cp's OEIS Frontend

A296356 a(n) = A296354(n) - A296355(n).

%I A296356 #26 Dec 25 2017 09:28:14
%S A296356 0,0,5,3,21,19,23,11,65,53,59,72,74,81,70,31,169,182,166,176,183,148,
%T A296356 202,188,210,202,180,228,218,216,185,79,441,345,411,467,433,458,416,
%U A296356 475,449,489,436,461,516,374,509,462,538,487,537,505,522,503,577,560
%N A296356 a(n) = A296354(n) - A296355(n).
%C A296356 This is the binary "early-birdness" of n (cf. A116700, A296364).
%C A296356 Theorem: a(n) > 0 for all n > 1.
%C A296356 Proof. The claim is true for 2 <= n <= 7, so assume n >= 8, and let u = 1... denote the binary expansion of n. Let L denote the list of all binary vectors whose concatenation gives A076478.
%C A296356 To show a(n)>0 it is enough to exhibit a pair of successive binary vectors b, c in L whose concatenation contains a copy of u that begins in b and is such that b appears in L before u does. There are three cases.
%C A296356 (i) Suppose n is even, say u = 1x0. Take c = x00, and let b be the vector preceding c in L, so that b = y11, say. Then bc = y11x00 contains u.
%C A296356 (ii) Suppose n = 2^k-1, u = 1^k. Take b = 01^(k-1), c = 10^(k-1), so that bc = 0 1^k 0^(k-1).
%C A296356 (iii) Otherwise, n is an odd number whose binary expansion contains a 0, say u = 1^k 0x1. Take c = 0x10^k, and let b be the vector preceding c in L, so that b = y1^k, say, and bc = y1^k 0x10^k.
%C A296356 In each case we need to verify that b does appear in L before u, but we leave this easy verification to the reader.  QED
%H A296356 Rémy Sigrist, <a href="/A296356/b296356.txt">Table of n, a(n) for n = 0..16384</a>
%Y A296356 Cf. A076478, A296354, A295355, A116700, A296364.
%K A296356 nonn,base,look
%O A296356 0,3
%A A296356 _N. J. A. Sloane_, Dec 14 2017, corrected and extended Dec 17 2017
%E A296356 More terms from _Rémy Sigrist_, Dec 19 2017