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A361929
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a(1) = 2; for n > 1, a(n) is the smallest positive integer > 1 not to share a factor with terms a(n-c .. n-1) where c = gcd(n-1,a(n-1)).
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2
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2, 3, 2, 3, 2, 3, 5, 2, 3, 7, 2, 3, 5, 2, 3, 7, 2, 3, 5, 2, 3, 7, 2, 3, 5, 11, 2, 3, 2, 3, 5, 2, 3, 7, 2, 3, 5, 2, 3, 7, 2, 3, 5, 2, 3, 7, 2, 3, 5, 2, 3, 7, 2, 3, 5, 11, 2, 3, 2, 3, 5, 2, 3, 7, 2, 3, 5, 2, 3, 7, 11, 2, 3, 2, 5, 7, 2, 3, 5, 2, 3, 7, 2, 3, 5, 11
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OFFSET
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1,1
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COMMENTS
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Conjecture: All primes will occur. See A361931 for the first occurrences of primes.
a(p+1) = 2 or 3 when p is prime.
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LINKS
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EXAMPLE
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For a(26), we see a(25) = 5. Then gcd(25,5) = 5, so a(26) must not share a factor with any of the previous 5 terms. The previous 5 terms {a(21), a(22), a(23), a(24), a(25)} = {3, 7, 2, 3, 5}, and the least positive number not to share a factor with {3, 7, 2, 3, 5} is 11, so a(26) = 11.
The first terms, alongside gcd(n,a(n)):
n a(n) gcd(n,(a(n))
- ---- ----
1 2 1
2 3 1
3 2 1
4 3 1
5 2 1
6 3 3
7 5 1
8 2 2
9 3 3
10 7 1
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MATHEMATICA
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K = {2}; While[Length@K < 86, p = 2; While[MemberQ[K[[Length@K - GCD[Length@K, Last@K] + 1 ;; Length@K]], p], p = NextPrime[p]]; AppendTo[K, p]]; Print[K]
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PROG
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(PARI) isok(w, k) = for (i=1, #w, if (gcd(k, w[i]) > 1 , return(0)); ); 1;
lista(nn) = my(va = vector(nn)); va[1] = 2; for (n=2, nn, my(k=2, ok = 0, w = vector(gcd(n-1, va[n-1]), i, va[n-i])); while (!ok, ok = isok(w, k); if (!ok, k++); ); va[n] = k; ); va; \\ Michel Marcus, Mar 31 2023
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CROSSREFS
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KEYWORD
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nonn
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AUTHOR
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STATUS
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approved
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