cp's OEIS Frontend

Number of elements in all subsets of {1,2,...,n-1} with no consecutive integers. Example: a(5)=10 because the subsets of {1,2,3,4} that have no consecutive elements, i.e., {}, {1}, {2}, {3}, {4}, {1,3}, {1,4}, {2,4}, the total number of elements is 10. - Emeric Deutsch, Dec 10 2003

If g is either of the real solutions to x^2-x-1=0, g'=1-g is the other one and phi is any 2 X 2-matricial solution to the same equation, not of the form gI or g'I, then Sum'_{i+j=n-1} g^i phi^j = F_n + (A001629(n) - A001629(n-1)g')*(phi-g'I), where i,j >= 0, F_n is the n-th Fibonacci number and I is the 2 X 2 identity matrix... - Michele Dondi (blazar(AT)lcm.mi.infn.it), Apr 06 2004

Number of 3412-avoiding involutions containing exactly one subsequence of type 321.

Number of binary sequences of length n with exactly one pair of consecutive 1's. - George J. Schaeffer (gschaeff(AT)andrew.cmu.edu), Sep 02 2004

For this sequence the n-th term is given by (nF(n+1)-F(n)+nF(n-1))/5 where F(n) is the n-th Fibonacci number. - Mrs. J. P. Shiwalkar and M. N. Deshpande (dpratap_ngp(AT)sancharnet.in), Apr 20 2005

If an unbiased coin is tossed n times then there are 2^n possible strings of H and T. Out of these, number of strings with exactly one 'HH' is given by a(n) where a(n) denotes n-th term of this sequence. - Mrs. J. P. Shiwalkar and M. N. Deshpande (dpratap_ngp(AT)sancharnet.in), May 04 2005

a(n) is half the number of horizontal dominoes in all domino tilings of a horizontally aligned 2 X n rectangle; a(n+1) = the number of vertical dominoes in all domino tilings of a horizontally aligned 2 X n rectangle; thus 2*a(n)+a(n+1)=n*F(n+1) = the number of dominoes in all domino tilings of a 2 X n rectangle, where F=A000045, the Fibonacci sequence. - Roberto Tauraso, May 02 2005; Graeme McRae, Jun 02 2006

a(n+1) = (-i)^(n-1)*(d/dx)S(n,x)|A049310%20for%20the%20S-polynomials.%20-%20_Wolfdieter%20Lang">{x=i}, where i is the imaginary unit, n >= 1. First derivative of Chebyshev S-polynomials evaluated at x=i multiplied by (-i)^(n-1). See A049310 for the S-polynomials. - _Wolfdieter Lang, Apr 04 2007

For n >= 4, a(n) is the number of weak compositions of n-2 in which exactly one part is 0 and all other parts are either 1 or 2. - Milan Janjic, Jun 28 2010

For n greater than 1, a(n) equals the absolute value of (1 - (1/2 - i/2)*(1 + (-1)^(n + 1))) times the x-coefficient of the characteristic polynomial of the (n-1) X (n-1) tridiagonal matrix with i's along the main diagonal (i is the imaginary unit), 1's along the superdiagonal and the subdiagonal and 0's everywhere else (see Mathematica code below). - John M. Campbell, Jun 23 2011

For n > 0: a(n) = Sum_{k=1..n-1} (A039913(n-1,k)) / 2. - Reinhard Zumkeller, Oct 07 2012

The right-hand side of a binomial-coefficient identity [Gauthier]. - N. J. A. Sloane, Apr 09 2013

a(n) is the number of edges in the Fibonacci cube Gamma(n-1) (see the Klavzar 2005 reference, p. 149). Example: a(3)=2; indeed, the Fibonacci cube Gamma(2) is the path P(3) having 2 edges. - Emeric Deutsch, Aug 10 2014

a(n) is the number of c(i)'s, including repetitions, in p(n), where p(n)/q(n) is the n-th convergent p(n)/q(n) of the formal infinite continued fraction [c(0), c(1), ...]; e.g., the number of c(i)'s in p(3) = c(0)*c(1)*c(2)*c(3) + c(0)*c(1) + c(0)*c(3) + c(2)*c(3) + 1 is a(5) = 10. - Clark Kimberling, Dec 23 2015

Also the number of maximal and maximum cliques in the (n-1)-Fibonacci cube graph. - Eric W. Weisstein, Sep 07 2017

a(n+1) is the total number of fixed points in all permutations p on 1, 2, ..., n such that |k-p(k)| <= 1 for 1 <= k <= n. - Katharine Ahrens, Sep 03 2019

From Steven Finch, Mar 22 2020: (Start)

a(n+1) is the total binary weight (cf. A000120) of all A000045(n+2) binary sequences of length n not containing any adjacent 1's.

