A169810 a(n) = n XOR n^2.
0, 0, 6, 10, 20, 28, 34, 54, 72, 88, 110, 114, 156, 164, 202, 238, 272, 304, 342, 378, 388, 428, 498, 518, 600, 616, 702, 706, 780, 852, 922, 990, 1056, 1120, 1190, 1258, 1332, 1404, 1410, 1494, 1640, 1720, 1742, 1810, 1980, 1988, 2154, 2190, 2352, 2384, 2550, 2586
Offset: 0
Examples
a(5) = 28: ..101 <- 5 11001 <- 25 ----- <- XOR 11100 -> 28
Links
- N. J. A. Sloane, Table of n, a(n) for n = 0..10000
Crossrefs
Programs
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Haskell
import Data.Bits (xor) a169810 n = n ^ 2 `xor` n :: Integer -- Reinhard Zumkeller, Dec 27 2012
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Maple
f:=proc(n) local i,t0,t1,t2,ts,tl,n1,n2; t1:=convert(n,base,2); t2:=convert(n^2,base,2); n1:=nops(t1); n2:=nops(t2); if n1 < n2 then ts:= t1; tl:=t2; else ts:=t2; tl:=t1; fi; t0:=[]; for i from 1 to nops(ts) do t0:=[op(t0), (ts[i] + tl[i]) mod 2 ]; od: for i from nops(ts)+1 to nops(tl) do t0:=[op(t0), tl[i]]; od: add(2^(i-1)*t0[i], i=1..nops(t0)); end; # second Maple program: a:= n-> Bits[Xor](n, n^2): seq(a(n), n=0..100); # Alois P. Heinz, Mar 29 2018
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Mathematica
a[n_]:=BitXor[n, n^2]; Array[a, 60, 0] (* Robert G. Wilson v, Jun 09 2010 *)
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PARI
A169810(n)=bitxor(n^2,n) \\ M. F. Hasler, May 07 2023
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Python
A169810=lambda n:n**2^n # M. F. Hasler, May 07 2023
Comments