A297405 Binary "cubes"; numbers whose binary representation consists of three consecutive identical blocks.
7, 42, 63, 292, 365, 438, 511, 2184, 2457, 2730, 3003, 3276, 3549, 3822, 4095, 16912, 17969, 19026, 20083, 21140, 22197, 23254, 24311, 25368, 26425, 27482, 28539, 29596, 30653, 31710, 32767, 133152, 137313, 141474, 145635, 149796, 153957, 158118, 162279, 166440, 170601, 174762, 178923, 183084, 187245
Offset: 1
Examples
42 in base 2 is 101010, which consists of three copies of the block "10".
Links
- Robert Israel, Table of n, a(n) for n = 1..10000
- Daniel M. Kane, Carlo Sanna, and Jeffrey Shallit, Waring's Theorem for Binary Powers, arXiv:1801.04483 [math.NT], 2018.
- Index entries for sequences related to binary expansion of n
Programs
-
Maple
a:= n-> (p-> n*(1+2^p+4^p))(1+ilog2(n)): seq(a(n), n=1..50); # Alois P. Heinz, Dec 29 2017
-
Mathematica
bc[n_]:=FromDigits[Join[n,n,n],2]; Flatten[Table[bc/@Select[Tuples[ {1,0},n],#[[1]] == 1&],{n,6}]]//Union (* Harvey P. Dale, Oct 09 2021 *)
-
PARI
a(n) = n=binary(n); fromdigits(concat([n, n, n]) , 2) \\ Iain Fox, Jul 04 2022
-
Python
def a(n): return int(bin(n)[2:]*3, 2) print([a(n) for n in range(1, 46)]) # Michael S. Branicky, Jul 04 2022 # Alternative: def A297405(n): p = n.bit_length() return n * (1 + 2**p + 4**p) print([A297405(n) for n in range(1, 46)]) # Peter Luschny, Jul 05 2022
Formula
a(n) = n*(1 + 2^p + 4^p) with p = 1 + floor(log_2(n)). - Alois P. Heinz, Dec 29 2017
G.f.: (7*x + Sum_{n>=1} (4^n + 3*8^n + (2^n + 2*4^n - 3*8^n)*x)*x^(2^n))/(1-x)^2. - Robert Israel, Dec 31 2017
Comments