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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

Table of n, a(n) for n=0..38.

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.)