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A084928
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If the numbers 1 to n^3 are arranged in a cubic array, a(n) is the minimum number of primes in each row of the n^2 rows in the "east-west view" that can have primes.
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3
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0, 1, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 1, 0, 1, 1, 1, 0, 0, 0, 0, 0, 0, 0, 0, 1, 0, 1, 0, 0, 0, 1, 0, 1, 1, 0, 0, 1, 0, 0, 1, 0, 1, 1, 1, 1, 0, 1, 1, 1, 1, 1, 1, 1, 1, 1, 0, 0, 1, 1, 1, 1, 1, 1, 1, 0, 2
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OFFSET
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1,121
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COMMENTS
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This is a three-dimensional generalization of A083382.
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REFERENCES
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See A083382 for references and links to the two-dimensional case.
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LINKS
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EXAMPLE
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For the case n=3, the numbers are arranged in a cubic array as follows:
1..2..3........10.11.12........19.20.21
4..5..6........13.14.15........22.23.24
7..8..9........16.17.18........25.26.27
The first row is (1,2,3), the second is (4,5,6), etc. Surprisingly, a(n) = 0 for all n from 3 to 66. It appears that a(n) > 0 for n > 128. This has been confirmed up to n = 1000.
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MATHEMATICA
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Table[minP=n; Do[s=0; Do[If[PrimeQ[n*(c-1)+r], s++ ], {r, n}]; minP=Min[s, minP], {c, n^2}]; minP, {n, 100}]
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PROG
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(PARI) A084928(n) = { my(m=-1); for(i=0, (n^2)-1, my(s=sum(j=(i*n), ((i+1)*n)-1, isprime(1+j))); if((m<0) || (s < m), m = s)); (m); }; \\ Antti Karttunen, Jan 01 2019
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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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