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A159272
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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.
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1
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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
(list; graph; refs; listen; history; internal format)
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OFFSET
| 0,5
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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=1-1 resp. 2=1+1, cf. 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).
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LINKS
| Several users at mersenneforum.org, A well-known puzzle..., February 2009.
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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, n-1, factor(m*(m+n))[, 1]==P & return(m); factor(m*(n-m))[, 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*(m-n))[, 1]==P & return(-m))}
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CROSSREFS
| Sequence in context: A072994 A052126 A094521 * A098372 A177236 A138567
Adjacent sequences: A159269 A159270 A159271 * A159273 A159274 A159275
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KEYWORD
| sign
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AUTHOR
| M. F. Hasler (www.univ-ag.fr/~mhasler), Apr 09 2009
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