0,2

Thus a(n)/a(m) = d_1 + 1/(d_2 + 1/(d_3 + ... + 1/d_k)) where m = n - 2^floor(log(n)/log2) + 1 and where d_j = b_j - b_(j+1) are the differences of the binary exponents b_j > b_(j+1) defined by: 4*n = 2^b_1 + 2^b_2 + 2^b_3 + ... 2^b_k.

All the rationals are uniquely represented by this sequence - compare Stern's diatomic sequence A002487.

This sequence lists the rationals >= 1 in order by the sum of the terms of their continued fraction expansions. For example, the numerators generated from partitions of 5 that do not end with 1 are listed together as 5, 7, 7, 8, 5, 7, 7, 8, since: 5/1 = [5]; 7/2 = [3;2]; 7/3 = [2;3]; 8/3 = [2;1,2]; 5/4 = [1;4]; 7/5 = [1;2,2]; 7/4 = [1;1,3]; 8/5 = [1;1,1,2].

a(2^k + 2^j + m) = (k-j)*a(2^j + m) + a(m) when 2^k > 2^j > m >=0.

a(0) = 1, a(2^k) = k + 2, a(2^k + 1) = 2*k + 1 (k>0), a(2^k + 2) = 3*k - 2 (k>1), a(2^k + 3) = 3*k - 1 (k>1), a(2^k + 4) = 4*k - 7 (k>2), a(2^k - 1) = the (k+1)th Fibonacci number; sum{m=0, 1, 2, ..., 2^(k-1)-1} a(2^k + m) = 3^k (k>0).

a(37)=17 as it is the numerator of 17/5 = 3 + 1/(2 + 1/2), which is a continued fraction that can be derived from the binary expansion of 4*37 = 2^7 + 2^4 + 2^2; the binary exponents are {7, 4, 2}, thus the differences of these exponents are {3, 2, 2}; giving the continued fraction expansion of 17/5=[3,2,2]. Also 3^3 = a(2^3) + a(2^3 + 1) + a(2^3 + 2) + a(2^3 + 3) = 5 + 7 + 7 + 8.

Cf. A071766.

Sequence in context: A131282 A114544 A154726 this_sequence A106500 A120245 A120246

Adjacent sequences: A071582 A071583 A071584 this_sequence A071586 A071587 A071588

easy,nice,nonn,frac

Paul D. Hanna (pauldhanna(AT)juno.com), Jun 01 2002