A265747 Numbers written in Jacobsthal greedy base.

0, 1, 2, 10, 11, 100, 101, 102, 110, 111, 200, 1000, 1001, 1002, 1010, 1011, 1100, 1101, 1102, 1110, 1111, 10000, 10001, 10002, 10010, 10011, 10100, 10101, 10102, 10110, 10111, 10200, 11000, 11001, 11002, 11010, 11011, 11100, 11101, 11102, 11110, 11111, 20000, 100000, 100001, 100002, 100010, 100011, 100100

Offset: 0

Author

Antti Karttunen, Dec 17 2015

Keywords

• nonn
• base

Comments

These are called "Jacobsthal Representation Numbers" in Horadam's 1996 paper.

Sum_{i=0..} digit(i)*A001045(2+digit(i)) recovers n from such representation a(n), where digit(0) stands for the least significant digit (at the right), and A001045(k) gives the k-th Jacobsthal number.

No larger digits than 2 will occur, which allows representing the same sequence in a more compact form by base-3 coding in A265746.

Sequence A197911 gives the terms with no digit "2" in their representation, while its complement A003158 gives the terms where "2" occurs at least once.

Numbers beginning with digit "2" in this representation are given by A020988(n) [= 2*A002450(n) = 2*A001045(2n)].

Examples

			For n=7, when selecting the terms of A001045 with the greedy algorithm, we need terms A001045(4) + A001045(2) + A001045(2) = 5 + 1 + 1, thus a(7) = "102".
For n=10, we need A001045(4) + A001045(4) = 5+5, thus a(10) = "200".
		

Links

• Antti Karttunen, Table of n, a(n) for n = 0..10923
• A. F. Horadam, Jacobsthal Representation Numbers, Fib Quart. 34 (1996), 40-54. (See especially page 50, which is page 11 in PDF.)

Crossrefs

Cf. A001045, A002450, A020988, A007089, A197911, A003158.

Cf. A265745 (sum of digits).

Cf. A265746 (same numbers interpreted in base-3, then shown in decimal).

Cf. A084639 (positions of repunits).

Cf. A007961, A014417, A014418, A244159 for analogous sequences.

Programs

Mathematica

jacob[n_] := (2^n - (-1)^n)/3; maxInd[n_] := Floor[Log2[3*n + 1]]; A265747[n_] := A265747[n] = 10^(maxInd[n] - 2) + A265747[n - jacob[maxInd[n]]]; A265747[0] = 0; Array[A265747, 100, 0] (* Amiram Eldar, Jul 21 2023 *)

PARI

A130249(n) = floor(log(3*n + 1) / log(2));
A001045(n) = (2^n - (-1)^n) / 3;
A265747(n) = {if(n==0, 0, my(d=n - A001045(A130249(n))); 10^(A130249(n)-2) + if(d == 0, 0, A265747(d)));} \\ Amiram Eldar, Jul 21 2023

Python

def greedyJ(n): m = (3*n+1).bit_length() - 1; return (m, (2**m-(-1)**m)//3)
def a(n):
    if n == 0: return 0
    place, value = greedyJ(n)
    return 10**(place-2) + a(n - value)
print([a(n) for n in range(49)]) # Michael S. Branicky, Jul 11 2021

Formula

a(0) = 0; for n >= 1, a(n) = 10^(A130249(n)-2) + a(n-A001045(A130249(n))).

a(n) = A007089(A265746(n)).