

A159272


Smallest integer m, in absolute value, such that m(n+m) has as prime factors exactly all primes <= sqrt(n); zero if no such m exists.


1



1, 0, 1, 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, 60, 1, 2, 3, 4, 5, 6, 3, 8, 6, 10, 4, 12, 2, 6, 15, 6, 2, 12, 21, 70, 21, 10, 7, 30, 15, 70, 15, 12, 14, 150, 9, 20, 42, 6, 30, 60, 3, 30, 15, 140, 15, 30, 3, 10, 30
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OFFSET

0,5


COMMENTS

For n<4 there are no primes <= sqrt(n). For n=1 and n=3 no a(n) would give 1 thus a(n)=0 is the only possibility. The given values for a(0) and a(2) correspond to writing 0=11 resp. 2=1+1, see below, but one might have chosen to put them zero, too.
The first unknown value is a(397).
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.)
If n is a prime, this value constitutes a proof of its primeness, since then either a or b, but not both, and thus n, are nonzero (mod p) for each prime < sqrt(n).


LINKS

Table of n, a(n) for n=0..75.
Several users at mersenneforum.org, A wellknown puzzle..., February 2009.


PROG

(PARI) A159272(n)={ local(P=vector(primepi(sqrtint(n)+!n), i, prime(i))~, M); P  return(!n(n==2)); M=P[ #P]; for( m=1, n1, factor(m*(m+n))[, 1]==P && return(m); factor(m*(nm))[, 1]==P && return(m)); for( m=1+n, 9e9, vecmax(factor(m)[, 1])>M & next; factor(m*(m+n))[, 1]==P && return(m); factor(m*(mn))[, 1]==P && return(m))}


CROSSREFS

Sequence in context: A052126 A094521 A321757 * A098372 A177236 A138567
Adjacent sequences: A159269 A159270 A159271 * A159273 A159274 A159275


KEYWORD

sign


AUTHOR

M. F. Hasler, Apr 09 2009


STATUS

approved



