# [seqfan] SImplified description of A161373 and A161374

Hagen von Eitzen hagen at von-eitzen.de
Fri Jun 12 08:29:47 CEST 2009

```Hello seqfans,

looking at the above sequences of early bird vs. punctual binary
numbers, as imple pattern emerges.
The proof is not difficult, but could someone please crosscheck if I
didn't miss a case or subcase or subsubsubcase ... ?

1) n = 2^k is punctual for all k>=0:
If a number n is early bird, then either the leading 1 of its binary
representation starts a number having less bits than n has; or another 1
starts a number <= n.
If n=2^k, there is only one 1 and the number starting there must be at
least k+1 digits long.
Therefore n is not early bird

2)  n = 2^(2k+1) + 2^k is punctual for all k>=1:
The leading 1 cannot start a shorter number - it would have to be 2^k,
followed by 2^k+1, not matching the last bit.
If a number m<=n starts at the other 1, then it is preceded by m-1,
hence  m ends in a 1 followed by (k-1) 0's followed by 1, esp. m>2^k and
m = 2^k +1 mod 2^(k+1).
Then m must begin with a 1, k 0's and then followed by at least k+1
digits, hence m >= 2^(2k+1).
Together with m = 2^k +1 mod 2^(k+1) we conclude m>n.
Hence n is not early bird.

3) All n with at least 3 1's in binary are early bird:
Write n in binary as 1.a.1.b.1.c where '.' denotes concatenation, a,b,c
are (possibly empty) strings and a and c consist only of 0's (i.e. we
have focused on the first, the second and the last 1).
3.1) Assume b contains at least one 0.
Then ingenearal x.b +1 = x.d for a string d with length(d)=length(b).
3.1.1) Assume length(c) > length(a).
Let m = 1.c.1.a.1.b. Then  m < n and m+1 = 1.c.1.a.1.d, hence n occurs
early at the junction of m and m+1.
3.1.2) Assume length(c) <= length(a).
Let m = 1.a.1.b < n, then m+1 =1.a.1.d begins with 1.c, hence n occurs
early at the junction of m and m+1.
3.2) Assume b consists only of 1's
3.2.1) Assume length(c)>=length(a).
Let m = 1.c.0.a.1.b, then m<n and m+1 begins with 1.c, hence n occurs
early at the junction of m and m+1.
3.2.2) Assume length(c) < length(a), esp. length(a)>0
Let m = 1.a.1.b < n, then m+1 begins with 1.c, hence n occurs early at
the junction of m and m+1.

4) n = 2^m + 2^k with  m>k and m != 2k+1 is early bird:
Write n = 1.a.1.c where a,c consist only of 0's.
4.1) Assume length(c)>length(a)>0.
Then m = 1.c.1.a <n and m+1 begins with 1.c, hence n occurs early at the
junction of m and m+1.
4.2) Assume length(c)<length(a).
Let m = 1.a. Then m+1 begins with 1.c, hence n occurs early at the
junction of m and m+1.
4.3) Assume length(c)>length(a)=0
Then n = 11.c. Let m = 1.c -1. Then m < n and odd, hence n occurs early
at the junction of m and m+1.

Summary:
A number n>3 is punctual iff it is of the form 2^k or 2^(2k+1) + 2^k.

Hagen

```