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A354773
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For terms of A354169 that are the sum of two distinct powers of 2, the exponent of the smaller power of 2.
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7
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0, 2, 0, 1, 2, 3, 8, 0, 1, 5, 12, 2, 3, 7, 16, 8, 9, 0, 10, 1, 4, 11, 24, 12, 13, 2, 14, 3, 6, 15, 32, 16, 17, 8, 18, 9, 19, 0, 20, 10, 21, 1, 22, 4, 5, 11, 48, 24, 25, 12, 26, 13, 27, 2, 28, 14, 29, 3, 30, 6, 7, 15, 64, 32, 33, 16, 34, 17, 35, 8, 36, 18, 37, 9, 38, 19, 39, 0, 40, 20, 41, 10, 42, 21, 43, 1, 44, 22, 45, 4, 46
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OFFSET
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1,2
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LINKS
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Michael De Vlieger, Thomas Scheuerle, Rémy Sigrist, N. J. A. Sloane, and Walter Trump, The Binary Two-Up Sequence, arXiv:2209.04108 [math.CO], Sep 11 2022.
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FORMULA
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The following is a conjectured recurrence for a(n). Basically a(n) = a(n/2-1) if n is even, and a(n) = (n+1)/2 if n is odd, except that there are four types of n which have a different formula, and there are 19 exceptional values for small n. Note that a(n) does not depend on earlier values when n is odd.
Here is the formula, which agrees with the first 10000 terms.
There are exceptional values as far out as n=61, so we take care of them first.
Initial conditons:
If n is on the list
[1, 2, 3, 4, 5, 6, 7, 9, 10, 11, 13, 14, 15, 21, 22, 29, 30, 45, 61]
then a(n) is given by the n-th term of the following list:
[0, 2, 0, 1, 2, 3, 8, 0, 1, 5, 12, 2, 3, 7, 16, 8, 9, 0, 10,
1, 4, 11, 24, 12, 13, 2, 14, 3, 6, 15, 32, 16, 17, 8, 18, 9,
19, 0, 20, 10, 21, 1, 22, 4, 5, 11, 48, 24, 25, 12, 26, 13,
27, 2, 28, 14, 29, 3, 30, 6, 7].
Otherwise, if n is even, a(n) = a(n/2-1).
Otherwise n is odd and is not one of the exceptions.
(I) If n = 3*2^k-3, k >= 5, then a(n) = (n-1)/4.
(II) If n = 2^k-3, k >= 4 then a(n) = (n-1)/4.
(III) If n = 3*2^k-1, k >= 2 then a(n) = n+1.
(IV) If n = 2^k-1, k >= 3 then a(n) = n+1.
(V) Otherwise a(n) = (n+1)/2.
(End)
The conjecture is now known to be true. See De Vlieger et al. (2022). - N. J. A. Sloane, Aug 29 2022
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PROG
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(Python)
from itertools import count, islice
from collections import deque
from functools import reduce
from operator import or_
def A354773_gen(): # generator of terms
aset, aqueue, b, f = {0, 1, 2}, deque([2]), 2, False
while True:
for k in count(1):
m, j, j2, r, s = 0, 0, 1, b, k
while r > 0:
r, q = divmod(r, 2)
if not q:
s, y = divmod(s, 2)
m += y*j2
j += 1
j2 *= 2
if s > 0:
m += s*2**b.bit_length()
if m not in aset:
if (s := bin(m)[:1:-1]).count('1') == 2:
yield s.index('1')
aset.add(m)
aqueue.append(m)
if f: aqueue.popleft()
b = reduce(or_, aqueue)
f = not f
break
(C++) See Links section.
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CROSSREFS
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KEYWORD
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base,nonn
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AUTHOR
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STATUS
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approved
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