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A242707
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Least number k > 2 such that (k!+n)/(k+n) is an integer, or 0 if no such k exists.
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1
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4, 0, 0, 0, 12, 4, 5032, 6, 2990, 329881, 6, 10, 1720, 9, 6, 4, 56, 5, 634, 18, 68, 12, 51848, 22, 124, 1671, 12, 6, 30, 28, 756, 30
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OFFSET
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1,1
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COMMENTS
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The entries a(n) = 0 are based on checking values of k <= 20000. - Derek Orr, May 21 2014
If a(n) > n, a(n) + n is necessarily prime.
For n = 33..100, the sequence continues with a(33), 12, 36, 6, 16, 36, 12, 12, 42, 40, 54, 42, 6, 16, 340730, 46, 774, 99, 90196, 16, 44, 52, 18, 12, 12, 1249, a(59), 9, a(61), 60, 30, 169, 43346, 12, 22, 38, 70, a(70), 2352, 70, a(73), 72, 36, 123183, a(77), 1283, a(79), 12, 118, a(82), 84, 82, a(85), 23, 42, 28, 110, 12, a(91), a(92), 883964, a(94), 14888, 6, 726, 96, 232, 10. For each n in {33, 59, 61, 70, 73, 77, 79, 82, 85, 91, 92, 94}, a(n) > 2*10^6.
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LINKS
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EXAMPLE
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(3!+1)/(3+1) = 7/4 is not an integer. (4!+1)/(4+1) = 25/5 = 5 is an integer. So a(1) = 4.
a(2) = 0, because k+2 can never divide k!+2: If k+2 = p > 3 is a prime, then k!+2 = (p-2)!+2 == (1+2) (mod p), using the fact that (p-2)! == 1 (mod p) for all primes p. If k+2 > 4 is composite, let k+2 = p*m for some prime p and some m > 2. Then p*m-2 >= m and also p*m-2 > p. Thus, k! = (p*m-2)! is divisible by p*m. (If m = p this still holds because then p*m-2 >= 2*m.) Therefore, k!+2 == 2 (mod k+2) in this case.
a(3) = 0 for similar reasons: Consider s = (k!+3) mod (k+3). If k+3 is prime, s = (k+8)/2 and if k+3 is composite, s = 3. So s is never 0.
a(4) = 0 as well: Let s = (k!+4) mod (k+4). If k+4 is prime, there are two cases. If k == 1 (mod 6), then s = (k+29)/6. If k == 3 (mod 6), then s = (5*k+45)/6 except at the following finite number of points (k,s): (3,3), (9,2), and (15,1). If k+4 is composite, s = 4 except at (k,s) = (5,7). Thus, s is never 0. - Jon E. Schoenfield, Oct 28 2014
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PROG
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(PARI) a(n)=k=3; while((k!+n)%(k+n)&&k<2e6, k++); k \\ program improved by Derek Orr, Oct 23 2014
(PARI) a(n)=for(k=3, n, if((k!+n)%(k+n)==0, return(k))); forprime(kn=2*n, 2e6+n, if(prod(i=2, kn-n, i, Mod(1, kn))==-n, return(kn-n))); 0 \\ Charles R Greathouse IV, Oct 28 2014
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CROSSREFS
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KEYWORD
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nonn,hard,more
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AUTHOR
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EXTENSIONS
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STATUS
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approved
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