cp's OEIS Frontend

Sequence also gives values of x satisfying 3*y^2 - x^2 = 2, the corresponding y being given by A001835(n+1). Moreover, quadruples(p, q, r, s) satisfying p^2 + q^2 + r^2 = s^2, where p = q and r is either p+1 or p-1, are termed nearly isosceles Pythagorean and are given by p = {x + (-1)^n}/3, r = p-(-1)^n, s = y for n > 1. - Lekraj Beedassy, Jul 19 2002

a(n)= A002531(1+2*n). - Anton Vrba (antonvrba(AT)yahoo.com), Feb 14 2007

361 written in base A001835(n+1) - 1 is the square of a(n). E.g., a(12) = 2672279, A001835(13) - 1 = 1542840. We have 361_(1542840) = 3*1542840 + 6*1542840 + 1 = 2672279^2. - Richard Choulet, Oct 04 2007

The lower principal convergents to 3^(1/2), beginning with 1/1, 5/3, 19/11, 71/41, comprise a strictly increasing sequence; numerators=A001834, denominators=A001835. - Clark Kimberling, Aug 27 2008

General recurrence is a(n) = (a(1) - 1)*a(n-1) - a(n-2), a(1) >= 4, lim_{n->infinity} a(n) = x*(k*x + 1)^n, k = (a(1) - 3), x = (1 + sqrt((a(1) + 1)/(a(1) - 3)))/2. Examples in OEIS: a(1) = 4 gives A002878, primes in it A121534. a(1) = 5 gives A001834, primes in it A086386. a(1) = 6 gives A030221, primes in it A299109. a(1) = 7 gives A002315, primes in it A088165. a(1) = 8 gives A033890, primes in it not in OEIS (do there exist any?). a(1) = 9 gives A057080, primes in {71, 34649, 16908641, ...}. a(1) = 10 gives A057081, primes in it {389806471, 192097408520951, ...}. - Ctibor O. Zizka, Sep 02 2008

Inverse binomial transform of A030192. - Philippe Deléham, Nov 19 2009

For positive n, a(n) equals the permanent of the (2*n) X (2*n) tridiagonal matrix with sqrt(6)'s along the main diagonal, and i's along the superdiagonal and the subdiagonal (i is the imaginary unit). - John M. Campbell, Jul 08 2011

x-values in the solution to 3x^2 + 6 = y^2 (see A082841 for the y-values). - Sture Sjöstedt, Nov 25 2011

Pisano period lengths: 1, 1, 2, 4, 3, 2, 8, 4, 6, 3, 10, 4, 12, 8, 6, 8, 18, 6, 5, 12, ... - R. J. Mathar, Aug 10 2012

The aerated sequence (b(n))A100047%20for%20a%20connection%20with%20Chebyshev%20polynomials.%20-%20_Peter%20Bala">{n>=1} = [1, 0, 5, 0, 19, 0, 71, 0, ...] is a fourth-order linear divisibility sequence; that is, if n | m then b(n) | b(m). It is the case P1 = 0, P2 = -2, Q = -1 of the 3-parameter family of divisibility sequences found by Williams and Guy. See A100047 for a connection with Chebyshev polynomials. - _Peter Bala, Mar 22 2015

Yong Hao Ng has shown that for any n, a(n) is coprime with any member of A001835 and with any member of A001075. - René Gy, Feb 26 2018

From Wolfdieter Lang, Oct 15 2020: (Start)

((-1)^n)*a(n) = X(n) = (-1)^n*(S(n, 4) + S(n-1, 4)) and Y(n) = X(n-1) gives all integer solutions (modulo sign flip between X and Y) of X^2 + Y^2 + 4*X*Y = +6, for n = -oo..+oo, with Chebyshev S polynomials (see A049310), with S(-1, x) = 0, and S(-|n|, x) = - S(|n|-2, x), for |n| >= 2.

This binary indefinite quadratic form of discriminant 12, representing 6, has only this family of proper solutions (modulo sign flip), and no improper ones.

This comment is inspired by a paper by Robert K. Moniot (private communication). See his Oct 04 2020 comment in A027941 related to the case of x^2 + y^2 - 3*x*y = -1 (special Markov solutions). (End)

Floretion Algebra Multiplication Program, FAMP Code: A001834 = (4/3)vesseq[ - .25'i + 1.25'j - .25'k - .25i' + 1.25j' - .25k' + 1.25'ii' + .25'jj' - .75'kk' + .75'ij' + .25'ik' + .75'ji' - .25'jk' + .25'ki' - .25'kj' + .25e], apart from initial term

Fill an infinity X infinity matrix with numbers so that 1..n^2 appear in the top left n X n corner for all n; write down the minimal elements in the rows and columns and sort into increasing order; maximize this list in the lexicographic order.

