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
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
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
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))}
(Python)
from sympy import primefactors, resultant, nthroot_mod
from sympy.abc import x
def A255852(n):
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
return k # Chai Wah Wu, May 08 2024
CROSSREFS
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