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 A076271 a(1) = 1, a(2) = 2, and for n > 2, a(n) = a(n-1) + gpf(a(n-1)), where gpf = greatest prime factor = A006530. 11
 1, 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 (list; graph; refs; listen; history; text; internal format)
 OFFSET 1,2 COMMENTS a(n+1) is the smallest number such that the largest prime divisor of a(n) is the highest common factor of a(n) and a(n+1). - Amarnath Murthy, Oct 17 2002 Essentially the same as A036441(n) = a(n+1) and A180107(n) = a(n-1) (n > 1). The equivalent sequence with A020639 = spf instead of A006530 = gpf begins a(1) = 1, a(2) = 2, and from then on we get all even numbers: a(n) = a(2) + 2*(n-2), n > 1. - M. F. Hasler, Apr 08 2015 From David James Sycamore, Apr 27 2017: (Start) The sequence contains only one prime; a(2)=2, all other terms (excluding a(1)=1) being composite, since if a(n) for some n>2 is assumed to be the first prime after 2, then a(n) = a(n-1) + gpf(a(n-1))= m*q+q = q*(m+1) for some integer m>1 and some prime q. This number is  composite; contradiction. Terms after a(3)=4 alternate between even and odd values since each is created by addition of a prime (odd term). All terms a(n) arise as consecutive multiples of consecutive primes occurring in their natural ascending order, 2,3,5,7.... (A000040). The number of (consecutive) terms which arise as multiples of p(n)= A000040(n) is 1 + p(n+1)- p(n-1), namely n-th term of the sequence:2,4,5,7,7,7,7,7,11, etc. Example: Number of multiples of 17, the 7th prime, is 1+p(8)-p(6) = 1+19-13 = 7. For any pair of consecutive primes, p,q (p

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Last modified August 20 07:48 EDT 2019. Contains 326143 sequences. (Running on oeis4.)