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A255869 Least m > 0 such that gcd(m^n+19, (m+1)^n+19) > 1, or 0 if there is no such m. 20

%I #19 Oct 16 2018 17:24:09

%S 1,0,3,2408,1,3976,608,28,1,88,23,464658,1,319924724,3,7,1,1628,138,

%T 2219409,1,6,5,594,1,872,3,92,1,392,65,2278155,1,3755866,4793,13,1,

%U 7873,3,614294,1,448812437,5

%N Least m > 0 such that gcd(m^n+19, (m+1)^n+19) > 1, or 0 if there is no such m.

%C See A118119, which is the main entry for this class of sequences.

%C a(43) <= 8153777984244162781089834. - _Max Alekseyev_, Aug 06 2015

%F 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).

%e For n=0 and n=4, see formula with k=0 resp. k=1.

%e For n=1, gcd(m^n+19, (m+1)^n+19) = gcd(m+19, m+20) = 1, therefore a(1)=0.

%e 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.

%t 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 *)

%o (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))}

%Y Cf. A118119, A255832, A255852-A255868

%K nonn,hard

%O 0,3

%A _M. F. Hasler_, Mar 09 2015

%E a(13)-a(40) from _Hiroaki Yamanouchi_, Mar 12 2015

%E a(41)-a(42) from _Max Alekseyev_, Aug 06 2015

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Last modified April 25 06:49 EDT 2024. Contains 371964 sequences. (Running on oeis4.)