|
%I #32 Apr 27 2024 17:21:15
%S 3,3,4,5,7,7,10,13,14,14,19,23,23,31,34,34,46,50,60,65,73,79,88,92,
%T 107,113,126,139,149,168,182,198,210,227,244,265,276,292,317,340,369,
%U 384,408,436,444,480,516,540,565,606,628,669,704,735,759,810,829,895,925
%N a(n) is the smallest number such that for n-bonacci constant c_n satisfies round(c_n^prime(m)) == 1 (mod 2*p_m) for every m>=a(n).
%C The n-bonacci constant is a unique root x_1>1 of the equation x^n-x^(n-1)-...-x-1=0. So, for n=2 we have Fibonacci constant phi or golden ratio (A001622); for n=3 we have tribonacci constant (A058265); for n=4 we have tetranacci constant (A086088), for n=5 (A103814), for n=6 (A118427), etc.
%H S. Litsyn and V. Shevelev, <a href="http://dx.doi.org/10.1142/S1793042105000339">Irrational Factors Satisfying the Little Fermat Theorem</a>, International Journal of Number Theory, vol.1, no.4 (2005), 499-512.
%H V. Shevelev, <a href="http://list.seqfan.eu/oldermail/seqfan/2014-March/012750.html">A property of n-bonacci constant</a>, Seqfan (Mar 23 2014)
%e Let n=2, then c_2 = phi (Fibonacci constant). We have round(c_2^2)=3 is not == 1 (mod 4), round(c_2^3)=4 is not == 1 (mod 6), while round(c_2^5)=11 == 1 (mod 10) and one can prove that for p>=5, we have round(c_2^p) == 1 (mod 2*p). Since 5=prime(3), then a(2)=3.
%Y Cf. A001622, A007619, A007663, A058265, A086088, A103814, A118427, A118428, A238693, A238697, A238698, A238700, A239502, A239544, A239564, A239565, A239566.
%K nonn
%O 2,1
%A _Vladimir Shevelev_ and _Peter J. C. Moses_, Mar 23 2014
|
