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 A137613 Omit the 1s from Rowland's sequence f(n) - f(n-1) = gcd(n,f(n-1)), where f(1) = 7. 10
 5, 3, 11, 3, 23, 3, 47, 3, 5, 3, 101, 3, 7, 11, 3, 13, 233, 3, 467, 3, 5, 3, 941, 3, 7, 1889, 3, 3779, 3, 7559, 3, 13, 15131, 3, 53, 3, 7, 30323, 3, 60647, 3, 5, 3, 101, 3, 121403, 3, 242807, 3, 5, 3, 19, 7, 5, 3, 47, 3, 37, 5, 3, 17, 3, 199, 53, 3, 29, 3, 486041, 3, 7, 421, 23 (list; graph; refs; listen; history; text; internal format)
 OFFSET 1,1 COMMENTS Rowland proves that each term is prime. He says it is likely that all odd primes occur. In the first 5000 terms, there are 965 unique primes and 397 is the least odd prime that does not appear. - T. D. Noe, Mar 01 2008 In the first 10000 terms, the least odd prime that does not appear is 587, according to Rowland. [Jonathan Sondow, Aug 14 2008] Removing duplicates from this sequence yields A221869. The duplicates are A225487. - Jonathan Sondow, May 03 2013 LINKS T. D. Noe, Table of n, a(n) for n = 1..5000 Jean-Paul Delahaye, Déconcertantes conjectures, Pour la science, 5 (2008), 92-97. [broken link] Brian Hayes, Pumping the Primes, bit-player, 19 August 2015. John Moyer, Source code in C and C++ to print this sequence or sorted and unique values from this sequence. [From John Moyer (jrm(AT)rsok.com), Nov 06 2009] Ivars Peterson, A New Formula for Generating Primes, The Mathematical Tourist. Eric S. Rowland, A simple prime-generating recurrence, Abstracts Amer. Math. Soc. 29 (No. 1, 2008), p. 50 (Abstract 1035-11-986). arXiv:0710.3217 Eric S. Rowland, A natural prime-generating recurrence, J. of Integer Sequences 11 (2008), Article 08.2.8. Eric Rowland, A simple recurrence that produces complex behavior ..., A New Kind of Science blog. Eric Rowland, Prime-Generating Recurrence, Wolfram Demonstrations Project, 2008. Jeffrey Shallit, Rutgers Graduate Student Finds New Prime-Generating Formula, Recursivity blog. V. Shevelev, Generalizations of the Rowland theorem, arXiv 2009 Wikipedia, Formula for primes FORMULA Denote by Lpf(n) the least prime factor of n. Then a(n) = Lpf(6-n+sum{i=1,...,n-1}a(i)). [Vladimir Shevelev, Mar 03 2010] EXAMPLE f(n) = 7, 8, 9, 10, 15, 18, 19, 20, ..., so f(n) - f(n-1) = 1, 1, 1, 5, 3, 1, 1, ... and a(n) = 5, 3, ... . We have a(1) = Lpf(6-1) = 5; a(2) = Lpf(6-2+5) = 3; a(3) = Lpf(6-3+5+3) = 11; a(4) = Lpf(6-4+5+3+11) = 3; a(5) = Lpf(6-5+5+3+11+3) = 23. [Vladimir Shevelev, Mar 03 2010] MAPLE A137613_list := proc(n) local a, c, k, L; L := NULL; a := 7; for k from 2 to n do     c := igcd(k, a);     a := a + c;     if c > 1 then L:=L, c fi; od; L end: A137613_list(500000); # Peter Luschny, Nov 17 2011 MATHEMATICA f[1] = 7; f[n_] := f[n] = f[n - 1] + GCD[n, f[n - 1]]; DeleteCases[Differences[Table[f[n], {n, 10^6}]], 1] (* Alonso del Arte, Nov 17 2011 *) PROG (Haskell) a137613 n = a137613_list !! (n-1) a137613_list =  filter (> 1) a132199_list -- Reinhard Zumkeller, Nov 15 2013 (PARI) ub=1000; n=3; a=9; while(n

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