cp's OEIS Frontend

Also the q-Stirling2 numbers at q = -1. - Peter Luschny, Mar 09 2020

Row sums give the Fibonacci sequence. So do the alternating row sums.

Triangle of coefficients of polynomials defined by p(-1,x) = p(0,x) = 1, p(n, x) = x*p(n-1, x) + p(n-2, x), for n >= 1. - Benoit Cloitre, May 08 2005 [rewritten with correct offset. - Wolfdieter Lang, Feb 18 2020]

Another version of triangle in A103631. - Philippe Deléham, Jan 01 2009

The T(n,k) coefficients appear in appendix 2 of Parks's remarkable article "A new proof of the Routh-Hurwitz stability criterion using the second method of Liapunov" if we assume that the b(n) coefficients are all equal to 1 and ignore the first column. The complete version of this triangle including the first column is A103631. - Johannes W. Meijer, Aug 11 2011

Signed ++--++..., the roots are chaotic using f(x) --> x^2 - 2 with cycle lengths shown in A003558 by n-th rows. Example: given row 3, x^3 + x^2 - 2x - 1; the roots are (a = 1.24697, ...; b = -0.445041, ...; c = -1.802937, ...). Then (say using seed b with x^2 - 2) we obtain the trajectory -0.445041, ... -> -1.80193, ... -> 1.24697, ...; matching the entry "3" in A003558(3). - Gary W. Adamson, Sep 06 2011

From Gary W. Adamson, Aug 25 2019: (Start)

Roots to the polynomials and terms in A003558 can all be obtained from the numbers below using a doubling series mod N procedure as follows: (more than one row may result). Any row ends when the trajectory produces a term already used. Then try the next higher odd term not used as the leftmost term, then repeat.

For example, for N = 11, we get: (1, 2, 4, 3, 5), showing that when confronted with two choices after the 4: (8 and -3), pick the smaller (abs) term, = 3. Then for the next row pick 7 (not used) and repeat the algorithm; succeeding only if the trajectory produces new terms. But 7 is also (-4) mod 11 and 4 was used. Therefore what I call the "r-t table" (for roots trajectory) has only one row: (1, 2, 4, 3, 5). Conjecture: The numbers of terms in the first row is equal to A003558 corresponding to N, i.e., 5 in this case with period 2.

Now for the roots to the polynomials. Pick N = 7. The polynomial is x^3 - x^2 - 2x + 1 = 0, with roots 1.8019..., -1.2469... and 0.445... corresponding to 2*cos(j*Pi/N), N = 7, and j = (1, 2, and 3). The terms (1, 2, 3) are the r-t terms for N = 7. For 11, the r-t terms are (1, 2, 4, 3, 5). This implies that given any roots of the corresponding polynomial, they are cyclic using f(x) --> x^2 - 2 with cycle lengths shown in A003558. The terms thus generated are 2*cos(j*Pi), with j = (1, 2, 4, 3, 5). Check: Begin with 2*j*Pi/N, with j = 1 (1.9189...). The other trajectory terms are: --> 1.6825..., --> 0.83083..., -1.3097...; 545...; (a 5 period and cyclic since we can begin with any of the constants). The r-t table for odd N begins as follows:

3...............1

5...............1, 2

7...............1, 2, 3

9...............1, 2, 4

...............3 (singleton terms reduce to "1") (9 has two rows)

11...............1, 2, 4, 3, 5

13...............1, 2, 4, 5, 3, 6

15...............1, 2, 4, 7

................3, 6 (dividing through by the gcd gives (1, 2))

................5. (singleton terms reduce to "1")

The result is that 15 has 3 factors (since 3 rows), and the values of those factors are the previous terms "N", corresponding to the r-t terms in each row. Thus, the first row is new, the second (1, 2), corresponds to N = 5, and the "1" in row 3 corresponds to N = 3. The factors are those values apart from 15 and 1. Note that all of the unreduced r-t terms in all rows for N form a complete set of the terms 1 through (N-1)/2 without duplication. (End)

From Gary W. Adamson, Sep 30 2019: (Start)

The 3 factors of the 7th degree polynomial for 15: (x^7 - x^6 - 6x^5 + 5x^4 + 10x^3 - 6x^2 - 4x + 1) can be determined by getting the roots for 2*cos(j*Pi/1), j = (1, 2, 4, 7) and finding the corresponding polynomial, which is x^4 + x^3 - 4x^2 - 4x + 1. This is the minimal polynomial for N = 15 as shown in Table 2, p. 46 of (Lang). The degree of this polynomial is 4, corresponding to the entry in A003558 for 15, = 4. The trajectories (3, 6) and (5) are j values for 2*cos(j*Pi/15) which are roots to x^2 - x - 1 (relating to the pentagon), and (x - 1), relating to the triangle. (End)

