cp's OEIS Frontend

-2, -4, -6, -8, -10, -12, -14, ... are the trivial zeros of the Riemann zeta function. - Vivek Suri (vsuri(AT)jhu.edu), Jan 24 2008

If a 2-set Y and an (n-2)-set Z are disjoint subsets of an n-set X then a(n-2) is the number of 2-subsets of X intersecting both Y and Z. - Milan Janjic, Sep 19 2007

A134452(a(n)) = 0; A134451(a(n)) = 2 for n > 0. - Reinhard Zumkeller, Oct 27 2007

Omitting the initial zero gives the number of prime divisors with multiplicity of product of terms of n-th row of A077553. - Ray Chandler, Aug 21 2003

A059841(a(n))=1, A000035(a(n))=0. - Reinhard Zumkeller, Sep 29 2008

(APSO) Alternating partial sums of (a-b+c-d+e-f+g...) = (a+b+c+d+e+f+g...) - 2*(b+d+f...), it appears that APSO(A005843) = A052928 = A002378 - 2*(A116471), with A116471=2*A008794. - Eric Desbiaux, Oct 28 2008

A056753(a(n)) = 1. - Reinhard Zumkeller, Aug 23 2009

Twice the nonnegative numbers. - Juri-Stepan Gerasimov, Dec 12 2009

The number of hydrogen atoms in straight-chain (C(n)H(2n+2)), branched (C(n)H(2n+2), n > 3), and cyclic, n-carbon alkanes (C(n)H(2n), n > 2). - Paul Muljadi, Feb 18 2010

For n >= 1; a(n) = the smallest numbers m with the number of steps n of iterations of {r - (smallest prime divisor of r)} needed to reach 0 starting at r = m. See A175126 and A175127. A175126(a(n)) = A175126(A175127(n)) = n. Example (a(4)=8): 8-2=6, 6-2=4, 4-2=2, 2-2=0; iterations has 4 steps and number 8 is the smallest number with such result. - Jaroslav Krizek, Feb 15 2010

For n >= 1, a(n) = numbers k such that arithmetic mean of the first k positive integers is not integer. A040001(a(n)) > 1. See A145051 and A040001. - Jaroslav Krizek, May 28 2010

Union of A179082 and A179083. - Reinhard Zumkeller, Jun 28 2010

a(k) is the (Moore lower bound on and the) order of the (k,4)-cage: the smallest k-regular graph having girth four: the complete bipartite graph with k vertices in each part. - Jason Kimberley, Oct 30 2011

For n > 0: A048272(a(n)) <= 0. - Reinhard Zumkeller, Jan 21 2012

Let n be the number of pancakes that have to be divided equally between n+1 children. a(n) is the minimal number of radial cuts needed to accomplish the task. - Ivan N. Ianakiev, Sep 18 2013

For n > 0, a(n) is the largest number k such that (k!-n)/(k-n) is an integer. - Derek Orr, Jul 02 2014

a(n) when n > 2 is also the number of permutations simultaneously avoiding 213, 231 and 321 in the classical sense which can be realized as labels on an increasing strict binary tree with 2n-1 nodes. See A245904 for more information on increasing strict binary trees. - Manda Riehl Aug 07 2014

It appears that for n > 2, a(n) = A020482(n) + A002373(n), where all sequences are infinite. This is consistent with Goldbach's conjecture, which states that every even number > 2 can be expressed as the sum of two prime numbers. - Bob Selcoe, Mar 08 2015

Number of partitions of 4n into exactly 2 parts. - Colin Barker, Mar 23 2015

Number of neighbors in von Neumann neighborhood. - Dmitry Zaitsev, Nov 30 2015

Unique solution b( ) of the complementary equation a(n) = a(n-1)^2 - a(n-2)*b(n-1), where a(0) = 1, a(1) = 3, and a( ) and b( ) are increasing complementary sequences. - Clark Kimberling, Nov 21 2017

