|
%I #25 Feb 18 2024 02:56:01
%S 0,0,0,0,-2,-1,2,1,-4,3,2,1,6,-1,-2,3,2,1,6,-1,-2,3,2,1,-6,5,4,3,2,1,
%T 30,-1,-2,-3,-4,-5,-6,3,-8,6,-10,4,-12,2,6,15,-6,-2,12,21,70,-21,-10,
%U 7,30,15,70,-15,12,-14,150,9,-20,42,6,-30,60,3,30,15,140,-15,-30,-3,10
%N Smallest integer m, in absolute value, such that |m(n+m)| has as prime factors exactly all primes <= sqrt(n); take m > 0 if m and -m both are solutions; a(n) = 0 if no such m exists.
%C As of today, the first unknown value is a(397).
%C For all n = 4, ..., 396, we know a way to write n = a +/- b where the prime factors of a*b are exactly all primes <= sqrt(n). This sequence lists min(a,b), with a minus sign if n = a + b (to have a(n) > 0 for most n).
%C If n is a prime, this value constitutes a proof of its primality, since then either a or b, but not both, and thus n, are nonzero (mod p) for each prime < sqrt(n).
%C Are a(152) = -42 and a(167) = -35 the last two negative values ? - _M. F. Hasler_, Feb 17 2024
%H M. F. Hasler, <a href="/A159272/b159272.txt">Table of n, a(n) for n = 0..315</a>, Feb 17 2024
%H Several users at mersenneforum.org, <a href="http://www.mersenneforum.org/showthread.php?p=168693">A well-known puzzle...</a>, February 2009.
%e For n < 4, a(n) = 0 by convention, or with the justification that there are no primes <= sqrt(n), and 0*(n+0) = 0 also has no prime factors (by convention).
%e For all larger n, |m*(n+m)| must have at least the prime factor 2, so m = 0 can't be a solution as above for n < 4, nor can it be m = -n.
%e For n = 4, a(4) = -2 because 2*(4-2) = 2*2 has only 2 as prime factor, and this is also the only prime <= sqrt(n) = 2. One easily checks that m = +-1 and m = 2 is not a solution.
%e For n = 5, a(5) = -1 because 1*(5-1) = 4 = 2*2 has only 2 as prime factor, which is also the only prime up to sqrt(5) ~ 2.236.
%e a(30) = 30 because 30*(30+30) = 1800 = 2^3 * 3^2 * 5^2 has exactly all primes < sqrt(30) ~ 5.5 as prime factors, and one can check that no smaller m yields a solution.
%o (PARI) A159272(n, oo=oo, start=1)={n>3 || return(0); my(p=primes([1,sqrtint(n)])~, P=p[#p]); for(m=start, oo, m>1 && vecmax(factor(m)[,1]) > P && next; factor(m*(m+n))[, 1]==p && return(m); factor(m*abs(m-n))[, 1]==p && return(-m))} \\ Edited to simplify, speed up & fix a bug. - _M. F. Hasler_, Feb 17 2024
%K sign
%O 0,5
%A _M. F. Hasler_, Apr 09 2009
%E a(30) corrected (thanks to a remark from Don Reble) and a(n) changed to 0 for n < 4 by _M. F. Hasler_, Feb 17 2024
|
