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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A233280 Permutation of nonnegative integers: a(n) = A003188(A054429(n)).

Original entry on oeis.org

0, 1, 2, 3, 4, 5, 7, 6, 8, 9, 11, 10, 14, 15, 13, 12, 16, 17, 19, 18, 22, 23, 21, 20, 28, 29, 31, 30, 26, 27, 25, 24, 32, 33, 35, 34, 38, 39, 37, 36, 44, 45, 47, 46, 42, 43, 41, 40, 56, 57, 59, 58, 62, 63, 61, 60, 52, 53, 55, 54, 50, 51, 49, 48, 64, 65, 67, 66
Offset: 0

Views

Author

Antti Karttunen, Dec 18 2013

Keywords

Comments

This permutation transforms the enumeration system of positive irreducible fractions A071766/A229742 (HCS) into the enumeration system A007305/A047679 (Stern-Brocot), and the enumeration system A245325/A245326 into A162909/A162910 (Bird). - Yosu Yurramendi, Jun 09 2015

Crossrefs

Inverse permutation: A233279.
Similarly constructed permutation pairs: A003188/A006068, A135141/A227413, A232751/A232752, A233275/A233276, A233277/A233278, A193231 (self-inverse).

Programs

  • Python
    from sympy import floor
    def a003188(n): return n^(n>>1)
    def a054429(n): return 1 if n==1 else 2*a054429(floor(n/2)) + 1 - n%2
    def a(n): return 0 if n==0 else a003188(a054429(n)) # Indranil Ghosh, Jun 11 2017
  • R
    maxrow <- 8 # by choice
    a <- 1
    for(m in 0:maxrow) for(k in 0:(2^m-1)){
    a[2^(m+1)+    k] <- a[2^m+      k] + 2^m
    a[2^(m+1)+2^m+k] <- a[2^(m+1)-1-k] + 2^(m+1)
    }
    a
    # Yosu Yurramendi, Apr 05 2017
    
  • Scheme
    (define (A233280 n) (A003188 (A054429 n)))
    ;; Alternative version, based on entangling even & odd numbers with odious and evil numbers:
    (definec (A233280 n) (cond ((< n 2) n) ((even? n) (A000069 (+ 1 (A233280 (/ n 2))))) (else (A001969 (+ 1 (A233280 (/ (- n 1) 2)))))))
    

Formula

a(n) = A003188(A054429(n)).
a(n) = A063946(A003188(n)).
a(n) = A054429(A154436(n)).
a(0)=0, a(1)=1, and otherwise, a(2n) = A000069(1+a(n)), a(2n+1) = A001969(1+a(n)). [A recurrence based on entangling even & odd numbers with odious and evil numbers]
a(n) = A258746(A180201(n)) = A180201(A117120(n)), n > 0. - Yosu Yurramendi, Apr 10 2017

A268087 a(n) = A162909(n) + A162910(n).

Original entry on oeis.org

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

Views

Author

Yosu Yurramendi, Jan 26 2016

Keywords

Comments

If the terms (n>0) are written as an array (in a left-aligned fashion) with rows of length 2^m, m >= 0:
2,
3, 3,
5, 4, 4, 5,
8, 7, 5, 7, 7, 5, 7, 8,
13,11, 9,12, 9, 6,10,11,11,10,6, 9,12, 9,11,13,
21,18,14,19,16,11,17,19,14,13,7,11,17,13,15,18,18,15,13,17,11,7,13,14,19,17,11,16, ...
a(n) is palindromic in each level m >= 0 (ranks between 2^m and 2^(m+1)-1), because in each level m >= 0 A162910 is the reverse of A162909:
a(2^m + k) = a(2^(m+1) - 1 - k), m >= 0, 0 <= k < 2^m.
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).
a(2^m + k) = A162909(2^(m+2) + k), a(2^m + k) = A162909(2^(m+1)+ 2^m + k), a(2^m + k) = A162910(2^(m+1) + k), m >= 0, 0 <= k < 2^m (empirical observations).
a(n) = A162911(n) + A162912(n), where A162911(n)/A162912(n) is the bit reversal permutation of A162909(n)/A162910(n) in each level m >= 0 (empirical observations).
a(n) = A162911(2n+1), a(n) = A162912(2n) for n > 0 (empirical observations). 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. A162909(n)/A162910(n) is also an enumeration system of all positive rationals (Bird system), 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), and A086592 (A020650+A020651).

