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A088430
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a(n) = the least positive d such that for p=prime(n), the numbers p+0d, p+1d, p+2d, ..., p+(p-1)d are all primes.
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11
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OFFSET
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1,2
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COMMENTS
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Problem discussed by Russell E. Rierson: starting with given p, find the least d such that the arithmetic progression p,p+d,p+2d,... contains only primes. Obviously, the maximum number of prime terms is p and to reach that maximum, d must be a multiple of all smaller primes. For example, a(5) is a multiple of 2*3*5*7.
There can be other maximum-length prime progressions starting at p, with larger d. (Zak Seidov found d=4911773580 for p=11.)
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LINKS
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Phil Carmody, a(7), NMBRTHRY Nov 2001
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FORMULA
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EXAMPLE
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n AP Last term
--------------
1 2+i 3
2 3+2*i 7
3 5+6*i 29
4 7+150*i 907
5 11+1536160080*i 15361600811
6 13+9918821194590*i 119025854335093
7 17+341976204789992332560*i 5471619276639877320977
8 19+2166703103992332274919550*i 39000655871861980948551919
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MATHEMATICA
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p = Prime[n]; m = Product[Prime[i], {i, 1, n - 1}];
d = m;
While[! AllTrue[Table[p + i*d, {i, 1, p - 1}], PrimeQ], d = d + m];
Return[d];
];
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CROSSREFS
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See A113834 for last term in the progression, and A231017 for the 2nd term.
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KEYWORD
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more,nonn
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AUTHOR
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EXTENSIONS
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a(7) was found by Phil Carmody. - Don Reble, Nov 23 2003
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STATUS
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approved
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