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A265210
Irregular triangle read by rows in which row n lists the base 3 digits of 2^n in reverse order, n >= 0.
3
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, 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, 0, 1, 1, 1, 1, 2, 2, 2, 1, 1, 2, 2, 2, 2, 1, 1
OFFSET
0,2
COMMENTS
The length of row n is A020915(n) = 1 + A136409(n).
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.
EXAMPLE
n
0: 1
1: 2
2: 1 1
3: 2 2
4: 1 2 1
5: 2 1 0 1
6: 1 0 1 2
7: 2 0 2 1 1
8: 1 1 1 0 0 1
9: 2 2 2 0 0 2
10: 1 2 2 1 0 1 1
11: 2 1 2 0 1 2 2
12: 1 0 2 1 2 1 2 1
13: 2 0 1 0 2 0 2 0 1
14: 1 1 2 0 1 1 1 1 2
15: 2 2 1 1 2 2 2 2 1 1
MATHEMATICA
(* Replace Flatten with Grid to display the triangle: *)
Flatten[Table[Reverse[IntegerDigits[2^n, 3]], {n, 0, 15}]]
PROG
(PARI) A265210_row(n)=Vecrev(digits(2^n, 3)) \\ M. F. Hasler, Dec 05 2015
CROSSREFS
Cf. A000079 (powers of 2), A004642 (powers of 2 written in base 3), A008776 (2*3^n).
Cf. A265209 (base 3 digits of 2^n).
Cf. A264980 (row n read as ternary number).
Cf. A037096 (numbers constructed from the inverse case, base 2 digits of 3^n).
Sequence in context: A210682 A293433 A177025 * A023396 A091221 A106495
KEYWORD
nonn,tabf,base
AUTHOR
L. Edson Jeffery, Dec 04 2015
STATUS
approved