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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A072726 Numerator of the rationals >= 1 whose continued fractions consist of only even terms, in ascending order by the sum of the continued fraction terms and descending by lowest order continued fraction terms to highest.

Original entry on oeis.org

1, 2, 4, 5, 6, 9, 9, 12, 8, 13, 17, 22, 13, 20, 22, 29, 10, 17, 25, 32, 25, 38, 40, 53, 17, 28, 38, 49, 32, 49, 53, 70, 12, 21, 33, 42, 37, 56, 58, 77, 33, 54, 72, 93, 58, 89, 97, 128, 21, 36, 54, 69, 56, 85, 89, 118, 42, 69, 93, 120, 77, 118, 128, 169
Offset: 0

Views

Author

Paul D. Hanna, Jul 09 2002

Keywords

Examples

			n: a(n)/A072727 has continued fraction:
0: 1/0 = [infinity]
1: 2/1 = [2]
2: 4/1 = [4]
3: 5/2 = [2;2]
4: 6/1 = [6]
5: 9/2 = [4;2]
6: 9/4 = [2;4]
7: 12/5 = [2;2,2]
8: 8/1 = [8]
9: 13/2 = [6;2]
10: 17/4 = [4;4]
11: 22/5 = [4;2,2]
12: 13/6 = [2;6]
13: 20/9 = [2;4,2]
14: 22/9 = [2;2,4]
15: 29/12= [2;2,2,2]
		

Crossrefs

Programs

  • Mathematica
    a[0] = 1; a[n_] := a[n] = Which[IntegerQ[k = Log[2, n]], 2 (k + 1), IntegerQ[k = Log[2, n - 1]], 4 k + 1, IntegerQ[k = Log[2, n + 1]], Fibonacci[k + 1, 2], True, Clear[k]; Hold[2*(k - j)*a[2^j + m] + a[m]] /. ToRules[Reduce[2^k > 2^j > m >= 0 && n == 2^k + 2^j + m, {k, j, m}, Integers]] // ReleaseHold];
    Table[Print["a(", n, ") = ", a[n]]; a[n], {n, 0, 63}] (* Jean-François Alcover, Jul 13 2016 *)

Formula

a(2^k + 2^j + m) = 2(k-j)*a(2^j + m) + a(m) when 2^k > 2^j > m >=0. a(0) = 1, a(2^k) = 2(k+1), a(2^k + 1) = 4*k + 1 (k>0), a(2^k - 1) = the (k+1)-th Pell number.

A072727 Denominator of the rationals >= 1 whose continued fractions consist of only even terms, in ascending order by the sum of the continued fraction terms and descending by lowest order continued fraction terms to highest.

Original entry on oeis.org

0, 1, 1, 2, 1, 2, 4, 5, 1, 2, 4, 5, 6, 9, 9, 12, 1, 2, 4, 5, 6, 9, 9, 12, 8, 13, 17, 22, 13, 20, 22, 29, 1, 2, 4, 5, 6, 9, 9, 12, 8, 13, 17, 22, 13, 20, 22, 29, 10, 17, 25, 32, 25, 38, 40, 53, 17, 28, 38, 49, 32, 49, 53, 70
Offset: 0

Views

Author

Paul D. Hanna, Jul 09 2002

Keywords

Crossrefs

Formula

a(n) = A072726(m) where m = n - 2^(floor(log_2(n))). a(0) = 0, a(2^k) = 1, a(2^k + 1) = 2, a(2^k - 1) = the k-th Pell number. [Corrected by Sean A. Irvine, Oct 22 2024]

A273493 a(n) = A245327(n) + A245328(n).

Original entry on oeis.org

2, 3, 3, 5, 5, 4, 4, 8, 8, 7, 7, 7, 7, 5, 5, 13, 13, 11, 11, 12, 12, 9, 9, 11, 11, 10, 10, 9, 9, 6, 6, 21, 21, 18, 18, 19, 19, 14, 14, 19, 19, 17, 17, 16, 16, 11, 11, 18, 18, 15, 15, 17, 17, 13, 13, 14, 14, 13, 13, 11, 11, 7, 7, 34, 34, 29, 29, 31, 31, 23, 23, 30, 30, 27, 27, 25, 25, 17, 17, 31, 31, 26, 26, 29, 29, 22, 22, 25
Offset: 1

