cp's OEIS Frontend

a(0)=0; thereafter if n occurs as a term of A005187, a(n)=its position in A005187, otherwise zero. This works as an "inverse" function for A005187 in a sense that a(A005187(n)) = n for all n.

a(n)*A234017(n) = 0 for all n.

The ES1 matrix coefficients are defined by ES1[2*m-1,n] = 2^(2*m-1) * int(y^(2*m-1)/(cosh(y))^(2*n),y=0..infinity)/(2*m-1)! for m = 1, 2, 3, .. and n = 1, 2, 3 .. .

This definition leads to ES1[2*m-1,n=1] = 2*eta(2*m-1) and the recurrence relation ES1[2*m-1,n] = ((2*n-2)/(2*n-1))*(ES1[2*m-1,n-1] - ES1[2*m-3,n-1]/(n-1)^2) which we used to extend our definition of the ES1 matrix coefficients to m = 0, -1, -2, .. . We discovered that ES1[ -1,n] = 0.5 for n = 1, 2, .. . As usual eta(m) = (1-2^(1-m))*zeta(m) with eta(m) the Dirichlet eta function and zeta(m) the Riemann zeta function.

The coefficients in the columns of the ES1 matrix, for m = 1, 2, 3, .. , and n = 2, 3, 4 .. , can be generated with the polynomials GF(z,n) for which we found the following general expression GF(z;n) = ((-1)^(n-1)*r(n)*CFN1(z,n)*GF(z;n=1) + ETA(z,n))/p(n).

The CFN1(z,n) polynomials depend on the central factorial numbers A008955.

The ETA(z,n) are the Eta polynomials which lead to the Eta triangle.

The zero patterns of the Eta polynomials resemble a UFO. These patterns resemble those of the Zeta, Beta and Lambda polynomials, see A160474, A160480 and A160487.

The first Maple algorithm generates the coefficients of the Eta triangle. The second Maple algorithm generates the ES1[2*m-1,n] coefficients for m= 0, -1, -2, -3, .. .

The M(n) sequence, see the second Maple algorithm, leads to Gould's sequence A001316 and a sequence that resembles the denominators of the Taylor series for tan(x), A156769(n).

Some of our results are conjectures based on numerical evidence, see especially A160466.

a(k) is the distance between k and the largest power of 2 not exceeding k, where k = n + 1. [Consider the sequence of even numbers <= k; after sending the first term to the last position delete all odd-indexed terms; the final term that remains after iterating the process is the a(k)-th even number.] - Lekraj Beedassy, May 26 2005

Triangle read by rows in which row n lists the first 2^(n-1) positive integers, n >= 1; see the example. - Omar E. Pol, Sep 10 2013

a(n+1) is the second-order survivor of the n-person Josephus problem where every second person is marked until only one remains, who is then eliminated; the process is repeated from the beginning until all but one is eliminated. a(n) is first a power of 2 when n is three times a power of 2. For example, the first appearances of 2, 4, 8 and 16 are at positions 3, 6, 12 and 24, or (3*1),(3*2),(3*4) and (3*8). Eugene McDonnell (eemcd(AT)aol.com), Jan 19 2002, reporting on work of Boyko Bantchev (Bulgaria).

Appears to coincide with following sequence: Let n >= 1. Start with a bag B containing n 1's. At each step, replace the two least elements x and y in B with the single element x+y. Repeat until B contains 2 or fewer elements. Let a(n) be the largest element remaining in B at this point. - David W. Wilson, Jul 01 2003

Hsien-Kuei Hwang, S Janson, TH Tsai (2016) show that A078881 is the same sequence, apart from the offset. - N. J. A. Sloane, Nov 26 2017

a(1)=1, a(2)=2, a(3)=3; a(n) is least k such that no three terms of a(1), a(2), ..., a(n-1), k form an arithmetic progression, except for the first triple (1,2,3).

From Gus Wiseman, Dec 29 2023: (Start)

These are numbers whose binary indices are all powers of 2, where a binary index of n (row n of A048793) is any position of a 1 in its reversed binary expansion. For example, the terms together with their binary expansions and binary indices begin:

0: 0 ~ {}

1: 1 ~ {1}

2: 10 ~ {2}

3: 11 ~ {1,2}

8: 1000 ~ {4}

9: 1001 ~ {1,4}

10: 1010 ~ {2,4}

11: 1011 ~ {1,2,4}

128: 10000000 ~ {8}

129: 10000001 ~ {1,8}

130: 10000010 ~ {2,8}

131: 10000011 ~ {1,2,8}

136: 10001000 ~ {4,8}

137: 10001001 ~ {1,4,8}

138: 10001010 ~ {2,4,8}

139: 10001011 ~ {1,2,4,8}

For powers of 3 we have A368531.

