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 1-4 of 4 results.

A246377 Permutation of natural numbers: a(1) = 1, a(p_n) = 2*a(n)+1, a(c_n) = 2*a(n), where p_n = n-th prime = A000040(n), c_n = n-th composite number = A002808(n).

Original entry on oeis.org

1, 3, 7, 2, 15, 6, 5, 14, 4, 30, 31, 12, 13, 10, 28, 8, 11, 60, 29, 62, 24, 26, 9, 20, 56, 16, 22, 120, 61, 58, 63, 124, 48, 52, 18, 40, 25, 112, 32, 44, 27, 240, 21, 122, 116, 126, 57, 248, 96, 104, 36, 80, 17, 50, 224, 64, 88, 54, 23, 480, 121, 42, 244, 232, 252, 114, 59, 496, 192, 208, 125, 72, 49, 160, 34, 100
Offset: 1

Views

Author

Antti Karttunen, Aug 27 2014

Keywords

Comments

This permutation is otherwise like Katarzyna Matylla's A135141, except that the role of even and odd numbers (or alternatively: primes and composites) has been swapped.
Because 2 is the only even prime, it implies that, apart from a(2)=3, odd numbers occur in odd positions only (along with many even numbers that also occur in odd positions).
This also implies that for each odd composite (A071904) there exists a separate infinite cycle in this permutation, apart from 9 and 15 which are in the same infinite cycle: (..., 23, 9, 4, 2, 3, 7, 5, 15, 28, 120, 82, 46, ...).

Crossrefs

Inverse: A246378.
Other related or similar permutations: A135141, A054429, A246201, A245703, A246376, A246379, A243071, A246681, A236854.
Differs from A237427 for the first time at n=19, where a(19) = 29, while A237427(19) = 62.

Formula

a(1) = 1, and for n > 1, if A010051(n) = 1 [i.e. when n is a prime], a(n) = 1+(2*a(A000720(n))), otherwise a(n) = 2*a(A065855(n)).
As a composition of related permutations:
a(n) = A054429(A135141(n)).
a(n) = A135141(A236854(n)).
a(n) = A246376(A246379(n)).
a(n) = A246201(A245703(n)).
a(n) = A243071(A246681(n)). [For n >= 1].
Other identities.
For all n > 1 the following holds:
A000035(a(n)) = A010051(n). [Maps primes to odd numbers > 1, and composites to even numbers, in some order. Permutations A246379 & A246681 have the same property].

A252735 a(1) = 0; for n > 1: a(2n) = a(n), a(2n+1) = 1 + a(A064989(n)).

Original entry on oeis.org

0, 0, 1, 0, 2, 1, 3, 0, 1, 2, 4, 1, 5, 3, 2, 0, 6, 1, 7, 2, 3, 4, 8, 1, 2, 5, 1, 3, 9, 2, 10, 0, 4, 6, 3, 1, 11, 7, 5, 2, 12, 3, 13, 4, 2, 8, 14, 1, 3, 2, 6, 5, 15, 1, 4, 3, 7, 9, 16, 2, 17, 10, 3, 0, 5, 4, 18, 6, 8, 3, 19, 1, 20, 11, 2, 7, 4, 5, 21, 2, 1, 12, 22, 3, 6, 13, 9, 4, 23, 2, 5, 8, 10, 14, 7, 1, 24, 3, 4, 2, 25, 6, 26, 5, 3, 15, 27, 1
Offset: 1

Views

Author

Antti Karttunen, Dec 21 2014

Keywords

Comments

Consider the binary tree illustrated in A005940: If we start from any n, computing successive iterations of A252463 until 1 is reached (i.e., we are traversing level by level towards the root of the tree, starting from that vertex of the tree where n is located at), a(n) gives the number of odd numbers > 1 encountered on the path (i.e., excluding the final 1 from the count but including the starting n if it was odd).

Crossrefs

Essentially one less than A061395.
Cf. also A246369.

