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
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]
Links
- T. D. Noe, Table of n, a(n) for n = 0..1023
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.
Comments