(End)

No two adjacent 1-bits. Permutation of A003714.

Replace 1 with 01 in binary. - Ralf Stephan, Oct 07 2003

Theorem: The following eight definitions are equivalent.

(P1) Numbers k such that the odd part of k (A000265(k)) is < sqrt(2k).

(P1) is the new definition, repeated here for convenience. Note that this is not the same as saying A000265(k) < A172471(k), since A172471(k) = floor(sqrt(2*k)).

(P2) Numbers k such that the odd divisors of k are < sqrt(2k).

(P2) and (P1) are obviously equivalent.

(P3) The numbers 1, S_0, S_1, S_2, ..., where

S_m = { 2^(m+1)*(2^m+i) : i = 0 .. 3*2^m - 1 }.

So S_0 = {2,4,6}, S_1 = {8,12,16,20,24,28}, S_2 = {32,40,48,...,120}, S_3 = {128,144,...,496}, ...

The proof that (P3) and (P1) are the same sequence is not difficult and will be added later. (P3) is equivalent to a formula stated without proof (it may have been only an empirical observation) in the original version of this entry.

(P4) Numbers k such that the odd part of k is <= A003056(k).

That is, the odd part of k is <= floor((sqrt(1+8*n)-1)/2). It is more difficult to show this is equivalent to (P1), but it is true.

(P5) Numbers k such that the odd divisors of k are <= A003056(k).

(P5) and (P4) are obviously equivalent.

(P6) Numbers k such that A001227(k) = A082647(k).

(P6) was the original definition. In words, it says that the number of odd divisors of k is equal to the number of ways to write k as a sum of an odd number of consecutive positive integers, or equivalently as a sum of d consecutive positive integers for some d dividing k. To show that (P6) is equivalent to (P1) one makes use of the Hirschhorn-Hirschhorn article.

(P7) Numbers k such that the odd part of k is <= the sum of divisors of the even part.

(P7) was contributed by Jaycob Coleman, Jun 21 2014. To show (P7) is equivalent to (P1), write k as 2^m*s where s is odd. Equality holds if and only if k is an even perfect number.

(P8) Numbers k such that A000265(k) <= A000203(A006519(k)) or also such that A000265(k) <= A038712(k).

(P8) was contributed by Michel Marcus, Aug 14 2014. It is a restatement of (P7).

(End of theorem)

A further equivalent property, (P9), follows at once from (P4). This was conjectured by Omar E. Pol, Apr 18 2017

(P9) These are the numbers k such that the sequence of successive widths in the symmetric representation of sigma(k) is unimodal.

Yet another equivalent property:

(P10) Numbers k >= 1 such if k = i + (i+1) + (i+2) + ... + (i+j-1) for some i >= 1 and j >= 1 then j is odd [Caballero, 2019]. - Michel Marcus, Jan 16 2020

This is a subsequence of A005153. - Jaycob Coleman, Jun 21 2014

The complement of this sequence is A281005. - Omar E. Pol, Apr 18 2017

Subsequence of A174973. - Omar E. Pol, Feb 01 2021

The sequence divides naturally into blocks of length 2^k, k = 0, 1, 2, ... On block k, let n go from 0 to 2^k-1, write n in binary using k bits and reverse the bits. - N. J. A. Sloane, Jun 11 2002

For example: the 3-bit strings are 000, 001, 010, 011, 100, 101, 110 and 111. When they are bit-reversed, we get 000, 100, 010, 110, 001, 101, 011, 111. Or, in decimal representation 0,4,2,6,1,5,3,7.

In other words: Given any n>1, the set of numbers A030109[i] for indexes i ranging from 2^n to 2^(n+1)-1 is a permutation of the set of consecutive integers {0,1,2,...,2^n-1}. Example: for n=2, we have the permutation of {0,2,1,3} of {0,1,2,3} This is important in the standard FFT algorithms requiring a starting (or ending) bit-reversal permutation of indices. - Stanislav Sykora, Mar 15 2012