Programs

Formula

a(1) = 0; for n > 1: a(2n) = a(n), a(2n+1) = 1 + a(A064989(n)).
a(n) = A080791(A156552(n)). [Number of nonleading 0-bits in A156552(n).]
Other identities:
For all n >= 2:
a(n) = A061395(n) - 1.
a(n) = A000120(A243071(n)) - 1. [One less than the binary weight of A243071(n).]
a(n) = A252464(n) - A252736(n) - 1.

A246348 a(1)=1, a(p_n) = 1 + a(n), a(c_n) = 1 + a(n), where p_n = n-th prime = A000040(n), c_n = n-th composite number = A002808(n); Also binary width of terms of A135141.

Original entry on oeis.org

1, 2, 3, 2, 4, 3, 3, 4, 3, 5, 5, 4, 4, 4, 5, 4, 4, 6, 5, 6, 5, 5, 4, 5, 6, 5, 5, 7, 6, 6, 6, 7, 6, 6, 5, 6, 5, 7, 6, 6, 5, 8, 5, 7, 7, 7, 6, 8, 7, 7, 6, 7, 5, 6, 8, 7, 7, 6, 5, 9, 7, 6, 8, 8, 8, 7, 6, 9, 8, 8, 7, 7, 6, 8, 6, 7, 9, 8, 6, 8, 7, 6, 5, 10, 8, 7, 9, 9, 6, 9, 8, 7, 10
Offset: 1

Views

Author

Antti Karttunen, Aug 27 2014

Keywords

Comments

If n = 1, the result is 1, otherwise, if n is prime, compute the result for that prime's index (A000720 or A049084) and add one, and if n is composite, compute the result for that composite's index (A065855) and add one.
a(n) tells how many calls (including the toplevel call) are required to compute A135141(n) or A246377(n) with a simple (nonmemoized) recursive algorithm as employed for example by Robert G. Wilson v's Mathematica-program of Feb 16 2008 in A135141 or Antti Karttunen's Scheme-proram in A246377.

Crossrefs

Programs

Formula

a(1) = 1, and for n >= 1, if A010051(n)=1 [that is, when n is prime], a(n) = 1 + a(A000720(n)), otherwise a(n) = 1 + a(A065855(n)). [A000720(n) and A065855(n) tell the number of primes, and respectively, composites <= n].
a(n) = A246369(n) + A246370(n).
a(n) = A070939(A135141(n)) = 1 + floor(log_2(A135141(n))). [Sequence gives also the binary width of terms of A135141].
a(n) = A070939(A246377(n)). [Also for 0/1-swapped version of that sequence].

A246370 a(1)=0, a(p_n) = 1 + a(n), a(c_n) = a(n), where p_n = n-th prime = A000040(n), c_n = n-th composite number = A002808(n); Also number of nonleading 0-bits in the binary representation of A135141(n).

Original entry on oeis.org

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

Views

Author

Antti Karttunen, Aug 27 2014

Keywords

Examples

			Consider n=30. It is the 19th composite number in A002808: 4, 6, 8, 9, 10, 12, 14, 15, 16, 18, 20, 21, 22, 24, 25, 26, 27, 28, 30, ...
Thus we consider next n=19, which is the 8th prime in A000040: 2, 3, 5, 7, 11, 13, 17, 19, ...
So we proceed with n=8, which is the 3rd composite number, and then with n=3, which is the 2nd prime, and then with n=2 which is the 1st prime, and we have finished.
All in all, it took us 5 steps (A246348(30) = 6 = 5+1) to reach 1, and on the journey, we encountered three primes, 19, 3 and 2, thus a(30) = 3.
		

Crossrefs

Formula

a(1) = 1, and for n >= 1, if A010051(n) = 1 [i.e. when n is prime], a(n) = 1 + a(A000720(n)), otherwise a(n) = a(A065855(n)). [A000720(n) and A065855(n) tell the number of primes, and respectively, composites <= n].
a(n) = A080791(A135141(n)). [a(n) tells also the number of nonleading zeros in binary representation of A135141(n)].
a(n) = A000120(A246377(n))-1. [Respectively, one less than the number of 1-bits in 0/1-swapped version of that sequence].
a(n) = A246348(n) - A246369(n) - 1.
Showing 1-4 of 4 results.