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A076546
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Let P = { p_1 = 3, p_2 = 5, ...} be the set of odd primes. If p_n in P can be written as p_n = q+r+s with q, r, s in P, let a(n) = largest such q, otherwise let a(n) = p_{n+1}.
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0
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5, 7, 11, 5, 7, 11, 13, 17, 23, 23, 31, 31, 37, 41, 47, 53, 53, 61, 61, 67, 73, 73, 83, 89, 89, 97, 101, 103, 107, 113, 113, 131, 131, 139, 139, 151, 157, 157, 167, 173, 173, 181, 181, 191, 193, 199, 211, 211, 223, 227, 233, 233, 241, 251, 257, 263, 263, 271, 271, 277
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OFFSET
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1,1
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COMMENTS
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11 belongs to a cycle of length 3 when a(n) is iterated. The paper by Janos studies the set of cycles.
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REFERENCES
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Ludvik Janos, On Vinagradov's 3-primes theorem, Abstracts Amer. Math. Soc., 25 (No. 2, 2002), p. 398, #01T-11-57.
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LINKS
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EXAMPLE
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3 has no such representation, so a(1) = 5. The 10th odd prime, 31, equals 23+5+3, with q=23 and no larger q exists, so a(1) = 23.
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PROG
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(PARI) {forprime(p=3, 300, b=0; q=precprime(p-1); while(b<1&&q>2, r=q; while(b<1&&r>2, s=r; while(b<1&&s>2, if(q+r+s==p, b=1; print1(q, ", "), s=precprime(s-1))); r=precprime(r-1)); q=precprime(q-1)); if(b<1, print1(nextprime(p+1), ", ")))}
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CROSSREFS
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KEYWORD
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nonn,easy
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AUTHOR
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EXTENSIONS
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STATUS
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approved
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