Also the maximum number of non-attacking bishops on an (n+1) X (n+1) board (n>0). (Cf. A000027 for rooks and queens (n>3), A008794 for kings or A030978 for knights.) - Martin Renner, Jan 26 2020

Integer k is even positive iff phi(2k) > phi(k), where phi is Euler's totient (A000010) [see reference De Koninck & Mercier]. - Bernard Schott, Dec 10 2020

Number of 3-permutations of n elements avoiding the patterns 132, 213, 312 and also number of 3-permutations avoiding the patterns 213, 231, 321. See Bonichon and Sun. - Michel Marcus, Aug 20 2022

a(n) gives the y-value of the integral solution (x,y) of the Pellian equation x^2 - (n^2 + 1)*y^2 = 1. The x-value is given by 2*n^2 + 1 (see Tattersall). - Stefano Spezia, Jul 24 2025

An n X n nonnegative matrix A is primitive (see A070322) iff every element of A^k is > 0 for some power k. If A is primitive then the power which should have all positive entries is <= n^2 - 2n + 2 (Wielandt).

a(n) = Phi_4(n), where Phi_k is the k-th cyclotomic polynomial.

As the positive solution to x=2n+1/x is x=n+sqrt(a(n)), the continued fraction expansion of sqrt(a(n)) is {n; 2n, 2n, 2n, 2n, ...}. - Benoit Cloitre, Dec 07 2001

a(n) is one less than the arithmetic mean of its neighbors: a(n) = (a(n-1) + a(n+1))/2 - 1. E.g., 2 = (1+5)/2 - 1, 5 = (2+10)/2 - 1. - Amarnath Murthy, Jul 29 2003

Equivalently, the continued fraction expansion of sqrt(a(n)) is (n;2n,2n,2n,...). - Franz Vrabec, Jan 23 2006

Number of {12,1*2*,21}-avoiding signed permutations in the hyperoctahedral group.

The number of squares of side 1 which can be drawn without lifting the pencil, starting at one corner of an n X n grid and never visiting an edge twice is n^2-2n+2. - Sébastien Dumortier, Jun 16 2005

Also, numbers m such that m^3 - m^2 is a square, (n*(1 + n^2))^2. - Zak Seidov

1 + 2/2 + 2/5 + 2/10 + ... = Pi*coth Pi [Jolley], see A113319. - Gary W. Adamson, Dec 21 2006

For n >= 1, a(n-1) is the minimal number of choices from an n-set such that at least one particular element has been chosen at least n times or each of the n elements has been chosen at least once. Some games define "matches" this way; e.g., in the classic Parker Brothers, now Hasbro, board game Risk, a(2)=5 is the number of cards of three available types (suits) required to guarantee at least one match of three different types or of three of the same type (ignoring any jokers or wildcards). - Rick L. Shepherd, Nov 18 2007

Positive X values of solutions to the equation X^3 + (X - 1)^2 + X - 2 = Y^2. To prove that X = n^2 + 1: Y^2 = X^3 + (X - 1)^2 + X - 2 = X^3 + X^2 - X - 1 = (X - 1)(X^2 + 2X + 1) = (X - 1)*(X + 1)^2 it means: (X - 1) must be a perfect square, so X = n^2 + 1 and Y = n(n^2 + 2). - Mohamed Bouhamida, Nov 29 2007

{a(k): 0 <= k < 4} = divisors of 10. - Reinhard Zumkeller, Jun 17 2009

Appears in A054413 and A086902 in relation to sequences related to the numerators and denominators of continued fractions convergents to sqrt((2*n)^2/4 + 1), n=1, 2, 3, ... . - Johannes W. Meijer, Jun 12 2010

For n > 0, continued fraction [n,n] = n/a(n); e.g., [5,5] = 5/26. - Gary W. Adamson, Jul 15 2010