From Gary W. Adamson, Aug 21 2019: (Start)

Matrices M of the form: (1's in the main diagonal, -1's in the subdiagonal, and the rest zeros) are chaotic if we replace (f(x) --> x^2 - 2) with f(x) --> M^2 - 2I, where I is the Identity matrix [1, 0, 0; 0, 1, 0; 0, 0, 1]. For example, with the 3 X 3 matrix M: [0, 0, 1; 0, 1, -1; 1, -1, 0]; the f(x) trajectory is:

....M^2 - 2I: [-1, -1, 0; -1, 0, -1; 0, -1, 0], then for the latter,

....M^2 - 2I: [0, 1, 1; 1, 0, 0; 1, 0, -1]. The cycle ends with period 3 since the next matrix is (-1) * the seed matrix. As in the case with f(x) --> x^2 - 2, the eigenvalues of the 3 chaotic matrices are (abs) 1.24697, 0.44504... and 1.80193, ... Also, the characteristic equations of the 3 matrices are the same as or variants of row 4 of the triangle below: (x^3 + x - 2x - 1) with different signs. (End)

Received from Herb Conn, Jan 2004: (Start)

Let x = 2*cos(2A) (A = Angle); then

sin(A)/sin A = 1

sin(3A)/sin A = x + 1

sin(5A)/sin A = x^2 + x - 1

sin(7A)/sin A = x^3 + x - 2x - 1

sin(9A)/sin A = x^4 + x^3 - 3x^2 - 2x + 1

... (signed ++--++...). (End)

Or Pascal's triangle (A007318) with duplicated diagonals. Also triangle of coefficients of polynomials defined by P_0(x) = 1 and for n>=1, P_n(x) = F_n(x) + F_(n+1)(x), where F_n(x) is Fibonacci polynomial (cf. A049310): F_n(x) = Sum_{i=0..floor((n-1)/2)} C(n-i-1,i)*x^(n-2*i-1). - Vladimir Shevelev, Apr 12 2012

The matrix inverse is given by

1;

1, 1;

0, -1, 1;

0, 1, -2, 1;

0, 0, 1, -2, 1;

0, 0, -1, 3, -3, 1;

0, 0, 0, -1, 3, -3, 1;

0, 0, 0, 1, -4, 6, -4, 1;

0, 0, 0, 0, 1, -4, 6, -4, 1;

... apart from signs the same as A124645. - R. J. Mathar, Mar 12 2013

Nonnegative values of y satisfying x^2 - 21*y^2 = 4; values of x are in A003501. - Wolfdieter Lang, Nov 29 2002

a(n) is equal to the permanent of the (n-1) X (n-1) Hessenberg matrix with 5's along the main diagonal, i's along the superdiagonal and the subdiagonal (i is the imaginary unit), and 0's everywhere else. - John M. Campbell, Jun 09 2011

For n >= 1, a(n) equals the number of 01-avoiding words of length n-1 on alphabet {0,1,2,3,4}. - Milan Janjic, Jan 25 2015

From Klaus Purath, Jul 26 2024: (Start)

For any three consecutive terms (x, y, z), y^2 - xz = 1 always applies.

a(n) = (t(i+2n) - t(i))/(t(i+n+1) - t(i+n-1)) where (t) is any recurrence t(k) = 4t(k-1) + 4t(k-2) - t(k-3) or t(k) = 5t(k-1) - t(k-2) without regard to initial values.

In particular, if the recurrence (t) of the form (4,4,-1) has the same three initial values as the current sequence, a(n) = t(n) applies.

a(n) = (t(k+1)*t(k+n) - t(k)*t(k+n+1))/(y^2 - xz) where (t) is any recurrence of the current family with signature (5,-1) and (x, y, z) are any three consecutive terms of (t), for integer k >= 0. (End)

Joe Keane (jgk(AT)jgk.org) observes that this sequence (beginning at 4) is "size of raises in pot-limit poker, one blind, maximum raising."

It appears that this sequence is the BinomialMean transform of A002315 - see A075271. - John W. Layman, Oct 02 2002

Number of (s(0), s(1), ..., s(2n+3)) such that 0 < s(i) < 8 and |s(i) - s(i-1)| = 1 for i = 1,2,...,2n+3, s(0) = 1, s(2n+3) = 4. - Herbert Kociemba, Jun 11 2004

a(n) = number of distinct matrix products in (A+B+C+D)^n where commutators [A,B]=[C,D]=0 but neither A nor B commutes with C or D. - Paul D. Hanna and Joshua Zucker, Feb 01 2006

The n-th term of the sequence is the entry (1,2) in the n-th power of the matrix M=[1,-1;-1,3]. - Simone Severini, Feb 15 2006