From Donald S. McDonald, Jan 09 2003: (Start)

Numbers of the form n^2 + 1 or n^2 + n + 1.

Locations of right angle turns in Ulam square spiral. (End)

a(n-1) (for n >= 1) is also the number u of unique Fibonacci/Lucas type sequences generated (the total number t of these sequences being a triangular number). Sum(n+1)=t. Then u=Sum((n+1/2) minus 0.5 for odd terms) except for the initial term. E.g., u=13: (n=6)+1 = 7; then 7/2 - 0.5 =3. So u = Sum(1, 1, 1, 2, 2, 3, 3) = 13. - Marco Matosic, Mar 11 2003

Number of (3412,123)-avoiding involutions in S_n.

Schur's Theorem (1905): the maximum number of mutually commuting linearly independent complex matrices of order n is floor((n^2)/4) + 1. - Jonathan Vos Post, Apr 03 2007

Let A be the Hessenberg n X n matrix defined by: A[1,j]=j mod 2, A[i,i]:=1, A[i,i-1]=-1, and A[i,j]=0 otherwise. Then, for n >= 1, a(n)=(-1)^(n-1)*coeff(charpoly(A,x),x). - Milan Janjic, Jan 24 2010

Except for the initial two terms, A033638 gives iterates of the nonsquare function: c(n) = f(c(n-1)), where f(n) = A000037(n) = n + floor(1/2 + sqrt(n)) = n-th nonsquare, starting with c(1)=2. - Clark Kimberling, Dec 28 2010

For n >= 1: for all permutations of [0..n-1]: number of distinct values taken by Sum_{k=0..n-1} (k mod 2) * pi(k). - Joerg Arndt, Apr 22 2011

First differences are A110654. - Jon Perry, Sep 12 2012

Number of (weakly) unimodal compositions of n with maximal part <= 2, see example. - Joerg Arndt, May 10 2013

Construct an infinite triangular matrix with 1's in the leftmost column and the natural numbers in all other columns but shifted down twice. Square the triangle and the sequence is the leftmost column vector. - Gary W. Adamson, Jan 27 2014

Equals the sum of terms in upward sloping diagonals of an infinite lower triangle with 1's in the leftmost column and the natural numbers in all other columns. - Gary W. Adamson, Jan 29 2014

a(n) is the number of permutations of length n avoiding both 213 and 321 in the classical sense which are breadth-first search reading words of increasing unary-binary trees. For more details, see the entry for permutations avoiding 231 at A245898. - Manda Riehl, Aug 05 2014

Number of partitions of n with no more than 2 parts > 1. - Wouter Meeussen, Feb 22 2015, revised Apr 24 2023

Number of possible values for the area of a polyomino whose perimeter is 2n + 4. - Luc Rousseau, May 10 2018

a(n) is the number of 231-avoiding even Grassmannian permutations of size n+1. - Juan B. Gil, Mar 10 2023

For n > 0, a(n) is the smallest number that requires n iterations of the map k -> k - floor(sqrt(k)) to reach 0. - Jon E. Schoenfield, Jun 24 2023

a(n) agrees with the lower matching number of the (n + 1) X (n + 1) black bishop graph from n = 1 up to at least n = 14. - Eric W. Weisstein, Dec 23 2024

For n > 0, obtain a positive integer a(n+1) recursively from a(n) by minimizing a(n+1) > a(n) so that each gap between a(k) and a(k+1) for 1 <= k <= n is used at most twice. - Gerold Jäger, Jun 04 2025

From Roger Ford, May 19 2025: (Start)

a(n) = the number of different total arch lengths for the top arches of semi-meanders with n+2 arches.

Example: Each arch length equals 1 + the number of covered arches.

For semi-meanders with 5 top arches there are 3 different values.

/\

//\\ /\ /\

///\\\ //\\ /\ / \

////\\\\ /\ ///\\\ //\\ //\/\\ /\ /\

Total arch lengths: 4+3+2+1 +1= 11 3+2+1 2+1= 9 3+1+1 +1 +1= 7, so a(3) = 3.

