A264980 -id:A264980 - cp's OEIS frontend

A263273 Bijective base-3 reverse: a(0) = 0; for n >= 1, a(n) = A030102(A038502(n)) * A038500(n).

0, 1, 2, 3, 4, 7, 6, 5, 8, 9, 10, 19, 12, 13, 22, 21, 16, 25, 18, 11, 20, 15, 14, 23, 24, 17, 26, 27, 28, 55, 30, 37, 64, 57, 46, 73, 36, 31, 58, 39, 40, 67, 66, 49, 76, 63, 34, 61, 48, 43, 70, 75, 52, 79, 54, 29, 56, 33, 38, 65, 60, 47, 74, 45, 32, 59, 42, 41, 68, 69, 50, 77, 72, 35, 62, 51, 44, 71, 78, 53, 80, 81

Offset: 0

Antti Karttunen, Dec 05 2015

Here the base-3 reverse has been adjusted so that the maximal suffix of trailing zeros (in base-3 representation A007089) stays where it is at the right side, and only the section from the most significant digit to the least significant nonzero digit is reversed, thus making this sequence a self-inverse permutation of nonnegative integers.
Because successive powers of 3 and 9 modulo 2, 4 and 8 are always either constant 1, 1, 1, ... or alternating 1, -1, 1, -1, ... it implies similar simple divisibility rules for 2, 4 and 8 in base 3 as e.g. 3, 9 and 11 have in decimal base (see the Wikipedia-link). As these rules do not depend on which direction they are applied from, it means that this bijection preserves the fact whether a number is divisible by 2, 4 or 8, or whether it is not. Thus natural numbers are divided to several subsets, each of which is closed with respect to this bijection. See the Crossrefs section for permutations obtained from these sections.
When polynomials over GF(3) are encoded as natural numbers (coefficients presented with the digits of the base-3 expansion of n), this bijection works as a multiplicative automorphism of the ring GF(3)[X]. This follows from the fact that as there are no carries involved, the multiplication (and thus also the division) of such polynomials could be as well performed by temporarily reversing all factors (like they were seen through mirror). This implies also that the sequences A207669 and A207670 are closed with respect to this bijection.

			For n = 15, A007089(15) = 120. Reversing this so that the trailing zero stays at the right yields 210 = A007089(21), thus a(15) = 21 and vice versa, a(21) = 15.

Antti Karttunen, Table of n, a(n) for n = 0..6561
Wikipedia, Divisibility rule
Index entries for sequences that are permutations of the natural numbers

Bisections: A264983, A264984.
Cf. A000035, A007089, A010873, A030102, A038500, A038502, A207669, A207670.
Cf. also A057889, A264965, A264966, A264980.
Permutations induced by various sections: A263272 (a(2n)/2), A264974 (a(4n)/4), A264978 (a(8n)/8), A264985, A264989.
Cf. also A004488, A140263, A140264, A246207, A246208 (other base-3 related permutations).

r[n_] := FromDigits[Reverse[IntegerDigits[n, 3]], 3]; b[n_] := n/ 3^IntegerExponent[n, 3]; c[n_] := n/b[n]; a[0]=0; a[n_] := r[b[n]]*c[n]; Table[a[n], {n, 0, 80}] (* Jean-François Alcover, Dec 29 2015 *)

from sympy import factorint
from sympy.ntheory.factor_ import digits
from operator import mul
def a030102(n): return 0 if n==0 else int(''.join(map(str, digits(n, 3)[1:][::-1])), 3)
def a038502(n):
    f=factorint(n)
    return 1 if n==1 else reduce(mul, [1 if i==3 else i**f[i] for i in f])
def a038500(n): return n/a038502(n)
def a(n): return 0 if n==0 else a030102(a038502(n))*a038500(n) # Indranil Ghosh, May 22 2017

(define (A263273 n) (if (zero? n) n (* (A030102 (A038502 n)) (A038500 n))))

a(0) = 0; for n >= 1, a(n) = A030102(A038502(n)) * A038500(n).
Other identities. For all n >= 0:
a(3*n) = 3*a(n).
A000035(a(n)) = A000035(n). [This permutation preserves the parity of n.]
A010873(a(n)) = 0 if and only if A010873(n) = 0. [See the comments section.]

A265345 Square array A(row,col): For row=0, A(0,col) = A265341(col), for row > 0, A(row,col) = A265342(A(row-1,col)).

Original entry on oeis.org

1, 3, 2, 7, 6, 4, 5, 10, 12, 8, 9, 22, 20, 24, 16, 21, 18, 28, 40, 48, 64, 13, 30, 36, 56, 80, 192, 32, 19, 26, 60, 72, 112, 160, 96, 184, 25, 14, 52, 120, 144, 224, 640, 552, 352, 11, 46, 76, 208, 240, 576, 448, 320, 1056, 704, 15, 58, 68, 136, 104, 480, 288, 1720, 1600, 2112, 1408

Offset: 1

Antti Karttunen, Dec 18 2015

Square array A(row,col) is read by downwards antidiagonals as: A(0,0), A(0,1), A(1,0), A(0,2), A(1,1), A(2,0), A(0,3), A(1,2), A(2,1), A(3,0), ...
All the terms in the same column are either all divisible by 3, or none of them are.
Reducing A265342 to its constituent sequences gives A265342(n) = A263273(2*A263273(n)). Iterating this function k times starting from n reduces to (because A263273 is an involution, so pairs of them are canceled) to A263273((2^k)*A263273(n)).