Hankel transform of this sequence is [1,-2,0,0,0,0,0,0,0,0,0,...]. - Philippe Deléham, Nov 21 2007

A204089 convolved with A000225, e.g., a(4) = 164 = (1*31 + 1*15 + 4*7 + 14*3 + 48*1) = (31 + 15 + 28 + 42 + 48). - Gary W. Adamson, Dec 23 2008

Equals INVERT transform of A000225: (1, 3, 7, 15, 31, ...). - Gary W. Adamson, May 03 2009

For n>=1, a(n-1) is the number of generalized compositions of n when there are 2^i-1 different types of the part i, (i=1,2,...). - Milan Janjic, Sep 24 2010

Binomial transform of A078057. - R. J. Mathar, Mar 28 2011

Pisano period lengths: 1, 1, 8, 1, 24, 8, 6, 1, 24, 24, 120, 8, 168, 6, 24, 1, 8, 24, 360, 24, ... . - R. J. Mathar, Aug 10 2012

a(n) is the diagonal of array A228405. - Richard R. Forberg, Sep 02 2013

From Wolfdieter Lang, Oct 01 2013: (Start)

a(n) appears together with A106731, both interspersed with zeros, in the representation of nonnegative powers of the algebraic number rho(8) = 2*cos(Pi/8) = A179260 of degree 4, which is the length ratio of the smallest diagonal and the side in the regular octagon.

The minimal polynomial for rho(8) is C(8,x) = x^4 - 4*x^2 + 2, hence rho(8)^n = A(n+1)*1 + A(n)*rho(8) + B(n+1)*rho(8)^2 + B(n)*rho(8)^3, n >= 0, with A(2*k) = 0, k >= 0, A(1) = 1, A(2*k+1) = A106731(k-1), k >= 1, and B(2*k) = 0, k >= 0, B(1) = 0, B(2*k+1) = a(k-1), k >= 1. See also the P. Steinbach reference given under A049310. (End)

The ratio a(n)/A006012(n) converges to 1+sqrt(2). - Karl V. Keller, Jr., May 16 2015

From Tom Copeland, Dec 04 2015: (Start)

An aerated version of this sequence is given by the o.g.f. = 1 / (1 - 4 x^2 + 2 x^4) = 1 / [x^4 a_4(1/x)] = 1 / determinant(I - x M) = exp[-log(1 -4 x + 2 x^4)], where M is the adjacency matrix for the simple Lie algebra B_4 given in A265185 with the characteristic polynomial a_4(x) = x^4 - 4 x^2 + 2 = 2 T_4(x/2) = A127672(4,x), where T denotes a Chebyshev polynomial of the first kind.

A133314 relates a(n) to the reciprocal of the e.g.f. 1 - 4 x + 4 x^2/2!. (End)

a(n) is the number of vertices of the Minkowski sum of n simplices with vertices e_(2*i+1), e_(2*i+2), e_(2*i+3), e_(2*i+4) for i=0,...,n-1, where e_i is a standard basis vector. - Alejandro H. Morales, Oct 03 2022

Also T(2n+1,n+1), T given by A027935. Also first row of Inverse Stolarsky array.

Third diagonal of array defined by T(i, 1)=T(1, j)=1, T(i, j)=Max(T(i-1, j)+T(i-1, j-1); T(i-1, j-1)+T(i, j-1)). - Benoit Cloitre, Aug 05 2003

Number of Schroeder paths of length 2(n+1) having exactly one up step starting at an even height (a Schroeder path is a lattice path starting from (0,0), ending at a point on the x-axis, consisting only of steps U=(1,1) (up steps), D=(1,-1) (down steps) and H=(2,0) (level steps) and never going below the x-axis). Schroeder paths are counted by the large Schroeder numbers (A006318). Example: a(1)=4 because among the six Schroeder paths of length 4 only the paths (U)HD, (U)UDD, H(U)D, (U)DH have exactly one U step that starts at an even height (shown between parentheses). - Emeric Deutsch, Dec 19 2004

Also: smallest number not writeable as the sum of fewer than n positive Fibonacci numbers. E.g., a(5)=88 because it is the smallest number that needs at least 5 Fibonacci numbers: 88 = 55 + 21 + 8 + 3 + 1. - Johan Claes, Apr 19 2005 [corrected for offset and clarification by Mike Speciner, Sep 19 2023] In general, a(n) is the sum of n positive Fibonacci numbers as a(n) = Sum_{i=1..n} A000045(2*i). See A001076 when negative Fibonacci numbers can be included in the sum. - Mike Speciner, Sep 24 2023

Except for first term, numbers a(n) that set a new record in the number of Fibonacci numbers needed to sum up to n. Position of records in sequence A007895. - Ralf Stephan, May 15 2005

