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A332877
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Arrange the first n primes in a circle in any order. a(n) is the minimum value of the largest product of two consecutive primes out of all possible orders.
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6
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6, 15, 21, 35, 55, 77, 91, 143, 187, 221, 253, 323, 391, 493, 551, 667, 713, 899, 1073, 1189, 1271, 1517, 1591, 1763, 1961, 2183, 2419, 2537, 2773, 3127, 3233, 3599, 3953, 4189, 4331, 4757, 4897, 5293, 5723, 5963, 6499, 6887, 7171, 7663, 8051, 8633, 8989, 9797, 9991, 10403, 10807, 11303
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OFFSET
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2,1
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COMMENTS
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It might appear that all terms are either the product of two consecutive primes or two primes with a prime in between (A333747). However, 253=11*23 is the first term that is not in that sequence.
The easiest optimal permutation of n primes is probably {p_1, p_n, p_2, p_n-1, …, p_ceiling(n/2)}. - Ivan N. Ianakiev, Apr 20 2020
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LINKS
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FORMULA
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Probably a(n) = A332765(n+1) for n > 4.
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EXAMPLE
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Here are the different ways to arrange the first 4 primes in a circle.
2-3
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5-7
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2-3
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7-5
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2-5
| | Products: 10, 15, 21, 14. Largest product: 21.
7-3
The minimum largest product is 21, so a(4)=21.
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MATHEMATICA
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primes[n_]:=Prime/@Range[n];
partition[n_]:=Partition[primes[n], UpTo[Ceiling[n/2]]];
riffle[n_]:=Riffle[partition[n][[1]], Reverse[partition[n][[2]]]];
a[n_]:=Max[Table[riffle[n][[i]]*riffle[n][[i+1]], {i, 1, n-1}]];
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PROG
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(PARI) a(n) = {my(x = oo); for (k=1, (n-1)!, my(vp = Vec(numtoperm(n, k-1))); vp = apply(x->prime(x), vp); x = min(x, max(vp[1]*vp[n-1], vecmax(vector(n-1, j, vp[j]*vp[j+1])))); ); x; } \\ Michel Marcus, Apr 14 2020
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CROSSREFS
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KEYWORD
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nonn
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AUTHOR
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EXTENSIONS
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STATUS
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approved
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