A157752 Smallest positive integer m such that m == prime(i) (mod prime(i+1)) for all 1<=i<=n.

2, 8, 68, 1118, 2273, 197468, 1728998, 1728998, 447914738, 10152454583, 1313795640428, 97783391392958, 5726413266646343, 38433316595821418, 15103232990013860963, 943894249589930135768, 52858423703753671390658, 932521283899305953765183, 8790842834979573009644273

Offset

1,1

Comments

Suggested by Chinese Remainder Theorem.

a(n) is prime for n = 1, 5, 10, 23, 30.

Links

Harvey P. Dale, Table of n, a(n) for n = 1..349

Maple

A157752 := proc(n)
    local lrem, leval, i ;
    lrem := [] ;
    leval := [] ;
    for i from 1 to n do
        lrem := [op(lrem), ithprime(i+1)] ;
        leval := [op(leval), ithprime(i)] ;
    end do:
    chrem(leval, lrem) ;
end proc: # R. J. Mathar, Apr 14 2016

Mathematica

a[n_] := ChineseRemainder[Prime[Range[n]], Prime[Range[2, n + 1]]] a[ # ] & /@ Range[30]
Table[With[{pr=Prime[Range[n]]}, ChineseRemainder[Most[pr], Rest[pr]]], {n, 2, 30}] (* Harvey P. Dale, Jun 11 2017 *)

Prog

(PARI) x=Mod(1, 1); for(i=1, 20, x=chinese(x, Mod(prime(i), prime(i+1))); print1(component(x, 2), ", "))
(Python)
from sympy.ntheory.modular import crt
from sympy import prime
def A157752(n): return int(crt((s:=[prime(i+1) for i in range(1, n)])+[prime(n+1)], [2]+s)[0]) # Chai Wah Wu, May 02 2023

Crossrefs

Cf. A053664, A071057, A121934.

Sequence in context: A262479 A372315 A192550 * A055547 A113087 A322495

Adjacent sequences: A157749 A157750 A157751 * A157753 A157754 A157755

Keyword

nonn

Author

Zak Seidov, Mar 05 2009

Extensions

Edited by Charles R Greathouse IV, Oct 28 2009

Status

approved