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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A269172 Permutation of natural numbers: a(1) = 1, a(2n) = 2*a(n), a(2n+1) = A250469(a(A269380(2n+1))).

Original entry on oeis.org

1, 2, 3, 4, 5, 6, 7, 8, 9, 10, 11, 12, 13, 14, 15, 16, 17, 18, 25, 20, 21, 22, 19, 24, 23, 26, 27, 28, 29, 30, 49, 32, 33, 34, 35, 36, 31, 50, 39, 40, 37, 42, 41, 44, 45, 38, 43, 48, 55, 46, 51, 52, 47, 54, 121, 56, 57, 58, 77, 60, 53, 98, 63, 64, 65, 66, 59, 68, 69, 70, 61, 72, 169, 62, 75, 100, 67, 78, 85, 80, 81
Offset: 1

Views

Author

Antti Karttunen, Mar 03 2016

Keywords

Crossrefs

Inverse: A269171.
Related or similar permutations: A260741, A260742, A269356, A269358, A255422.
Cf. also A269394 (a(3n)/3) and A269396.
Differs from A255408 for the first time at n=38, where a(38) = 50, while A255408(38) = 38.

Formula

a(1) = 1, then after for even n, a(n) = 2*a(n/2), and for odd n, A250469(a(A269380(n))).
a(1) = 1, for n > 1, a(n) = A083221(A260738(n), a(A260739(n))).
As a composition of other permutations:
a(n) = A252755(A269386(n)).
a(n) = A252753(A269388(n)).
Other identities. For all n >= 1:
A000035(a(n)) = A000035(n). [This permutation preserves the parity of n.]
a(A003309(n)) = A008578(n). [Maps Ludic numbers to noncomposites.]

A253555 a(1) = 0, a(2n) = 1 + a(n), a(2n+1) = 1 + a(A250470(2n+1)); also binary width of terms of A252754 and A252756.

Original entry on oeis.org

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

Views

Author

Antti Karttunen, Jan 12 2015

Keywords

Comments

a(n) tells how many iterations of A253554 are needed before 1 is reached, i.e., the distance of n from 1 in binary trees like A252753 and A252755.

Crossrefs

Differs from A252464 for the first time at n=21, where a(21) = 4, while A252463(21) = 5.

Formula

a(1) = 0; for n > 1: a(n) = 1 + a(A253554(n)).
a(n) = A029837(1+A252754(n)) = A029837(1+A252756(n)).
a(n) = A253556(n) + A253557(n).
Other identities.
For all n >= 1:
a(A000079(n)) = n. [I.e., a(2^n) = n.]
a(A000040(n)) = n.
a(A001248(n)) = n+1.
For n >= 2, a(n) = A253558(n) + A253559(n).

A269865 Permutation of natural numbers: a(1) = 1, a(2n) = 2*a(n), a(2n+1) = A250469(1+a(n)).

Original entry on oeis.org

1, 2, 3, 4, 5, 6, 9, 8, 7, 10, 15, 12, 11, 18, 27, 16, 25, 14, 21, 20, 13, 30, 45, 24, 17, 22, 33, 36, 23, 54, 81, 32, 19, 50, 75, 28, 35, 42, 63, 40, 55, 26, 39, 60, 37, 90, 135, 48, 49, 34, 51, 44, 29, 66, 99, 72, 41, 46, 69, 108, 91, 162, 243, 64, 85, 38, 57, 100, 125, 150, 225, 56, 31, 70, 105, 84, 47, 126, 189, 80, 43, 110, 165, 52
Offset: 1

Views

Author

Antti Karttunen, Mar 12 2016

Keywords

Comments

This sequence can be represented as a binary tree. When the parent contains n, the left hand child contains 2n, while the value of right hand child is obtained by applying A250469(1+n):
1
|
................../ \..................
2 3
4......../ \........5 6......../ \........9
/ \ / \ / \ / \
/ \ / \ / \ / \
/ \ / \ / \ / \
8 7 10 15 12 11 18 27
16 25 14 21 20 13 30 45 24 17 22 33 36 23 54 81
etc.
Note how all nodes with odd n have a right hand child with value 3n.

Crossrefs

Inverse: A269866.
Cf. A250469.
Related or similar permutations: A269359, A269863, A269864, A269867, A246375, A249814, A252755, A270195.

Formula

a(1) = 1, a(2n) = 2*a(n), a(2n+1) = A250469(1+a(n)).

A253554 a(1) = 1, a(2n) = n, a(2n+1) = A250470(2n+1).

Original entry on oeis.org

1, 1, 2, 2, 3, 3, 5, 4, 4, 5, 7, 6, 11, 7, 6, 8, 13, 9, 17, 10, 8, 11, 19, 12, 9, 13, 10, 14, 23, 15, 29, 16, 12, 17, 15, 18, 31, 19, 14, 20, 37, 21, 41, 22, 16, 23, 43, 24, 25, 25, 18, 26, 47, 27, 21, 28, 20, 29, 53, 30, 59, 31, 22, 32, 27, 33, 61, 34, 24, 35, 67, 36, 71, 37, 26, 38, 35, 39, 73, 40, 28, 41, 79, 42, 33, 43, 30
Offset: 1

Views

Author

Antti Karttunen, Jan 12 2015

Keywords

Comments

Divide the even numbers by two, and for odd numbers n >= 3, a(n) = A078898(n)-th number k for which A055396(k) = A055396(n)-1.
For any number n >= 2 in binary trees A252753 and A252755, a(n) gives the number which is the parent of n.