The only real solution of the form f(x) = A*x^p with negative p which satisfies f^(m)(x) = f^[-1](x), x >= 0, m >= 1, with f^(m) the m-th derivative and f^[-1] the compositional inverse of f, is obtained for m=2*n, p=p(n)= -(sqrt(a(n))-n) and A=A(n)=(fallfac(p(n),2*n))^(-p(n)/(p(n)+1)), with fallfac(x,k):=Product_{j=0..k-1} (x-j) (falling factorials). See the T. Koshy reference, pp. 263-4 (there are also two solutions for positive p, see the corresponding comment in A087475). - Wolfdieter Lang, Oct 21 2010

n + sqrt(a(n)) = [2*n;2*n,2*n,...] with the regular continued fraction with period 1. This is the even case. For the general case see A087475 with the Schroeder reference and comments. For the odd case see A078370.

a(n-1) counts configurations of non-attacking bishops on a 2 X n strip [Chaiken et al., Ann. Combin. 14 (2010) 419]. - R. J. Mathar, Jun 16 2011

Also numbers k such that 4*k-4 is a square. Hence this sequence is the union of A053755 and A069894. - Arkadiusz Wesolowski, Aug 02 2011

a(n) is also the Moore lower bound on the order, A191595(n), of an (n,5)-cage. - Jason Kimberley, Oct 17 2011

Left edge of the triangle in A195437: a(n+1) = A195437(n,0). - Reinhard Zumkeller, Nov 23 2011

If h (5,17,37,65,101,...) is prime is relatively prime to 6, then h^2-1 is divisible by 24. - Vincenzo Librandi, Apr 14 2014

The identity (4*n^2+2)^2 - (n^2+1)*(4*n)^2 = 4 can be written as A005899(n)^2 - a(n)*A008586(n)^2 = 4. - Vincenzo Librandi, Jun 15 2014

a(n) is also the number of permutations simultaneously avoiding 213 and 321 in the classical sense which can be realized as labels on an increasing strict binary tree with 2n-1 nodes. See A245904 for more information on increasing strict binary trees. - Manda Riehl, Aug 07 2014

a(n-1) is the maximum number of stages in the Gale-Shapley algorithm for finding a stable matching between two sets of n elements given an ordering of preferences for each element (see Gura et al.). - Melvin Peralta, Feb 07 2016

Because of Fermat's little theorem, a(n) is never divisible by 3. - Altug Alkan, Apr 08 2016

For n > 0, if a(n) points are placed inside an n X n square, it will always be the case that at least two of the points will be a distance of sqrt(2) units apart or less. - Melvin Peralta, Jan 21 2017

Also the limit as q->1^- of the unimodal polynomial (1-q^(n*k+1))/(1-q) after making the simplification k=n. The unimodal polynomial is from O'Hara's proof of unimodality of q-binomials after making the restriction to partitions of size <= 1. See G_1(n,k) from arXiv:1711.11252. As the size restriction s increases, G_s->G_infinity=G: the q-binomials. Then substituting k=n and q=1 yields the central binomial coefficients: A000984. - Bryan T. Ek, Apr 11 2018

a(n) is the smallest number congruent to both 1 (mod n) and 2 (mod n+1). - David James Sycamore, Apr 04 2019

a(n) is the number of permutations of 1,2,...,n+1 with exactly one reduced decomposition. - Richard Stanley, Dec 22 2022

From Klaus Purath, Apr 03 2025: (Start)

The odd prime factors of these terms are always of the form 4*k + 1.

All a(n) = D satisfy the Pell equation (k*x)^2 - D*y^2 = -1. The values for k and the solutions x, y can be calculated using the following algorithm: k = n, x(0) = 1, x(1) = 4*D - 1, y(0) = 1, y(1) = 4*D - 3. The two recurrences are of the form (4*D - 2, -1). The solutions x, y of the Pell equations for n = {1 ... 14} are in OEIS.

It follows from the above that this sequence is a subsequence of A031396. (End)

Number of balanced strings of length n: let d(S) = #(1's) - #(0's), # == count in S, then S is balanced if every substring T of S has -2 <= d(T) <= 2.

