OFFSET
0,1
COMMENTS
Based on a 2 for zero and three for one substitution, this apparently unique
Goedelization of the binary numbers uses the Fibonacci sequence to make
the digit ordering unique.
FORMULA
The sum of digits is weighted by the Fibonacci sequence to give:
a(n)=6^(sum[n]*Fibonacci[n])*2^(Count[n,2]+PosititionSum[n,2])*3^(Count[n,3]+PosititionSum[n,3])/3
MATHEMATICA
Table[6^(Sum[Table[((Reverse[IntegerDigits[n, 2]]) /. 0 -> 2) /. 1 -> 3, {n, 0, 50}][[n]][[ m]]*Fibonacci[m], {m, 1, Length[ Table[((Reverse[IntegerDigits[n, 2]]) /. 0 -> 2) /. 1 -> 3, {n, 0, 50}][[n]]]}])*2^(-(Count[ Table[((IntegerDigits[n, 2]) /. 0 -> 2) /. 1 -> 3, {n, 0, 50}][[n]], 2] + Apply[Plus, Flatten[ Position[Table[((Reverse[IntegerDigits[n, 2]]) /. 0 -> 2) /. 1 -> 3, {n, 0, 50}][[n]], 2]]]))*3^(-( Count[Table[((IntegerDigits[n, 2]) /. 0 -> 2) /. 1 -> 3, {n, 0, 50}][[ n]], 3] + Apply[Plus, Flatten[Position[Table[(( Reverse[IntegerDigits[n, 2]]) /. 0 -> 2) /. 1 -> 3, { n, 0, 50}][[n]], 3]]]))/3, {n, 1, 51}]
CROSSREFS
KEYWORD
nonn,uned
AUTHOR
Roger L. Bagula, Apr 28 2010
STATUS
approved