The only three 2-bitstrings without adjacent 1's are 00, 01 and 10. The bitsums of these are 0, 1 and 1. Adding these give a(3)=2.

The only five 3-bitstrings without adjacent 1's are 000, 001, 010, 100 and 101. The bitsums of these are 0, 1, 1, 1 and 2. Adding these give a(4)=5.

The only eight 4-bitstrings without adjacent 1's are 0000, 0001, 0010, 0100, 1000, 0101, 1010 and 1001. The bitsums of these are 0, 1, 1, 1, 1, 2, 2, and 2. Adding these give a(5)=10. (End)

Number of tilings of a 1 X n strip with monominoes (1 X 1 squares) and at least one domino (1 X 2 rectangles), where exactly one of the dominoes is colored gold. - Greg Dresden and Jiachen Weng, Jul 31 2025

For some applications it is better to start this sequence with an extra 1 at the beginning: 1, 1, 2, 37, 266, 2431, 27007, 353522, 5329837, ... (again with offset 0). This sequence now has its own entry - see A144301.

Number of partitions of {1,...,k}, n <= k <= 2n, into n blocks with no more than 2 elements per block. Restated, number of ways to use the elements of {1,...,k}, n <= k <= 2n, once each to form a collection of n sets, each having 1 or 2 elements. - Bob Proctor, Apr 18 2005, Jun 26 2006. E.g., for n=2 we get: (k=2): {1,2}; (k=3): {1,23}, {2,13}, {3,12}; (k=4): {12,34}, {13,24}, {14,23}, for a total of a(2) = 7 partitions.

Equivalently, number of sequences of n unlabeled items such that each item occurs just once or twice (cf. A105749). - David Applegate, Dec 08 2008

Numerator of (n+1)-th convergent to 1+tanh(1). - Benoit Cloitre, Dec 20 2002

The following Maple lines show how this sequence and A144505, A144498, A001514, A144513, A144506, A144514, A144507, A144301 are related.

f0:=proc(n) local k; add((n+k)!/((n-k)!*k!*2^k),k=0..n); end; [seq(f0(n),n=0..10)];

# that is this sequence

f1:=proc(n) local k; add((n+k+1)!/((n-k)!*k!*2^k),k=0..n); end; [seq(f1(n),n=0..10)];

# that is A144498

f2:=proc(n) local k; add((n+k+2)!/((n-k)!*k!*2^k),k=0..n); end; [seq(f2(n),n=0..10)];

# that is A144513; divided by 2 gives A001514

f3:=proc(n) local k; add((n+k+3)!/((n-k)!*k!*2^k),k=0..n); end; [seq(f3(n),n=0..10)];

# that is A144514; divided by 6 gives A144506

f4:=proc(n) local k; add((n+k+4)!/((n-k)!*k!*2^k),k=0..n); end; [seq(f4(n),n=0..10)];

# that divided by 24 gives A144507

a(n) is also the numerator of the continued fraction sequence beginning with 2 followed by 3 and the remaining odd numbers: [2,3,5,7,9,11,13,...]. - Gil Broussard, Oct 07 2009

Also, number of scenarios in the Gift Exchange Game when a gift can be stolen at most once. - N. J. A. Sloane, Jan 25 2017

The row polynomials with exponents in increasing order (e.g., third row: 1+3x+3x^2) are Grosswald's y_{n}(x) polynomials, p. 18, Eq. (7).

Also called Bessel numbers of first kind.

