A273050 Numbers k such that ror(k) XOR rol(k) = k, where ror(x)=A038572(x) is x rotated one binary place to the right, rol(x)=A006257(x) is x rotated one binary place to the left, and XOR is the binary exclusive-or operator.
0, 5, 6, 45, 54, 365, 438, 2925, 3510, 23405, 28086, 187245, 224694, 1497965, 1797558, 11983725, 14380470, 95869805, 115043766, 766958445, 920350134, 6135667565, 7362801078, 49085340525, 58902408630, 392682724205, 471219269046
Offset: 1
Crossrefs
Programs
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Mathematica
ok[n_] := Block[{x = IntegerDigits[n, 2]}, x == BitXor @@@ Transpose@ {RotateLeft@ x, RotateRight@ x}]; Select[ Range[0, 10^5], ok] (* Giovanni Resta, May 14 2016 *) ok[n_] := Block[{x = IntegerDigits[n, 2]}, x == BitXor @@@ Transpose[ {RotateLeft[x], RotateRight[x]}]]; Select[LinearRecurrence[{0, 9, 0, -8}, {0, 5, 6, 45}, 100], ok] (* Jean-François Alcover, May 22 2016, after Giovanni Resta *) -
Python
def ROR(n): # returns A038572(n) BL = len(bin(n))-2 return (n>>1) + ((n&1) << (BL-1)) def ROL(n): # returns A006257(n) for n>0 BL = len(bin(n))-2 return (n*2) - (1<
Formula
Conjectures from Colin Barker, May 22 2016: (Start)
a(n) = (-11+(-1)^n+2^(-1/2+(3*n)/2)*(3-3*(-1)^n+5*sqrt(2)+5*(-1)^n*sqrt(2)))/14.
a(n) = 5*(2^(3*n/2)-1)/7 for n even.
a(n) = 3*(2^((3*n)/2-1/2)-2)/7 for n odd.
a(n) = 9*a(n-2)-8*a(n-4) for n>4.
G.f.: x^2*(5+6*x) / ((1-x)*(1+x)*(1-8*x^2)).
(End)
Extensions
a(19)-a(27) from Giovanni Resta, May 14 2016