Views

Author

Yosu Yurramendi, May 23 2016

Keywords

Comments

The terms (n>0) may be written as a left-justified array with rows of length 2^m, m >= 0:
2,
3, 3,
5, 5, 4, 4,
8, 8, 7, 7, 7, 7, 5, 5,
13,13,11,11,12,12, 9, 9,11,11,10,10, 9, 9, 6, 6,
21,21,18,18,19,19,14,14,19,19,17,17,16,16,11,11,18,18,15,15,17,17,13,13,14,14,...
All columns have the Fibonacci sequence property: a(2^(m+2) + k) = a(2^(m+1) + k) + a(2^m + k), m >= 0, 0 <= k < 2^m (empirical observations).
The terms (n>0) may also be written as a right-justified array with rows of length 2^m, m >= 0:
2,
3, 3,
5, 5, 4, 4,
8, 8, 7, 7, 7, 7, 5, 5,
13,13,11,11,12,12, 9, 9,11,11,10,10, 9, 9, 6, 6,
...,19,19,17,17,16,16,11,11,18,18,15,15,17,17,13,13,14,14,13,13,11,11, 7, 7,
Each column is an arithmetic sequence. The differences of the arithmetic sequences repeat the sequence A071585: a(2^(m+2) -1 - 2k) - a(2^(m+1) -1 - 2k) = A071585(k-1), m > 0, 0 <= k < 2^m ; a(2^(m+2) -1 - 2k - 1) - a(2^(m+1) -1 - 2k - 1) = A071585(k-1), m > 0, 0 <= k < 2^m .
n>1 occurs in this sequence phi(n) = A000010(n) times, as it occurs in A007306 (Franklin T. Adams-Watters' comment), that is the sequence obtained by adding numerator and denominator in the Calkin-Wilf enumeration system of positive rationals. A245327(n)/A245328(n) is also an enumeration system of all positive rationals, and in each level m >= 0 (ranks between 2^m and 2^(m+1)-1) rationals are the same in both systems. Thus a(n) has the same terms in each level as A007306.
a(n) = A273494(A059893(n)), a(A059893(n)) = A273494(n), n > 0. - Yosu Yurramendi, May 30 2017

Crossrefs

Programs

  • PARI
    b(n) = my(b=binary(n)); fromdigits(concat(b[1], Vecrev(vector(#b-1, k, b[k+1]))), 2); \\ from A059893
    a(n) = my(n=b(n), x=1, y=1); for(i=0, logint(n, 2), if(bittest(n, i), [x, y]=[x+y, y], [x, y]=[y, x+y])); x \\ Mikhail Kurkov, Mar 11 2023

Formula

a(n) = A007306(A284459(n)), n > 0. - Yosu Yurramendi, Aug 23 2021

A273494 a(n) = A245325(n) + A245326(n).

Original entry on oeis.org

2, 3, 3, 5, 4, 5, 4, 8, 7, 7, 5, 8, 7, 7, 5, 13, 11, 12, 9, 11, 10, 9, 6, 13, 11, 12, 9, 11, 10, 9, 6, 21, 18, 19, 14, 19, 17, 16, 11, 18, 15, 17, 13, 14, 13, 11, 7, 21, 18, 19, 14, 19, 17, 16, 11, 18, 15, 17, 13, 14, 13, 11, 7, 34, 29, 31, 23, 30, 27, 25, 17, 31, 26, 29, 22, 25, 23, 20, 13, 29, 25, 26, 19, 27, 24, 23, 16, 23
Offset: 1

Views

Author

Yosu Yurramendi, May 23 2016

Keywords

Comments

The terms (n>0) may be written as a left-justified array with rows of length 2^m, m >= 0:
2,
3, 3,
5, 4, 5, 4,
8, 7, 7, 5, 8, 7, 7, 5,
13,11,12, 9,11,10, 9, 6,13,11,12, 9,11,10, 9, 6,
21,18,19,14,19,17,16,11,18,15,17,13,14,13,11, 7,21,18,19,14,19,17,...
All columns have the Fibonacci sequence property: a(2^(m+2) + k) = a(2^(m+1) + k) + a(2^m + k), m >= 0, 0 <= k < 2^m (empirical observations).
The terms (n>0) may also be written as a right-justified array with rows of length 2^m, m >= 0:
2,
3, 3,
5, 4, 5, 4,
8, 7, 7, 5, 8, 7, 7, 5,
13,11,12, 9,11,10, 9, 6,13,11,12, 9,11,10, 9, 6,
..., 18,15,17,13,14,13,11, 7,21,18,19,14,19,17,16,11,18,15,17,13,14,13,11, 7,
Each column is an arithmetic sequence. The differences of the arithmetic sequences give the sequence A071585: a(2^(m+1)-1-k) - a(2^m-1-k) = A071585(k), m >= 0, 0 <= k < 2^m.
n > 1 occurs in this sequence phi(n) = A000010(n) times, as it occurs in A007306 (Franklin T. Adams-Watters's comment), which is the sequence obtained by adding numerator and denominator in the Calkin-Wilf enumeration system of positive rationals. A245325(n)/A245326(n) is also an enumeration system of all positive rationals, and in each level m >= 0 (ranks between 2^m and 2^(m+1)-1) rationals are the same in both systems. Thus a(n) has the same terms in each level as A007306.
The same property occurs in all numerator+denominator sequences of enumeration systems of positive rationals, as, for example, A007306 (A007305+A047679), A071585 (A229742+A071766), A086592 (A020650+A020651), A268087 (A162909+A162910).

