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%I #12 Mar 31 2022 16:59:57
%S 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,
%T 168,12,13,33,14,15,38,39,40,98,17,18,45,19,20,50,51,52,127,53,54,132,
%U 55,56,137,138,139,337,58,59,144,60,61,149,150,151,366,63,64
%N 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).
%C This sequence is to Pell numbers what A048680 is to Fibonacci numbers.
%C This sequence is a permutation of the nonnegative integers, with inverse A352417.
%H Rémy Sigrist, <a href="/A352416/b352416.txt">Table of n, a(n) for n = 0..10000</a>
%H A. F. Horadam, <a href="https://www.fq.math.ca/Scanned/32-3/horadam2.pdf">Maximal representations of positive integers by Pell numbers</a>, The Fibonacci Quarterly, Vol. 32, No. 3 (1994), pp. 240-244.
%H <a href="/index/Per#IntegerPermutation">Index entries for sequences that are permutations of the natural numbers</a>
%e For n = 7:
%e - the ternary expansion of 7 is "21",
%e - after appending 0's, we obtain "201",
%e - so a(7) = 2*A000129(2+1) + 0*A000129(1+1) + 1*A000129(0+1) = 2*5 + 1*1 = 11.
%o (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))) }
%Y Cf. A000129, A048680, A317204, A352417 (inverse).
%K nonn,base
%O 0,3
%A _Rémy Sigrist_, Mar 15 2022
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