OFFSET
1,2
COMMENTS
A245325(n)/a(n) enumerates all the reduced nonnegative rational numbers exactly once.
If the terms (n>0) are written as an array (in a left-aligned fashion) with rows of length 2^m, m = 0,1,2,3,...
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,
13,11,12,9,11,10,9,6,13,11,12,9,11,10,9,6,8,7,7,5,8,7,7,5,5,4,5,4,3,3,2,1,
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 for the first terms of those sequences.
If the rows are written in a right-aligned fashion:
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,
13,11,12,9,11,10,9,6,13,11,12,9,11,10,9,6,8,7,7,5,8,7,7,5,5,4,5,4,3,3,2,1,
then each column is constant and the terms are from A071585 (a(2^m-1-k) = A071585(k), k = 0,1,2,...).
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 the 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.
FORMULA
PROG
(R)
blocklevel <- 6 # arbitrary
a <- 1
for(m in 0:blocklevel) for(k in 0:(2^(m-1)-1)){
a[2^(m+1)+k] <- a[2^m+k] + a[2^m+2^(m-1)+k]
a[2^(m+1)+2^(m-1)+k] <- a[2^(m+1)+k]
a[2^(m+1)+2^m+k] <- a[2^m+k]
a[2^(m+1)+2^m+2^(m-1)+k] <- a[2^m+2^(m-1)+k]
}
a
(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
CROSSREFS
KEYWORD
nonn,frac
AUTHOR
Yosu Yurramendi, Jul 18 2014
STATUS
approved