A352416 A permutation related to minimal Pell representations: append a 0 after each 2 in the ternary expansion of n, and then replace each place value, say 3^k with k >= 0, by A000129(k+1).
0, 1, 4, 2, 3, 9, 10, 11, 28, 5, 6, 16, 7, 8, 21, 22, 23, 57, 24, 25, 62, 26, 27, 67, 68, 69, 168, 12, 13, 33, 14, 15, 38, 39, 40, 98, 17, 18, 45, 19, 20, 50, 51, 52, 127, 53, 54, 132, 55, 56, 137, 138, 139, 337, 58, 59, 144, 60, 61, 149, 150, 151, 366, 63, 64
Offset: 0
Examples
For n = 7: - the ternary expansion of 7 is "21", - after appending 0's, we obtain "201", - so a(7) = 2*A000129(2+1) + 0*A000129(1+1) + 1*A000129(0+1) = 2*5 + 1*1 = 11.
Links
- Rémy Sigrist, Table of n, a(n) for n = 0..10000
- A. F. Horadam, Maximal representations of positive integers by Pell numbers, The Fibonacci Quarterly, Vol. 32, No. 3 (1994), pp. 240-244.
- Index entries for sequences that are permutations of the natural numbers
Programs
-
PARI
a(n) = { my (v=0, t=0, d); for (k=0, oo, if (n, d=n%3; n\=3; if (d==2, t++); if (d, v+=d*([2, 1; 1, 0]^(k+1+t))[2, 1]), return (v))) }
Comments