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Search: id:A036441
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| A036441 |
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a(n+1) = next number having largest prime dividing a(n) as a factor, with a(1) = 2. |
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+0
3
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| 2, 4, 6, 9, 12, 15, 20, 25, 30, 35, 42, 49, 56, 63, 70, 77, 88, 99, 110, 121, 132, 143, 156, 169, 182, 195, 208, 221, 238, 255, 272, 289, 306, 323, 342, 361, 380, 399, 418, 437, 460, 483, 506, 529, 552, 575, 598, 621, 644, 667, 696, 725, 754, 783, 812, 841, 870
(list; graph; listen)
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OFFSET
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1,1
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COMMENT
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a(2,n) satisfies the following inequality: 1/4*(n^2+3*n+1)<=a(n)<=1/4*(n-2)^2. Also a(a(r,k), n)=a(r,n+k-1), for all n,k in N\{0} and all r in N\{0,1}; a(prime(k), n)=a(prime(i), n+prime(k)-prime(i)), for all k,i,n e N\{0}, with k >= i, n >= prime(k-1) and with prime(x) := x-th prime.
Essentially the same as A076271.
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FORMULA
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a(m, n) := a(m, n-1) + lp(a(m, n-1)), a(m, 1) := m; with lp(x) := "largest prime factor of x"
a(n) = p(m)*(n+2-p(m)), where p(k) is the k-th prime and m is the smallest index such that n+2 <= p(m) + p(m+1). [From Max Alekseyev (maxale(AT)gmail.com), Oct 21 2008]
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EXAMPLE
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a(2,2)=4 because 2+ lp(2)= 2+2=4; a(2,3)=6 because 4 + lp(4)= 4+2=6
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CROSSREFS
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Cf. A006530.
Sequence in context: A130025 A145802 A076271 this_sequence A134678 A135146 A053096
Adjacent sequences: A036438 A036439 A036440 this_sequence A036442 A036443 A036444
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KEYWORD
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eigen,nice,nonn
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AUTHOR
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Frederick Magata (fmagata(AT)smail.uni-koeln.de)
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EXTENSIONS
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Better description from Reinhard Zumkeller (reinhard.zumkeller(AT)gmail.com), Feb 04, 2002
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