For semi-meanders with 6 top arches there are 5 values: 8, 10, 12, 14, 16, so a(4) = 5. (End)

Matrix inverse of A124645.

Let L(n,x) = Sum_{k=0..n} T(n,k)*x^(n-k) and Pi=3.14...:

L(n,x) = Product_{k=1..n} (x - 2*cos((2*k-1)*Pi/(2*n+1)));

Sum_{k=0..n} T(n,k) = L(n,1) = A010892(n+1);

Sum_{k=0..n} abs(T(n,k)) = A000045(n+2);

abs(T(n,k)) = A065941(n,k), T(n,k) = A065941(n,k)*A087960(k);

T(2*n,k) + T(2*n+1,k+1) = 0 for 0 <= k <= 2*n;

T(n,0) = A000012(n) = 1; T(n,1) = -1 for n > 0;

T(n,2) = -(n-1) for n > 1; T(n,3) = A000027(n)=n for n > 2;

T(n,4) = A000217(n-3) for n > 3; T(n,5) = -A000217(n-4) for n > 4;

T(n,6) = -A000292(n-5) for n > 5; T(n,7) = A000292(n-6) for n > 6;

T(n,n-3) = A058187(n-3)*(-1)^floor(n/2) for n > 2;

T(n,n-2) = A008805(n-2)*(-1)^floor((n+1)/2) for n > 1;

T(n,n-1) = A008619(n-1)*(-1)^floor(n/2) for n > 0;

T(n,n) = L(n,0) = (-1)^floor((n+1)/2);

L(n,1) = A010892(n+1); L(n,-1) = A061347(n+2);

L(n,2) = 1; L(n,-2) = A005408(n)*(-1)^n;

L(n,3) = A001519(n); L(n,-3) = A002878(n)*(-1)^n;

L(n,4) = A001835(n+1); L(n,-4) = A001834(n)*(-1)^n;

L(n,5) = A004253(n); L(n,-5) = A030221(n)*(-1)^n;

L(n,6) = A001653(n); L(n,-6) = A002315(n)*(-1)^n;

L(n,7) = A049685(n); L(n,-7) = A033890(n)*(-1)^n;

L(n,8) = A070997(n); L(n,-8) = A057080(n)*(-1)^n;

L(n,9) = A070998(n); L(n,-9) = A057081(n)*(-1)^n;

L(n,10) = A072256(n+1); L(n,-10) = A054320(n)*(-1)^n;

L(n,11) = A078922(n+1); L(n,-11) = A097783(n)*(-1)^n;

L(n,12) = A077417(n); L(n,-12) = A077416(n)*(-1)^n;

L(n,13) = A085260(n);

L(n,14) = A001570(n); L(n,-14) = A028230(n)*(-1)^n;

L(n,n) = A108366(n); L(n,-n) = A108367(n).

Row n of the matrix inverse (A124645) has g.f.: x^floor(n/2)*(1-x)^(n-floor(n/2)). - Paul D. Hanna, Jun 12 2005

From L. Edson Jeffery, Mar 12 2011: (Start)

Conjecture: Let N=2*n+1, with n > 2. Then T(n,k) (0 <= k <= n) gives the k-th coefficient in the characteristic function p_N(x)=0, of degree n in x, for the n X n tridiagonal unit-primitive matrix G_N (see [Jeffery]) of the form

G_N=A_{N,1}=

(0 1 0 ... 0)

(1 0 1 0 ... 0)

(0 1 0 1 0 ... 0)

...

(0 ... 0 1 0 1)

(0 ... 0 1 1),

with solutions phi_j = 2*cos((2*j-1)*Pi/N), j=1,2,...,n. For example, for n=3,

G_7=A_{7,1}=

(0 1 0)

(1 0 1)

(0 1 1).

We have {T(3,k)}=(1,-1,-2,1), while the characteristic function of G_7 is p(x) = x^3-x^2-2*x+1 = 0, with solutions phi_j = 2*cos((2*j-1)*Pi/7), j=1,2,3. (End)

The triangle sums, see A180662 for their definitions, link A108299 with several sequences, see the crossrefs. - Johannes W. Meijer, Aug 08 2011

