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.

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

0,2

Links

T. D. Noe, Table of n, a(n) for n = 0..1023

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.

Example

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]

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 *)

Crossrefs

Cf. A072727, A071585, A071766, A072728, A072729, A000129.

Sequence in context: A104704 A169884 A341260 * A343585 A153242 A143070

Adjacent sequences: A072723 A072724 A072725 * A072727 A072728 A072729

Keyword

easy,frac,nice,nonn

Author

Paul D. Hanna, Jul 09 2002

Status

approved