Number of "fold lines" seen when a rectangular piece of paper is folded n+1 times along alternate orthogonal directions and then unfolded. - Quim Castellsaguer (qcastell(AT)pie.xtec.es), Dec 30 1999

Also the number of binary strings with the property that, when scanning from left to right, once the first 1 is seen in position j, there must be a 1 in positions j+2, j+4, ... until the end of the string. (Positions j+1, j+3, ... can be occupied by 0 or 1.) - Jeffrey Shallit, Sep 02 2002

a(n-1) is also the Moore lower bound on the order of a (3,n)-cage. - Eric W. Weisstein, May 20 2003 and Jason Kimberley, Oct 30 2011

Partial sums of A016116. - Hieronymus Fischer, Sep 15 2007

Equals row sums of triangle A152201. - Gary W. Adamson, Nov 29 2008

From John P. McSorley, Sep 28 2010: (Start)

a(n) = DPE(n+1) is the total number of k-double-palindromes of n up to cyclic equivalence. See sequence A180918 for the definitions of a k-double-palindrome of n and of cyclic equivalence. Sequence A180918 is the 'DPE(n,k)' triangle read by rows where DPE(n,k) is the number of k-double-palindromes of n up to cyclic equivalence. For example, we have a(4) = DPE(5) = DPE(5,1) + DPE(5,2) + DPE(5,3) + DPE(5,4) + DPE(5,5) = 0 + 2 + 2 + 1 + 1 = 6.

The 6 double-palindromes of 5 up to cyclic equivalence are 14, 23, 113, 122, 1112, 11111. They come from cyclic equivalence classes {14,41}, {23,32}, {113,311,131}, {122,212,221}, {1112,2111,1211,1121}, and {11111}. Hence a(n)=DPE(n+1) is the total number of cyclic equivalence classes of n containing at least one double-palindrome.

(End)

From Herbert Eberle, Oct 02 2015: (Start)

For n > 0, there is a red-black tree of height n with a(n-1) internal nodes and none with less.

In order a red-black tree of given height has minimal number of nodes, it has exactly 1 path with strictly alternating red and black nodes. All nodes outside this height defining path are black.

Consider:

mrbt5 R

/ \

/ \

/ \

/ B

/ / \

mrbt4 B / B

/ \ B E E

/ B E E

mrbt3 R E E

/ \

/ B

mrbt2 B E E

/ E

mrbt1 R

E E

(Red nodes shown as R, blacks as B, externals as E.)

Red-black trees mrbt1, mrbt2, mrbt3, mrbt4, mrbt5 of respective heights h = 1, 2, 3, 4, 5; all minimal in the number of internal nodes, namely 1, 2, 4, 6, 10.

Recursion (let n = h-1): a(-1) = 0, a(n) = a(n-1) + 2^floor(n/2), n >= 0.

(End)

Also the number of strings of length n with the digits 1 and 2 with the property that the sum of the digits of all substrings of uneven length is not divisible by 3. An example with length 8 is 21221121. - Herbert Kociemba, Apr 29 2017

a(n-2) is the number of achiral n-bead necklaces or bracelets using exactly two colors. For n=4, the four arrangements are AAAB, AABB, ABAB, and ABBB. - Robert A. Russell, Sep 26 2018

Partial sums of powers of 2 repeated 2 times, like A200672 where is 3 times. - Yuchun Ji, Nov 16 2018

Also the number of binary words of length n with cuts-resistance <= 2, where, for the operation of shortening all runs by one, cuts-resistance is the number of applications required to reach an empty word. Explicitly, these are words whose sequence of run-lengths, all of which are 1 or 2, has no odd-length run of 1's sandwiched between two 2's. - Gus Wiseman, Nov 28 2019

Also the number of up-down paths with n steps such that the height difference between the highest and lowest points is at most 2. - Jeremy Dover, Jun 17 2020

Also the number of non-singleton integer compositions of n + 2 with no odd part other than the first or last. Including singletons gives A052955. This is an unsorted (or ordered) version of A351003. The version without even (instead of odd) interior parts is A001911, complement A232580. Note that A000045(n-1) counts compositions without odd parts, with non-singleton case A077896, and A052952/A074331 count non-singleton compositions without even parts. Also the number of compositions y of n + 1 such that y_i = y_{i+1} for all even i. - Gus Wiseman, Feb 19 2022

Apart from leading term (which should really be 3/2), same as A055010.

