login
Smallest k > 1 dividing tau(k*2^n) (cf. A000005).
1

%I #10 Feb 06 2019 17:18:58

%S 2,4,2,10,2,3,2,6,2,4,2,3,2,4,2,12,2,3,2,5,2,4,2,3,2,4,2,7,2,3,2,6,2,

%T 4,2,3,2,4,2,5,2,3,2,6,2,4,2,3,2,4,2,12,2,3,2,6,2,4,2,3,2,4,2,10,2,3,

%U 2,6,2,4,2,3,2,4,2,12,2,3,2,5,2,4,2,3,2,4,2,11,2,3,2,6,2,4,2,3,2,4,2,5,2,3

%N Smallest k > 1 dividing tau(k*2^n) (cf. A000005).

%C Sequence contains the values 2,3,4,5,6,7,10,11,12 only. Is there any pattern ? a(m) = 4 for m =1,9,13,21,25,33,37,45,49,...This subsequence has the recurrence b(1)=1, b(2k)=b(2k-1)+8, b(2k+1)=b(2k)+4. If a(m) = 10 then m == 0 (mod 3). If m>n, a(n)=10, a(m)=10 then m-n == 0 (mod 60).

%H Antti Karttunen, <a href="/A072866/b072866.txt">Table of n, a(n) for n = 0..16384</a>

%o (PARI) A072866(n) = { n = (1<<n); for(k=2,oo,if(!(numdiv(k*n)%k), return(k))); }; \\ _Antti Karttunen_, Feb 06 2019

%Y Cf. A000005.

%K easy,nonn

%O 0,1

%A _Benoit Cloitre_, Jul 27 2002