

A122706


Smallest prime p such that p^n is equal to the sum of 3 consecutive primes.


4



23, 7, 11, 29, 79, 29, 509, 53, 467, 1571, 61, 7, 1553, 31, 1097, 11, 397, 11, 163, 677, 23, 103, 1723, 11, 1759, 67, 433, 149, 919, 2879, 293, 9907, 1103, 1153, 179, 6199, 2683, 1877, 4373, 4679, 953, 2341, 8069, 3779, 3691, 28463, 991, 1061, 2447, 5471
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OFFSET

1,1


COMMENTS

Corresponding numbers k such that a(n)^n = p(k) + p(k+1) + p(k+2) are given by A157197.
It is not known if a(n) exists for all n.


LINKS



FORMULA

For m = (p^n)/3 (not an integer), if q,r are largest primes and s,t are smallest primes such that q < r < m < s < t, then p^n must equal either q+r+s or r+s+t.  Robert G. Wilson v


EXAMPLE

a(1) = 23 because A070934(1) = p(3) + p(4) + p(5) = 5 + 7 + 11 = 23 is prime, but p(1) + p(2) + p(3) = 2 + 3 + 5 = 10 is composite and p(2) + p(3) + p(4) = 3 + 5 + 7 = 15 is composite.
a(2) = 7 because A122654(1) = 7 is prime and p(6) + p(7) + p(8) = 13 + 17 + 19 = 49 = 7^2, but p(k) + p(k+1) + p(k+2) are not squares for 0 < k < 6.


MATHEMATICA

PrevPrim[n_] := Block[{k = n  1}, While[ !PrimeQ[k], k ]; k]; NextPrim[n_] := Block[{k = n + 1}, While[ !PrimeQ[k], k++ ]; k]; f[n_] := Block[{p = If[n < 2, 5, 3]}, While[r = PrevPrim@ Floor[p^n/3]; q = PrevPrim@r; s = NextPrim@r; t = NextPrim@s; p^n != q + r + s && p^n != r + s + t, p = NextPrim@p]; p]; Array[f, 50] (* Robert G. Wilson v *)


CROSSREFS



KEYWORD

nonn


AUTHOR



EXTENSIONS



STATUS

approved



