THEOREM 2 . Let k, 1, n be integers

THEOREM 2 . Let k, 1, n be integers such that k
>_ 3, 1 >_ 2
and n + k >_ p (k) ,
where
p(k)
is the least prime satisfying p (k)
>_
k. Then there is a prime p >_ k for
which
a p
# 0 (mod
1),
where
ap
is the power of p dividing (n + 1) . . . (n + k) .

結果 (英語) 1:

THEOREM 2. Let k, 1, and n be integers such that k> _ > 1, 3 _ 2and n+k > _ p(k),wherep(k)is the least prime satisfying p(k)>_Then there is a prime p > k. _ k forwhichp# 0 (mod1),whereAPis the power of p dividing (n+1). (n+k).

結果 (英語) 2:

THEOREM 2. Let K, 1, N Be Integers Such That K
> _ 3, 1> _ 2
And N Tasu K> _ P (K),
Where
P (K)
Is The Least Prime Satisfying P (K)
> _
K . Then There Is A Prime P> _ K For
Which
A P
# 0 (Mod
1),
Where
Ap
Is The Power Of P Dividing (N Tasu 1)... (N Tasu K).

結果 (英語) 3:

THEOREM Let K 1 2,, be integers n k such thatA-yp-10, 1 2 3N, P and K, a-yp-10WhereP KIs the least prime satisfying P (k)A-yp-10There is a Then K K for a-yp-10 prime pWhichA p# 0 Mod(1)WhereAPIs the power of the P (n + 1) dividing 0.0.0.0 (n + k).