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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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6
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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
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OFFSET
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1,1
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COMMENTS
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a(n) satisfies the following inequality: (1/4)*(n^2 + 3*n + 1) <= a(n) <= (1/4)*(n+2)^2. [Corrected by M. F. Hasler, Apr 08 2015]
The present sequence is the special case a(n) = a(2,n) with a more general a(m, n) := a(m, n-1) + gpf(a(m, n-1)), a(m, 1) := m, where gpf(x) := "greatest prime factor of x" = A006530(x). 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 in N\{0}, with k >= i, n >= prime(k-1) and with prime(x) := x-th prime.
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LINKS
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FORMULA
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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). - Max Alekseyev, Oct 21 2008
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EXAMPLE
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a(2,2) = 4 because 2 + gpf(2) = 2 + 2 = 4;
a(2,3) = 6 because 4 + gpf(4) = 4 + 2 = 6.
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MATHEMATICA
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NestList[#+FactorInteger[#][[-1, 1]]&, 2, 60] (* Harvey P. Dale, Dec 02 2012 *)
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PROG
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(Haskell)
a036441 n = a036441_list !! (n-1)
a036441_list = tail a076271_list
(PARI) a(n)=(n+2-if(n\2+1<(p=nextprime(n\2+1))&&n+1<p+n=precprime(p-1), p=n, p))*p \\ M. F. Hasler, Apr 08 2015
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CROSSREFS
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KEYWORD
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eigen,nice,nonn
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AUTHOR
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Frederick Magata (frederick.magata(AT)uni-muenster.de)
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EXTENSIONS
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STATUS
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approved
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