cp's OEIS Frontend

This is a front-end for the Online Encyclopedia of Integer Sequences, made by Christian Perfect. The idea is to provide OEIS entries in non-ancient HTML, and then to think about how they're presented visually. The source code is on GitHub.

A348853 Delete any least significant 0's from the Zeckendorf representation of n, leaving its "odd" part.

Original entry on oeis.org

1, 1, 1, 4, 1, 6, 4, 1, 9, 6, 4, 12, 1, 14, 9, 6, 17, 4, 19, 12, 1, 22, 14, 9, 25, 6, 27, 17, 4, 30, 19, 12, 33, 1, 35, 22, 14, 38, 9, 40, 25, 6, 43, 27, 17, 46, 4, 48, 30, 19, 51, 12, 53, 33, 1, 56, 35, 22, 59, 14, 61, 38, 9, 64, 40, 25, 67, 6, 69, 43, 27, 72
Offset: 1

Views

Author

Kevin Ryde, Nov 14 2021

Keywords

Comments

Terms are odd Zeckendorfs A003622 and the fixed points are where n is odd already so that a(n) = n iff n is in A003622.
A139764(n) is the least significant "10..00" part of n so Zeckendorf multiplication n = A101646(a(n), A139764(n)).
The equivalent delete least significant 0's in binary is A000265 so that conversion to Fibbinary (A003714) and back gives a(n) = A022290(A000265(A003714(n))).
a(n) = 1 iff n is a Fibonacci number >= 1 (A000045) since they are Zeckendorf 100..00.
a(n) = 4 iff n is a Lucas number >= 4 (A000032) since they are Zeckendorf 10100..00 which reduces to 101.
In the Wythoff array A035513, a(n) is the term in column 0 of the row containing n, and hence the formula below using row number A019586 to select which of the odds (column 0) is a(n).

Examples

			n    = 81 = Zeckendorf 101001000.
a(n) = 19 = Zeckendorf 101001.
		

Crossrefs

Cf. A189920 (Zeckendorf digits), A003622 (odds), A003849 (final digit), A005206, A319433 (shift down).
Cf. A000045 (Fibonacci), A000032 (Lucas).
Cf. A035513 (Wythoff array), A019586 (row number).
Cf. A003714 (Fibbinary), A022290 (its inverse).
In other bases: A000265 (binary), A004151 (decimal).

Programs

  • PARI
    my(phi=quadgen(5)); a(n) = my(q,r); while([q,r]=divrem(n+2,phi); r<1, n=q-1); n;

Formula

a(n) = n if A003849(n)=1, otherwise a(n) = a(A005206(n)) = a(A319433(n)).
a(n) = A003622(A019586(n) + 1).
Sum_{k=1..n} a(k) ~ n^2/(2*phi), where phi is the golden ratio (A001622). - Amiram Eldar, Feb 17 2024