Find next higher number with same number of 1's in binary form

AND

x&(-x) => gives last set bit from right .

ADDITION

x + ( x & - x ) => sets first unset bit after first continuous set bits from right to left.

XOR

x ^ ( x + ( x & - x ) ) => gives first continuous set bits from right to left with 1 extra set bit after continuous set bits.

DIVISION

( x ^ ( x + ( x & - x ) ) ) / ( x& -x ) => brings continuous set bits to right.

SHIFT RIGHT

( x ^ ( x + ( x & - x ) ) ) / ( x& -x ) >>2 => removes 1 extra set bits and 1 more as single 1 has to be at higher position.

OR

((x+(x&-x))|(((x^(x+(x&-x))))/(x&(-x))>>2)) => gives next higher number .