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%I #18 Dec 06 2015 11:17:57
%S 1,2,1,1,2,2,1,2,1,2,1,0,1,1,0,1,2,2,0,2,1,1,1,1,1,0,0,1,2,2,2,0,0,2,
%T 1,2,2,1,0,1,1,2,1,2,0,1,2,2,1,0,2,1,2,1,2,1,2,0,1,0,2,0,2,0,1,1,1,2,
%U 0,1,1,1,1,2,2,2,1,1,2,2,2,2,1,1
%N Irregular triangle read by rows in which row n lists the base 3 digits of 2^n in reverse order, n >= 0.
%C The length of row n is A020915(n) = 1 + A136409(n).
%C Conjecture 1: The sequence in column k is periodic, with period p(k) = 2*3^(k-1) = A008776(k-1), k >= 1, and in which the numbers 0,1,2 appear with equal frequency, for each k>1.
%e n
%e 0: 1
%e 1: 2
%e 2: 1 1
%e 3: 2 2
%e 4: 1 2 1
%e 5: 2 1 0 1
%e 6: 1 0 1 2
%e 7: 2 0 2 1 1
%e 8: 1 1 1 0 0 1
%e 9: 2 2 2 0 0 2
%e 10: 1 2 2 1 0 1 1
%e 11: 2 1 2 0 1 2 2
%e 12: 1 0 2 1 2 1 2 1
%e 13: 2 0 1 0 2 0 2 0 1
%e 14: 1 1 2 0 1 1 1 1 2
%e 15: 2 2 1 1 2 2 2 2 1 1
%t (* Replace Flatten with Grid to display the triangle: *)
%t Flatten[Table[Reverse[IntegerDigits[2^n, 3]], {n, 0, 15}]]
%o (PARI) A265210_row(n)=Vecrev(digits(2^n,3)) \\ _M. F. Hasler_, Dec 05 2015
%Y Cf. A000079 (powers of 2), A004642 (powers of 2 written in base 3), A008776 (2*3^n).
%Y Cf. A265209 (base 3 digits of 2^n).
%Y Cf. A264980 (row n read as ternary number).
%Y Cf. A037096 (numbers constructed from the inverse case, base 2 digits of 3^n).
%K nonn,tabf,base
%O 0,2
%A _L. Edson Jeffery_, Dec 04 2015
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