A036215 -id:A036215 - cp's OEIS frontend

A246200 Self-inverse permutation of natural numbers: a(n) = A057889(3*n) / 3.

0, 1, 2, 3, 4, 5, 6, 7, 8, 9, 10, 11, 12, 19, 14, 15, 16, 17, 18, 13, 20, 21, 22, 27, 24, 35, 38, 23, 28, 39, 30, 31, 32, 33, 34, 25, 36, 41, 26, 29, 40, 37, 42, 43, 44, 75, 54, 59, 48, 67, 70, 51, 76, 83, 46, 55, 56, 71, 78, 47, 60, 79, 62, 63, 64, 65, 66, 49, 68, 81, 50, 57, 72, 73, 82, 45, 52, 77, 58, 61, 80, 69

Offset: 0

Antti Karttunen, Aug 27 2014

In binary system, 3 ("11" in binary), has a similar shortcut rule for divisibility as eleven has in decimal system. This rule doesn't depend on which end of the number representation it is applied from, thus, if we reverse the number 3*n with "balanced bit-reverse" (A057889), the result should still be divisible by 3. Moreover, because the reversing operation is itself a self-inverse involution, and the prime factorization of any natural number is unique, we get a self-inverse permutation of nonnegative integers when we divide the bit-reversed result with 3.

Antti Karttunen, Table of n, a(n) for n = 0..10921
Index entries for sequences that are permutations of the natural numbers

Cf. A036215, A057889, A003714, A048724, A083822, A083824.

def a057889(n):
    x=bin(n)[2:]
    y=x[::-1]
    return int(str(int(y))+(len(x) - len(str(int(y))))*'0', 2)
def a(n): return a057889(3*n)//3
print([a(n) for n in range(101)]) # Indranil Ghosh, Jun 11 2017

(define (A246200 n) (/ (A057889 (* 3 n)) 3))

a(n) = A057889(3*n) / 3.

A264980 Base-3 reversal of 2^n: a(n) = A030102(A000079(n)).

Original entry on oeis.org

1, 2, 4, 8, 16, 64, 32, 184, 352, 704, 1408, 1880, 2824, 14032, 10328, 56128, 100576, 145784, 189472, 370304, 731752, 4388248, 2924096, 11175712, 15965704, 31930448, 63861880, 383165344, 255439712, 1021772344, 510875648, 2550188248, 5619691648, 9689861048, 17830350904, 79068724264, 34109913224, 192259976368, 133338241880

Offset: 0

Antti Karttunen, Dec 05 2015

			2^5 = 32 in base 3 = "1012" (= A007089(32)) as 1*27 + 1*3 + 2*1 = 32. 2^6 = 64 in base 3 = "2101" as 2*27 + 1*9 + 1*1 = 64. "1012" reversed is "2101" and vice versa, thus a(5) = 64 and a(6) = 32.

Antti Karttunen, Table of n, a(n) for n = 0..512

Leftmost column of A265345.
Cf. A000079, A007089, A030102, A263273, A265210, A265342.
Cf. also A036215.

base(n) = {my(a=[n%3]); while(0Altug Alkan, Dec 29 2015

a(n) = A030102(A000079(n)) = A263273(A000079(n)).

a(0) = 1, for n >= 1, a(n) = A265342(a(n-1)).

A036214 Bit-reversing masks for 2*n bits.

Original entry on oeis.org

0, 18, 4740, 17966088, 1136090292240, 1171507928472027168, 19496308761789043518734400, 5212738348288268369644435170918528, 22344471816287582119092726913736555148345600, 1533995044405866391626076022957811770200509055768723968

Offset: 0

Antti Karttunen

R. Schroeppel: DECsystem-10/20 Processor Reference Manual AA-H391A-TK, Chapter 2, User Operations, section 2.15: Programming Examples: Reversing Order of Digits.

M. Beeler, R. W. Gosper, and R. Schroeppel, A Bit-Reversing Example in HAKMEM (Item 167).
Bitsavers, DECsystem-10 DECSYSTEM-20 Processor Reference Manual AA-H391A-TK, Fifth Edition, July 1980 (Updated, June 1982). (Section 2.15, Programming Examples, example "Reversing Order of Digits" on p. 2-116, page 177 of PDF)
Antti Karttunen, A Simple C program Demonstrating Bit Reversals.
Index entries for sequences related to binary expansion of n.

Cf. A036213, A036215.

Table[2^n*(2^(2*n^2 + 3*n + 1) + 2^(2*n^2 + 2*n) - 2^(3*n + 1) - 1)/(2^(2*n + 1) - 1), {n, 0, 10}] (* Wesley Ivan Hurt, Jun 10 2024 *)

A036214(n) = 2^n * ( 2^(2*n^2+3*n+1) + 2^(2*n^2+2*n) - 2^(3*n+1) - 1 ) / (2^(2*n+1) - 1); \\ Antti Karttunen, Jan 14 2024

a(n) = 2^n * ( 2^(2*n^2+3*n+1) + 2^(2*n^2+2*n) - 2^(3*n+1) - 1 ) / (2^(2*n+1) - 1).

log(a(n)) ~ log(4) * n ^ 2. - Bill McEachen, Jul 13 2024

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A246200 Self-inverse permutation of natural numbers: a(n) = A057889(3*n) / 3.

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A264980 Base-3 reversal of 2^n: a(n) = A030102(A000079(n)).

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A036214 Bit-reversing masks for 2*n bits.

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