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A119447
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Leading diagonal of triangle A119446, as described in A100461, except with a(1,n) = prime(n) instead of 2^(n-1).
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2
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2, 2, 3, 3, 3, 3, 3, 3, 3, 7, 7, 7, 7, 7, 7, 7, 7, 7, 7, 7, 7, 7, 7, 7, 13, 13, 7, 7, 7, 7, 13, 13, 13, 13, 13, 13, 13, 13, 13, 13, 13, 13, 13, 13, 13, 13, 13, 13, 13, 13, 13, 13, 13, 13, 13, 13, 13, 13, 13, 13, 13, 13, 13, 13, 13, 13, 19, 19, 19, 19, 19, 19, 19, 19, 19, 19, 19, 19, 19
(list; graph; refs; listen; history; internal format)
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OFFSET
| 1,1
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COMMENTS
| a(181) = 27 is the first term greater than 19. This is because prime(181)/181 > 6 for the first time. In general this sequence is determined by prime(n)/n: the pattern for each row of the triangle is that it ends with prime(n), preceded by multiples of k = prime(n)/n down to k^2, then the largest multiple of k-1 less than k^2 and the largest multiple of k-2 less than that and so on. This sequence gives the multiple of 1. See A000960 for the sequence that gives the ending value for each starting k.
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FORMULA
| With k = prime(n)/n, a(n) = A000960(k).
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CROSSREFS
| Cf. A100461 for powers of 2, A119444 for Fibonacci and A119446 for triangle corresponding to this diagonal.
Sequence in context: A130147 A096143 A025792 * A157720 A077463 A084556
Adjacent sequences: A119444 A119445 A119446 * A119448 A119449 A119450
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KEYWORD
| nonn
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AUTHOR
| Joshua Zucker (joshua.zucker(AT)stanfordalumni.org), May 20 2006
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