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Square array A(n,k) [where n is row and k is column] is read by descending antidiagonals: A(1,1), A(1,2), A(2,1), A(1,3), A(2,2), A(3,1), etc.

For n >= 3, each row n lists the numbers that appear n times in A004001. See also A051135.

Square array read by descending antidiagonals: A(1,1), A(1,2), A(2,1), A(1,3), A(2,2), A(3,1), etc.

The topmost row (row 1) of the array is A000079 (powers of 2), and in general each row 2^k contains the sequence (2^n - k), starting from the term (2^(k+1) - k). This follows from the properties (3) and (4) of A004001 given on page 227 of Kubo & Vakil paper (page 3 in PDF).

Moreover, each row 2^k - 1 (for k >= 2) contains the sequence 2^n - n - (k-2), starting from the term (2^(k+1) - (2k-1)). To see why this holds, consider the definitions of sequences A162598 and A265332, the latter which also illustrates how the frequency counts Q_n for A004001 are recursively constructed (in the Kubo & Vakil paper).

Numbers n for which A004001(n-1) < A004001(n) < A004001(n+1).

On Jul 15 1988 during a colloquium talk at Bell Labs, John Conway stated that he could prove that a(n)/n -> 1/2 as n approached infinity, but that the proof was extremely difficult. He therefore offered $100 to someone who could find an n_0 such that for all n >= n_0, we have |a(n)/n - 1/2| < 0.05, and he offered $10,000 for the least such n_0. I took notes (a scan of my notebook pages appears below), plus the talk - like all Bell Labs Colloquia at that time - was recorded on video. John said afterwards that he meant to say $1000, but in fact he said $10,000. I was in the front row. The prize was claimed by Colin Mallows, who agreed not to cash the check. - N. J. A. Sloane, Oct 21 2015

a(n) - a(n-1) = 0 or 1 (see the D. Newman reference). - Emeric Deutsch, Jun 06 2005

a(A188163(n)) = n and a(m) < n for m < A188163(n). - Reinhard Zumkeller, Jun 03 2011

From Daniel Forgues, Oct 04 2019: (Start)

Conjectures:

a(n) = n/2 iff n = 2^k, k >= 1.

a(n) = 2^(k-1): k times, for n = 2^k - (k-1) to 2^k, k >= 1. (End)

Complement of A088359; A051135(a(n)) > 1. [Reinhard Zumkeller, Jun 03 2011]

From Antti Karttunen, Jan 18 2016: (Start)

This set of numbers is closed with respect to A004001, see A266188.

After 1, one more than the positions of zeros in A093879.

(End)

Out of the first one million terms (a(10^6) = 510403), 258661 occur only once.

Complement of A087686; A051135(a(n)) = 1. - Reinhard Zumkeller, Jun 03 2011

From Antti Karttunen, Jan 18 2016: (Start)

In general, out of the first 2^(n+1) terms of A004001, 2^(n-1) - 1 terms (a quarter) occur only once. See also illustration in A265332.

One more than the positions of ones in A093879.

(End)

From Antti Karttunen, Jan 10 2016: (Start)

This is the sequence b(n) mentioned on page 229 (page 5 of PDF) in Kubo & Vakil paper, but using starting offset 1 instead of 2.

The recursive sum formula for A004001, a(n) = a(a(n-1)) + a(n-a(n-1)) can be written also as a(n) = a(a(n-1)) + a(A080677(n-1)).

This is the least monotonic left inverse for sequence A087686. Proof: Taking the first differences of this sequence yields the characteristic function for the complement of A188163, because A188163 gives the positions where A004001 increases, and this sequence increases by one whenever A004001 does not increase (and vice versa). Sequence A188163 is also 1 followed by A088359 (see comment in former), whose complement A087686 is, thus A087686 is also the complement of A188163, apart from the initial one. Note also how A087686 is closed with respect to A004001 (see A266188).

(End)

All the terms are 0 or 1: it is easy to show that if {b(n)} = A004001, b(n)>=b(n-1) and b(n)Benoit Cloitre, Jun 05 2004