The roots to the polynomials are chaotic using iterates of the operation (x^2 - 2), with cycle lengths L and initial seeds returning to the same term or (-1)* the seed. Periodic cycle lengths L are shown in A003558 such that for the polynomial represented by row r, the cycle length L is A003558(r-1). The matrices corresponding to the rows as characteristic polynomials are likewise chaotic [cf. Kappraff et al., 2005] with the same cycle lengths but substituting 2*I for the "2" in (x^2 - 2), where I = the Identity matrix. For example, the roots to x^3 - x^2 - 2x + 1 = 0 are 1.801937..., -1.246979..., and 0.445041... With 1.801937... as the initial seed and using (x^2 - 2), we obtain the 3-period trajectory of 8.801937... -> 1.246979... -> -0.445041... (returning to -1.801937...). We note that A003558(2) = 3. The corresponding matrix M is: [0,1,0; 1,0,1; 0,1,1,]. Using seed M with (x^2 - 2*I), we obtain the 3-period with the cycle completed at (-1)*M. - Gary W. Adamson, Feb 07 2012

(x,y) = (a(n), a(n+1)) are solutions of (x+y)^2/(1+xy)=9, the other solutions are in A033888. - Floor van Lamoen, Dec 10 2001

This sequence consists of the odd-indexed terms of A001906 (whose terms are the values of x such that 5*x^2 + 4 is a square). The even-indexed terms of A001906 are in A033888. Limit_{n->infinity} a(n)/a(n-1) = phi^4 = (7 + 3*sqrt(5))/2. - Gregory V. Richardson, Oct 13 2002

General recurrence is a(n) = (a(1)-1)*a(n-1) - a(n-2), a(1) >= 4, lim_{n->infinity} a(n) = x*(k*x+1)^n, k = a(1) - 3, x = (1 + sqrt((a(1)+1)/(a(1)-3)))/2. Examples in OEIS: a(1)=4 gives A002878. a(1)=5 gives A001834. a(1)=6 gives A030221. a(1)=7 gives A002315. a(1)=8 gives A033890. a(1)=9 gives A057080. a(1)=10 gives A057081. - Ctibor O. Zizka, Sep 02 2008

Indices of square numbers which are also 12-gonal. - Sture Sjöstedt, Jun 01 2009

For positive n, a(n) equals the permanent of the (2n) X (2n) tridiagonal matrix with 3's along the main diagonal, and i's along the superdiagonal and the subdiagonal (i is the imaginary unit). - John M. Campbell, Jul 08 2011

If we let b(0) = 0 and, for n >= 1, b(n) = A033890(n-1), then the sequence b(n) will be F(4n-2) and the first difference is L(4n) or A056854. F(4n-2) is also the ratio of golden spiral length (rounded to the nearest integer) after n rotations. L(4n) is also the pitch length ratio. See illustration in links. - Kival Ngaokrajang, Nov 03 2013

The aerated sequence (b(n))n>=1 = [1, 0, 8, 0, 55, 0, 377, 0, ...] is a fourth-order linear divisibility sequence; that is, if n | m then b(n) | b(m). It is the case P1 = 0, P2 = -5, Q = -1 of the 3-parameter family of divisibility sequences found by Williams and Guy. See A100047. - Peter Bala, Mar 22 2015

Solutions y of Pell equation x^2 - 5*y^2 = 4; corresponding x values are in A342710 (see A342709). - Bernard Schott, Mar 19 2021

For any numbers, A and B, both appearing in the sequence, if gcd(A,B)=1, then A*B is also in the sequence. - Andrew Weimholt, Jul 01 2008

The primes in this sequence are the NSW primes (A088165). For the terms less than 2^31, the only powers greater than 1 appearing in the prime factorization of numbers are 3^3 and 13^2. It appears that all terms are +-1 (mod 8). See A224988 for even numbers. - T. D. Noe, Jul 06 2008, Apr 25 2013

A basis for this sequence is given by A002315. This can be considered as the convergents of quasiregular continued fractions or a special 6-ary numeration system (see A. S. Fraenkel) which gives the characterization of positions of some heap or Wythoff game. What is the Sprague-Grundy function of this game?