Examples

			m = 3, k = 6: a(38) = 17, a(22) = 10, a(14) = 7.
		

Crossrefs

Programs

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

Formula

a(2^(m+2)+k) = a(2^(m+1)+k) + a(2^m+k) with m = 0, 1, 2, ... and 0 <= k < 2^m (empirical observation).
a(A059893(n)) = a(n) for n > 0. - Yosu Yurramendi, May 30 2017
From Yosu Yurramendi, May 14 2019: (Start)
Take the smallest m > 0 such that 0 <= k < 2^(m-1), and choose any M >= m,
a((1/3)*( A016921(2^(m-1)+k)*4^(M-m)-1)) = 2*a(2^(m-1)+k)*(M-m) + a(2^m+2*k ).
a((1/3)*(2*A016921(2^(m-1)+k)*4^(M-m)-2)) = 2*a(2^(m-1)+k)*(M-m) + a(2^m+2*k ) + a(2^(m-1)+k).
a((1/3)*( A016969(2^(m-1)+k)*4^(M-m)-2)) = 2*a(2^(m-1)+k)*(M-m) + a(2^m+2*k+1).
a((1/3)*(2*A016969(2^(m-1)+k)*4^(M-m)-1)) = 2*a(2^(m-1)+k)*(M-m) + a(2^m+2*k+1) + a(2^(m-1)+k). (End)
a(n) = A007306(A258996(n)), n > 0. - Yosu Yurramendi, Jun 23 2021

A052913 a(n+2) = 5*a(n+1) - 2*a(n), with a(0) = 1, a(1) = 4.

Original entry on oeis.org

1, 4, 18, 82, 374, 1706, 7782, 35498, 161926, 738634, 3369318, 15369322, 70107974, 319801226, 1458790182, 6654348458, 30354161926, 138462112714, 631602239718, 2881086973162, 13142230386374, 59948977985546, 273460429154982, 1247404189803818, 5690100090709126
Offset: 0

Views

Author

encyclopedia(AT)pommard.inria.fr, Jan 25 2000

Keywords

Comments

Main diagonal of the array: m(1,j)=3^(j-1), m(i,1)=1; m(i,j) = m(i-1,j) + m(i,j-1): 1 3 9 27 81 ... / 1 4 13 40 ... / 1 5 18 58 ... / 1 6 24 82 ... - Benoit Cloitre, Aug 05 2002
a(n) is also the number of 3 X n matrices of integers for which the upper-left hand corner is a 1, the rows and columns are weakly increasing, and two adjacent entries differ by at most 1. - Richard Stanley, Jun 06 2010
a(n) is the number of compositions of n when there are 4 types of 1 and 2 types of other natural numbers. - Milan Janjic, Aug 13 2010
If a Stern's sequence based enumeration system of positive irreducible fractions is considered (for example, A007305/A047679, or A162909/A162910, or A071766/A229742, or A245325/A245326, ...), and if it is organized by blocks or levels (n) with 2^n terms (n>=0), and the products numerator*denominator, term by term, are summed at each level n, then the resulting sequence of integers is a(n). - Yosu Yurramendi, May 23 2015
Number of 1’s in the substitution system {0 -> 110, 1 -> 11110} at step n from initial string "1" (1 -> 11110 -> 11110111101111011110110 -> ...) . - Ilya Gutkovskiy, Apr 10 2017

Crossrefs

Cf. A007482 (inverse binomial transform).

