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Repeating k-th ternary repunit (A003462) 2^k times, k >= 0.
4

%I #16 Nov 07 2024 21:58:45

%S 0,1,1,4,4,4,4,13,13,13,13,13,13,13,13,40,40,40,40,40,40,40,40,40,40,

%T 40,40,40,40,40,40,121,121,121,121,121,121,121,121,121,121,121,121,

%U 121,121,121,121,121,121,121,121,121,121,121,121,121,121,121,121,121,121,121

%N Repeating k-th ternary repunit (A003462) 2^k times, k >= 0.

%C a(n) is the greatest ternary repunit that is not greater than the n-th number with no 2 in ternary representation.

%F A032924(n) = a(n) + A107681(n);

%F A081604(A107681(n)) <= A081604(a(n)) = A081604(A032924(n)) = A000523(n+1).

%F a(n) = A003462(A000523(n+1)).

%e k=1: A003462(1) = (3^1-1)/2 = 1, therefore a(1) = a(2^1) = 1;

%e k=2: A003462(2) = (3^2-1)/2 = 4, therefore a(2+1) = a(2+2) =

%e a(2+3) = a(2+2^2) = 4.

%t With[{nn=5},Flatten[Table[#[[1]],{#[[2]]}]&/@Thread[{Table[FromDigits[ PadRight[{},n,1],3],{n,nn}],2^Range[nn]}]]] (* _Harvey P. Dale_, Jan 04 2013 *)

%o (PARI) apply( {A107680(n)=3^exponent(n+1)\2}, [0..66]) \\ _M. F. Hasler_, Jun 22 2020

%o (Python)

%o def A107680(n): return 3**((n+1).bit_length()-1)-1>>1 # _Chai Wah Wu_, Nov 07 2024

%Y Cf. A007089, A003462 (repunits in base 3), A000523 (number of digits in binary representation of n).

%K nonn,changed

%O 0,4

%A _Reinhard Zumkeller_, May 20 2005

%E Corrected by _T. D. Noe_, Oct 25 2006

%E Extended to a(0) = 0 by _M. F. Hasler_, Jun 23 2020