A181901 - OEIS

A181901

a(n) = least positive integer m such that 2(s_k)^2 for k=1,...,n are pairwise distinct modulo m where s_k = Sum_{j=1..k} (-1)^(k-j)*p_j and p_j is the j-th prime.

1, 4, 7, 9, 13, 17, 19, 23, 25, 31, 37, 41, 43, 47, 53, 59, 61, 67, 71, 73, 79, 83, 89, 97, 101, 103, 107, 109, 113, 127, 131, 137, 139, 149, 151, 157, 163, 167, 173, 179, 181, 191, 193, 197, 199, 211, 223, 227, 229, 233

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OFFSET

1,2

COMMENTS

On Mar 28 2012, Zhi-Wei Sun conjectured that a(n) is the (n+1)-th prime p_{n+1} with the only exceptions being a(1)=1, a(2)=4, a(4)=9 and a(9)=25. He has shown that 2(s_k)^2 (k=1,...,n) are indeed pairwise distinct modulo p_{n+1} and hence a(n) does not exceed p_{n+1}.

Note that the sequence 0,s_1,s_2,s_3,... is A008347 introduced by N. J. A. Sloane and J. H. Conway.

Compare a(n) with the sequence A210640.

The conjecture was verified for n up to 2*10^5 by the author in 2018, and for n up to 3*10^5 by Chang Zhang (a student at Nanjing Univ.) in June 2020. - Zhi-Wei Sun, Jun 22 2020

LINKS

Zhi-Wei Sun, Table of n, a(n) for n = 1..600

Zhi-Wei Sun, An amazing recurrence for primes, a message to Number Theory List, March 31, 2012.

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

EXAMPLE

We have a(4)=9 since 2(s_1)^2=8, 2(s_2)^2=2, 2(s_3)^2=32, 2(s_4)^2=18 are pairwise distinct modulo 9 but not pairwise distinct modulo any of 1,...,8.

MATHEMATICA

s[n_]:=Sum[(-1)^k*Prime[k], {k, 1, n}]

f[n_]:=2*s[n]^2

R[n_, m_]:=Union[Table[Mod[f[k], m], {k, 1, n}]]

Do[Do[If[Length[R[n, m]]==n, Print[n, " ", m]; Goto[aa]], {m, 1, Prime[n+1]}];

Print[n]; Label[aa]; Continue, {n, 1, 600}]