Programs

  • GAP
    a:=[1,4];; for n in [3..30] do a[n]:=5*a[n-1]-2*a[n-2]; od; a; # G. C. Greubel, Oct 16 2019
    
  • Magma
    I:=[1,4]; [n le 2 select I[n] else 5*Self(n-1)-2*Self(n-2): n in [1..35]]; // Vincenzo Librandi, May 24 2015
    
  • Magma
    R:=PowerSeriesRing(Integers(), 25); Coefficients(R!((1-x)/(1-5*x+2*x^2))); // Marius A. Burtea, Oct 16 2019
  • Maple
    spec := [S,{S=Sequence(Union(Prod(Sequence(Z),Union(Z,Z)),Z,Z))},unlabeled]: seq(combstruct[count](spec,size=n), n=0..20);
    seq(coeff(series((1-x)/(1-5*x+2*x^2), x, n+1), x, n), n = 0..30); # G. C. Greubel, Oct 16 2019
  • Mathematica
    Transpose[NestList[{Last[#],5Last[#]-2First[#]}&, {1,4},20]][[1]] (* Harvey P. Dale, Mar 12 2011 *)
    LinearRecurrence[{5, -2}, {1, 4}, 25] (* Jean-François Alcover, Jan 08 2019 *)
  • PARI
    Vec((1-x)/(1-5*x+2*x^2) + O(x^30)) \\ Michel Marcus, Mar 05 2015
    
  • Sage
    def A052913_list(prec):
        P. = PowerSeriesRing(ZZ, prec)
        return P((1-x)/(1-5*x+2*x^2)).list()
    A052913_list(30) # G. C. Greubel, Oct 16 2019
    

Formula

G.f.: (1-x)/(1-5*x+2*x^2).
a(n) = Sum_{alpha=RootOf(1 - 5*z + 2*z^2)} (1/17)*(3+alpha)*alpha^(-1-n).
a(n) = ((17+3*sqrt(17))/34)*((5+sqrt(17))/2)^n + ((17-3*sqrt(17))/34)*((5-sqrt(17))/2)^n. - N. J. A. Sloane, Jun 03 2002
a(n) = A107839(n) - A107839(n-1). - R. J. Mathar, May 21 2015
a(n) = 2*A020698(n-1), n>1. - R. J. Mathar, Nov 23 2015
E.g.f.: (1/17)*exp(5*x/2)*(17*cosh(sqrt(17)*x/2) + 3*sqrt(17)*sinh(sqrt(17)*x/2)). - Stefano Spezia, Oct 16 2019
a(n) = 3*A107839(n-1) + (-1)^n*A152594(n) with A107839(-1) = 0. - Klaus Purath, Jul 29 2020

Extensions

Typo in definition corrected by Bruno Berselli, Jun 07 2010

A086593 Bisection of A086592, denominators of the left-hand half of Kepler's tree of fractions.

Original entry on oeis.org

2, 3, 4, 5, 5, 7, 7, 8, 6, 9, 10, 11, 9, 12, 11, 13, 7, 11, 13, 14, 13, 17, 15, 18, 11, 16, 17, 19, 14, 19, 18, 21, 8, 13, 16, 17, 17, 22, 19, 23, 16, 23, 24, 27, 19, 26, 25, 29, 13, 20, 23, 25, 22, 29, 26, 31, 17, 25, 27, 30, 23, 31, 29, 34, 9, 15, 19, 20, 21, 27, 23, 28, 21, 30
Offset: 1

Views

Author

Antti Karttunen, Aug 28 2003

Keywords

Comments

Also denominator of alternate fractions in Kepler's tree as shown in A294442. - N. J. A. Sloane, Nov 20 2017

