A036214 -id:A036214 - cp's OEIS frontend

A036215 Binary reversal of 3^n.

1, 3, 9, 27, 69, 207, 621, 3345, 4275, 25497, 38247, 229173, 589185, 1669443, 5205897, 14045019, 34319397, 102566511, 307313949, 1843835217, 2312645619, 13776780249, 20417442711, 112792132341, 290155405761, 847524815523, 2611222884297, 7627711248315

Offset: 0

Antti Karttunen

Compute 3^n in binary, reverse the bits, from 0 to the most significant bit of the power.

Vincenzo Librandi, Table of n, a(n) for n = 0..1000

Cf. A036213 and A036214.

Table[FromDigits[Reverse[IntegerDigits[3^n, 2]], 2], {n, 0, 30}] (* Vincenzo Librandi, Sep 09 2013 *)

a(n) = subst(Polrev(binary(3^n)), x, 2); \\ Michel Marcus, Sep 08 2013

a(n) = A030101(A000244(n)). - Michel Marcus, Sep 08 2013

More terms from Michel Marcus, Sep 08 2013

A036213 Duplicating binary multipliers; i.e., n+1 1-bits placed 2n bits from each other.

Original entry on oeis.org

1, 5, 273, 266305, 4311810305, 1127000493261825, 4723519685917965029377, 316931994050834867150735294465, 340287559297026369749534115703797383169, 5846028850153881119687907085637645039610972340225, 1606939576755992644461949257743820820735113393327883823349761

Offset: 0

Antti Karttunen

A 2n-bit binary number can be reversed by multiplying it first by 2 and the n-th element of this sequence, masking it (bit and) with n-th element of A036214 and taking remainder of the division by (2^(2n + 2) - 1).

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

Vincenzo Librandi, Table of n, a(n) for n = 0..40
M. Beeler, R. W. Gosper, and R. Schroeppel, A Bit-Reversing Example in HAKMEM (Item 167).
A. Karttunen, A Simple C program Demonstrating Bit Reversals.

[1] cat [((2^((2*(n^2))+2*(n)))-1)/((2^(2*n))-1): n in [1..10]]; // Vincenzo Librandi, Aug 03 2017

Join[{1}, Table[((2^((2 (n^2)) + 2 (n))) - 1) / ((2^(2 n)) - 1), {n, 20}]] (* Vincenzo Librandi, Aug 03 2017 *)

a(n) = if (n==0, 1, ((2^((2*(n^2))+2*(n)))-1)/((2^(2*n))-1)) \\ Michel Marcus, Jun 07 2013

a(0) = 1, a(n) = (2^(2*n^2+2*n)-1) / (2^(2*n)-1).

cp's OEIS Frontend

A036215 Binary reversal of 3^n.

Views

Author

Keywords

Comments

Links

Crossrefs

Programs

Mathematica

PARI

Formula

Extensions