Crossrefs

Bisections: A000027 and A250472.
Cf. A253555 (the number of iterations needed to reach 1 from n).
Differs from A252463 for the first time at n=21, where a(21) = 8, while A252463(21) = 10.

Programs

  • Scheme
    (define (A253554 n) (cond ((<= n 1) n) ((even? n) (/ n 2)) (else (A250470 n))))

Formula

a(1) = 1, a(2n) = n, a(2n+1) = A250470(2n+1).

A253556 a(1) = 0; after which, a(2n) = a(n), a(2n+1) = 1 + a(A250470(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, 1, 4, 8, 1, 2, 5, 3, 3, 9, 2, 10, 0, 2, 6, 3, 1, 11, 7, 4, 2, 12, 1, 13, 4, 1, 8, 14, 1, 3, 2, 2, 5, 15, 3, 2, 3, 3, 9, 16, 2, 17, 10, 5, 0, 4, 2, 18, 6, 2, 3, 19, 1, 20, 11, 6, 7, 4, 4, 21, 2, 4, 12, 22, 1, 3, 13, 3, 4, 23, 1, 3, 8, 1, 14, 5, 1, 24, 3, 7, 2, 25
Offset: 1

Views

Author

Antti Karttunen, Jan 12 2015

Keywords

Comments

Consider the binary tree illustrated in A252753 and A252755: If we start from any n, computing successive iterations of A253554 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

One less than A253558.
Powers of two, A000079, gives the positions of zeros.
Differs from A252735 for the first time at n=21, where a(21) = 1, while A252735(21) = 3.

Formula

a(1) = 0; after which, a(2n) = a(n), a(2n+1) = 1 + a(A250470(n)).
a(n) = A253555(n) - A253557(n).
a(n) = A253558(n) - 1.
a(n) = A080791(A252754(n)). [Number of nonleading 0-bits in A252754(n).]
Other identities. For all n >= 2:
a(n) = A000120(A252756(n)) - 1. [One less than the binary weight of A252756(n).]

A253559 a(1) = 0; for n>1: a(n) = A253557(n) - 1.

Original entry on oeis.org

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

Views

Author

Antti Karttunen, Jan 12 2015

Keywords

Comments

Consider the binary trees illustrated in A252753 and A252755: If we start from any n, computing successive iterations of A253554 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), a(n) gives the number of even numbers > 2 encountered on the path (i.e., excluding the 2 from the count but including the starting n if it was even).

Crossrefs

Essentially, one less than A253557.
A008578 gives the positions of zeros.
Differs from A252736 for the first time at n=21, where a(21) = 2, while A252736(21) = 1.

Programs

Formula

a(n) = A080791(A252756(n)). [Number of nonleading 0-bits in A252756(n).]
a(1) = 0; for n>1: a(n) = A253557(n) - 1.
Other identities. For all n >= 2:
a(n) = A000120(A252754(n)) - 1. [One less than the binary weight of A252754(n).]
a(n) = A253555(n) - A253558(n).

A269375 Tree of Lucky sieve, mirrored: a(0) = 1, a(1) = 2; after which a(2n) = 2*a(n), a(2n+1) = A269369(a(n)).

Original entry on oeis.org

1, 2, 4, 3, 8, 5, 6, 7, 16, 17, 10, 19, 12, 11, 14, 9, 32, 41, 34, 61, 20, 23, 38, 27, 24, 29, 22, 39, 28, 35, 18, 13, 64, 89, 82, 145, 68, 95, 122, 91, 40, 53, 46, 81, 76, 107, 54, 45, 48, 65, 58, 103, 44, 59, 78, 57, 56, 77, 70, 123, 36, 47, 26, 15, 128, 185, 178, 313, 164, 239, 290, 217, 136, 197, 190, 333, 244, 359, 182, 147, 80
Offset: 0

Views

Author

Antti Karttunen, Mar 01 2016

Keywords

Comments

Permutation of natural numbers obtained from the Lucky sieve. Note the indexing: Domain starts from 0, range from 1.
This sequence can be represented as a binary tree. Each left hand child is obtained by doubling the parent's contents, and each right hand child is obtained by applying A269369 to the parent's contents:
1
|
...................2...................
4 3
8......../ \........5 6......../ \........7
/ \ / \ / \ / \
/ \ / \ / \ / \
/ \ / \ / \ / \
16 17 10 19 12 11 14 9
32 41 34 61 20 23 38 27 24 29 22 39 28 35 18 13
etc.
Sequence A269377 is obtained from the mirror image of the same tree.

Crossrefs

Inverse: A269376.
Cf. A000959 (with 2 inserted between 1 and 3 forms the right edge of the tree).
Related or similar permutations: A163511, A260742, A269377.
Cf. also A252755, A269385.

