cp's OEIS Frontend

a(n) = A006997(3^n-1).

It appears that, aside from the first term, this is the (L)-sieve transform of A016789 ={2,5,8,11,...,3n+2....}. This has been verified up to a(30)=311071. See A152009 for the definition of the (L)-sieve transform. - John W. Layman, Nov 20 2008

a(n) is also the largest-index square reachable in n jumps if we start at square 0 of the Infinite Sidewalk. - Jose Villegas, Mar 27 2023

The RUNS transform of this sequence appears to be yet another version of the Fibonacci word (cf. A001468, A001950, A076662). - N. J. A. Sloane, Mar 29 2025

Fibonacci base shift right: for n >= 0, a(n+1) = Sum_{k in A_n} F_{k-1}, where n = Sum_{k in A_n} F_k (unique) expression of n as a sum of "noncontiguous" Fibonacci numbers (with index >=2). - Michele Dondi (bik.mido(AT)tiscalenet.it), Dec 30 2001 [corrected, and aligned with sequence offset by Peter Munn, Jan 10 2018]

Numerators a(n) of fractions slowly converging to phi, the golden ratio: let a(1) = 0, b(n) = n - a(n); if (a(n) + 1) / b(n) < (1 + sqrt(5))/2, then a(n+1) = a(n) + 1, else a(n+1) = a(n). a(n) + b(n) = n and as n -> +infinity, a(n) / b(n) converges to (1 + sqrt(5))/2. For all n, a(n) / b(n) < (1 + sqrt(5))/2. - Robert A. Stump (bee_ess107(AT)msn.com), Sep 22 2002

a(10^n) gives the first few digits of phi=(sqrt(5)-1)/2.

Comment corrected, two alternative ways, by Peter Munn, Jan 10 2018: (Start)

(a(n) = a(n+1) or a(n) = a(n-1)) if and only if a(n) is in A066096.

a(n+1) = a(n+2) if and only if n is in A003622.

(End)

From Wolfdieter Lang, Jun 28 2011: (Start)

a(n+1) counts for n >= 1 the number of Wythoff A-numbers not exceeding n.

a(n+1) counts also the number of Wythoff B-numbers smaller than A(n+2), with the Wythoff A- and B-sequences A000201 and A001950, respectively.

a(n+1) = Sum_{j=1..n} A005614(j-1) for n >= 1 (no rounding problems like in the above definition, because the rabbit sequence A005614(n-1) for n >= 1, can be defined by a substitution rule).

a(n+1) = A(n+1)-(n+1) (serving, together with the last equation, as definition for A(n+1), given the rabbit sequence).

a(n+1) = A005206(n), n >= 0.

(End)

Let b(n) = floor((n+1)/phi). Then b(n) + b(b(n-1)) = n [Granville and Rasson]. - N. J. A. Sloane, Jun 13 2014

Most of the natural numbers are members. Conjecture: there are infinitely many nonmembers. Is there an estimate for a(k)/k ?

A118164(n) = number of representations of a(n) as sum of consecutive earlier terms. - Reinhard Zumkeller, Apr 13 2006

F(n) is not equal to M(n) if and only if n+1 is a Fibonacci number (A000045); a(n) = A005379(n) + A192687(n). - Reinhard Zumkeller, Jul 12 2011

Differs from A098294 at indices n = 0, 17, 20, 22, 25, 27, 29, 30, ... - M. F. Hasler, Jun 29 2014

M(n) is not equal to F(n) if and only if n+1 is a Fibonacci number (A000045); a(n)=A005379(n)-A192687(n). [Reinhard Zumkeller, Jul 12 2011]

Equals its own "second derivative" (cf. A006337).

Presumably this is the same as the following sequence from Hofstadter's book: the number of triangular numbers between each successive pair of squares. More precisely, a(n) is the number of triangular numbers T such that n^2 <= T < (n+1)^2. E.g., a(3) = 2 because 3^2 <= T < 4^2 permits T(4) = 10 and T(5) = 15 and no other triangular number. - Hugo van der Sanden, May 03 2005.

a(n) = A214848(n) = A022846(n+1) - A022846(n). - Reinhard Zumkeller, Mar 03 2014

A003056(n) gives the number of odd terms in the first n terms of this sequence. Modulo 2, this sequence becomes A023531. - T. D. Noe, Jul 24 2007

The present definition was the original definition of this sequence. It was later changed to "Sequence formed from rows of triangle A046936", but this seems less satisfactory. - N. J. A. Sloane, Oct 26 2014

Equals its own "derivative", which is formed by counting the strings of 1's that lie between 2's.

Conjecture: A006340 = continued fraction expansion of (2.729967741... = sup{f(n,1)}), where f(1,x) = x + 1 and thereafter f(n,x) = x + 1 if n is in the lower Wythoff sequence (A000201), else f(n,x) = 1/x. The first 12 values of f(n,1) are given in Example at A245216. - Clark Kimberling, Jul 14 2014

From Michel Dekking, Mar 05 2018: (Start)

The description of this sequence is not correct, since the derivative of a equals

a' = 1,2,2,1,2,2,1,2,1,2,2,1,2,1,2,2,...

The claim by Hofstadter in formula (4) in the 1977 letter to Sloane is also not correct, since the second derivative of a is equal to

a'' = 2,2,1,2,1,2,2,1,2,1,2,2,1,...

so a is not equal to its own second derivative.

Nevertheless, this sequence has a self-similarity property: if one replaces every chunk 212 with 1 and every chunk 21212 with 2, then one obtains back the original sequence. In other words, (a(n)) is the unique fixed point of the morphism sigma given by sigma: 1->212, 2->21212.

This can be proved following the ideas of Chapter 2 in Lothaire's book and Section 4 of my paper "Substitution invariant Sturmian words and binary trees".

To comply with these references change the alphabet to {0,1}. This changes sigma into the morphism 0->101, 1->10101.

The fractional part {tau} of tau is larger than 1/2; as it is convenient to have it smaller than 1/2 we change to beta = 1-tau = (3-sqrt(5))/2.

This changes the morphism 0->101, 1->10101 to its mirror image psi given by 0->01010, 1->010.

Let psi_1 and psi_2 be the elementary Sturmian morphisms given by

psi_1(0)=01 , psi_1(1)=1, psi_2(0)=10, psi_2(1)=0.

Then psi = psi_2^2 psi_1.

This already shows that psi generates a Sturmian sequence with certain parameters alpha and rho: s(alpha,rho) = ([(n+1)*alpha+rho]-[n*alpha+rho]).

Since psi is the composition psi_2^2psi_1, the parameters of s(alpha,rho) are given by the composition T:=T_2^2T_1 of the fractional linear maps

T_1(x,y) = ((1-x)/(2-x),(1-y)/(2-x)),

T_2(x,y) = ((1-x)/(2-x), (2-x-y)/(2-x)).

Since one can verify that T(beta,1/2)=(beta,1/2), it follows that

alpha = beta, and rho = 1/2.

(End)

Suggested by A006336 and A007604.