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

1,3

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

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

Formula

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

Example

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.

Prog

(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

Crossrefs

Aside from the first term, subsequence of A174659.

Cf. A000120, A037301, A065363.

Sequence in context: A219956 A331276 A230597 * A124867 A199991 A161598

Adjacent sequences: A330901 A330902 A330903 * A330905 A330906 A330907

Keyword

nonn,base

Author

Thomas König, May 02 2020

Extensions

Offset corrected by Thomas König, Jul 09 2020

Status

approved