cp's OEIS Frontend

For n >= 3, also the number of maximal independent vertex sets (and minimal vertex covers) in the n-prism graph. - Eric W. Weisstein, Mar 30 and Aug 07 2017

From Petros Hadjicostas, Jul 08 2018: (Start)

Let q and m be positive integers. We denote by f1(m,q,n) the number of (marked) cyclic q-ary strings of length n that contain no runs of lengths > m when no wrapping around is allowed, and by f2(m,q,n) when wrapping around is allowed.

It is clear that f1(m,q,n) = f2(m,q,n) for n > m, but f1(m,q,n) = q^n and f2(m,q,n) = q^n - q when 1 <= n <= m.

Burstein and Wilf (1997) and Edlin and Zeilberger (2000) considered f1(m,q,n) while Hadjicostas and Zhang considered f2(m,q,n).

Let g(m, q, x) = (m+1-m*q*x)/(1-q*x+(q-1)*x^(m+1)) - (m+1)/(1-x^(m+1)).

By generalizing Moser (1991, 1993), Burstein and Wilf (1997) proved that the g.f. of the numbers f1(m,q,n) is F1(m,q,x) = ((1-x^m)/(1-x))*(q*x + (q-1)*x* g(m, q, x)).

Using the above formula by Burstein and Wilf (1997), Hadjicostas and Zhang (2018) proved that the g.f. of the numbers f2(m,q,n) is F2(m,q,x) = ((q-1)*x*(1-x^m)/(1-x))*g(m, q, x).

A necklace is an unmarked cyclic string. If f3(m,q,n) is the number of q-ary necklaces of length n with no runs of length > m (and wrapping around is allowed), then f3(m,q,n) = (1/n)*Sum_{d|n} phi(n/d)*f2(m,q,d), where phi(.) is Euler's totient function. Using this formula and F2(m,q,x), Hadjicostas and Zhang (2018) proved that the g.f. of the numbers f3(m,q,n) is given by F3(m,q,x) = -(q-1)*x*(1-x^m)/((1-x)*(1-x^(m+1))) - Sum_{s>=1} (phi(s)/s)*log(1 - (q-1)*(x^s - x^(s*(m+1)))/(1-x^s)).

For the current sequence, we have q = 2 and m = 2. We have a(n) = f1(m=2, q=2, n) = f2(m=2, q=2, n) for n >= 3, but for a(1) and a(2) it is unclear what approach the author of the sequence is following. He has a(1) = q^1 = 2, but a(2) = q^2 - q = 2^2 - 2 = 2. (Note that, for q = m = 2, we have f1(m=2, q=2, 1) = 2, f1(m=2, q=2, 2) = 4, f2(m=2, q=2, 1) = 0, and f2(m=2, q=2, 2) = 2.)

If A(x) is the g.f. of the current sequence, we have A(x) = F1(m=2,q=2, x) - 2*x^2 = F2(m=2, q=2, x) + 2*x.

When m = 1 and q = 3, we have f1(m=1, q=3, n) = number of marked cyclic words on three letters with no two consecutive like letters. We have f1(m=1, q=3, n) = A092297(n) for n >= 2. This was first stated in the comments of that sequence by G. Critzer.

When m = 1 and q = 4, we have f1(m=1, q=4, n) = number of marked cyclic words on four letters with no two consecutive like letters. We have f1(m=1, q=4, n) = A218034(n) for n >= 1. This was first stated in the comments of that sequence by J. Arndt.

A generalization of the above formula by Burstein and Wilf (1997) was given by Taylor (2014) in Section 5 of his paper. (End)

For n>=1, the set {A008747(6n+-1)} is the set of numbers of the form a^2 + 5*(a+1)^2 for -inf < a < inf. Furthermore the set A008747(6n) is A033581(n). - Kieren MacMillan, Dec 19 2007

For n>1, a(n-1) is the number of aperiodic necklaces (Lyndon words) with k<=3 black beads and n-k white beads. For n=4 we have for example a(3)=3 aperiodic necklaces: BWWW, BBWW and BBBW. BWBW is periodic and is not counted. - Herbert Kociemba, Oct 23 2016

From Klaus Brockhaus, May 30 2010: (Start)

Periodic sequence: Repeat [1, 1, 1, 3].

Continued fraction expansion of (9+sqrt(165))/14.

Decimal expansion of 371/3333. (End)

Final nonzero digit of n^n in base 4. - José María Grau Ribas, Jan 19 2012

A Chebyshev transform of A001045(n+1): if A(x) is the g.f. of a sequence, map it to ((1-x^2)/(1+x^2))*A(x/(1+x^2)).

Also decimal expansion of 1111/9990. - Elmo R. Oliveira, Feb 18 2024

Also partial quotients of the continued fraction expansion of sqrt(5/2). - Hugo Pfoertner, Jan 10 2025

This sequence was investigated in cooperation with Paul Barry.

Generating floretion: - 0.5'i - 0.5'k - 0.5j' - 0.5'ii' + 0.5'jj' - 0.5'kk' + 0.5'ik' - 0.5'ki' ("tes").

From Joshua P. Bowman, Sep 28 2023: (Start)

a(n) is equal to the number of circular binary sequences of length n+1 with an even number of 0's and no consecutive 1's. A circular binary sequence is a finite sequence of 0's and 1's for which the first and last digits are considered to be adjacent. Rotations are distinguished from each other.

a(n) is also equal to the number of matchings in the cycle graph C_{n+1} for which the number of edges plus the number of unmatched vertices is even.

a(n) is also equal to the number of circular compositions of n+1 into an even number of 1's and 2's. (End)

Apply the Riordan array (1/(1-x^4),x) to floor((n+3)/3). - Paul Barry, Jan 20 2006

Number of partitions of n into parts 1, 3, and 4. - David Neil McGrath, Aug 30 2014

Also, a(n-4) is equal to the number of partitions mu of n of length 3 such that mu_1-mu_2 is even and mu_2-mu_3 is odd or vice versa (see below example). - John M. Campbell, Jan 29 2016

With four 0's prepended and offset 0, a(n) is the number of partitions of n into four parts whose 2nd and 3rd largest parts are equal. - Wesley Ivan Hurt, Jan 05 2021

Setting m=2 in

log(m) = Sum_{n>0} (n mod m - (n-1) mod m)/n [1]

yields the sum

log(2) = (1 -1/2) +(1/3 -1/4) +(1/5 -1/6)+...

Substituting every -1/d by 1/d - 2/d we obtain

log(2) = (1+1/2-1)+(1/3+1/4-1/2)+(1/5+1/6-1/3)+...

a(n) is the sequence of denominators of this modified sum with unit numerators, so

Sum_{k>0} 1/a(k) = log(2)

Substituting -1/d by -2/d + 1/d would yield another permutation (one positive, one negative, one positive) with the same sum of inverses.

Similar sequences (m positives, one negative) may be obtained for the logarithm of any integer m>0. A001057 is the case m=1, with sum of inverses log(1).

Equation [1] is a result of expanding log( Sum_{0<=k<=m-1} x^k ) at x=1 (see comment to A061347.)

Nonsimple continued fraction expansion of 1+1/sqrt(3) = 1 + A020760. - R. J. Mathar, Mar 08 2012

Apart from the top row, the same as A177121.

Sum_{k>=1} T(n,k)/k = log(n); this has been pointed out by Jaume Oliver Lafont in A061347 and A002162.