Crossrefs

Programs

  • Mathematica
    (* b = A020650 *) b[1] = 1; b[2] = 2; b[3] = 1; b[n_] := b[n] = Switch[ Mod[n, 4], 0, b[n/2 + 1] + b[n/2], 1, b[(n - 1)/2 + 1], 2, b[(n - 2)/2 + 1] + b[(n - 2)/2], 3, b[(n - 3)/2]]; a[1] = 2; a[n_] := b[4 n - 4]; Array[a, 100] (* Jean-François Alcover, Jan 22 2016, after Yosu Yurramendi's formula for A020650 *)
  • R
    maxlevel <- 15
    d <- c(1,2)
    for(m in 0:maxlevel)
     for(k in 1:2^m) {
       d[2^(m+1)    +k] <- d[k] + d[2^m+k]
       d[2^(m+1)+2^m+k] <- d[2^(m+1)+k]
    }
    a <- vector()
    for(m in 0:maxlevel) for(k in 0:(2^m-1)) a[2^m+k] <- d[2^(m+1)+k]
      a[1:63]
    # Yosu Yurramendi, May 16 2018

Formula

a(n) = A086592(2n-1) = A020650(4n-2).
a(n+1) = A071585(n) + A071766(n), n >= 0. - Yosu Yurramendi, Jun 30 2014
From Yosu Yurramendi, Jan 04 2016: (Start)
a(2^(m+1)+k+1) - a(2^m+k+1) = A071585(k), m >= 0, 0 <= k < 2^m.
a(2^(m+2)-k) = a(2^(m+1)-k) + a(2^m-k), m > 0, 0 <= k < 2^m-1.
(End)
a(2^n) = A000045(n+3). - Antti Karttunen, Jan 29 2016, based on above.
a(n) = A020651(4n-1), a(n+1) = A020651(4n+1), n > 0. - Yosu Yurramendi, May 08 2018
a(2^m+k) = A071585(2^(m+1)+k), m >= 0, 0 <= k < 2^m. - Yosu Yurramendi, May 16 2018

A180201 Inverse permutation to A180200.

Original entry on oeis.org

0, 1, 2, 3, 5, 4, 6, 7, 11, 10, 8, 9, 13, 12, 14, 15, 23, 22, 20, 21, 17, 16, 18, 19, 27, 26, 24, 25, 29, 28, 30, 31, 47, 46, 44, 45, 41, 40, 42, 43, 35, 34, 32, 33, 37, 36, 38, 39, 55, 54, 52, 53, 49, 48, 50, 51, 59, 58, 56, 57, 61, 60, 62, 63, 95, 94, 92, 93, 89, 88, 90, 91, 83
Offset: 0

Views

Author

Reinhard Zumkeller, Aug 15 2010

Keywords

Comments

A180199(n) = a(a(n));
a(A180198(n)) = A180198(a(n)) = A180200(n);
a(A075427(n)) = A075427(n).
This permutation transforms the enumeration system of positive irreducible fractions A245325/A245326 into the enumeration system A007305/A047679 (Stern-Brocot), and enumeration system A071766/A229742 (HCS) into A162909/A162910 (Bird). - Yosu Yurramendi, Jun 09 2015

Programs

  • R
    #
    maxn <- 63 # by choice
    a <- 1
    for(n in 1:maxn){
    a[2*n  ] <- 2*a[n] + (n%%2 == 0)
    a[2*n+1] <- 2*a[n] + (n%%2 != 0)}
    a <- c(0, a)
    # Yosu Yurramendi, May 23 2020

Formula

a(n) = A233280(A258746(n)) = A117120(A233280(n)), n > 0. - Yosu Yurramendi, Apr 10 2017 [Corrected by Yosu Yurramendi, Mar 14 2025]
a(0) = 0, a(1) = 1, for n > 0 a(2*n) = 2*a(n) + [n even], a(2*n + 1) = 2*a(n) + [n odd]. - Yosu Yurramendi, May 23 2020
From Alan Michael Gómez Calderón, Mar 04 2025: (Start)
a(n) = A054429(n) XOR floor(n/2) for n > 0.
a(n) = A054429(A003188(n)) for n > 0. (End)
a(n) = A154436(A054429(n)), n > 0. - Yosu Yurramendi, Mar 11 2025

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]

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

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

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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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