A080940 is easy

[hv at crypt.org]

(Noticed while looking at the "Collatz related sequence" problem.)

Make the comment more forceful, and add the cross-reference. I'm not
convinced the second comment is particularly illuminating, but I guess
it doesn't hurt.

  change from:
%C A080940 a(n)=0 iff n=2^k; if a(n)>0 then a(n)=2^k for some k;
%Y A080940 Cf. A007088, A000079, A080941, A080942.
%K A080940 nonn,base
  to:
%C A080940 by definition, identical to A006519 except that a(2^k) = 0 for all k
%Y A080940 Cf. A007088, A000079, A080941, A080942, A006519.
%K A080940 nonn,base,easy
  giving:

%I A080940
%S A080940 0,0,1,0,1,2,1,0,1,2,1,4,1,2,1,0,1,2,1,4,1,2,1,8,1,2,1,4,1,2,1,0,1,2,1,
%T A080940 4,1,2,1,8,1,2,1,4,1,2,1,16,1,2,1,4,1,2,1,8,1,2,1,4,1,2,1,0,1,2,1,4,1,2,
%U A080940 1,8,1,2,1,4,1,2,1,16,1,2,1,4,1,2,1,8,1,2,1,4,1,2,1,32,1,2,1,4,1,2,1,8
%N A080940 Smallest divisor of n which is less n and also a suffix of n in binary representation; a(n)=0 if no such divisor exists.
%C A080940 by definition, identical to A006519 except that a(2^k) = 0 for all k
%C A080940 a(3*2^k)=2^k and a(m)<2^k for m<3*2^k (see A007283).
%e A080940 n=6='110', divisors<6: 1='1', 2='10', and 3='11', therefore a(6)=2='10';
%e A080940 n=7='111', divisors<7: 1='1', therefore a(7)=1;
%e A080940 n=8='1000', divisors<8: 1='1', 2='10', and 4='100', therefore a(8)=0.
%Y A080940 Cf. A007088, A000079, A080941, A080942, A006519.
%Y A080940 Adjacent sequences: A080937 A080938 A080939 this_sequence A080941 A080942 A080943
%Y A080940 Sequence in context: A087991 A076092 A080468 this_sequence A080941 A053616 A094718
%K A080940 nonn,base,easy
%O A080940 1,6
%A A080940 Reinhard Zumkeller (reinhard.zumkeller(AT)lhsystems.com), Feb 25 2003

Hugo