A324338 a(n) = A002487(1+A006068(n)).

1, 1, 1, 2, 1, 3, 3, 2, 1, 4, 5, 3, 4, 3, 2, 5, 1, 5, 7, 4, 7, 5, 3, 8, 5, 4, 3, 7, 2, 7, 8, 5, 1, 6, 9, 5, 10, 7, 4, 11, 9, 7, 5, 12, 3, 11, 13, 8, 6, 5, 4, 9, 3, 10, 11, 7, 2, 9, 12, 7, 11, 8, 5, 13, 1, 7, 11, 6, 13, 9, 5, 14, 13, 10, 7, 17, 4, 15, 18, 11, 11, 9, 7, 16, 5, 17, 19, 12, 3, 14, 19, 11, 18, 13, 8, 21, 7, 6, 5, 11, 4, 13, 14, 9, 3, 13

Offset: 0

Author

Antti Karttunen, Feb 23 2019

Keywords

• nonn

Comments

Like in A324337, a few terms preceding each 2^k-th term (here always 1) seem to consist of a batch of nearby Fibonacci numbers (A000045) in some order. For example, a(65533) = 987, a(65534) = 610 and a(65535) = 1597.

For all n > 0 A324338(n)/A324337(n) constitutes an enumeration system of all positive rationals. For all n > 0 A324338(n) + A324337(n) = A071585(n). - Yosu Yurramendi, Oct 22 2019

Links

• Antti Karttunen, Table of n, a(n) for n = 0..16384
• Antti Karttunen, Data supplement: n, a(n) computed for n = 0..65537
• Index entries for sequences related to binary expansion of n
• Index entries for sequences related to Stern's sequences

Crossrefs

Cf. A000045, A002487, A006068, A324288, A324337.

Programs

PARI

A006068(n)= { my(s=1, ns); while(1, ns = n >> s; if(0==ns, break()); n = bitxor(n, ns); s <<= 1; ); return (n); } \\ From A006068
A002487(n) = { my(s=sign(n), a=1, b=0); n = abs(n); while(n>0, if(bitand(n, 1), b+=a, a+=b); n>>=1); (s*b); };
A324338(n) = A002487(1+A006068(n));

R

maxlevel <- 6 # by choice #
b <- 0; A324338 <- 1; A324337 <- 1
for(i in 1:2^maxlevel) {
  b[2*i  ] <-     b[i]
  b[2*i+1] <- 1 - b[i]
  A324338[2*i  ] <- A324338[i]          + A324337[i]*   b[i]
  A324338[2*i+1] <- A324338[i]          + A324337[i]*(1-b[i])
  A324337[2*i  ] <- A324338[i]*(1-b[i]) + A324337[i]
  A324337[2*i+1] <- A324338[i]*   b[i]  + A324337[i]}
#
A324338[1:127]; A324337[1:127]
# Yosu Yurramendi, Oct 22 2019

Formula

a(n) = A002487(1+A006068(n)).

a(2^n) = 1 for all n >= 0.

From Yosu Yurramendi, Oct 22 2019: (Start)

a(2^m+2^(m-1)+k) = A324337(2^m+ k), m > 0, 0 <= k < 2^(m-1)

a(2^m+ k) = A324337(2^m+2^(m-1)+k), m > 0, 0 <= k < 2^(m-1). (End)

a(n) = A324337(A063946(n)), n > 0. Yosu Yurramendi, Nov 04 2019

a(n) = A002487(A233279(n)), n > 0. Yosu Yurramendi, Nov 08 2019

From Yosu Yurramendi, Nov 28 2019: (Start)

a(2^(m+1)+k) - a(2^m+k) = A324337(k), m >= 0, 0 <= k < 2^m.

a(A059893(2^(m+1)+A000069(k+1))) - a(A059893(2^m+A000069(k+1))) = A071585(k), m >= 1, 0 <= k < 2^(m-1).

a(A059893(2^m+ A001969(k+1))) = A071585(k), m >= 0, 0 <= k < 2^(m-1). (End)

From Yosu Yurramendi, Nov 29 2019: (Start)

For n > 0:

A324338(n) + A324337(n) = A071585(n).

A324338(2*A001969(n) )-A324337(2*A001969(n) ) = A071585(n-1)

A324338(2*A001969(n)+1)-A324337(2*A001969(n)+1) = -A324337(n-1)

A324338(2*A000069(n) )-A324337(2*A000069(n) ) = -A071585(n-1)

A324338(2*A000069(n)+1)-A324337(2*A000069(n)+1) = A324338(n-1) (End)

a(n) = A002487(A233279(n)). Yosu Yurramendi, Dec 27 2019