Crossrefs

Programs

  • PARI
    a(n) = my(x=1, y=1); for(i=0, logint(n, 2), if(bittest(n, i), [x, y]=[x+y, y], [x, y]=[y, x+y])); x \\ Mikhail Kurkov, Mar 10 2023

Formula

a(n) = A273493(A059893(n)), a(A059893(n)) = A273493(n), n > 0. - Yosu Yurramendi, May 30 2017
a(n) = A007306(A059893(A180200(n))) = A007306(A059894(A154435(n))). - Yosu Yurramendi, Sep 20 2021

A324337 a(n) = A002487(A006068(n)).

Original entry on oeis.org

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

Views

Author

Antti Karttunen, Feb 23 2019

Keywords

Comments

Like in A324338, a few terms preceding each position n = 2^k seem to be a batch of nearby Fibonacci numbers in some order.
For all n > 0 A324338(n)/A324337(n) constitutes an enumeration system of all positive rationals. For all n > 0 A324338(n) + A324337(n) = A071585(n). - Yosu Yurramendi, Oct 22 2019

Crossrefs

Programs

Formula

From Yosu Yurramendi, Oct 22 2019: (Start)
a(2^m+ k) = A324338(2^m+2^(m-1)+k), m > 0, 0 <= k < 2^(m-1)
a(2^m+2^(m-1)+k) = A324338(2^m+ k), m > 0, 0 <= k < 2^(m-1). (End)
a(n) = A324338(A063946(n)), n > 0. Yosu Yurramendi, Nov 04 2019
a(n) = A002487(A248663(A283477(n))). - Antti Karttunen, Nov 06 2019
a(n) = A002487(1+A233279(n)). - Yosu Yurramendi, Nov 08 2019
From Yosu Yurramendi, Nov 28 2019: (Start)
a(2^(m+1)+k) - a(2^m+k) = A324338(k), m >= 0, 0 <= k < 2^m.
a(A059893(2^(m+1)+A001969(k+1))) - a(A059893(2^m+A001969(k+1))) = A071585(k), m >= 0, 0 <= k < 2^(m-1).
a(A059893(2^(m+1)+ A000069(k+1))) = A071585(k), m >= 1, 0 <= k < 2^(m-1). (End)
From Yosu Yurramendi, Nov 29 2019: (Start)
For n > 0:
A324338(n) + A324337(n) = A071585(n).
A324338(2*A001969(n) )-A324337(2*A001969(n) ) = A071585(n-1)
A324338(2*A001969(n)+1)-A324337(2*A001969(n)+1) = -A324337(n-1)
A324338(2*A000069(n) )-A324337(2*A000069(n) ) = -A071585(n-1)
A324338(2*A000069(n)+1)-A324337(2*A000069(n)+1) = A324338(n-1) (End)
a(n) = A002487(1+A233279(n)). - Yosu Yurramendi, Dec 27 2019

A324338 a(n) = A002487(1+A006068(n)).

Original entry on oeis.org

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

Views

Author

Antti Karttunen, Feb 23 2019

Keywords

Comments

Like in A324337, a few terms preceding each 2^k-th term (here always 1) seem to consist of a batch of nearby Fibonacci numbers (A000045) in some order. For example, a(65533) = 987, a(65534) = 610 and a(65535) = 1597.
For all n > 0 A324338(n)/A324337(n) constitutes an enumeration system of all positive rationals. For all n > 0 A324338(n) + A324337(n) = A071585(n). - Yosu Yurramendi, Oct 22 2019

