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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.

%I A255869 #24 May 08 2024 02:28:27
%S A255869 1,0,3,2408,1,3976,608,28,1,88,23,464658,1,319924724,3,7,1,1628,138,
%T A255869 2219409,1,6,5,594,1,872,3,92,1,392,65,2278155,1,3755866,4793,13,1,
%U A255869 7873,3,614294,1,448812437,5
%N A255869 Least m > 0 such that gcd(m^n+19, (m+1)^n+19) > 1, or 0 if there is no such m.
%C A255869 See A118119, which is the main entry for this class of sequences.
%C A255869 a(43) <= 8153777984244162781089834. - _Max Alekseyev_, Aug 06 2015
%F A255869 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 A255869 For n=0 and n=4, see formula with k=0 resp. k=1.
%e A255869 For n=1, gcd(m^n+19, (m+1)^n+19) = gcd(m+19, m+20) = 1, therefore a(1)=0.
%e A255869 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 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 A255869 (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))}
%o A255869 (Python)
%o A255869 from sympy import primefactors, resultant, nthroot_mod
%o A255869 from sympy.abc import x
%o A255869 def A255869(n):
%o A255869     if n == 0: return 1
%o A255869     k = 0
%o A255869     for p in primefactors(resultant(x**n+19,(x+1)**n+19)):
%o A255869         for d in (a for a in nthroot_mod(-19,n,p,all_roots=True) if pow(a+1,n,p)==-19%p):
%o A255869             k = min(d,k) if k else d
%o A255869     return k # _Chai Wah Wu_, May 07 2024
%Y A255869 Cf. A118119, A255832, A255852-A255868
%K A255869 nonn,hard
%O A255869 0,3
%A A255869 _M. F. Hasler_, Mar 09 2015
%E A255869 a(13)-a(40) from _Hiroaki Yamanouchi_, Mar 12 2015
%E A255869 a(41)-a(42) from _Max Alekseyev_, Aug 06 2015