A210186 - OEIS

A210186

a(n) = least integer m>1 such that m divides none of P_i + P_j with 0<i<j<=n where P_k is the product of the first k primes.

2, 3, 5, 7, 11, 19, 23, 23, 23, 47, 59, 61, 71, 71, 71, 101, 101, 101, 101, 101, 101, 113, 113, 113, 113, 113, 113, 113, 113, 113, 223, 223, 223, 223, 223, 223, 223, 223, 223, 223, 223, 223, 223, 223, 223, 223, 223, 223, 487, 487, 661, 661, 661, 661, 661, 661, 661, 661, 661, 719, 719, 719, 719, 719, 719, 811, 811, 811, 811, 811, 811, 811, 811, 811, 811

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OFFSET

1,1

COMMENTS

Conjecture: all the terms are primes and a(n) < n^2 for all n > 1.

LINKS

Zhi-Wei Sun, Table of n, a(n) for n = 1..258

Romeo Meštrović, Euclid's theorem on the infinitude of primes: a historical survey of its proofs (300 BC--2012) and another new proof, arXiv preprint arXiv:1202.3670 [math.HO], 2012-2018. - From N. J. A. Sloane, Jun 13 2012

Zhi-Wei Sun, A function taking only prime values, message to Number Theory List, Feb. 21, 2012.

Zhi-Wei Sun, On functions taking only prime values, J. Number Theory, Vol. 133, No. 8 (2013), pp. 2794-2812.

EXAMPLE

We have a(3)=5 since 2+2*3, 2+2*3*5, 2*3+2*3*5 are pairwise distinct modulo m=5 but not pairwise distinct modulo m=2,3,4.

MATHEMATICA

P[n_]:=Product[Prime[k], {k, 1, n}]

R[n_, m_]:=Product[If[Mod[P[k]+P[j], m]==0, 0, 1], {k, 2, n}, {j, 1, k-1}]

Do[Do[If[R[n, m]==1, Print[n, " ", m]; Goto[aa]], {m, 2, Max[2, n^2]}]; Print[n]; Label[aa]; Continue, {n, 1, 300}]

CROSSREFS

Cf. A000040, A210144, A208494, A208643, A207982.

Sequence in context: A081889 A078139 A347337 * A120628 A322471 A262837

Adjacent sequences: A210183 A210184 A210185 * A210187 A210188 A210189

KEYWORD

nonn

AUTHOR

Zhi-Wei Sun, Mar 18 2012

STATUS

approved