
Extensions
Edited and extended by Max Alekseyev, Nov 11 2009
Comments