Sequence generalized: sigma_r-numbers are numbers n for which sigma_r(n)/sigma_0(n) = c^r. Sigma_r(n) denotes sum of r-th powers of divisors of n; c,r positive integers. This sequence are sigma_2-numbers, A003601 are sigma_1-numbers. In a weaker form we have sigma_r(n)/sigma_0(n) = c^t; t is an integer from <1,r>. - Ctibor O. Zizka, Jul 14 2008

The primes in this sequence are prime numerators with an odd index in A001333. The RMS values (A141812) of prime RMS numbers (this sequence) are prime Pell numbers (A000129) with an odd index. - Ctibor O. Zizka, Aug 13 2008

From Ctibor O. Zizka, Aug 30 2008: (Start)

The set of RMS numbers n could be split into subsets according to the number and form of divisors of n. By definition, RMS(n) = sqrt(sigma_2(n) / sigma_0(n)) should be an integer. Now consider some examples. For n prime number, n has 2 divisors [1,n] and we have to solve Pell's equation n^2 = 2*C^2 - 1; C positive integer. The solution is a prime n of the form u(i) = 6*u(i-1) - u(i-2), i >= 2, u(0)=1, u(1)=7, known as an NSW prime (A088165). For n = p_1*p_2, p_1 and p_2 primes, n has 4 divisors {1; p_1; p_2; p_1*p_2}. There are 2 possible cases. Firstly p^2 = (2*C)^2 - 1 which does not hold for any prime p; secondly p_1^2 = 2*C_1^2 - 1 and p_2^2 = 2*C_2^2 - 1; C_1 and C_2 positive integers.

The solution is that p_1 and p_2 are different NSW primes. If n = p^3, divisors of n are {1; p; p^2; p^3} and we have to solve the Diophantine equation (p^8 - 1)/(p - 1) = (2*C)^2. This equation has no solution for any prime p. RMS numbers n with 4 divisors are only of the form n = p_1*p_2, with p_1 and p_2 NSW primes. The general case is n = p_1*...*p_t, n has 2^t divisors, and for t >= 3, NSW primes are not the only solution. If some of the prime divisors are equals p_i = p_j = ... = p_k, the general case n = p_1*...*p_t is "degenerate" because of the multiplicity of prime factors and therefore n has fewer than 2^t divisors. (End)

General recurrence is a(n) = (a(1)-1)*a(n-1) - a(n-2), a(1) >= 4, lim_{n->infinity} a(n) = x*(k*x+1)^n, k = a(1) - 3, x = (1 + sqrt((a(1)+1)/(a(1)-3)))/2. Examples in OEIS: a(1)=4 gives A002878, whose prime terms give A121534. a(1)=5 gives A001834, whose prime terms give A086386. a(1)=6 gives A030221, whose prime terms {29, 139, 3191, ...} are not a sequence on the OEIS. a(1)=7 gives A002315, whose prime terms give A088165. a(1)=8 gives A033890; the OEIS does not have its prime terms as a sequence (do there exist any prime terms?). a(1)=9 gives A057080, whose prime terms {71, 34649, 16908641, ...} are not a sequence in the OEIS. a(1)=10 gives A057081, whose prime terms {389806471, 192097408520951, ...} are not a sequence in the OEIS. - Ctibor O. Zizka, Sep 02 2008

16 of the first 1660 terms are even (the smallest is 2217231104). The first 16 even terms are all divisible by 30976. - Donovan Johnson, Apr 16 2013

All the 83 even terms up to 10^13 (see A224988) are divisible by 30976. - Giovanni Resta, Oct 29 2019

Riordan array ((1+x)/(1-x)^2, x/(1-x)^2). Row sums are A002878. Diagonal sums are A003945. Inverse is A113187. An interesting factorization is (1/(1-x), x/(1-x))(1+2*x, x*(1+x)). - Paul Barry, Oct 17 2005

Central coefficients of rows with odd numbers of term are A052227.

From Wolfdieter Lang, Jun 26 2011: (Start)

T(k,s) appears as T_s(k) in the Knuth reference, p. 285.

This triangle is related to triangle A156308(n,m), appearing in this reference as U_m(n) on p. 285, by T(k,s) - T(k-1,s) = A156308(k,s), k>=s>=1 (identity on p. 286). T(k,s) = A156308(k+1,s+1) - A156308(k,s+1), k>=s>=0 (identity on p. 286).

(End)

A111125 is jointly generated with A208513 as an array of coefficients of polynomials v(n,x): initially, u(1,x)= v(1,x)= 1; for n>1, u(n,x)= u(n-1,x) +x*(x+1)*v(n-1) and v(n,x)= u(n-1,x) +x*v(n-1,x) +1. See the Mathematica section. The columns of A111125 are identical to those of A208508. Here, however, the alternating row sums are periodic (with period 1,2,1,-1,-2,-1). - Clark Kimberling, Feb 28 2012

