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a(n) = A296354(n) - A296355(n).
4

%I #26 Dec 25 2017 09:28:14

%S 0,0,5,3,21,19,23,11,65,53,59,72,74,81,70,31,169,182,166,176,183,148,

%T 202,188,210,202,180,228,218,216,185,79,441,345,411,467,433,458,416,

%U 475,449,489,436,461,516,374,509,462,538,487,537,505,522,503,577,560

%N a(n) = A296354(n) - A296355(n).

%C This is the binary "early-birdness" of n (cf. A116700, A296364).

%C Theorem: a(n) > 0 for all n > 1.

%C 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 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 (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 (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 (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 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 Rémy Sigrist, <a href="/A296356/b296356.txt">Table of n, a(n) for n = 0..16384</a>

%Y Cf. A076478, A296354, A295355, A116700, A296364.

%K nonn,base,look

%O 0,3

%A _N. J. A. Sloane_, Dec 14 2017, corrected and extended Dec 17 2017

%E More terms from _Rémy Sigrist_, Dec 19 2017