cp's OEIS Frontend

The floor of the arithmetic mean of the first n+1 positive integers. - Cino Hilliard, Sep 06 2003

Number of partitions of n into powers of 2 where no power is used more than three times, or 4th binary partition function (see A072170).

Number of partitions of n in which the greatest part is at most 2. - Robert G. Wilson v, Jan 11 2002

Number of partitions of n into at most 2 parts. - Jon Perry, Jun 16 2003

a(n) = #{k=0..n: k+n is even}. - Paul Barry, Sep 13 2003

Number of symmetric Dyck paths of semilength n+2 and having two peaks. E.g., a(6)=4 because we have UUUUUUU*DU*DDDDDDD, UUUUUU*DDUU*DDDDDD, UUUUU*DDDUUU*DDDDD and UUUU*DDDDUUUU*DDDD, where U=(1,1), D=(1,-1) and * indicates a peak. - Emeric Deutsch, Jan 12 2004

Smallest positive integer whose harmonic mean with another positive integer is n (for n > 0). For example, a(6)=4 is already given (as 4 is the smallest positive integer such that the harmonic mean of 4 (with 12) is 6) - but the harmonic mean of 2 (with -6) is also 6 and 2 < 4, so the two positive integer restrictions need to be imposed to rule out both 2 and -6.

Second outermost diagonal of Losanitsch's triangle (A034851). - Alonso del Arte, Mar 12 2006

Arithmetic mean of n-th row of A080511. - Amarnath Murthy, Mar 20 2003

a(n) is the number of ways to pay n euros (or dollars) with coins of one and two euros (respectively dollars). - Richard Choulet and Robert G. Wilson v, Dec 31 2007

Inverse binomial transform of A045623. - Philippe Deléham, Dec 30 2008

Coefficient of q^n in the expansion of (m choose 2)_q as m goes to infinity. - Y. Kelly Itakura (yitkr(AT)mta.ca), Aug 21 2002

This Itakura comment follows from a partial fraction decomposition (m choose 2)q = [(1-q^(2m-2))/(1+q) + (1-q^(2m-2))/(1-q) +2 (1-q^(m-1))^2/(1-q)^2]/4. Interpreted as generating functions in q, they have convolution structures; the first term in the numerator creates +1,-1,+1,-1 etc, the 2nd term creates +1,+1,+1,+1 etc., the 3rd term 2,4,6,8, etc. as m->infinity. - _R. J. Mathar, Sep 25 2008

Binomial transform of (-1)^n*A034008(n) = [1,0,1,-2,4,-8,16,-32,...]. - Philippe Deléham, Nov 15 2009

From Jon Perry_, Nov 16 2010: (Start)

Column sums of:

1 1 1 1 1 1...

1 1 1 1...

1 1...

..............

--------------

1 1 2 2 3 3... (End)

This sequence is also the half-convolution of the powers of 1 sequence A000012 with itself. For the definition of half-convolution see a comment on A201204, where also the rule for the o.g.f. is given. - Wolfdieter Lang, Jan 09 2012

a(n) is also the number of roots of the n-th Bernoulli polynomial in the right half-plane for n>0. - Michel Lagneau, Nov 08 2012

a(n) is the number of symmetry-allowed, linearly-independent terms at n-th order in the series expansion of the Exe vibronic perturbation matrix, H(Q) (cf. Viel & Eisfeld). - Bradley Klee, Jul 21 2015

a(n) is the number of distinct integers in the n-th row of Pascal's triangle. - Melvin Peralta, Feb 03 2016

a(n+1) for n >= 3 is the diameter of the Generalized Petersen Graph G(n, 1). - Nick Mayers, Jun 06 2016

The arithmetic function v_1(n,2) as defined in A289198. - Robert Price, Aug 22 2017

Also, this sequence is the second column in the triangle of the coefficients of the sum of two consecutive Fibonacci polynomials F(n+1, x) and F(n, x) (n>=0) in ascending powers of x. - Mohammad K. Azarian, Jul 18 2018

a(n+2) is the least k such that given any k integers, there exist two of them whose sum or difference is divisible by n. - Pablo Hueso Merino, May 09 2020

Column k = 2 of A051159. - John Keith, Jun 28 2021

From David Callan, Nov 02 2006: (Start)

a(n) = number of (unlabeled, rooted) ordered trees on n-1 vertices in which all outdegrees are <= 2 and, for each vertex of outdegree 2, the sizes of its two subtrees are weakly increasing left to right (n >= 2). The number b(n) of such trees on n vertices satisfies the recurrence b[1]=1; b[n_]/;n>=2 := b[n] = b[n-1] + Sum_{i=1..floor((n-1)/2)} b[i]b[n-1-i], the first term counting trees whose root has outdegree 1 and the sum counting trees whose root has outdegree 2 by size of the left subtree. This recurrence generates b(n) = a(n+1), n >= 1. For example, the a(5)=3 such trees are:

.|....|...../\..

