A036213 - OEIS

%I #34 Jan 14 2024 04:36:48

%S 1,5,273,266305,4311810305,1127000493261825,4723519685917965029377,

%T 316931994050834867150735294465,

%U 340287559297026369749534115703797383169,5846028850153881119687907085637645039610972340225,1606939576755992644461949257743820820735113393327883823349761

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

%C 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).

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

%H Vincenzo Librandi, <a href="/A036213/b036213.txt">Table of n, a(n) for n = 0..40</a>

%H M. Beeler, R. W. Gosper, and R. Schroeppel, <a href="http://www.inwap.com/pdp10/hbaker/hakmem/hacks.html#item167">A Bit-Reversing Example in HAKMEM (Item 167)</a>.

%H A. Karttunen, <a href="/A036213/a036213.txt">A Simple C program Demonstrating Bit Reversals</a>.

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

%t Join[{1}, Table[((2^((2 (n^2)) + 2 (n))) - 1) / ((2^(2 n)) - 1), {n, 20}]] (* _Vincenzo Librandi_, Aug 03 2017 *)

%o (PARI) a(n) = if (n==0, 1, ((2^((2*(n^2))+2*(n)))-1)/((2^(2*n))-1)) \\ _Michel Marcus_, Jun 07 2013

%o (Magma) [1] cat [((2^((2*(n^2))+2*(n)))-1)/((2^(2*n))-1): n in [1..10]]; // _Vincenzo Librandi_, Aug 03 2017

%K nonn,base

%O 0,2

%A _Antti Karttunen_