Formula

a(0) = 1, a(1) = 2; after which, a(2n) = 2*a(n), a(2n+1) = A269369(a(n)).
As a composition of related permutations:
a(n) = A260742(A269385(n)).
Other identities. For all n >= 2:
A000035(a(n)) = A000035(n). [This permutation preserves the parity of n from a(2)=4 onward.]

A269385 Tree of Ludic sieve, mirrored: a(0) = 1, a(1) = 2; after which, a(2n) = 2*a(n), a(2n+1) = A269379(a(n)).

Original entry on oeis.org

1, 2, 4, 3, 8, 9, 6, 5, 16, 21, 18, 19, 12, 15, 10, 7, 32, 45, 42, 49, 36, 51, 38, 31, 24, 33, 30, 35, 20, 27, 14, 11, 64, 93, 90, 109, 84, 123, 98, 85, 72, 105, 102, 125, 76, 111, 62, 55, 48, 69, 66, 79, 60, 87, 70, 59, 40, 57, 54, 65, 28, 39, 22, 13, 128, 189, 186, 229, 180, 267, 218, 191, 168, 249, 246, 305, 196, 291, 170, 151, 144
Offset: 0

Views

Author

Antti Karttunen, Mar 01 2016

Keywords

Comments

Permutation of natural numbers obtained from the Ludic sieve. Note the indexing: Domain starts from 0, range from 1.
This sequence can be represented as a binary tree. Each left hand child is obtained by doubling the parent's contents, and each right hand child is obtained by applying A269379 to the parent's contents:
1
|
...................2...................
4 3
8......../ \........9 6......../ \........5
/ \ / \ / \ / \
/ \ / \ / \ / \
/ \ / \ / \ / \
16 21 18 19 12 15 10 7
32 45 42 49 36 51 38 31 24 33 30 35 20 27 14 11
etc.
Sequence A269387 is obtained from the mirror image of the same tree.

Crossrefs

Inverse: A269386.
Cf. A003309 (right edge of the tree).
Related or similar permutations: A163511, A260741, A269387, A269171.
Cf. also A252755, A269375.

Formula

a(0) = 1, a(1) = 2; after which, a(2n) = 2*a(n), a(2n+1) = A269379(a(n)).
As a composition of related permutations:
a(n) = A269171(A252755(n)).
a(n) = A260741(A269375(n)).
Other identities. For all n >= 2:
A000035(a(n)) = A000035(n). [This permutation preserves the parity of n from a(2)=4 onward.]

A253558 a(n) = A253556(n) + 1.

Original entry on oeis.org

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

Views

Author

Antti Karttunen, Jan 12 2015

Keywords

Comments

Consider the binary trees illustrated in A252753 and A252755: If we start from any n, computing successive iterations of A253554 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 encountered on the path (i.e., including both the final 1 and the starting n if it was odd).

Crossrefs

One more than A253556.
Powers of two, A000079, gives the positions of ones.
After n=1, differs from A061395 for the first time at n=21, where a(21) = 2, while A061395(21) = 4.

Programs

Formula

a(n) = A253556(n) + 1.
a(n) = A080791(A252754(n)) + 1. [One more than the number of nonleading 0-bits in A252754(n).]
Other identities.
For all n >= 1:
a(A000040(n)) = n.
For all n >= 2:
a(n) = A000120(A252756(n)). [Binary weight of A252756(n).]
a(n) = A253555(n) - A253559(n).

A269851 a(0) = 1, a(A087686(1+n)) = 2*a(n), a(A088359(n)) = A250469(a(n)), where A088359 and A087686 = numbers that occur only once (resp. more than once) in A004001.

Original entry on oeis.org

1, 2, 4, 3, 8, 9, 5, 6, 16, 21, 25, 7, 18, 15, 10, 12, 32, 45, 55, 49, 11, 42, 51, 35, 50, 27, 14, 36, 33, 30, 20, 24, 64, 93, 115, 91, 121, 13, 90, 123, 125, 77, 110, 147, 65, 98, 39, 22, 84, 105, 85, 102, 87, 70, 100, 57, 54, 28, 72, 69, 66, 60, 40, 48, 128, 189, 235, 203, 187, 169, 17, 186, 267, 305, 217, 143, 230
Offset: 0

Views

Author

Antti Karttunen, Mar 07 2016

Keywords

Comments

Permutation of natural numbers obtained from the sieve of Eratosthenes, combined with the permutation obtained from Hofstadter-Conway $10000 sequence (A004001). Note the indexing: Domain starts from 0, range from 1.

Crossrefs

Inverse: A269852.
Related or similar permutations: A252755, A267111, A269855.

Formula

a(0) = 1, a(1) = 2, for n > 1, if A093879(n-1) = 0 [when n is in A087686], a(n) = 2*a(n - A004001(n)), otherwise [when n is in A088359], a(n) = A250469(a(A004001(n)-1)).
As a composition of related permutations:
a(n) = A252755(A267111(n)).
Other identities. For all n >= 0:
a(2^n) = 2^(n+1).
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