A179016 The infinite trunk of binary beanstalk: The only infinite sequence such that a(n-1) = a(n) - number of 1's in binary representation of a(n).

0, 1, 3, 4, 7, 8, 11, 15, 16, 19, 23, 26, 31, 32, 35, 39, 42, 46, 49, 53, 57, 63, 64, 67, 71, 74, 78, 81, 85, 89, 94, 97, 101, 104, 109, 112, 116, 120, 127, 128, 131, 135, 138, 142, 145, 149, 153, 158, 161, 165, 168, 173, 176, 180, 184, 190, 193, 197, 200, 205, 209

Offset: 0

Author

Carl R. White, Jun 24 2010

Keywords

• easy
• nice
• nonn
• base

Comments

a(n) tells in what number we end in n steps, when we start climbing up the infinite trunk of the "binary beanstalk" from its root (zero). The name "beanstalk" is due to Antti Karttunen.

There are many finite sequences such as 0,1,2; 0,1,3,4,7,9; etc. obeying the same condition (see A218254) and as the length increases, so (necessarily) does the similarity to this infinite sequence.

Links

• Alois P. Heinz and Antti Karttunen, Table of n, a(n) for n = 0..16405 (first 1000 terms from Alois P. Heinz)
• Paul Tek, Illustration of the first terms

Crossrefs

Cf. A000120, A010062, A011371, A213710, A213711, A213717, A213730, A213731, A218600, A218616, A218789, A233271, A218602, A054429. First differences: A213712, complement: A213713.

A subsequence of A005187, i.e., a(n) = A005187(A213715(n)). For all n,

A071542(a(n)) = n, and furthermore A213708(n) <= a(n) <= A173601(n). (Cf. A218603, A218604).

Rows of A218254, when reversed, converge towards this sequence.

Cf. A276623, A219648, A219666, A255056, A276573, A276583, A276613 for analogous constructions, and also A259934.

Programs

Mathematica

TakeWhile[Reverse@ NestWhileList[# - DigitCount[#, 2, 1] &, 10^3, # > 0 &], # <= 209 &] (* Michael De Vlieger, Sep 12 2016 *)

Formula

a(0)=0, a(1)=1, and for n > 1, if n = A218600(A213711(n)) then a(n) = (2^A213711(n)) - 1, and in other cases, a(n) = a(n+1) - A213712(n+1). (This formula is based on Carl White's observation that this iterated/converging path must pass through each (2^n)-1. However, it would be very interesting to know whether the sequence admits more traditional recurrence(s), referring to previous, not to further terms in the sequence in their definition!) - Antti Karttunen, Oct 26 2012

a(n) = A218616(A218602(n)). - Antti Karttunen, Mar 04 2013

a(n) = A054429(A233271(A218602(n))). - Antti Karttunen, Dec 12 2013

Extensions

Starting offset changed from 1 to 0 by Antti Karttunen, Nov 05 2012