cp's OEIS Frontend

This is a front-end for the Online Encyclopedia of Integer Sequences, made by Christian Perfect. The idea is to provide OEIS entries in non-ancient HTML, and then to think about how they're presented visually. The source code is on GitHub.

Showing 1-2 of 2 results.

A277813 a(n) = A115384(A277812(n)) = index of the row where n is located in array A277880.

Original entry on oeis.org

1, 2, 1, 3, 2, 1, 4, 5, 3, 2, 6, 1, 7, 8, 4, 9, 5, 3, 10, 2, 11, 12, 6, 1, 13, 14, 7, 15, 8, 4, 16, 17, 9, 5, 18, 3, 19, 20, 10, 2, 21, 22, 11, 23, 12, 6, 24, 1, 25, 26, 13, 27, 14, 7, 28, 29, 15, 8, 30, 4, 31, 32, 16, 33, 17, 9, 34, 5, 35, 36, 18, 3, 37, 38, 19, 39, 20, 10, 40, 2, 41, 42, 21, 43, 22, 11, 44, 45, 23, 12, 46, 6, 47, 48, 24, 1, 49, 50
Offset: 1

Views

Author

Antti Karttunen, Nov 03 2016

Keywords

Comments

a(n) = index of the first odious number which is encountered when we start from k = n and iterate the map k -> A003188(A006068(k)/2).
Ordinal transform of A277822.
This works as a left inverse of A000069 like also does A115384.

Crossrefs

Cf. A003945 (the positions of 1's in this sequence).

Programs

Formula

a(n) = A115384(A277812(n)).
Other identities. For all n >= 1:
a(A000069(n)) = n.
a(n) = a(floor(n/2)) when A010060(n) = 0.

A277808 a(n) = number of iterations of map k -> A003188(A006068(k)/2) that are required (when starting from k = n) until k is an odious number.

Original entry on oeis.org

0, 0, 1, 0, 1, 2, 0, 0, 1, 2, 0, 3, 0, 0, 1, 0, 1, 2, 0, 3, 0, 0, 1, 4, 0, 0, 1, 0, 1, 2, 0, 0, 1, 2, 0, 3, 0, 0, 1, 4, 0, 0, 1, 0, 1, 2, 0, 5, 0, 0, 1, 0, 1, 2, 0, 0, 1, 2, 0, 3, 0, 0, 1, 0, 1, 2, 0, 3, 0, 0, 1, 4, 0, 0, 1, 0, 1, 2, 0, 5, 0, 0, 1, 0, 1, 2, 0, 0, 1, 2, 0, 3, 0, 0, 1, 6, 0, 0, 1, 0, 1, 2, 0, 0, 1, 2, 0, 3, 0, 0, 1, 0, 1, 2, 0, 3, 0, 0, 1, 4
Offset: 1

Views

Author

Antti Karttunen, Nov 03 2016

Keywords

Crossrefs

One less than A277822.
A left inverse of A003945.
Cf. A277812 (gives the odious number where such an iteration is finished at when starting from k=n).

Formula

a(n) = A010059(n) * A001511(n).
If A010060(n) = 1 [when n is one of the odious numbers, A000069], then a(n) = 0, otherwise a(n) = 1 + a(A003188(A006068(n)/2)).
Other identities:
For all n >= 0, a(A003945(n)) = n.
Showing 1-2 of 2 results.