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A255852 Least k > 0 such that gcd(k^n+2, (k+1)^n+2) > 1, or 0 if there is no such k.

%I A255852 #33 May 10 2024 02:33:04
%S A255852 1,0,1,51,1,40333,1,434,1,16,1,1234,1,78607,1,8310,1,817172,1,473,1,
%T A255852 116,1,22650,1,736546059,1,22,1,1080982,1,252,1,7809,1,644,1,
%U A255852 1786225573,1
%N A255852 Least k > 0 such that gcd(k^n+2, (k+1)^n+2) > 1, or 0 if there is no such k.
%C A255852 See A118119, which is the main entry for this class of sequences.
%C A255852 a(39) <= 8105110304875691067. - _Max Alekseyev_, Aug 06 2015
%C A255852 a(41) = 34290868, a(49) <= 2002111070, a(47) = 32286649814088452353414982038778088771611290478685407234712300075870593693164721\
%C A255852 99455164873287615636327176797646292254029648497024652505965417768073756378034012\
%C A255852 80883965289152013363422286845290874810700297549641281106223286199677401563701715\
%C A255852 56997846264124867393209579875386439424082082891813462700417531719383529314983727. - _Hiroaki Yamanouchi_, Mar 10 2015
%C A255852 a(43) = 3585, a(45) = 5, a(51) = 16, a(57) = 22, a(59) = 4495, a(63) = 1291, a(65) = 108, a(67) = 220, a(69) = 218039, a(71) = 2112. - _Chai Wah Wu_, May 08 2024
%F A255852 a(2k)=1 for k>=0, because gcd(1^(2k)+2,2^(2k)+2) = gcd(3,4^k-1) = 3.
%F A255852 a(2k+1) = A255832(k).
%e A255852 For n=1, gcd(k^n+2,(k+1)^n+2) = gcd(k+2,k+3) = 1, therefore a(1)=0.
%e A255852 For n=2k, see formula.
%e A255852 For n=3, we have gcd(51^3+2,52^3+2) = 109, and the pair (k,k+1)=(51,52) is the smallest which yields a GCD > 1, therefore a(3)=51.
%t A255852 A255852[n_] := Module[{m = 1}, While[GCD[m^n + 2, (m + 1)^n + 2] <= 1, m++]; m];
%t A255852 Join[{1, 0}, Table[A255852[n], {n, 2, 24}]]
%o A255852 (PARI) a(n, c=2, L=10^7, S=1)={n!=1 && for(a=S, L, gcd(a^n+c, (a+1)^n+c)>1&&return(a))}
%o A255852 (Python)
%o A255852 from sympy import primefactors,resultant, nthroot_mod
%o A255852 from sympy.abc import x
%o A255852 def A255852(n):
%o A255852     if n == 0: return 1
%o A255852     k = 0
%o A255852     for p in primefactors(resultant(x**n+2,(x+1)**n+2)):
%o A255852         for d in (a for a in sorted(nthroot_mod(-2,n,p,all_roots=True)) if pow(a+1,n,p)==-2%p):
%o A255852             k = min(d,k) if k else d
%o A255852             break
%o A255852     return k # _Chai Wah Wu_, May 08 2024
%Y A255852 Cf. A118119, A255832, A255853-A255869
%K A255852 nonn,hard,more
%O A255852 0,4
%A A255852 _M. F. Hasler_, Mar 08 2015
%E A255852 a(25),a(37),a(41),a(47) conjectured by _Hiroaki Yamanouchi_, Mar 10 2015; confirmed by _Max Alekseyev_, Aug 06 2015