The triangle a(n,k) has factorization [C(n,k)][C(k,n-k)]Diag((2n-1)!!) The triangle a(n-k,k) is A100861, which gives coefficients of scaled Hermite polynomials. - Paul Barry, May 21 2005

Related to k-matchings of the complete graph K_n by a(n,k)=A100861(n+k,k). Related to the Morgan-Voyce polynomials by a(n,k)=(2k-1)!!*A085478(n,k). - Paul Barry, Aug 17 2005

Related to Hermite polynomials by a(n,k)=(-1)^k*A060821(n+k, n-k)/2^n. - Paul Barry, Aug 28 2005

The row polynomials, the Bessel polynomials y(n,x):=Sum_{m=0..n} (a(n,m)*x^m) (called y_{n}(x) in the Grosswald reference) satisfy (x^2)*(d^2/dx^2)y(n,x) + 2*(x+1)*(d/dx)y(n,x) - n*(n+1)*y(n,x) = 0.

a(n-1, m-1), n >= m >= 1, enumerates unordered n-vertex forests composed of m plane (aka ordered) increasing (rooted) trees. Proof from the e.g.f. of the first column Y(z):=1-sqrt(1-2*z) (offset 1) and the Bergeron et al. eq. (8) Y'(z)= phi(Y(z)), Y(0)=0, with out-degree o.g.f. phi(w)=1/(1-w). See their remark on p. 28 on plane recursive trees. For m=1 see the D. Callan comment on A001147 from Oct 26 2006. - Wolfdieter Lang, Sep 14 2007

The asymptotic expansions of the higher order exponential integrals E(x,m,n), see A163931 for information, lead to the Bessel numbers of the first kind in an intriguing way. For the first four values of m these asymptotic expansions lead to the triangles A130534 (m=1), A028421 (m=2), A163932 (m=3) and A163934 (m=4). The o.g.f.s. of the right hand columns of these triangles in their turn lead to the triangles A163936 (m=1), A163937 (m=2), A163938 (m=3) and A163939 (m=4). The row sums of these four triangles lead to A001147, A001147 (minus a(0)), A001879 and A000457 which are the first four right hand columns of A001498. We checked this phenomenon for a few more values of m and found that this pattern persists: m = 5 leads to A001880, m=6 to A001881, m=7 to A038121 and m=8 to A130563 which are the next four right hand columns of A001498. So one by one all columns of the triangle of coefficients of Bessel polynomials appear. - Johannes W. Meijer, Oct 07 2009

a(n,k) also appear as coefficients of (n+1)st degree of the differential operator D:=1/t d/dt, namely D^{n+1}= Sum_{k=0..n} a(n,k) (-1)^{n-k} t^{1-(n+k)} (d^{n+1-k}/dt^{n+1-k}. - Leonid Bedratyuk, Aug 06 2010

a(n-1,k) are the coefficients when expanding (xI)^n in terms of powers of I. Let I(f)(x) := Integral_{a..x} f(t) dt, and (xI)^n := x Integral_{a..x} [ x_{n-1} Integral_{a..x_{n-1}} [ x_{n-2} Integral_{a..x_{n-2}} ... [ x_1 Integral_{a..x_1} f(t) dt ] dx_1 ] .. dx_{n-2} ] dx_{n-1}. Then: (xI)^n = Sum_{k=0..n-1} (-1)^k * a(n-1,k) * x^(n-k) * I^(n+k)(f)(x) where I^(n) denotes iterated integration. - Abdelhay Benmoussa, Apr 11 2025

Discussion of Central Factorial Numbers by N. J. A. Sloane, Feb 01 2011: (Start)

Here is Riordan's definition of the central factorial numbers t(n,k) given in Combinatorial Identities, Section 6.5:

For n >= 0, expand the polynomial

x^[n] = x*Product{i=1..n-1} (x+n/2-i) = Sum_{k=0..n} t(n,k)*x^k.

The t(n,k) are not always integers. The cases n even and n odd are best handled separately.

For n=2m, we have:

x^[2m] = Product_{i=0..m-1} (x^2-i^2) = Sum_{k=1..m} t(2m,2k)*x^(2k).

E.g. x^[8] = x^2(x^2-1^2)(x^2-2^2)(x^2-3^2) = x^8-14x^6+49x^4-36x^2,

which corresponds to row 4 of the present triangle.

So the m-th row of the present triangle gives the absolute values of the coefficients in the expansion of Product_{i=0..m-1} (x^2-i^2).

Equivalently, and simpler, the n-th row gives the coefficients in the expansion of Product_{i=1..n-1}(x+i^2), highest powers first.

