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A255869
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Least m > 0 such that gcd(m^n+19, (m+1)^n+19) > 1, or 0 if there is no such m.
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20
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1, 0, 3, 2408, 1, 3976, 608, 28, 1, 88, 23, 464658, 1, 319924724, 3, 7, 1, 1628, 138, 2219409, 1, 6, 5, 594, 1, 872, 3, 92, 1, 392, 65, 2278155, 1, 3755866, 4793, 13, 1, 7873, 3, 614294, 1, 448812437, 5
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OFFSET
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0,3
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COMMENTS
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See A118119, which is the main entry for this class of sequences.
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LINKS
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FORMULA
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a(4k) = 1 for k>=0, because gcd(1^(4k)+19, 2^(4k)+19) = gcd(20, 16^k-1) >= 5 since 16 = 1 (mod 5).
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EXAMPLE
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For n=0 and n=4, see formula with k=0 resp. k=1.
For n=1, gcd(m^n+19, (m+1)^n+19) = gcd(m+19, m+20) = 1, therefore a(1)=0.
For n=2, gcd(3^2+19, 4^2+19) = 7 and (m,m+1) = (3,4) is the smallest pair which yields a GCD > 1 here.
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MATHEMATICA
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A255869[n_] := Module[{m = 1}, While[GCD[m^n + 19, (m + 1)^n + 19] <= 1, m++]; m]; Join[{1, 0}, Table[A255869[n], {n, 2, 12}]] (* Robert Price, Oct 16 2018 *)
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PROG
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(PARI) a(n, c=19, L=10^7, S=1)={n!=1 && for(a=S, L, gcd(a^n+c, (a+1)^n+c)>1 && return(a))}
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CROSSREFS
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KEYWORD
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nonn,hard
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AUTHOR
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EXTENSIONS
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STATUS
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approved
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