cp's OEIS Frontend

n XOR n-1, i.e., nim-sum of a pair of consecutive numbers.

Denominator of quotient sigma(2*n)/sigma(n). - Labos Elemer, Nov 04 2003

a(n) = the Towers of Hanoi disc moved at the n-th move, using standard moves with discs labeled (1, 3, 7, 15, ...) starting from top (smallest = 1). - Gary W. Adamson, Oct 26 2009

Equals row sums of triangle A168312. - Gary W. Adamson, Nov 22 2009

In the binary expansion of n, delete everything left of the rightmost 1 bit, and set all bits to the right of it. - Ralf Stephan, Aug 22 2013

Every finite sequence of positive integers summing to n may be termwise dominated by a subsequence of the first n values in this sequence [see Bannister et al., 2013]. - David Eppstein, Aug 31 2013

Sum of powers of 2 dividing n. - Omar E. Pol, Aug 18 2019

Given the binary expansion of (n-1) as {b[k-1], b[k-2], ..., b[2], b[1], b[0]}, then the binary expansion of a(n) is {bitand(b[k-1], b[k-2], ..., b[2], b[1], b[0]), bitand(b[k-2], ..., b[2], b[1], b[0]), ..., bitand(b[2], b[1], b[0]), bitand(b[1], b[0]), b[0], 1}. Recursively stated - 0th bit (L.S.B) of a(n), a(n)[0] = 1, a(n)[i] = bitand(a(n)[i-1], (n-1)[i-1]), where n[i] = i-th bit in the binary expansion of n. - Chinmaya Dash, Jun 27 2020

In the prime factorization of n = p1^e1 * ... pk^ek, add together the number of trailing 1-bits in each exponent e when they are written in binary.

Restricted growth sequence transform of the ordered pair [A007814(n), A007949(n)].

For all i, j:

A305900(i) = A305900(j) => a(i) = a(j),

a(i) = a(j) => A122841(i) = A122841(j),

a(i) = a(j) => A244417(i) = A244417(j),

a(i) = a(j) => A322316(i) = A322316(j) => A072078(i) = A072078(j).

If and only if a(k) > a(i) for all k > i then k is in A003586, - David A. Corneth, Dec 03 2018

That is, A003586 gives the positions of records (1, 2, 3, 4, 5, ...) in this sequence.

Sequence A126760 (without its initial zero) and this sequence are ordinal transforms of each other.

A276156(n) converts the binary expansion of n to a number whose primorial base representation has the same digits of 0's and 1's, thus each one of its terms is a unique sum of distinct primorial numbers. In this sequence that sum is then divided by the largest primorial that divides it, which only depends on the position of the least significant 1-bit in the binary expansion of the original n, that is, the 2-adic valuation of n.

To compute a(n) we shift its binary representation right as many steps k as necessary that the result were an odd number. Then one is added to that odd number, and the prime factorization of the resulting even number is shifted the same k number of steps towards larger primes, whose product is then decremented by one to get the final result.

In the essence, a(n) tells which number in array A246275 is at the same position where n is in the array A135764. As the topmost row in both arrays is A005408 (odd numbers), they are fixed, i.e. a(2n+1) = 2n+1 for all n.

Equally: a(n) tells which number in array A246273 is at the same position where n is in the array A054582, as they are the transposes of above two arrays.

Restricted growth sequence transform of A286161, of the ordered pair [A007814(n), A046523(n)].

For all i, j: a(i) = a(j) => A291761(i) = A291761(j).

This is Guy Steele's sequence GS(1, 6) (see A135416).

See the comments in A246676. This is otherwise similar permutation, except that after having reached an odd number 2m-1 when we have shifted the binary representation of n right k steps, here, in contrary to A246676, we don't shift the primes in the prime factorization of the even number 2m, but instead of an even number (2*a(m)), shifting it the same number (k) of positions towards larger primes, whose product is then decremented by one to get the final result.

From Antti Karttunen, Jan 18 2015: (Start)

This can be viewed as an entanglement or encoding permutation where the complementary pairs of sequences to be interwoven together are even and odd numbers (A005843/A005408) which are entangled with another complementary pair: even numbers in the order they appear in A253885 and odd numbers in their usual order: (A253885/A005408).

From the above follows also that this sequence can be represented as a binary tree. Each child to the left is obtained by doubling the parent and subtracting one, and each child to the right is obtained by applying A253885 to the parent:

1

|

...................2...................

3 4

5......../ \........8 7......../ \........6

/ \ / \ / \ / \

/ \ / \ / \ / \

/ \ / \ / \ / \

9 14 15 24 13 26 11 10

17 20 27 34 29 80 47 48 25 32 51 124 21 44 19 12

(End)

a(n) is the constant term of the reduction by x^2->x+1 of the polynomial encoded in the prime factorization of n. (Assuming here only polynomials with nonnegative integer coefficients, see e.g. A206296 for the details of the encoding).

Completely additive with a(prime(k)) = F(k-2), where F(k) denotes the k-th Fibonacci number, A000045(k) for k >= 0, or A039834(-k) for k <= 0. - Peter Munn, Apr 05 2021, incorporating comment by Antti Karttunen, Dec 15 2015

All positive integers eventually reach 1 in the Collatz problem iff all nonnegative integers eventually reach 0 with repeated application of this map, i.e., if for all n, the sequence n, a(n), a(a(n)), a(a(a(n))), ... eventually hits 0.

0 <= a(n) <= (3n+1)/2, with the upper bound being achieved for all odd n.

The positions of the zeros are given by A020988 = (2/3)*(4^n-1). This is because if n = (2/3)*(4^k-1), then m = 2n+1 = (1/3)*(4^(k+1)-1), and 3m+1 = 4^(k+1) is a power of 4. - Howard A. Landman, Mar 14 2010

Subsequence of A025480, a(n) = A025480(3n+1), i.e., A025480 = 0,[0],1,0,[2],1,3,[0],4,2,[5],1,6,[3],7,0,[8],4,9,[2],10,5,[11],1,12,[6],13,3,[14],... with elements of A173732 in brackets. - Paul Tarau, Mar 21 2010

A204418(a(n)) = 1. - Reinhard Zumkeller, Apr 29 2012

Original name: "A compression of the Collatz (or 3x+1) sequence considered as a map from odd numbers to odd numbers." - Michael De Vlieger, Oct 07 2019