Binomial transform of A040001. Inverse binomial transform of A053156.

a(n) = A105728(n+1,2). - Reinhard Zumkeller, Apr 18 2005

Row sums of triangle A133567. - Gary W. Adamson, Sep 16 2007

Row sums of triangle A135226. - Gary W. Adamson, Nov 23 2007

a(n) = number of partitions Pi of [n+1] (in standard increasing form) such that the permutation Flatten[Pi] avoids the patterns 2-1-3 and 3-1-2. Example: a(3)=11 counts all 15 partitions of [4] except 13/24, 13/2/4 which contain a 2-1-3 and 14/23, 14/2/3 which contain a 3-1-2. Here "standard increasing form" means the entries are increasing in each block and the blocks are arranged in increasing order of their first entries. - David Callan, Jul 22 2008

An elephant sequence, see A175654. For the corner squares four A[5] vectors, with decimal values 42, 138, 162, 168, lead to this sequence. For the central square these vectors lead to the companion sequence A003945. - Johannes W. Meijer, Aug 15 2010

The binary representation of a(n) has n+1 digits, where all digits are 1's except digit n-1. For example: a(4) = 23 = 10111 (2). - Omar E. Pol, Dec 02 2012

Row sums of triangle A209561. - Reinhard Zumkeller, Dec 26 2012

If a Stern's sequence based enumeration system of positive irreducible fractions is considered (for example, A007305/A047679, A162909/A162910, A071766/A229742, A245325/A245326, ...), and if it is organized by blocks or levels (n) with 2^n terms (n >= 0), and the fractions, term by term, are summed at each level n, then the resulting sequence of integers is a(n) + 1/2, apart from leading term (which should be 1/2). - Yosu Yurramendi, May 23 2015

For n >= 2, A083329(n) in binary representation is a string [101..1], also 10 followed with (n-1) 1's. For n >= 3, A036563(n) in binary representation is a string [1..101], also (n-2) 1's followed with 01. Thus A083329(n) is a reflection of the binary representation of A036563(n+1). Example: A083329(5) = 101111 in binary, A036563(6) = 111101 in binary. - Ctibor O. Zizka, Nov 06 2018

For n > 0, a(n) is the minimum number of turns in (n+1)-dimensional Euclidean space needed to visit all 2^(n+1) vertices of the (n+1)-cube (e.g., {0,1}^(n+1)) and return to the starting point, moving along straight-line segments between turns (turns may occur elsewhere in R^(n+1)). - Marco Ripà, Aug 14 2025

a(n+1) is the reflection of a(n) through a(n-1) on the numberline. - Floor van Lamoen, Aug 31 2004

If a zero is added as the (new) a(0) in front, the sequence represents the inverse binomial transform of A001045. Partial sums are in A077898. - R. J. Mathar, Aug 30 2008

a(n) = A077953(2*n+3). - Reinhard Zumkeller, Oct 07 2008

Related to the Fibonacci sequence by an INVERT transform: if A(x) = 1+x^2*g(x) is the generating function of the a(n) prefixed with 1, 0, then 1/A(x) = 2+(x+1)/(x^2-x+1) is the generating function of 1, 0, -1, 1, -2, 3, ..., the signed Fibonacci sequence A000045 prefixed with 1. - Gary W. Adamson, Jan 07 2011

