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A255852
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Least k > 0 such that gcd(k^n+2, (k+1)^n+2) > 1, or 0 if there is no such k.
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21
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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
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OFFSET
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0,4
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COMMENTS
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See A118119, which is the main entry for this class of sequences.
a(41) = 34290868, a(49) <= 2002111070, a(47) = 32286649814088452353414982038778088771611290478685407234712300075870593693164721\
99455164873287615636327176797646292254029648497024652505965417768073756378034012\
80883965289152013363422286845290874810700297549641281106223286199677401563701715\
56997846264124867393209579875386439424082082891813462700417531719383529314983727. - Hiroaki Yamanouchi, Mar 10 2015
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
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LINKS
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FORMULA
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a(2k)=1 for k>=0, because gcd(1^(2k)+2,2^(2k)+2) = gcd(3,4^k-1) = 3.
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EXAMPLE
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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.
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MATHEMATICA
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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}]]
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PROG
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(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))}
(Python)
from sympy import primefactors, resultant, nthroot_mod
from sympy.abc import x
if n == 0: return 1
k = 0
for p in primefactors(resultant(x**n+2, (x+1)**n+2)):
for d in (a for a in sorted(nthroot_mod(-2, n, p, all_roots=True)) if pow(a+1, n, p)==-2%p):
k = min(d, k) if k else d
break
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CROSSREFS
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KEYWORD
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nonn,hard,more
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AUTHOR
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EXTENSIONS
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STATUS
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approved
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