A317204 -id:A317204 - cp's OEIS frontend

A286667 Positions of 1 in A286665; complement of A286666.

2, 4, 5, 7, 9, 10, 12, 14, 16, 17, 19, 21, 22, 24, 26, 28, 29, 31, 33, 34, 36, 38, 39, 41, 43, 45, 46, 48, 50, 51, 53, 55, 57, 58, 60, 62, 63, 65, 67, 68, 70, 72, 74, 75, 77, 79, 80, 82, 84, 86, 87, 89, 91, 92, 94, 96, 98, 99, 101, 103, 104, 106, 108, 109

Offset: 1

Clark Kimberling, May 13 2017

a(n) - a(n-1) is in {2,3} for n>=2.  Conjecture: a(n)/n -> 1 + sqrt(1/2).
This conjecture follows easily from the fact that (a(n)) is a Beatty sequence, see my comments in A286665. - Michel Dekking, Mar 11 2018
Numbers with a positive number of trailing 0's in their minimal representation in terms of the positive Pell numbers (A317204). - Amiram Eldar, Mar 16 2022

			As a word, A286665 = 010110101101010110101101..., in which 1 is in positions 2,4,5,7,9,...

Clark Kimberling, Table of n, a(n) for n = 1..10000

Cf. A171588, A286665, A286666, A317204.

s = Nest[Flatten[# /. {0 -> {0, 0, 1}, 1 -> {0}}] &, {0}, 6] (* A171588 *)
w = StringJoin[Map[ToString, s]]
w1 = StringReplace[w, {"0" -> "01"}]
st = ToCharacterCode[w1] - 48 ; (* A286665 *)
p0 = Flatten[Position[st, 0]];  (* A286666 *)
p1 = Flatten[Position[st, 1]];  (* A286667 *)

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).

Original entry on oeis.org

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

Rémy Sigrist, Mar 15 2022

This sequence is to Pell numbers what A048680 is to Fibonacci numbers.

This sequence is a permutation of the nonnegative integers, with inverse A352417.

			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.

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

Cf. A000129, A048680, A317204, A352417 (inverse).

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))) }

A352417 Inverse permutation to A352416.

Original entry on oeis.org

0, 1, 3, 4, 2, 9, 10, 12, 13, 5, 6, 7, 27, 28, 30, 31, 11, 36, 37, 39, 40, 14, 15, 16, 18, 19, 21, 22, 8, 81, 82, 84, 85, 29, 90, 91, 93, 94, 32, 33, 34, 108, 109, 111, 112, 38, 117, 118, 120, 121, 41, 42, 43, 45, 46, 48, 49, 17, 54, 55, 57, 58, 20, 63, 64, 66

Offset: 0

Rémy Sigrist, Mar 15 2022

			A352416(7) = 11, so a(11) = 7.

Cf. A000129, A317204, A352416.

a(n) = { my (p=[1, 2]); for (k=2, oo, if (n<=p[k], my (v=0, d); while (n, v+=3^k*d=n\p[k]; if (d==2, v/=3); n-=d*p[k]; k--); return (v/3), p = concat(p, 2*p[k]+p[k-1]))) }

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A286667 Positions of 1 in A286665; complement of A286666.

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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).

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A352417 Inverse permutation to A352416.

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