Also: Gaussian binomial coefficients [n+1,1], or q-integers, for q=-2, diagonal k=1 in the triangular (or column r=1 in the square) array A015109. - M. F. Hasler, Nov 04 2012

With a leading zero, 0, 1, -1, 3, -5, 11, -21, 43, -85, 171, -341, 683, ... we obtain the Lucas U(-1,-2) sequence. - R. J. Mathar, Jan 08 2013

Let m = a(n). Then 18*m^2 - 12*m + 1 = A000225(2n+3). - Roderick MacPhee, Jan 17 2013

Draw n ellipses in the plane (n > 0); sequence gives maximum number of regions into which the plane is divided (cf. A014206, A386480).

Least k such that Z(k,2) <= Z(n,3) where Z(m,s) = Sum_{i>=m} 1/i^s = zeta(s) - Sum_{i=1..m-1} 1/i^s. - Benoit Cloitre, Nov 29 2002

For n > 2, third diagonal of A154685. - Vincenzo Librandi, Aug 06 2010

a(k) is also the Moore lower bound A198300(k,6) on the order A054760(k,6) of a (k,6)-cage. Equality is achieved if and only if there exists a finite projective plane of order k - 1. A sufficient condition for this is that k - 1 be a prime power. - Jason Kimberley, Oct 17 2011 and Jan 01 2013

From Jess Tauber, May 20 2013: (Start)

For neutron shell filling in spherical atomic nuclei, this sequence shows numerical differences between filled spin-split suborbitals sharing all quantum numbers except the principal quantum number n, and here all n's must differ by 1. Only a small handful of exceptions exist.

This sequence consists of summed pairs of every other doubled triangular number. It also can be created by taking differences between nuclear magic numbers from the harmonic oscillator (HO)(doubled tetrahedral) set and the spin-orbit (SO) set (2,6,14,28,50,82,126,184,...), with either set being larger. So SO-HO: 2-0=2, 6-0=6, 14-0=14, 28-2=26, 50-8=42, 82-20=62, 126-40=86, 184-70=114, and HO-SO: 2-0=2, 8-2=6, 20-6=14, 40-14=26, 70-28=42, 112-50=62, 168-82=86, 240-126=114. From the perspective of idealized HO periodic structure, with suborbitals in order from largest to smallest spin, alternating by parity, the HO-SO set is spaced two period analogs PLUS one suborbital, while the SO-HO set is spaced two period analogs MINUS one suborbital. (End)

The known values of f(k,6) and F(k,6) in Brown (1967), Table 1, closely match this sequence. - N. J. A. Sloane, Jul 09 2015

Numbers k such that 2*k - 3 is a square. - Bruno Berselli, Nov 08 2017

Numbers written 222 in number base B, including binary with 'digit' 2: 222(2)=14, 222(3)=26, ... - Ron Knott, Nov 14 2017

WARNING: The offset of this sequence has been changed from 0 to 1 without correcting the formulas and programs, many of them correspond to the original indexing a(0)=0, a(1)=1, ... - M. F. Hasler, Oct 06 2014

Numbers n such that no entry in n-th row of Pascal's triangle is divisible by 3, i.e., such that A062296(n) = 0.

The base 3 representation of these numbers is 222...222 or 122...222.

a(n+1) is the smallest number with ternary digit sum = n: A053735(a(n+1)) = n and A053735(m) <> n for m < a(n+1). - Reinhard Zumkeller, Sep 15 2006

A138002(a(n)) = 0. - Reinhard Zumkeller, Feb 26 2008

Also, number of terms in S(n), where S(n) is defined in A114482. - N. J. A. Sloane, Nov 13 2014

a(n+1) is also the Moore lower bound on the order of a (4,g)-cage. - Jason Kimberley, Oct 30 2011

This sequence is closed under multiplication by any integer k > 0. The primitive elements of the sequence (those not divisible by any smaller element) are the primes, A000040. - Franklin T. Adams-Watters, May 22 2006

