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A249241
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a(n) = p - prime(n)!/prime(n)#, where p is the smallest prime number > prime(n)!/prime(n)#+1.
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0
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2, 2, 3, 5, 11, 7, 29, 17, 17, 397, 47, 67, 23, 41, 31, 157, 409, 31, 151, 109, 199, 191, 131, 61, 103, 547, 179, 269, 389, 317, 181, 331, 307, 173, 1259, 1289, 619, 131, 223, 683, 139, 241, 191, 101, 1039, 1367, 1153, 241, 1187, 479, 149, 181, 487, 1093, 571, 1151, 809, 199, 823, 491, 191, 151, 1321, 197, 163, 337, 467, 659, 673, 877, 487, 743, 313, 673, 857, 677, 1021
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OFFSET
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1,1
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COMMENTS
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Conjecture: All terms are prime.
While Fortune's conjecture (A005235) uses products of primes, this sequence uses products of composite numbers (more exactly: of nonprimes, because 1 belongs to them). It looks like all multiples of prime(n)# (except some powers) lead to a sequence which contains only prime numbers.
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LINKS
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EXAMPLE
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n = 1; prime(1)!/prime(1)# = 2/2 = 1; p = nextprime(1+1) = 3; a(1) = 3-1 = 2.
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PROG
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(MuPAD) q:=1; p:=1; for i from 1 to 100 do q:=nextprime(q+1); p:=p*q; N:=nextprime((fact(q)/p)+2)-fact(q)/p; print(i, N); end_for:
(PARI) A092435(n)=prime(n)!/prod(i=1, n, prime(i))
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CROSSREFS
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KEYWORD
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nonn
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AUTHOR
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STATUS
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approved
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