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

0,2

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)

Links

Table of n, a(n) for n=0..55.

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)

Example

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.

Prog

(Magma) fb:=func<n|(Multiplicity(Intseq(n, 2), 1)) eq 3>; 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

Crossrefs

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

Cf. A014311, A340068.

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

Sequence in context: A026147 A032785 A096887 * A268574 A023631 A080641

Adjacent sequences: A340158 A340159 A340160 * A340162 A340163 A340164

Keyword

nonn,easy,base

Author

Marius A. Burtea, Dec 30 2020

Status

approved