Possible sums of the final scores of completed Chicago Bears football games. 1 point only is an impossible score in American football. But with the safety 2 and the field goal 3, we can construct the set of integers greater than 1. We can prove this by noting that if a score is even, we can build it with a series of safeties. Of course the other allowed scorings of 3, 6, and 1 after a touchdown, could also be used. Now if a score is odd it is of the form 2k+3. So for any odd number 2m+1, we subtract 3 (or 1 field goal) from it to make it even and divide by 2 to get the number of safeties we need to add back to the field goal. Symbolically, let the odd number be 2m+1; then (2m+1 - 3)/2 = m-1 safeties are needed. Add this to 3 and you will have the number. For example, say we want a score of 99. 99 = 2m+1 and m = 49. So m-1 = 48 safeties + 1 field goal = 99 points. - Cino Hilliard, Feb 03 2006

Possible nonnegative values of (a*b-c*d) where a,b,c and d are distinct positive integers and a+b=c+d. All positive values >=2 are possible: for even values 2n take a=m+n, b=m-n+1, c=m+n+1, d=m-n, where m>n; for odd values 2n+1 take a=m+n, b=m-n, c=m+n+1, d=m-n-1, where m>n+1. Elementary algebra shows that the values 0 and 1 are not possible without violating the assumption that a,b,c and d are distinct. - John Grint, Sep 26 2011

Also numbers n such that a semiprime is equal to the sum of n primes. Bachraoui proved that there is a prime between 2n and 3n for every n > 1, so every n > 1 is in this sequence since any number in that range is the sum of n integers each of which is either 2 or 3. - Charles R Greathouse IV, Oct 27 2011

From Jason Kimberley, Oct 30 2011: (Start)

Moore lower bound on the order of a (k,g) cage: A198300 (square); rows: A000027 (k=2), A027383 (k=3), A062318 (k=4), A061547 (k=5), A198306 (k=6), A198307 (k=7), A198308 (k=8), A198309 (k=9), A198310 (k=10), A094626 (k=11); columns: this sequence (g=3), A005843 (g=4), A002522 (g=5), A051890 (g=6), A188377 (g=7).

Digit string 12 read in base n-1 (for n>3 or by extending notation). (End)

Positive integers whose number of divisors is not 1. - Omar E. Pol, Aug 11 2012

Positive integers where the number of parts function on the set of 2-ary partitions is equidistributed mod 2. - Tom Edgar, Apr 26 2016

This sequence is also the Pierce Expansion of 1/exp(1). - G. C. Greubel, Nov 15 2016

Natural numbers with at least one prime factor. - Michal Bozon, Apr 24 2017

a(n) is the reciprocal of the Integral_{x=0..1} x^n dx. - Felix Huber, Aug 19 2023

If the Fibonacci pairs are kept in the natural order (F(n),F(n+1)), it appears that the first term of the pair occurs in A002487 at the index given by A061547(n).

Equals row sums of triangle A177954. - Gary W. Adamson, May 15 2010

Starting at n=3, begin subtracting from (2^(n-1)-1)/2^(n-1): 3/4 - 1/2 = 1/4 with 1+4=5=a(3); 7/8 - 1/4 = 5/8 with 5+8=13=a(4); 15/16 - 5/8 = 5/16 with 5+16=21= a(5); 31/32 - 5/16 = 21/32 with 21+32=53=a(6); 63/64 - 21/32 = 21/64 with 21+64=85=a(7) and so on. For n odd in the first fraction (2^(n-1)-1)/2^(n-1), the result approaches 1/3, and for n even in the first fraction, the result approaches 2/3. - J. M. Bergot, May 08 2015

Also, the decimal representation of the x-axis, from the left edge to the origin, of the n-th stage of growth of the two-dimensional cellular automaton defined by "Rule 678", based on the 5-celled von Neumann neighborhood, initialized with a single black (ON) cell at stage zero. See A283641. - Robert Price, Mar 12 2017