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%I #26 Feb 20 2023 03:40:15
%S 1,2,1,3,3,2,1,5,4,5,4,3,3,2,1,8,7,7,5,8,7,7,5,5,4,5,4,3,3,2,1,13,11,
%T 12,9,11,10,9,6,13,11,12,9,11,10,9,6,8,7,7,5,8,7,7,5,5,4,5,4,3,3,2,1,
%U 21,18,19,14,19,17,16,11,18,15,17,13,14,13,11,7,21,18,19,14,19,17,16,11,18,15,17,13,14,13,11,7,13,11,12,9,11
%N Denominators of an enumeration system of the reduced nonnegative rational numbers.
%C A245325(n)/a(n) enumerates all the reduced nonnegative rational numbers exactly once.
%C If the terms (n>0) are written as an array (left-aligned fashion) with rows of length 2^m, m = 0,1,2,3,...
%C 1,
%C 2, 1,
%C 3, 3, 2,1,
%C 5, 4, 5,4, 3, 3,2,1,
%C 8, 7, 7,5, 8, 7,7,5, 5, 4, 5,4, 3, 3,2,1,
%C 13,11,12,9,11,10,9,6,13,11,12,9,11,10,9,6,8,7,7,5,8,7,7,5,5,4,5,4,3,3,2,1,
%C then the sum of the m-th row is 3^m (m = 0,1,2,), and each column k is a Fibonacci sequence. These Fibonacci sequences are equal to Fibonacci sequences from A.... except the first terms of those sequences.
%C If the rows are written in a right-aligned fashion:
%C 1,
%C 2,1,
%C 3,3,2,1,
%C 5,4,5,4,3,3,2,1,
%C 8,7,7,5,8,7,7,5,5,4,5,4,3,3,2,1,
%C 13,11,12,9,11,10,9,6,13,11,12,9,11,10,9,6,8,7,7,5,8,7,7,5,5,4,5,4,3,3,2,1,
%C then each column is constant and the terms are from A071585 ( a(2^m-1-k) = A071585(k) , k = 0,1,2,...).
%C If the sequence is considered by blocks of length 2^m, m = 0,1,2,..., the blocks of this sequence are permutations of terms of blocks from A002487 (Stern's diatomic series or Stern-Brocot sequence), and, more precisely, the reverses of blocks of A071766 ( a(2^m+k) = A071766(2^(m+1)-1-k), m = 0,1,2,..., k = 0,1,2,...,2^m-1). Moreover, each block is the bit-reversed permutation of the corresponding block of A245328.
%H <a href="/index/Fo#fraction_trees">Index entries for fraction trees</a>
%F a(n) = A002487(1+A059893(A180200(n))) = A002487(A059893(A154435(n))). - _Yosu Yurramendi_, Sep 20 2021
%o (R)
%o blocklevel <- 6 # arbitrary
%o a <- 1
%o for(m in 0:blocklevel) for(k in 0:(2^(m-1)-1)){
%o a[2^(m+1)+k] <- a[2^m+k] + a[2^m+2^(m-1)+k]
%o a[2^(m+1)+2^(m-1)+k] <- a[2^(m+1)+k]
%o a[2^(m+1)+2^m+k] <- a[2^m+k]
%o a[2^(m+1)+2^m+2^(m-1)+k] <- a[2^m+2^(m-1)+k]
%o }
%o a
%o (PARI) a(n) = my(A=1); for(i=0, logint(n, 2), if(bittest(2*n, i), A++, A=(A+1)/A)); denominator(A) \\ _Mikhail Kurkov_, Feb 20 2023
%Y Cf. A245325, A002487, A071585, A071766, A273494.
%Y Cf. A002487, A059893, A154435.
%K nonn,frac
%O 1,2
%A _Yosu Yurramendi_, Jul 18 2014
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