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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
Sequence in context: A111402 A174732 A087408 * A160474 A317620 A317415
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 March 19 04:58 EDT 2024. Contains 370952 sequences. (Running on oeis4.)