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A329423
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Each term is visited once according to the rule: go left (resp. right) a(n) places if a(n) is prime (resp. not prime), starting at a(1) = 1. Choose a(n) along this journey as the largest possible prime, or else the smallest possible composite, not occurring earlier and compatible with these rules.
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2
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1, 4, 6, 8, 9, 3, 12, 14, 5, 21, 25, 7, 20, 10, 51, 16, 26, 22, 11, 15, 34, 2, 28, 17, 63, 42, 48, 33, 36, 32, 13, 18, 23, 60, 19, 24, 52, 91, 45, 29, 69, 55, 27, 39, 66, 58, 116, 78, 85, 37, 30, 105, 94, 90, 35, 75, 81, 77, 112, 43, 31, 41, 65, 106, 38, 50, 72, 44, 141, 47, 133, 180, 100, 150, 49, 121, 168, 128, 110
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OFFSET
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1,2
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COMMENTS
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More formally: from index i, move to index i - a(i), resp. to i + a(i), if a(i) is prime, resp. not prime.
The sequence is conjectured to be a permutation of the positive integers.
See A330154 for the lexicographic earliest variant, rather than choosing the largest possible prime (and choosing terms following the journey).
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LINKS
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EXAMPLE
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At index i = 1, a(1) = 1 is the smallest available number and possible since it drives to the right. This implies a move of a(1) = 1 place to the right, i.e., to index i = 2.
Then a(2) cannot be 2 or a larger prime, since this would imply a move too far to the left, so we choose the smallest available composite number, a(2) = 4. This takes us 4 places further to the right, to index 2 + 4 = 6.
Then a(6) can be equal to the prime 3, but not 5 which would drive back to a(1) and create a loop, preventing from visiting all terms. This leads to index 6 - 3 = 3.
Then a(3) cannot be 2 which would lead to already visited a(1), nor a larger prime which would take too far to the left. So a(3) = 6, the smallest available composite number. This leads to index 3 + 6 = 9.
Here a(9) can be a prime, the largest possible choice is a(9) = 5, leading to index 9 - 5 = 4.
Now a(4) cannot be 2 nor 3 nor a larger prime. Therefore a(4) = 8, the smallest available composite number. This leads to index 4 + 8 = 12.
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PROG
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(PARI) upto(N)={ my(U=[1], V=[2], f(k)=(-1)^isprime(k)*k, add(V, t)=if(V[1]+1<t, setunion(V, [t]), V[1]=t; while(#V>1&&V[2]==V[1]+1, V=V[^1]); V), A=Vec(1, N), done=1, n=1, k, t); while(n<=N||N>=n=valuation(done+1, 2), until(!A[n]|| N < n+=f(A[n]), bittest(done, n)&&(n=oo)&&break; done+=1<<n); n>N&& next; k=n-V[1]; while(U[1]<k=precprime(k-1), setsearch(U, k)|| setsearch(V, t=n-k)|| break); U[1]<k|| forcomposite(k=U[1]+1, oo, setsearch(U, k)|| setsearch(V, t=n+k)|| t<=V[1]|| break); U=add(U, A[n]=abs(t-n)); V=add(V, n=t)); print("Terms > "V[1]" may be incorrect."); A} \\ Use upto(600)[1..90] to get the displayed DATA, for smaller N, terms from a(72) = 180 on may be incorrect. - M. F. Hasler, Dec 05 2019
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CROSSREFS
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KEYWORD
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AUTHOR
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STATUS
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approved
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