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%I #19 Feb 20 2023 06:14:06
%S 1,3,5,9,11,15,17,27,121,187,191,275,277,573,831,14641,14653,109443,
%T 109451,131877,161183,249101,249103,254221,214710409,1603785503,
%U 3146151623077,23500268975459,23500268975497,352504034632455,352504034632459,675511501766876508493,8283939628810696270871857123
%N Two numbers k and l we call equivalent if they have the same vector of exponents with positive components in prime power factorization. Let a(1)=1, a(2)=3. If n>=3 is prime, then a(n) is the smallest prime greater than a(n-1); otherwise, a(n)>a(n-1) is the smallest number equivalent to n such that prime power factorization of a(n) contains only primes which already appeared in the sequence.
%C The sequence contains exactly 33 terms.
%e By the condition, a(12) should be more than a(11)=191. Since 12 has vector of positive exponents (2,1), then we seek already constructed prime terms p<q in the sequence and choose the smallest number of the form p^2*q>191. It is 275=5^2*11. Thus a(12)=275. Further, a(13) should be the nearest prime more than 275. It is 277.
%o (Sage)
%o @CachedFunction
%o def A178443(n):
%o if n <= 2: return {1:1, 2:3}[n]
%o if is_prime(n): return next_prime(A178443(n-1))
%o psig_n = list(m for p,m in factor(n))
%o primes_seen = sorted(set(filter(is_prime, map(A178443, range(2,n)))))
%o possibles = (prod(p**m for p,m in zip(pvec, psig_n)) for pvec in Combinations(primes_seen, len(psig_n)))
%o return min(p for p in possibles if p > A178443(n-1))
%o # _D. S. McNeil_, Jan 01 2011
%Y Cf. A178442, A172980, A172999
%K nonn,fini,full
%O 1,2
%A _Vladimir Shevelev_, Dec 22 2010
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