			The top left corner of the array:
    1,    3,    7,    5,    9,   21,   13,   19,   25,   11,   15,    39, .
    2,    6,   10,   22,   18,   30,   26,   14,   46,   58,   66,    78, .
    4,   12,   20,   28,   36,   60,   52,   76,   68,   44,   84,   156, .
    8,   24,   40,   56,   72,  120,  208,  136,   88,  232,  168,   624, .
   16,   48,   80,  112,  144,  240,  104,  200,  496,  424,  336,   312, .
   64,  192,  160,  224,  576,  480,  520,  256,  344,  608,  672,  1560, .
   32,   96,  640,  448,  288, 1920, 1144,  512, 1984,  736, 1344,  3432, .
  184,  552,  320, 1720, 1656,  960, 2072, 1024, 1376, 4384, 5160,  6216, .
  352, 1056, 1600,  824, 3168, 4800, 3712, 6040, 5344, 2936, 2472, 11136, .
  ...

Inverse: A265346.
Transpose: A265347.
Leftmost column: A264980.
Topmost row: A265341.
Row index: A265330 (zero-based), A265331 (one-based).
Column index: A265910 (zero-based), A265911 (one-based).
Cf. also A265342.
Related permutations: A263273, A265895.

(define (A265345 n) (A265345bi (A002262 (+ -1 n)) (A025581 (+ -1 n)))) ;; o=1.
(define (A265345bi row col) (A263273 (* (A000079 row) (A263273 (A265341 col))))) ;; Faster than below.
(define (A265345bi row col) (if (= 0 row) (A265341 col) (A265342 (A265345bi (- row 1) col)))) ;; row>=0, col>=0.

For row=0, A(0,col) = A265341(col), for row>0, A(row,col) = A265342(A(row-1,col)).

A(row, col) = A263273((2^row) * A263273(A265341(col))). [The above reduces to this.]

A265342 Permutation of even numbers: a(n) = 2 * A265351(n).

Original entry on oeis.org

0, 2, 4, 6, 8, 22, 12, 10, 16, 18, 20, 58, 24, 26, 76, 66, 64, 70, 36, 14, 40, 30, 28, 34, 48, 46, 52, 54, 56, 166, 60, 62, 184, 174, 172, 178, 72, 74, 220, 78, 80, 238, 228, 226, 232, 198, 68, 202, 192, 190, 196, 210, 208, 214, 108, 38, 112, 42, 44, 130, 120, 118, 124, 90, 32, 94, 84, 82, 88, 102, 100, 106, 144

Offset: 0

Antti Karttunen, Dec 07 2015

nonn

Iterating this sequence as 1, a(1), a(a(1)), a(a(a(1))), ... yields A264980.

Antti Karttunen, Table of n, a(n) for n = 0..9841

Cf. A265351.

Cf. also A265341, A263273, A264980.

from sympy import factorint
from sympy.ntheory.factor_ import digits
from operator import mul
def a030102(n): return 0 if n==0 else int(''.join(map(str, digits(n, 3)[1:][::-1])), 3)
def a038502(n):
    f=factorint(n)
    return 1 if n==1 else reduce(mul, [1 if i==3 else i**f[i] for i in f])
def a038500(n): return n/a038502(n)
def a263273(n): return 0 if n==0 else a030102(a038502(n))*a038500(n)
def a263272(n): return a263273(2*n)/2
def a(n): return 2*a263272(a263273(n)) # Indranil Ghosh, May 25 2017

Scheme
```
(define (A265342 n) (* 2 (A265351 n)))
```

a(n) = 2 * A265351(n).

A265210 Irregular triangle read by rows in which row n lists the base 3 digits of 2^n in reverse order, n >= 0.

Original entry on oeis.org

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

L. Edson Jeffery, Dec 04 2015

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.

			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

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).

(* Replace Flatten with Grid to display the triangle: *)
Flatten[Table[Reverse[IntegerDigits[2^n, 3]], {n, 0, 15}]]

A265210_row(n)=Vecrev(digits(2^n,3)) \\ M. F. Hasler, Dec 05 2015

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A263273 Bijective base-3 reverse: a(0) = 0; for n >= 1, a(n) = A030102(A038502(n)) * A038500(n).

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A265345 Square array A(row,col): For row=0, A(0,col) = A265341(col), for row > 0, A(row,col) = A265342(A(row-1,col)).

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A265342 Permutation of even numbers: a(n) = 2 * A265351(n).

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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.

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