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.

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A292603 Doudna-tree reduced modulo 4: a(n) = A005940(1+n) mod 4.

Original entry on oeis.org

1, 2, 3, 0, 1, 2, 1, 0, 3, 2, 3, 0, 1, 2, 3, 0, 3, 2, 1, 0, 3, 2, 1, 0, 1, 2, 3, 0, 1, 2, 1, 0, 1, 2, 1, 0, 3, 2, 3, 0, 1, 2, 1, 0, 3, 2, 3, 0, 1, 2, 3, 0, 1, 2, 1, 0, 3, 2, 3, 0, 1, 2, 3, 0, 1, 2, 3, 0, 1, 2, 3, 0, 3, 2, 1, 0, 3, 2, 1, 0, 3, 2, 3, 0, 1, 2, 3, 0, 3, 2, 1, 0, 3, 2, 1, 0, 1, 2, 3, 0, 1, 2, 1, 0, 3, 2, 3, 0, 1, 2, 3, 0, 3, 2, 1, 0, 3, 2, 1, 0, 1
Offset: 0

Views

Author

Antti Karttunen, Dec 01 2017

Keywords

Examples

			The first six levels of the binary tree (compare also to the illustrations given at A005940 and A292602):
                               1
                               |
                               2
                ............../ \..............
               3                               0
        ....../ \......                 ....../ \......
       1               2               1               0
      / \             / \             / \             / \
     /   \           /   \           /   \           /   \
    3     2         3     0         1     2         3     0
   / \   / \       / \   / \       / \   / \       / \   / \
  3   2 1   0     3   2 1   0     1   2 3   0     1   2 1   0
		

Crossrefs

Cf. A004767 (gives the positions of 0's), A016813 (of 2's).

Programs

Formula

a(n) = A010873(A005940(1+n)).
a(n) + 4*A292602(n) = A005940(1+n).
a(2n+1) = 2*a(n) mod 4.
a(A004767(n)) = 0.
a(A016813(n)) = 2.
a(2*A156552(A246261(n))) = 1.
a(2*A156552(A246263(n))) = 3.
a(n * 2^(1+A246271(A005940(1+n)))) = 1.

A331302 Number of 4k+3 composites encountered when traversing from n to the root of A005940-tree.

Original entry on oeis.org

0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 1, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 1, 0, 0, 1, 0, 0, 0, 0, 2, 0, 0, 0, 1, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 1, 0, 0, 1, 1, 0, 0, 0, 0, 1, 0, 0, 1, 0, 0, 0, 0, 0, 0, 2, 0, 0, 0, 0, 1, 0, 2, 1, 0, 0, 0, 0, 0, 0, 1, 0, 1, 0, 0, 0, 2, 0, 0, 0, 2, 0, 0, 0, 1, 0, 0, 1, 0, 0, 1
Offset: 1

Views

Author

Antti Karttunen, Feb 07 2020

Keywords

Comments

For numbers > 1, iterate the map x -> A252463(x) which divides even numbers by 2 and for odd numbers shifts every prime in the prime factorization one index step towards smaller primes with A064989. a(n) counts the composite numbers of the form 4k+3 (A091236) encountered until 1 has been reached, including in the count also n itself if it is of the same form.

Examples

			Here -> stands for transition x -> A252463(x):
For n = 35, 35 mod 4 = 3, 35 -> 15 and 15 mod 4 = 3 also, but then 15 -> 6 (with 6 mod 4 = 2), and 6 -> 3, a prime, after which only noncomposites occur in the trajectory -> 2 -> 1, thus a(35) = 2 as there were exactly two 4k+3 composites on the whole path.
		

Crossrefs

Subsequences of the indices of zeros: A093641, A028982 (see A292583 for the explanation of the latter).

Programs

Formula

a(1) = 0, a(p) = 0 for all primes, otherwise a(n) = [n == 3 (mod 4)] + a(A252463(n)).
a(2^k * p) [with k>=0, p prime] = a(n^2) = a(2*(n^2)) = 0, with zeros occurring also on some other positions.
a(n) <= A292377(n).
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