A038572 a(n) = n rotated one binary place to the right.
0, 1, 1, 3, 2, 6, 3, 7, 4, 12, 5, 13, 6, 14, 7, 15, 8, 24, 9, 25, 10, 26, 11, 27, 12, 28, 13, 29, 14, 30, 15, 31, 16, 48, 17, 49, 18, 50, 19, 51, 20, 52, 21, 53, 22, 54, 23, 55, 24, 56, 25, 57, 26, 58, 27, 59, 28, 60, 29, 61, 30, 62, 31, 63, 32, 96, 33, 97, 34, 98, 35, 99, 36, 100
Offset: 0
Examples
For n = 35, 35_10 = 100011_2, which after rotating one binary place to the right becomes 110001. Now, 110001_2 = 49_10. So, a(35) = 49. - _Indranil Ghosh_, Jan 21 2017
Links
- Indranil Ghosh, Table of n, a(n) for n = 0..20000 (first 1024 terms from T. D. Noe)
Programs
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Haskell
a038572 0 = 0 a038572 n = a053645 n * m + n' where (n', m) = divMod n 2 -- Reinhard Zumkeller, Dec 03 2012
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Maple
A038572 := proc(n) convert(n,base,2) ; ListTools[Rotate](%,1) ; add( op(i,%)*2^(i-1),i=1..nops(%)) ; end proc: # R. J. Mathar, May 20 2016
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Mathematica
Table[ FromDigits[ RotateRight[ IntegerDigits[n, 2]], 2], {n, 0, 80}] (* Robert G. Wilson v *) -
PARI
a(n)=if(n<2,return(n)); my(d=binary(n)); fromdigits(concat(d[#d], d[1..#d-1]),2) \\ Charles R Greathouse IV, Sep 02 2015
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Python
def A038572(n): x = bin(n)[2:] return int(x[-1]+x[:-1],2) # Indranil Ghosh, Jan 21 2017
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Python
def A038572(n): return (n>>1)+(1<
Formula
a(n) = A053645(n) * A000035(n) + A004526(n) = most significant bit(n) * least significant bit(n) + floor(n/2).
a(0)=0, a(1)=1, a(2n) = n, a(2n+1) = 2a(n) + 2a(n+1) - n. - Ralf Stephan, Oct 24 2003
Comments