Crossrefs

Programs

Formula

a(n) = A002487(1+A006068(n)).
a(2^n) = 1 for all n >= 0.
From Yosu Yurramendi, Oct 22 2019: (Start)
a(2^m+2^(m-1)+k) = A324337(2^m+ k), m > 0, 0 <= k < 2^(m-1)
a(2^m+ k) = A324337(2^m+2^(m-1)+k), m > 0, 0 <= k < 2^(m-1). (End)
a(n) = A324337(A063946(n)), n > 0. Yosu Yurramendi, Nov 04 2019
a(n) = A002487(A233279(n)), n > 0. Yosu Yurramendi, Nov 08 2019
From Yosu Yurramendi, Nov 28 2019: (Start)
a(2^(m+1)+k) - a(2^m+k) = A324337(k), m >= 0, 0 <= k < 2^m.
a(A059893(2^(m+1)+A000069(k+1))) - a(A059893(2^m+A000069(k+1))) = A071585(k), m >= 1, 0 <= k < 2^(m-1).
a(A059893(2^m+ A001969(k+1))) = A071585(k), m >= 0, 0 <= k < 2^(m-1). (End)
From Yosu Yurramendi, Nov 29 2019: (Start)
For n > 0:
A324338(n) + A324337(n) = A071585(n).
A324338(2*A001969(n) )-A324337(2*A001969(n) ) = A071585(n-1)
A324338(2*A001969(n)+1)-A324337(2*A001969(n)+1) = -A324337(n-1)
A324338(2*A000069(n) )-A324337(2*A000069(n) ) = -A071585(n-1)
A324338(2*A000069(n)+1)-A324337(2*A000069(n)+1) = A324338(n-1) (End)
a(n) = A002487(A233279(n)). Yosu Yurramendi, Dec 27 2019

A072728 Numerator of rationals >= 1 whose continued fractions consist only of 1's and 2's, in ascending order by the sum of the continued fraction terms and descending by lowest order continued fraction terms to highest.

Original entry on oeis.org

1, 2, 3, 5, 5, 8, 7, 8, 12, 13, 11, 12, 13, 19, 19, 21, 17, 18, 19, 19, 21, 29, 31, 30, 31, 34, 27, 26, 29, 29, 31, 30, 31, 34, 46, 45, 50, 46, 49, 49, 50, 55, 41, 44, 41, 43, 47, 46, 45, 50, 46, 49, 49, 50, 55
Offset: 0

Views

Author

Paul D. Hanna, Jul 09 2002

Keywords

Examples

			n: a(n)/A072729(n) has continued fraction:
0: 1/1 = [1]
1: 2/1 = [2]
2: 3/2 = [1;2]
3: 5/2 = [2;2]
4: 5/3 = [1;1,2]
5: 8/3 = [2;1,2]
6: 7/5 = [1;2,2]
7: 8/5 = [1;1,1,2]
8: 12/5 = [2;2,2]
9: 13/5 = [2;1,1,2]
10: 11/8 = [1;2,1,2]
11: 12/7 = [1;1,2,2]
12: 13/8 = [1;1,1,1,2]
13: 19/8 = [2;2,1,2]
14: 19/7 = [2;1,2,2]
15: 21/8 = [2;1,1,1,2]
16: 17/12= [1;2,2,2]
17: 18/13= [1;2,1,1,2]
18: 19/11= [1;1,2,1,2]
19: 19/12= [1;1,1,2,2]
20: 21/13= [1;1,1,1,1,2]
		

Crossrefs

Formula

a(F(n)+F(n-3)+m) = a(F(n-1)+m) + a(F(n-3)+m) when 02; a(F(n)+m) = 2*a(F(n-2)+m) + a(F(n-4)+m) when 03; where a(0)=1, a(F(n)-1) = F(n) = n-th Fibonacci number; a(F(2n-1)) = n-th Pell number.

A072729 Denominator of rationals >= 1 whose continued fractions consist only of 1's and 2's, in ascending order by the sum of the continued fraction terms and descending by lowest order continued fraction terms to highest.

Original entry on oeis.org

1, 1, 2, 2, 3, 3, 5, 5, 5, 5, 8, 7, 8, 8, 7, 8, 12, 13, 11, 12, 13, 12, 13, 11, 12, 13, 19, 19, 21, 17, 18, 19, 19, 21, 19, 19, 21, 17, 18, 19, 19, 21, 29, 31, 30, 31, 34, 27, 26, 29, 29, 31, 30, 31, 34
Offset: 0

Views

Author

Paul D. Hanna, Jul 09 2002

Keywords

Crossrefs

Formula

a(F(n)+F(n-3)+m) = a(F(n-1)+m) + a(F(n-3)+m) when 02; a(F(n)+m) = 2*a(F(n-2)+m) + a(F(n-4)+m) when 03; where a(0)=1, a(F(n+1)-1) = F(n) = n-th Fibonacci number; a(F(2n+1)) = n-th Pell number.
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