This triangle T(k,s) (with signs and columns scaled with powers of 5) appears in the expansion of Fibonacci numbers F=A000045 with multiples of odd numbers as indices in terms of odd powers of F-numbers. See the Jennings reference, p. 108, Theorem 1. Quoted as Lemma 3 in the Ozeki reference given in A111418. The formula is: F_{(2*k+1)*n} = Sum_{s=0..k} ( T(k,s)*(-1)^((k+s)*n)*5^s*F_{n}^(2*s+1) ), k >= 0, n >= 0. - Wolfdieter Lang, Aug 24 2012

From Wolfdieter Lang, Oct 18 2012: (Start)

This triangle T(k,s) appears in the formula x^(2*k+1) - x^(-(2*k+1)) = Sum_{s=0..k} ( T(k,s)*(x-x^(-1))^(2*s+1) ), k>=0. Prove the inverse formula (due to the Riordan property this will suffice) with the binomial theorem. Motivated to look into this by the quoted paper of Wang and Zhang, eq. (1.4).

Alternating row sums are A057079.

The Z-sequence of this Riordan array is A217477, and the A-sequence is (-1)^n*A115141(n). For the notion of A- and Z-sequences for Riordan triangles see a W. Lang link under A006232. (End)

The signed triangle ((-1)^(k-s))*T(k,s) gives the coefficients of (x^2)^s of the polynomials C(2*k+1,x)/x, with C the monic integer Chebyshev T-polynomials whose coefficients are given in A127672 (C is there called R). See the odd numbered rows there. This signed triangle is the Riordan array ((1-x)/(1+x)^2, x/(1+x)^2). Proof by comparing the o.g.f. of the row polynomials where x is replaced by x^2 with the odd part of the bisection of the o.g.f. for C(n,x)/x. - Wolfdieter Lang, Oct 23 2012

From Wolfdieter Lang, Oct 04 2013: (Start)

The signed triangle S(k,s) := ((-1)^(k-s))*T(k,s) (see the preceding comment) is used to express in a (4*(k+1))-gon the length ratio s(4*(k+1)) = 2*sin(Pi/4*(k+1)) = 2*cos((2*k+1)*Pi/(4*(k+1))) of a side/radius as a polynomial in rho(4*(k+1)) = 2*cos(Pi/4*(k+1)), the length ratio (smallest diagonal)/side:

s(4*(k+1)) = Sum_{s=0..k} ( S(k,s)*rho(4*(k+1))^(2*s+1) ).

This is to be computed modulo C(4*(k+1), rho(4*(k+1)) = 0, the minimal polynomial (see A187360) in order to obtain s(4*(k+1)) as an integer in the algebraic number field Q(rho(4*(k+1))) of degree delta(4*(k+1)) (see A055034). Thanks go to Seppo Mustonen for asking me to look into the problem of the square of the total length in a regular n-gon, where this formula is used in the even n case. See A127677 for the formula in the (4*k+2)-gon. (End)

From Wolfdieter Lang, Aug 14 2014: (Start)

The row polynomials for the signed triangle (see the Oct 23 2012 comment above), call them todd(k,x) = Sum_{s=0..k} ( (-1)^(k-s)*T(k,s)*x^s ) = S(k, x-2) - S(k-1, x-2), k >= 0, with the Chebyshev S-polynomials (see their coefficient triangle (A049310) and S(-1, x) = 0), satisfy the recurrence todd(k, x) = (-1)^(k-1)*((x-4)/2)*todd(k-1, 4-x) + ((x-2)/2)*todd(k-1, x), k >= 1, todd(0, x) = 1. From the Aug 03 2014 comment on A130777.

This leads to a recurrence for the signed triangle, call it S(k,s) as in the Oct 04 2013 comment: S(k,s) = (1/2)*(1 + (-1)^(k-s))*S(k-1,s-1) + (2*(s+1)*(-1)^(k-s) - 1)*S(k-1,s) + (1/2)*(-1)^(k-s)*Sum_{j=0..k-s-2} ( binomial(j+s+2,s)*4^(j+2)* S(k-1, s+1+j) ) for k >= s >= 1, and S(k,s) = 0 if k < s and S(k,0) = (-1)^k*(2*k+1). Note that the recurrence derived from the Riordan A-sequence A115141 is similar but has simpler coefficients: S(k,s) = sum(A115141(j)*S(k-1,s-1+j), j=0..k-s), k >= s >=1.

(End)

From Tom Copeland, Nov 07 2015: (Start)

Rephrasing notes here: Append an initial column of zeros, except for a 1 at the top, to A111125 here. Then the partial sums of the columns of this modified entry are contained in A208513. Append an initial row of zeros to A208513. Then the difference of consecutive pairs of rows of the modified A208513 generates the modified A111125. Cf. A034807 and A127677.

