cp's OEIS Frontend

For n > 1, a(n) is the expected number of tosses of a fair coin to get n-1 consecutive heads. - Pratik Poddar, Feb 04 2011

For n > 2, Sum_{k=1..a(n)} (-1)^binomial(n, k) = A064405(a(n)) + 1 = 0. - Benoit Cloitre, Oct 18 2002

For n > 0, the number of nonempty proper subsets of an n-element set. - Ross La Haye, Feb 07 2004

Numbers j such that abs( Sum_{k=0..j} (-1)^binomial(j, k)*binomial(j + k, j - k) ) = 1. - Benoit Cloitre, Jul 03 2004

For n > 2 this formula also counts edge-rooted forests in a cycle of length n. - Woong Kook (andrewk(AT)math.uri.edu), Sep 08 2004

For n >= 1, conjectured to be the number of integers from 0 to (10^n)-1 that lack 0, 1, 2, 3, 4, 5, 6 and 7 as a digit. - Alexandre Wajnberg, Apr 25 2005

Beginning with a(2) = 2, these are the partial sums of the subsequence of A000079 = 2^n beginning with A000079(1) = 2. Hence for n >= 2 a(n) is the smallest possible sum of exactly one prime, one semiprime, one triprime, ... and one product of exactly n-1 primes. A060389 (partial sums of the primorials, A002110, beginning with A002110(1)=2) is the analog when all the almost primes must also be squarefree. - Rick L. Shepherd, May 20 2005

From the second term on (n >= 1), the binary representation of these numbers is a 0 preceded by (n - 1) 1's. This pattern (0)111...1110 is the "opposite" of the binary 2^n+1: 1000...0001 (cf. A000051). - Alexandre Wajnberg, May 31 2005

The numbers 2^n - 2 (n >= 2) give the positions of 0's in A110146. Also numbers k such that k^(k + 1) = 0 mod (k + 2). - Zak Seidov, Feb 20 2006

Partial sums of A155559. - Zerinvary Lajos, Oct 03 2007

Number of surjections from an n-element set onto a two-element set, with n >= 2. - Mohamed Bouhamida, Dec 15 2007

It appears that these are the numbers n such that 3*A135013(n) = n*(n + 1), thus answering Problem 2 on the Mathematical Olympiad Foundation of Japan, Final Round Problems, Feb 11 1993 (see link Japanese Mathematical Olympiad).

Let P(A) be the power set of an n-element set A and R be a relation on P(A) such that for all x, y of P(A), xRy if x is a proper subset of y or y is a proper subset of x and x and y are disjoint. Then a(n+1) = |R|. - Ross La Haye, Mar 19 2009

The permutohedron Pi_n has 2^n - 2 facets [Pashkovich]. - Jonathan Vos Post, Dec 17 2009

First differences of A005803. - Reinhard Zumkeller, Oct 12 2011

For n >= 1, a(n + 1) is the smallest even number with bit sum n. Cf. A069532. - Jason Kimberley, Nov 01 2011

a(n) is the number of branches of a complete binary tree of n levels. - Denis Lorrain, Dec 16 2011

For n>=1, a(n) is the number of length-n words on alphabet {1,2,3} such that the gap(w)=1. For a word w the gap g(w) is the number of parts missing between the minimal and maximal elements of w. Generally for words on alphabet {1,2,...,m} with g(w)=g>0 the e.g.f. is Sum_{k=g+2..m} (m - k + 1)*binomial((k - 2),g)*(exp(x) - 1)^(k - g). a(3)=6 because we have: 113, 131, 133, 311, 313, 331. Cf. A240506. See the Heubach/Mansour reference. - Geoffrey Critzer, Apr 13 2014

For n > 0, a(n) is the minimal number of internal nodes of a red-black tree of height 2*n-2. See the Oct 02 2015 comment under A027383. - Herbert Eberle, Oct 02 2015

Conjecture: For n>0, a(n) is the least m such that A007814(A000108(m)) = n-1. - L. Edson Jeffery, Nov 27 2015

Actually this follows from the procedure for determining the multiplicity of prime p in C(n) given in A000108 by Franklin T. Adams-Watters: For p=2, the multiplicity is the number of 1 digits minus 1 in the binary representation of n+1. Obviously, the smallest k achieving "number of 1 digits" = k is 2^k-1. Therefore C(2^k-2) is divisible by 2^(k-1) for k > 0 and there is no smaller m for which 2^(k-1) divides C(m) proving the conjecture. - Peter Schorn, Feb 16 2020

For n >= 0, a(n) is the largest number you can write in bijective base-2 (a.k.a. the dyadic system, A007931) with n digits. - Harald Korneliussen, May 18 2019

The terms of this sequence are also the sum of the terms in each row of Pascal's triangle other than the ones. - Harvey P. Dale, Apr 19 2020

For n > 1, binomial(a(n),k) is odd if and only if k is even. - Charlie Marion, Dec 22 2020

For n >= 2, a(n+1) is the number of n X n arrays of 0's and 1's with every 2 X 2 square having density exactly 2. - David desJardins, Oct 27 2022

For n >= 1, a(n+1) is the number of roots of unity in the unique degree-n unramified extension of the 2-adic field Q_2. Note that for each p, the unique degree-n unramified extension of Q_p is the splitting field of x^(p^n) - x, hence containing p^n - 1 roots of unity for p > 2 and 2*(2^n - 1) for p = 2. - Jianing Song, Nov 08 2022

Given irregular one minute fuses, the shortest amount of time that can be measured over n minutes is time(n) = n + 1/2^(fuse(n)) minutes. For example, fuse(1)=3, for 9/8 = 1 + 1/(2^fuse(1)). Over 2 minutes, time(2) = 2 + 1/(2^fuse(2)) = 2049/1024 minutes. The value for fuse(3) is larger than 2↑↑↑↑↑↑↑↑↑16, in Knuth's up-arrow notation. - Ed Pegg Jr, Apr 03 2011; edited by Junyan Xu, Jan 04 2012

Suppose you have n ropes and a lighter. Each rope burns at a nonconstant rate but takes exactly one hour to burn completely from one end to the other. You can only light the ropes at either of their ends but can decide when to light each end as you see fit. If you're strategic in how you burn the ropes, how many specific lengths of time can you measure? (For example, if you had one rope, you could measure two lengths of time: one hour, by simply burning the entire rope from one end, and half an hour, by burning the rope from both ends and marking when the flames meet.)

In this sequence, the time intervals begin when the first rope (or safety fuse, or match cord) begins burning. Without this restriction, if we are allowed to burn some ropes beforehand, more time intervals are available for measuring off. - Andrey Zabolotskiy, Feb 28 2017