OFFSET
1,2
COMMENTS
From Davis Smith, May 09 2021: (Start)
For n > 2, a(n) cannot be a power of 2.
If A007088(n) (the binary expansion of n) contains a string of k zeros, then it contains A007088(2^m), where 0 <= m <= k, as a substring. Similarly, if A007088(n) contains a string of k ones, then it contains A007088(2^m - 1), where 1 <= m <= k. Strings of zeros and ones are the most compact way to have powers of 2 and powers of 2 minus 1 (respectively) as substrings in a binary expansion. This means that A007088(a(n)) will contain a string of A000523(n) ones and a string of A000523(n) zeros. The binary expansion of a(2^k - 1) will contain a string of k ones and a string of k - 1 zeros.
Conjecture: a(n) == 0 (mod A053644(n)), i.e., A007088(a(n)) ends with the longest string of zeros. It follows from this that a(2^k) = 2*a(2^k - 1). A conjecture related to this is that a(2^k - 1) = 2*a(2^k - 2) + 2^(k - 1), i.e., A007088(a(2^k - 1)) ends with the longest string of ones followed by the longest string of zeros. Ending with the longest string of ones followed by the longest string of zeros is not true for all A007088(a(n)), as some have a hiccup before starting their string of zeros, e.g., a(10), a(18), a(22), and a(34).
Conjecture: a(2^k + 1) = 2^(k + floor(log_2(a(2^k)))) + a(2^k), i.e., concatenate the binary expansion of 2^(k - 1) to the front of the binary expansion of a(2^k) in order to get the binary expansion of a(2^k + 1).
(End)
All terms belong to A261467. - Rémy Sigrist, May 11 2021
From Jon E. Schoenfield, Jun 03 2021: (Start)
Conjecture: the binary expansion of a(n) contains exactly ceiling(n/2) 1's iff 2^m - 7 <= n <= 2^m + 6 for some integer m >= 3. (See Links.)
Conjecture: for n > 1, the binary expansion of a(n) begins with that of 2^floor(log_2(n-1)) + 1.(End)
From Davis Smith, Jun 05 2021: (Start)
For a proof that a(n) == 2^floor(log_2(n)) (mod 2^(floor(log_2(n)) + 1)), see my second link (not the b-file). This also proves the conjecture from May 09 2021 which states that it is congruent to 0 (mod A053644(n)). A proof for the related conjecture would likely rely on an explanation of values of n such that a(n) is not congruent to (2^floor(log_2(n)) - 1)*2^floor(log_2(n)) (mod 2^(2*floor(log_2(n)))), i.e. the values of n such that A007088(a(n)) does not end with a string of floor(log_2(n)) ones followed immediately by a string of floor(log_2(n)) zeros. A proof for Jon E. Schoenfield's second conjecture on Jun 03 2021 would satisfy my more restricted second conjecture and it may follow necessarily from my proof, assuming that A007088(a(n)) must begin with either A007088(2^floor(log_2(n - 1)) + 1) or A007088(2^floor(log_2(n))). (End)
LINKS
Davis Smith, Table of n, a(n) for n = 1..64
David A. Corneth, substrings of a(33) and a(36) listed.
Jon E. Schoenfield, Conjecture on the number of 1's in the binary expansion of a(n).
Jon E. Schoenfield, Values for n = 1..64 in binary and decimal.
FORMULA
A144016(a(n)) >= n. - Rémy Sigrist, May 11 2021
EXAMPLE
a(6)=44 because 101100 (44 in base 2) is the smallest number that contains 1, 10, 11, 100, 101 and 110 (1 through 6 in base 2).
Terms begin as follows (see Links for a longer table):
.
a(n)
=========================
n decimal binary
-- ------- ----------------
1 1 1
2 2 10
3 6 110
4 12 1100
5 44 101100
6 44 101100
7 92 1011100
8 184 10111000
9 1208 10010111000
10 1256 10011101000
11 4792 1001010111000
12 4792 1001010111000
13 9912 10011010111000
14 9912 10011010111000
15 19832 100110101111000
16 39664 1001101011110000
PROG
(PARI)
A056744_vec(n)={
my(
L=List([1]), x=L[#L], Z=n+#L, B=binary(x),
A=setbinop((y, z)->fromdigits(B[y..z], 2), [1..#B])
);
while(#L<Z, while((#A<(#L+2))||(A[#L+2]!=#L+1),
B=binary(x++); A=setbinop((y, z)->fromdigits(B[y..z], 2), [1..#B])); listput(L, x)); Vec(L)
} \\ Davis Smith, May 09 2021
CROSSREFS
KEYWORD
base,nonn
AUTHOR
Fred J. Schalekamp, Aug 15 2000
EXTENSIONS
More terms from Naohiro Nomoto, Jul 20 2001
a(25)-a(31) from Ray Chandler, Nov 06 2008
a(32) from Davis Smith, May 10 2021
a(33) from Jon E. Schoenfield, May 11 2021
STATUS
approved