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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 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. 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: A326321 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 May 24 16:12 EDT 2024. Contains 372778 sequences. (Running on oeis4.)