.|.../.\.....|..

.|.............. (End)

From R. J. Mathar, Mar 27 2009: (Start)

The connection with the Rayleigh polynomials Phi(2n,x) of A158616 is that Phi(2n,x) = Sum_{i=1..a(n)} 2^(n_i) Product_{j=2..n-1} (x+j)^(n_ij), as described by Kishore.

So a(n) counts the terms in the representation of the polynomial Phi(2n,x) as a sum over these "base" polynomials.

For example, Phi(12,x) = 2^4*(x+2)^2*(x+3) + 2^2*(x+2)*(x+3)^2 + 2^3*(x+2)*(x+3)*(x+4) + 2^3*(x+2)*(x+3)*(x+5) + 2^2*(x+2)*(x+4)*(x+5) + 2*(x+3)^2*(x+5) has a(6)=6 terms. (End)

From Wolfdieter Lang, Jan 06 2012: (Start)

The o.g.f. G(x) := Sum_{n>=0} a(n)*x^n, with a(0)=0, satisfies the relation (G(x))^2 - 2*G(x) + G2(x^2) + 2*x = 0, with the o.g.f. G2(x) := Sum_{n>=0} a(n)^2*x^n of the squares. This can be proved from the connection to the half-convolution of the sequence with itself (for this notion see a comment on A201204, where also the rule for the o.g.f. is given). (End)

Limit_{n->infinity} a(n)^(1/n) = 2.49086422... . - Vaclav Kotesovec, Oct 15 2014

This sequence diverges from A001190 for n >= 8. A001190(n) gives the number of unlabeled binary trees with n leaves and n-1 internal nodes. - Andrew Howroyd, Apr 01 2023

A Catalan-like formula using consecutive odd numbers. Recall that Catalan numbers (A000108) are given by ((n+1)+(n)-1)!/((n+1)!(n)!).

From David Callan, Jun 01 2006: (Start)

a(n) = number of Dyck (2n)-paths (i.e., semilength = 2n) all of whose interior returns to ground level (if any) occur at or before the (2n-2)-nd step, that is, they occur strictly before the midpoint of the path.

For example, a(2)=7 counts UUUUDDDD, UUUDUDDD, UUDUUDDD, UUDUDUDD, UUUDDUDD, UD.UUUDDD, UD.UUDUDD ("." denotes an interior return to ground level).

This result follows immediately from an involution on Dyck paths, due to Emeric Deutsch, defined by E->E, UPDQ -> UQDP (where E is the empty Dyck path; U=upstep, D=downstep and P,Q are arbitrary Dyck paths), because the involution is fixed-point-free on Dyck (2n)-paths and contains one path of the type being counted in each orbit.

a(n) = Sum_{k=0..n-1} C(2n-1-2k)*C(2k). This identity has the following combinatorial interpretation:

a(n) is the number of odd-GL-marked Dyck (2n-1)-paths. An odd-GL vertex is a vertex at location (2i,0) for some odd i >= 1 (path starts at origin). An odd-GL-marked Dyck path is a Dyck path with one of its odd-GL vertices marked. For example, a(2)=7 counts UUUDDD*, UUDUDD*, UD*UUDD, UDUUDD*, UD*UDUD, UDUDUD*, UUDDUD* (the * denotes the marked odd-GL vertex). (End)

a(n+1) = Sum_{k=0..n} C(k)*C(2*n+1-k), n >= 0, with C(n) = A000108(n), also gives the odd part of the bisection of the half-convolution of the Catalan sequence A000108 with itself. For the definition of the half-convolution of a sequence with itself see a comment on A201204. There one also finds the rule for the o.g.f. given below in the formula section. The even part of this bisection is found under A201205. - Wolfdieter Lang, Jan 05 2012

From Peter Bala, Dec 01 2015: (Start)

Let x = p/q be a positive rational in reduced form with p,q > 0. Define Cat(x) = (1/(2*p + q))*binomial(2*p + q, p). Then Cat(n) = Catalan(n). This sequence is Cat(n + 1/2) = (1/(4*n + 4))*binomial(4*n + 4, 2*n + 1). Cf. A265101 (Cat(n + 1/3)), A265102 (Cat(n + 1/4)) and A265103 (Cat(n + 1/5)).

