A340161 a(n) is the smallest number k for which the set {k + 1, k + 2, ..., k + k} contains exactly n elements with exactly three 1-bits (A014311).

1, 4, 6, 7, 10, 11, 13, 18, 19, 21, 25, 34, 35, 37, 41, 49, 66, 67, 69, 73, 81, 97, 130, 131, 133, 137, 145, 161, 193, 258, 259, 261, 265, 273, 289, 321, 385, 514, 515, 517, 521, 529, 545, 577, 641, 769, 1026, 1027, 1029, 1033, 1041, 1057, 1089, 1153, 1281, 1537

Offset: 0

Author

Marius A. Burtea, Dec 30 2020

Keywords

• nonn
• easy
• base

Comments

When n = m*(m-1)/2 + 1, m >= 2 (A000124 \ {1}), then a(n) = k = 2^m+2, m >= 2 (A052548 \ {3, 4}), and only for these values of k, there exists only one set, {k+1, k+2, ..., 2k}, that contains exactly n elements whose binary representation has exactly three 1's (see A340068). - Bernard Schott, Jan 03 2021

From David A. Corneth, Jan 03 2021: (Start)

a(n) = A018900(n) + 1 for n >= 1.

Proof: Let T(k) be the number of values in {k+1, k+2, ..., k+k} that have exactly 3 ones in their binary expansion. Let h(k) be 1 if k has exactly 3 ones in its binary expansion and 0 otherwise and let w(k) be the binary weight of k (cf. A000120). Then T(k + 1) = T(k) + h(2*k + 1) + h(2*k + 2) - h(k + 1) = T(k) + h(2*k + 1) + h(2*(k + 1)) - h(k + 1) but as h(2^m * k) = h(k) two terms cancel and we have T(k + 1) = T(k) + h(2*k + 1). If w(2*k + 1) = w(k) + 1 = 3 then w(k) = 2 which holds for k in A018900. (End)

Examples

			For k in {1, 2, 3}, the sets are {1, 2}, {3, 4} and {4, 5, 6}, which do not contain numbers in A014311, so a(0) = 1.
For k = 4, the set is {5, 6, 7, 8} with 7 = A014311(1), so a(1) = 4.
For k = 6, the set {7, 8, 9, 10, 11, 12} contains the elements 7 = A014311(1) and 11 = A014311(2), so a(2) = 6.
		

Crossrefs

Essentially a duplicate of A018900 - N. J. A. Sloane, Jan 23 2021.

Cf. A014311, A340068.

Cf. also A000124, A052548 \ {3} (is a subsequence).

Programs

Magma

fb:=func; a:=[]; for n in [0..64] do k:=1; while #[s:s in [k+1..2*k]|fb(s)] ne n do k:=k+1; end while; Append(~a,k); end for; a;

PARI

first(n) = {my(res = vector(n), t = 1); res[1] = 1; for(i = 2, oo, if(hammingweight(2*i-1) == 3, t++; if(t > n, return(res)); res[t] = i))} \\ David A. Corneth, Jan 03 2021

Python

from math import isqrt, comb
def A340161(n): return 1+(1<<(m:=isqrt(n<<3)+1>>1))+(1<<(n-1-comb(m,2))) if n else 1 # Chai Wah Wu, Mar 10 2025

Formula

From Bernard Schott, Jan 03 2021: (Start)

a(m*(m-1)/2 + 1) = 2^m + 2 for m >= 2.

a(m*(m-1)/2 + 2) = 2^m + 3 for m >= 2.

a(n) = A018900(n) + 1 for n >= 1 (see A340068). (End)