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

0, 0, 5, 3, 21, 19, 23, 11, 65, 53, 59, 72, 74, 81, 70, 31, 169, 182, 166, 176, 183, 148, 202, 188, 210, 202, 180, 228, 218, 216, 185, 79, 441, 345, 411, 467, 433, 458, 416, 475, 449, 489, 436, 461, 516, 374, 509, 462, 538, 487, 537, 505, 522, 503, 577, 560

Offset: 0

Author

N. J. A. Sloane, Dec 14 2017, corrected and extended Dec 17 2017

Keywords

• nonn
• base
• look

Comments

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

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

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.

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.

(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.

(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).

(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.

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

Links

• Rémy Sigrist, Table of n, a(n) for n = 0..16384

Crossrefs

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

Extensions

More terms from Rémy Sigrist, Dec 19 2017