For relations among the characteristic polynomials of Cartan matrices of the Coxeter root groups, Chebyshev polynomials, cyclotomic polynomials, and the polynomials of this entry, see Damianou (p. 20 and 21) and Damianou and Evripidou (p. 7).

As suggested by the equations on p. 7 of Damianou and Evripidou, the signed row polynomials of this entry are given by (p(n,x))^2 = (A(2*n+1, x) + 2)/x = (F(2*n+1, (2-x), 1, 0, 0, ... ) + 2)/x = F(2*n+1, -x, 2*x, -3*x, ..., (-1)^n n*x)/x = -F(2*n+1, x, 2*x, 3*x, ..., n*x)/x, where A(n,x) are the polynomials of A127677 and F(n, ...) are the Faber polynomials of A263196. Cf. A127672 and A127677.

(End)

The row polynomials P(k, x) of the signed triangle S(k, s) = ((-1)^(k-s))*T(k, s) are given from the row polynomials R(2*k+1, x) of triangle A127672 by

P(k, x) = R(2*k+1, sqrt(x))/sqrt(x). - Wolfdieter Lang, May 02 2021

Number of (s(0), s(1), ..., s(2n+4)) such that 0 < s(i) < 10 and |s(i) - s(i-1)| = 1 for i = 1,2,...,2n+4, s(0) = 1, s(2n+4) = 5. - Herbert Kociemba, Jun 14 2004

Binomial transform of A002878. - Philippe Deléham, Oct 04 2005

Diagonal of square array A216219. - Philippe Deléham, Mar 15 2013

Lim_{n->oo} a(n+1)/a(n) = 2 + phi = A296184, where phi = A001622. - Wolfdieter Lang, Nov 16 2023~

This is the r=5 member in the r-family of sequences S_r(n) defined in A092184 where more information can be found.

Number of spanning trees of the wheel W_n on n+1 vertices. - Emeric Deutsch, Mar 27 2005

Also number of spanning trees of the n-helm graph. - Eric W. Weisstein, Jul 16 2011

a(n) is the smallest number requiring n terms when expressed as a sum of Lucas numbers (A000204). - David W. Wilson, Jan 10 2006

This sequence has a primitive prime divisor for all terms beyond the twelfth. - Anthony Flatters (Anthony.Flatters(AT)uea.ac.uk), Aug 17 2007

From Giorgio Balzarotti, Mar 11 2009: (Start)

Determinant of power series of gamma matrix with determinant 1:

a(n) = Determinant(A + A^2 + A^3 + A^4 + A^5 + ... + A^n)

where A is the submatrix A(1..2,1..2) of the matrix with factorial determinant

A = [[1,1,1,1,1,1,...],[1,2,1,2,1,2,...],[1,2,3,1,2,3,...],[1,2,3,4,1,2,...],

[1,2,3,4,5,1,...],[1,2,3,4,5,6,...],...]. Note: Determinant A(1..n,1..n)= (n-1)!.

See A158039, A158040, A158041, A158042, A158043, A158044, for sequences of matrix 2!,3!,... (End)

The previous comment could be rephrased as: a(n) = -det(A^n - I) where I is the 2 X 2 identity matrix and A = [1, 1; 1, 2]. - Peter Bala, Mar 20 2015

a(n) is also the number of points of Arnold's "cat map" that are on orbits of period n-1. This is a map of the two-torus T^2 into itself. If we regard T^2 as R^2 / Z^2, the action of this map on a two vector in R^2 is multiplication by the unit-determinant matrix A = [2, 1;1, 1], with the vector components taken modulo one. As such, an explicit formula for the n-th entry of this sequence is -det(I-A^n). - Bruce Boghosian, Apr 26 2009

7*a(n) gives the total number of vertices in a heptagonal hyperbolic lattice {7,3} with n total levels, in which an open heptagon is centered at the origin. - Robert M. Ziff, Apr 10 2011

The sequence is the case P1 = 5, P2 = 6, Q = 1 of the 3 parameter family of 4th-order linear divisibility sequences found by Williams and Guy. - Peter Bala, Apr 03 2014

Determinants of the spiral knots S(3,k,(1,-1)). a(k) = det(S(3,k,(1,-1))). These knots are also the weaving knots W(k,3) and the Turk's Head Links THK(3,k). - Ryan Stees, Dec 14 2014

