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 A061303 Given a prime p, let s(p,0)=p and let s(p,n+1) be the smallest prime == 1 (mod s(p,n)). Let S(p) be the sequence {s(p,n): n=0,1,...}. Let a(0)=2 and let a(n+1) be the smallest prime not in any of the sequences S(a(0)), ..., S(a(n)). 3
 2, 5, 13, 17, 19, 31, 37, 41, 43, 61, 67, 71, 73, 79, 89, 97, 101, 109, 113, 127, 131, 137, 139, 151, 157, 163, 181, 193, 197, 199, 211, 223, 229, 233, 239, 241, 251, 257, 271, 277, 281, 307, 313, 331, 337, 349, 353, 373, 379, 397, 401, 409, 419, 421, 431, 433 (list; graph; refs; listen; history; text; internal format)
 OFFSET 0,1 COMMENTS It is conjectured for primes p and q the sequences S(p) and S(q) are disjoint, unless one is contained in the other. Also values of n such that gcd(n! , phi(n!)) equals gcd((n-1)! , phi((n-1)!)), see proof by Don Reble. - Wouter Meeussen, Mar 18 2014 Primes p such that phi(p) divides phi(Product_{primes q <= p} phi(q)), where phi is A000010. - Richard R. Forberg, Sep 11 2024 REFERENCES Amarnath Murthy, On the divisors of Smarandache Unary Sequence. Smarandache Notions Journal, Vol. 11, No. 1-2-3, Spring 2000. Amarnath Murthy, Smarandache Prime Generator Sequence (to be published in Smarandache Notions Journal). LINKS Table of n, a(n) for n=0..55. Wouter Meeussen, Don Reble's equivalence proof EXAMPLE a(0)=2 so S(a(0))={2,3,7,29,...}, which is A061092. Hence a(1)=5 so S(a(1))={5,11,23,47,...}. Hence a(2)=13 so S(a(2))={13,53,107,643,...}, ... MATHEMATICA (* start *) s[p_, 0] := s[p, 0]=p; s[p_, n_] := s[p, n]=Module[{q}, For[q=s[p, n-1]+1, !PrimeQ[q], q+=s[p, n-1], Null]; q]; ins[q_, p_] := Module[{k}, For[k=0, s[p, k]<=q, k++, If[s[p, k]==q, Return[True]]]; False]; a[0]=2; a[n_] := a[n]=Module[{i, j, q}, For[i=1, True, i++, q=Prime[i]; For[j=0, j

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Last modified September 16 23:59 EDT 2024. Contains 375984 sequences. (Running on oeis4.)