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A123701
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Minimum number k > 0 such that abs(n^k - k^n) = A078202(n) is prime, or -1 if such k > 0 does not exist.
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1
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3, 5, 1, 1, 2, 1, 2, 1, 2, 3, 8, 1, 6, 1, 68, -1, 2, 1, 2, 1, 32
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OFFSET
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1,1
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COMMENTS
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A078202(n) is the smallest prime of the form abs(n^k - k^n), the absolute difference between n^k and k^n, or -1 if no such prime exists. A078202(n) = {2, 7, 2, 3, 7, 5, 79, 7, 431, 58049, 8375575711, 11, 13055867207, 13, 94233563770233419658037661865757455268745312881861761180195872329157714108064193, -1, 130783, 17, ...}. a(n) = -1 for n = {16,64,...} when A078202(n) = -1. a(n) = 1 for n = {3,4,6,8,12,14,18,20,...} = A008864(n) Primes + 1, when A078202(p+1) = p. Currently a(n) is not known for n = {22,28,33,36,37,39,40,46,55,56,57,59,...}. a(23)-a(27) = {60,1,12,5,-1}. a(29)-a(32) = {98,1,42,1}. a(34)-a(35) = {69,6}. a(38) = 1. a(41)-a(45) = {60,1,32,1,44}. a(47)-a(54) = {110,1,24,9,2,3,2,1}. a(58) = 93. a(60)-a(64) = {1,180,1,88,-1}.
Let x >= 2 and y >= 1 and k >= 1 and n == x^(xy). Then either (x,y,k) = (2,1,3) or (x,y,k) = (2,1,1) or abs(n^k - k^n) is composite. If we have (x,y) == (2,1), then n == 4, and we can check that a(4) == 1. Therefore, if n != 4 is a power of a number of the form x^x, then a(n) == -1. - Lucas A. Brown, Mar 25 2024
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LINKS
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MATHEMATICA
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f[n_] := Block[{k = If[EvenQ@n || n < 4, 1, 2]}; While[ ! PrimeQ@Abs[n^k - k^n], k += 2]; k] (* Robert G. Wilson v *)
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CROSSREFS
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KEYWORD
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hard,more,sign
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AUTHOR
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STATUS
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approved
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