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Numbers which, written in base 256, have all digits less than 16 and no two adjacent digits equal.
15

%I #30 Mar 10 2023 13:33:14

%S 1,2,3,4,5,6,7,8,9,10,11,12,13,14,15,256,258,259,260,261,262,263,264,

%T 265,266,267,268,269,270,271,512,513,515,516,517,518,519,520,521,522,

%U 523,524,525,526,527,768,769,770,772,773,774

%N Numbers which, written in base 256, have all digits less than 16 and no two adjacent digits equal.

%C Sequence A033014 consists of the numbers that have all base 16 digits repeated *exactly* twice. (This is equivalent to say that the base-256 digits are 0x00, 0x11, 0x22,... or 0xFF, in hex notation, and no two adjacent base-256 digits are equal.) Thus, these numbers are divisible by 0x11 = 17, and the result of the division is a number which has no other base-256 digits than 0x00, 0x01,... or 0x0F, and no two adjacent digits equal. Conversely, it is clear that exactly these numbers are terms of A033014 when multiplied by 17 = 0x11. - _M. F. Hasler_, Feb 05 2014

%H Vincenzo Librandi, <a href="/A043320/b043320.txt">Table of n, a(n) for n = 1..1800</a>

%F a(n) = A033014(n)/17. [This was initially the definition of the sequence. - _M. F. Hasler_, Feb 03 2014]

%t Select[Range[20000], Union[Length/@Split[IntegerDigits[#, 16]]]=={2}&]/17 (* _Vincenzo Librandi_, Feb 06 2014 *)

%o (PARI) is_A043320(n)={(n=[n])&&!until(!n[1], ((n=divrem(n[1], 256))[2]<16 && n[1]%16!=n[2])||return)} \\ _M. F. Hasler_, Feb 03 2014

%o (Python)

%o from itertools import count, islice, groupby

%o def A043320_gen(startvalue=1): # generator of terms >= startvalue

%o return filter(lambda n:set(len(list(g)) for k, g in groupby(hex(17*n)[2:]))=={2},count(max(startvalue,1)))

%o A043320_list = list(islice(A043320_gen(),20)) # _Chai Wah Wu_, Mar 10 2023

%Y Cf. A043307 - A043319, A043291, A033001 - A033014, A033015 - A033029.

%K nonn,base

%O 1,2

%A _Clark Kimberling_

%E New definition by _M. F. Hasler_, Feb 03 2014