Even-indexed Fibonacci numbers (1, 3, 8, 21, ...) convolved with (1, 2, 2, 2, 2, ...). - Gary W. Adamson, Aug 09 2016

a(n) is the number of ways to tile a bracelet of length n with 1-color square, 2-color dominos, 3-color trominos, etc. - Yu Xiao, May 23 2020

a(n) is the number of face-labeled unfoldings of a pyramid whose base is a simple n-gon. Cf. A103536. - Rick Mabry, Apr 17 2023

a(n) = L(n,-5)*(-1)^n, where L is defined as in A108299; see also A004253 for L(n,+5). - Reinhard Zumkeller, Jun 01 2005

General recurrence is a(n) = (a(1)-1)*a(n-1) - a(n-2), a(1) >= 4; lim_{n->oo} a(n) = x*(k*x+1)^n, k =(a(1)-3), x=(1+sqrt((a(1)+1)/(a(1)-3)))/2. Examples in OEIS: a(1)=4 gives A002878. a(1)=5 gives A001834. a(1)=6 gives the present sequence. a(1)=7 gives A002315. a(1)=8 gives A033890. a(1)=9 gives A057080. a(1)=10 gives A057081. - Ctibor O. Zizka, Sep 02 2008

The primes in this sequence are 29, 139, 3191, 15289, 350981, 1681651, ... - Ctibor O. Zizka, Sep 02 2008

Inverse binomial transform of A030240. - Philippe Deléham, Nov 19 2009

For positive n, a(n) equals the permanent of the (2n)X(2n) matrix with sqrt(7)'s along the main diagonal, and i's along the superdiagonal and the subdiagonal (i is the imaginary unit). - John M. Campbell, Jul 08 2011

The aerated sequence (b(n))n>=1 = [1, 0, 6, 0, 29, 0, 139, 0, ...] is a fourth-order linear divisibility sequence; that is, if n | m then b(n) | b(m). It is the case P1 = 0, P2 = -3, Q = -1 of the 3-parameter family of divisibility sequences found by Williams and Guy. See A100047 for a connection with Chebyshev polynomials. - Peter Bala, Mar 22 2015

From Wolfdieter Lang, Oct 26 2020: (Start)

((-1)^n)*a(n) = X(n) = ((-1)^n)*(S(n, 5) + S(n-1, 5)) and Y(n) = X(n-1) gives all integer solutions (modulo sign flip between X and Y) of X^2 + Y^2 + 5*X*Y = +7, for n = -oo..+oo, with Chebyshev S polynomials (see A049310), with S(-1, x) = 0, and S(-n, x) = - S(n-2, x), for n >= 2.

This binary indefinite quadratic form of discriminant 21, representing 7, has only this family of proper solutions (modulo sign flip), and no improper ones.

This comment is inspired by a paper by Robert K. Moniot (private communication). See his Oct 04 2020 comment in A027941 related to the case of x^2 + y^2 - 3*x*y = -1 (special Markov solutions). (End)

Next term a(9) is too large (99 digits) to include here. - Ray Chandler, Sep 21 2003

These primes are the prime RMS numbers (A140480): primes p such that (1+p^2)/2 is a square r^2. Then r is a Pell number, A000129. - T. D. Noe, Jul 01 2008

Also prime numerators with an odd index in A001333. - Ctibor O. Zizka, Aug 13 2008

r in the above note of T. D. Noe is a prime Pell number (A000129) with an odd index. - Ctibor O. Zizka, Aug 13 2008

General recurrence is a(n) = (a(1)-1)*a(n-1) - a(n-2), a(1) >= 4, lim_{n->infinity} a(n) = x*(k*x+1)^n, k = a(1)-3, x = (1+sqrt((a(1)+1)/(a(1)-3)))/2. Examples in the OEIS: a(1)=4 gives A002878, primes in it A121534. a(1)=5 gives A001834, primes in it A086386. a(1)=6 gives A030221, primes in it not in the OEIS {29, 139, 3191, ...}. a(1)=7 gives A002315, primes in it A088165. a(1)=8 gives A033890, primes in it not in the OEIS (do there exist any ?). a(1)=9 gives A057080, primes in it not in the OEIS {71, 34649, 16908641, ...}. a(1)=10 gives A057081, primes in it not in the OEIS {389806471, 192097408520951, ...}. - Ctibor O. Zizka, Sep 02 2008