A330904 Numbers m such that the number of 1's in the binary expansion of m equals the sum of the balanced ternary trits of m.

0, 1, 10, 12, 13, 34, 36, 37, 66, 67, 120, 121, 192, 193, 202, 264, 265, 272, 273, 282, 283, 354, 355, 360, 361, 514, 516, 517, 520, 526, 544, 576, 577, 688, 840, 841, 848, 849, 904, 928, 1026, 1027, 1028, 1029, 1032, 1033, 1038, 1039, 1062, 1063, 1074, 1075

Offset: 1

Author

Thomas König, May 02 2020

Keywords

• nonn
• base

Comments

If a(n) mod 6 = 0, then a(n+1) = a(n)+1.

a(41) = 1026, a(42) = 1027, a(43) = 1028 and a(44) = 1029 is the first time that four consecutive numbers appear in a(n). Conjecture: There is no occurrence of five or more consecutive numbers in a(n). Tested by exhaustive search up to 3^30. - Thomas König, Jul 19 2020

Examples

			34_10 = 11T1_bt = 10010_2, the sum of the digits is 1+1-1+1 = 2 for balanced ternary and 1+1 = 2 for base 2, so 34 is a term.
		

Links

• Thomas König, Table of n, a(n) for n = 1..20000
• Thomas König, C program
• Thomas König, Fortran program to test the conjecture that there are at most four consecutive numbers in sequence

Crossrefs

Aside from the first term, subsequence of A174659.

Cf. A000120, A037301, A065363.

Programs

PARI

bt(n)= if (n==0, return (0)); my(d=digits(n, 3), c=1); while(c, if(d[1]==2, d=concat(0, d)); c=0; for(i=2, #d, if(d[i]==2, d[i]=-1; d[i-1]+=1; c=1))); vecsum(d); \\ A065363
isok(m) = bt(m) == hammingweight(m); \\ Michel Marcus, Jun 07 2020

Formula

Integers m such that A065363(m) = A000120(m).

Extensions

Offset corrected by Thomas König, Jul 09 2020