Number of maximal faces of the rational associahedron Ass(2*n + 1, 2*n + 3). Number of lattice paths from (0, 0) to (2*n + 3, 2*n + 1) using steps of the form (1, 0) and (0, 1) and staying above the line y = (2*n + 1)/(2*n + 3)*x. See Armstrong et al. (End)

Also the number of ordered rooted trees with 2n nodes, most of which are leaves, i.e., the odd bisection of A358585. This follows from Callan's formula below. - Gus Wiseman, Nov 27 2022

a(n) = ((n+1)*F(2*n+3)+(2*n+3)*F(2*(n+1)))/5 with F(n)=A000045(n) (Fibonacci numbers). One half of odd-indexed A001629(n), n >= 2 (Fibonacci convolution).

Convolution of F(2n+1) (A001519) and F(2n+2) (A001906(n+1)). - Graeme McRae, Jun 07 2006

Number of reentrant corners along the lower contours of all directed column-convex polyominoes of area n+3 (a reentrant corner along the lower contour is a vertical step that is followed by a horizontal step). a(n) = Sum_{k=0..ceiling((n+1)/2)} k*A121466(n+3,k). - Emeric Deutsch, Aug 02 2006

From Wolfdieter Lang, Jan 02 2012: (Start)

a(n) = A024458(2*n), n >= 1 (bisection, even arguments).

a(n) is also the odd part of the bisection of the half-convolution of the sequence A000045(n+1), n >= 0, with itself. See a comment on A201204 for the definition of the half-convolution of a sequence with itself. There one also finds the rule for the o.g.f. which in this case is Chato(x)/2 with the o.g.f. Chato(x) = 2*(1-x)/(1-3*x+x^2)^2 of A001629(2*n+3), n >= 0.

(End)

From Wolfdieter Lang, Jan 02 2012: (Start)

chat(n):=a(n+1), n>=0, is the half-convolution of the sequence A000045(n+1), n>=0, with itself. For the definition of half-convolution see a comment on A201204, where also the rule to find the o.g.f. is given. Here the o.g.f. is obtained from (U(x)^2 + U2(x^2))/2 with U(x)=1/(1-x-x^2), the o.g.f. of A000045(n+1), n>=0, and U2(x):=(1-x)/((1+x)*(1-3*x+x^2)) the o.g.f. of A007598(n+1), n>=0. This coincides with the o.g.f. given below in the formula section after x has been divided.

For the bisection of this half-convolution see A027991(n+1) and A001870(n), n>=0.

(End)

For the definition of the half-convolution of a sequence with itself see a comment to A201204.

The odd part of this bisection is found under A065097.

The even part of this bisection of A099924 is found in A203573.

This is also the odd part of the bisection of A201207 (half-convolution of the Lucas sequence with itself). See a comment on A201204 for the definition of half-convolution of a sequence with itself. There the rule for the o.g.f. is given.

In general the exponential (also known as binomial) half-convolution of a sequence {b(n), n>=0} with itself is defined by

bhat(n) := Sum_{k=0..floor(n/2)} binomial(n,k)*b(k)*b(n-k), n>=0.

The e.g.f. of the sequence {bhat(n)} is Bhat(x) = ((B(x))^2 + B2(x^2))/2, with the e.g.f. B(x) of {b(n), n>=0} and the e.g.f. B2(x) := Sum_((b(n)^2/n!)*x^n/n!, n>=0) of the scaled squares. The proof runs along the same line as the one given for the ordinary half-convolution in a comment on A201204. In fact, bhat(n)/n! is the ordinary half-convolution of the sequence {b(n)/n!, n>=0} with itself.

Here b(n) = A000984(n) = binomial(2*n,n), n>=0, B(x) = exp(2*x)*BesselI(0,2*x) (see the Abramowitz-Stegun reference and link under A008277 for BesselI, p. 375, eq. 9.6.10) and B2(x) = hypergeometric([1/2,1/2],[1,1,1],16*x).

Half-convolution of factorials (A000142) with itself. For the definition of the half-convolution of a sequence with itself see a comment to A201204. - Vladimir Reshetnikov, Oct 05 2016

From Wolfdieter Lang, Jan 02 2012: (Start)

a(n) = A024458(2*n-1), n>=1 (bisection, odd arguments).

chate(n):=a(n+1), n>=0, is the even part of the bisection of the half-convolution of the sequence A000045(n+1), n>=0, with itself. See a comment on A201204 for the definition of half-convolution. There one finds also the rule for the o.g.f.s of the bisection. Here the o.g.f. of the sequence chate(n), n>=0, is Chate(x):= (Ce(x)+U2(x))/2 with Ce(x)=(1-x+x^2)/(1-3*x+x^2)^2, the o.g.f. of A054444(n), and

U2(x)=(1-x)/((1+x)*(1-3*x+x^2)), the o.g.f. of A007598(n+1), n>=0. This results (after multiplying with x) in the o.g.f. given below in the formula section. It is equivalent to the explicit formula given there, as can be seen after a partial fraction decomposition of the o.g.f.

(End)