cp's OEIS Frontend

In other words, let S(n) contain place values of 1's in the binary expansion of n, ordered greatest to least, where S(n,1) = floor(log_2(n+1)) = A000523(n+1) and the remaining terms in S strictly decrease. This sequence reads S(n,k)+1 instead as a multiplicity of prime(k) so as to produce a number with strictly decreasing prime exponents.

For n>0: first occurrence of n in row 2^(n-1), and when the table is seen as a flattened list at position n*2^(n-1)+1, cf. A005183. - Reinhard Zumkeller, Nov 16 2013

Row n lists the positions of 1's in the reversed binary expansion of n. Compare to triangles A112798 and A213925. - Gus Wiseman, Jul 22 2019

Length of n-th row given by A000120(n);

Min of n-th row given by A001511(n);

Sum of n-th row given by A029931(n);

Product of n-th row given by A096111(n);

Max of n-th row given by A113473(n);

Numerator of sum of reciprocals of n-th row given by A116416(n);

Denominator of sum of reciprocals of n-th row given by A116417(n);

LCM of n-th row given by A271410(n).

The first appearance of n is at A001787(n - 1).

n-th row begins at index A000788(n - 1) for n > 0.

Also the reversed positions of 1's in the reversed binary expansion of n. Also the reversed partial sums of the n-th composition in standard order (row n of A066099). Reversing rows gives A048793. - Gus Wiseman, Jan 17 2023

This sequence contains every increasing finite sequence. For example, the finite sequence {0,2,3,5} arises from n = 45.

Essentially A030308(n,k)*k, then entries removed where A030308(n,k)=0. - R. J. Mathar, Nov 30 2007

In the corresponding irregular triangle {a(n)+1}, the m-th row gives all positive integer roots m_i of polynomial {m,k}. - see link [Shevelev]; see also A264613. - Vladimir Shevelev, Dec 13 2015

Exponent of 2 in the superfactorials, i.e., a(n) = A007814(A000178(n)). - Ralf Stephan, Jan 03 2014

Original name: "Countdown sequences, allowing gaps."

Only digits 0 through 9 are used. The last term is 9876543210.

Equals A009995, sorted lexicographically. - Reinhard Zumkeller, Oct 14 2015

There are 2^k terms starting with digit k >= 0, they start at index 2^k. The countdown sequences, i.e., digits of the n-th term, are given in rows of A272011. - M. F. Hasler, Dec 11 2019

a(2^j)=0. Local extrema are a(2^j-1) = 2^j-j-1 (A000295).

Create a binary tree starting with x. To follow 0 from the root, apply f(x)=2x. To follow 1, apply g(x)=(2x-1)/3. For example, starting with x, the string 010 {also known as f(g(f(x)))}, you would get (8x-2)/3. These expressions represent the reverse Collatz function and will provide numbers whose Collatz path may include x. These expressions will all be of the form (2^a*x-b)/3^c. This sequence concerns b. What makes b interesting is that if you draw the tree, each level of the tree will have the same sequence of values for b. The root of the tree x, can be written as (2^0*x-0)/3^0, which has the first value for b. Each subsequent level contains twice as many values of b.

This sequence is 0 followed by a permutation of A213539, and therefore consists of 0 plus the elements of A116640 multiplied by 2^k, where k >= 0. E.g., 1, 5, 7, 19 becomes 0, 2^0*1, 2^1*1, 2^0*5, 2^2*1, 2^0*7, 2^1*5, 2^0*19, ... - Joe Slater, Dec 19 2016

When this sequence is arranged as an irregular triangle the sum of each row a(2^k)...a(2^(k+1)-1) equals A081039(2^k). The cumulative sum from a(0) to a(2^k-1) equals A002697(k). - Joe Slater, Apr 12 2018

In other words, the n-th row gives the numerical values of the runs of 1's in the binary expansion of n.

a(n) is a sharp lower bound of the greatest whole number k such that there is a hypergraph (V, H) with |V| = k having no isolated vertices and containing no partitions of size greater than n (see Brian & Larson link, i.e. Definition 3.1, Lemma 4.2 and Proof of Theorem 4.6).

Partial sums of A063787. - Robert Israel, Nov 28 2019