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A036213 Duplicating binary multipliers, i.e., n+1 1-bits placed 2n bits from each other. 3
1, 5, 273, 266305, 4311810305, 1127000493261825, 4723519685917965029377, 316931994050834867150735294465, 340287559297026369749534115703797383169, 5846028850153881119687907085637645039610972340225, 1606939576755992644461949257743820820735113393327883823349761 (list; graph; refs; listen; history; text; internal format)
OFFSET

0,2

COMMENTS

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).

REFERENCES

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

LINKS

Vincenzo Librandi, Table of n, a(n) for n = 0..40

Beeler, M., Gosper, R. W. and Schroeppel, R., A Bit-Reversing Example in HAKMEM (Item 167)

A. Karttunen, A Simple C-program Demonstrating Bit-Reversals

FORMULA

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

MATHEMATICA

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

PROG

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

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

CROSSREFS

Sequence in context: A153322 A234324 A066210 * A262548 A112901 A213958

Adjacent sequences:  A036210 A036211 A036212 * A036214 A036215 A036216

KEYWORD

nonn,base

AUTHOR

Antti Karttunen

STATUS

approved

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Last modified December 13 16:57 EST 2017. Contains 295959 sequences.