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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. 21
 1, 0, 1, 51, 1, 40333, 1, 434, 1, 16, 1, 1234, 1, 78607, 1, 8310, 1, 817172, 1, 473, 1, 116, 1, 22650, 1, 736546059, 1, 22, 1, 1080982, 1, 252, 1, 7809, 1, 644, 1, 1786225573, 1 (list; graph; refs; listen; history; text; internal format)
 OFFSET 0,4 COMMENTS See A118119, which is the main entry for this class of sequences. a(39) <= 8105110304875691067. - Max Alekseyev, Aug 06 2015 a(41) = 34290868, a(49) <= 2002111070, a(47) = 32286649814088452353414982038778088771611290478685407234712300075870593693164721\ 99455164873287615636327176797646292254029648497024652505965417768073756378034012\ 80883965289152013363422286845290874810700297549641281106223286199677401563701715\ 56997846264124867393209579875386439424082082891813462700417531719383529314983727. - Hiroaki Yamanouchi, Mar 10 2015 LINKS FORMULA a(2k)=1 for k>=0, because gcd(1^(2k)+2,2^(2k)+2) = gcd(3,4^k-1) = 3. a(2k+1) = A255832(k). EXAMPLE For n=1, gcd(k^n+2,(k+1)^n+2) = gcd(k+2,k+3) = 1, therefore a(1)=0. For n=2k, see formula. 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. MATHEMATICA A255852[n_] := Module[{m = 1}, While[GCD[m^n + 2, (m + 1)^n + 2] <= 1, m++]; m]; Join[{1, 0}, Table[A255852[n], {n, 2, 24}]] PROG (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))} CROSSREFS Cf. A118119, A255832, A255853-A255869 Sequence in context: A111402 A174732 A087408 * A160474 A317620 A317415 Adjacent sequences:  A255849 A255850 A255851 * A255853 A255854 A255855 KEYWORD nonn,hard,more AUTHOR M. F. Hasler, Mar 08 2015 EXTENSIONS a(25),a(37),a(41),a(47) conjectured by Hiroaki Yamanouchi, Mar 10 2015; confirmed by Max Alekseyev, Aug 06 2015 STATUS approved

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Last modified February 25 11:54 EST 2020. Contains 332233 sequences. (Running on oeis4.)