For n odd, n=2m+1, we have:

x^[2m+1] = x*Product_{i=0..m-1}(x^2-((2i+1)/2)^2) = Sum_{k=0..m} t(2m+1,2k+1)*x^(2k+1).

E.g. x^[5] = x(x^2-(1/2)^2)(x^2-(3/2)^2) = x^5-10x^3/4+9x/16,

which corresponds to row 2 of the triangle in A008956.

We now rescale to get integers by replacing x by x/2 and multiplying by 2^(2m+1) (getting 1, -10, 9 from the example).

The result is that row m of triangle A008956 gives the coefficients in the expansion of x*Product_{i=0..m} (x^2-(2i+1)^2).

Equivalently, and simpler, the n-th row of A008956 gives the coefficients in the expansion of Product_{i=0..n-1} (x+(2i+1)^2), highest powers first.

Note that the n-th row of A182867 gives the coefficients in the expansion of Product_{i=1..n} (x+(2i)^2), highest powers first.

(End)

Contribution from Johannes W. Meijer, Jun 18 2009: (Start)

We define Beta(n-z,n+z)/Beta(n,n) = Gamma(n-z)*Gamma(n+z)/Gamma(n)^2 = sum(EG2[2m,n]*z^(2m), m = 0..infinity) with Beta(z,w) the Beta function. The EG2[2m,n] coefficients are quite interesting, see A161739. Our definition leads to EG2[2m,1] = 2*eta(2m) and the recurrence relation EG2[2m,n] = EG2[2m,n-1] - EG2[2m-2,n-1]/(n-1)^2 for m = -2, -1, 0, 1, 2, ... and n = 2, 3, ... , with eta(m) = (1-2^(1-m))*zeta(m) with eta(m) the Dirichlet eta function and zeta(m) the Riemann zeta function. We found for the matrix coefficients EG2[2m,n] = sum((-1)^(k+n)*t1(n-1,k-1)*2*eta(2*m-2*n+2*k)/((n-1)!)^2,k=1..n) with the central factorial numbers t1(n,m) as defined above, see also the Maple program.

From the EG2 matrix we arrive at the ZG2 matrix, see A161739 for its odd counterpart, which is defined by ZG2[2m,1] = 2*zeta(2m) and the recurrence relation ZG2[2m,n] = ZG2[2m-2,n-1]/(n*(n-1))-(n-1)*ZG2[2m,n-1]/n for m = -2, -1, 0, 1, 2, ... and n = 2, 3, ... . We found for the ZG2[2m,n] = Sum_{k=1..n} (-1)^(k+1)*t1(n-1,k-1)* 2* zeta(2*m-2*n+2*k)/((n-1)!*(n)!), and we see that the central factorial numbers t1(n,m) once again play a crucial role.

(End)

Convolution of A001906(n), n >= 1 (even-indexed Fibonacci numbers) with itself.

A001787 and this sequence arise in counting ordered trees of height at most k where only the rightmost branch at the root actually achieves this height and the count is by the number of edges, with k = 3 for A001787 and k = 4 for this sequence.

Gives the number of 3412-avoiding permutations containing exactly one subsequence of type 321. - Dan Daly (ddaly(AT)du.edu), Apr 24 2008

From Wolfdieter Lang, Oct 06 2008: (Start)

a(n) is the denominator of the n-th approximant to the continued fraction 1^2/(6+3^2/(6+5^2/(6+... for Pi-3. W. Lang, Oct 06 2008, after an e-mail from R. Rosenthal. Cf. A142970 for the corresponding numerators.

The e.g.f. g(x)=(1+x)/(1-2*x)^(5/2) satisfies (1-4*x^2)*g''(x) - 2*(8*x+3)*g'(x) -9*g(x) = 0 (from the three term recurrence given below). Also g(x)=hypergeom([2,3/2],[1],2*x). (End)

Number of descents in all fixed-point-free involutions of {1,2,...,2(n+1)}. A descent of a permutation p is a position i such that p(i) > p(i+1). Example: a(1)=6 because the fixed-point-free involutions 2143, 3412, and 4321 have 2, 1, and 3 descents, respectively. - Emeric Deutsch, Jun 05 2009

First differences of A193651. - Vladimir Reshetnikov, Apr 25 2016

a(n-2) is the number of maximal elements in the absolute order of the Coxeter group of type D_n. - Jose Bastidas, Nov 01 2021

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)