

A307345


Numbers k such that every prime p <= sqrt(k) divides k*(k1).


1



1, 2, 3, 4, 5, 6, 7, 8, 9, 10, 12, 13, 15, 16, 18, 19, 21, 22, 24, 25, 30, 31, 36, 40, 45, 46, 70, 85, 91, 105, 106, 120
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OFFSET

1,2


COMMENTS

If k is in the sequence, the first Chebyshev function theta(sqrt(k)) = Sum_{p <= sqrt(k)} log(p) <= 2 log(k). Now it is known that theta(x) = x + O(x/log(x)), so this can't happen if k is sufficiently large. Thus the sequence is finite.
For x >= 2, theta(x) >= x  1.2323*x/log(x) (see Dusart, Theorem 5.2). Thus theta(sqrt(k)) > 2*log(k) for k >= 417. Since there are no other terms < 417, the largest term is 120.


LINKS



EXAMPLE

120 is in the sequence because all primes <= sqrt(120) (namely 2,3,5,7) divide 120*119.


MAPLE

Res:= NULL:
P:= 1:
q:= 2: t:= 4:
for n from 1 to 10^6 do
if n = t then P:= P*q; q:= nextprime(q); t:= q^2 fi;
if n*(n1) mod P = 0 then Res:= Res, n fi
od:
Res;


MATHEMATICA

seqQ[k_] := AllTrue[Select[Range@Floor@Sqrt@k, PrimeQ], Divisible[k (k  1), #] &]; Select[Range[120], seqQ] (* Amiram Eldar, Apr 03 2019 *)


PROG

(Sage)
def isA307345(k):
r = prime_range(isqrt(k)+1)
return all([p.divides(k*(k1)) for p in r])
print([n for n in (1..120) if isA307345(n)]) # Peter Luschny, Apr 03 2019
(PARI) isok(k) = forprime(p=1, sqrtint(k), if (k*(k1) % p, return(0))); return(1); \\ Michel Marcus, Apr 05 2019


CROSSREFS



KEYWORD

nonn,fini,full


AUTHOR



STATUS

approved