Successive extremal petal bends beta(n) = a(n-2). See the Ring Lemma of Rodin and Sullivan in K. Stephenson, Introduction to Circle Packing (Cambridge U. P., 2005), pp. 73-74 and 318-321. - David W. Cantrell (DWCantrell(AT)sigmaxi.net)

a(n+1)= AAB^(n)(1), n>=1, with compositions of Wythoff's complementary A(n):=A000201(n) and B(n)=A001950(n) sequences. See the W. Lang link under A135817 for the Wythoff representation of numbers (with A as 1 and B as 0 and the argument 1 omitted). E.g., 4=`110`, 12=`1100`, 33=`11000`, 88=`110000`, ..., in Wythoff code. AA(1)=1=a(1) but for uniqueness reason 1=A(1) in Wythoff code. - N. J. A. Sloane, Jun 29 2008

Start with n. Each n generates a sublist {n-1,n-1,n-2,..,1}. Each element of each sublist also generates a sublist. Add numbers in all terms. For example, 3->{2,2,1} and both 2->{1,1}, so a(3) = 3 + 2 + 2 + 1 + 1 + 1 + 1 + 1 = 12. - Jon Perry, Sep 01 2012

For n>0: smallest number such that the inner product of Zeckendorf binary representation and its reverse equals n: A216176(a(n)) = n, see also A189920. - Reinhard Zumkeller, Mar 10 2013

Also, numbers m such that 5*m*(m+2)+1 is a square. - Bruno Berselli, May 19 2014

Also, number of nonempty submultisets of multisets of weight n that span an initial interval of integers (see 2nd example). - Gus Wiseman, Feb 10 2015

From Robert K. Moniot, Oct 04 2020: (Start)

Including a(-1):=0, consecutive terms (a(n-1),a(n))=(u,v) or (v,u) give all points on the hyperbola u^2-u+v^2-v-4*u*v=0 with both coordinates nonnegative integers. Note that this follows from identifying (1,u+1,v+1) with the Markov triple (1,Fibonacci(2n-1),Fibonacci(2n+1)). See A001519 (comments by Robert G. Wilson, Oct 05 2005, and Wolfdieter Lang, Jan 30 2015).

Let T(n) denote the n-th triangular number. If i, j are any two successive elements of the above sequence then (T(i-1) + T(j-1))/T(i+j-1) = 3/5. (End)

From Hieronymus Fischer, Jan 02 2009: (Start)

Set c:=1+sqrt(2). Then the fractional part of c^n equals 1/c^n, if n odd. For even n, the fractional part of c^n is equal to 1-(1/c^n).

c:=1+sqrt(2) satisfies c-c^(-1)=floor(c)=2, hence c^n + (-c)^(-n) = round(c^n) for n>0, which follows from the general formula of A001622.

1/c = sqrt(2)-1.

See A001622 for a general formula concerning the fractional parts of powers of numbers x>1, which satisfy x-x^(-1)=floor(x).

Other examples of constants x satisfying the relation x-x^(-1)=floor(x) include A001622 (the golden ratio: where floor(x)=1) and A098316 (the "bronze" ratio: where floor(x)=3). (End)

In terms of continued fractions the constant c can be described by c=[2;2,2,2,...]. - Hieronymus Fischer, Oct 20 2010

Side length of smallest square containing five circles of diameter 1. - Charles R Greathouse IV, Apr 05 2011

Largest radius of four circles tangent to a circle of radius 1. - Charles R Greathouse IV, Jan 14 2013

An analog of Fermat theorem: for prime p, round(c^p) == 2 (mod p). - Vladimir Shevelev, Mar 02 2013

n*(1+sqrt(2)) is the perimeter of a 45-45-90 triangle with hypotenuse n. - Wesley Ivan Hurt, Apr 09 2016

This algebraic integer of degree 2, with minimal polynomial x^2 - 2*x - 1, is also the length ratio diagonal/side of the second largest diagonal in the regular octagon (not counting the side). The other two diagonal/side ratios are A179260 and A121601. - Wolfdieter Lang, Oct 28 2020

c^n = A001333(n) + A000129(n) * sqrt(2). - Gary W. Adamson, Apr 26 2023

c^n = c * A000129(n) + A000129(n-1), where c = 1 + sqrt(2). - Gary W. Adamson, Aug 30 2023

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

Denominators of continued fraction convergents to sqrt(3), for n >= 1.

Also denominators of continued fraction convergents to sqrt(3) - 1. See A048788 for numerators. - N. J. A. Sloane, Dec 17 2007. Convergents are 1, 2/3, 3/4, 8/11, 11/15, 30/41, 41/56, 112/153, ...

Consider the mapping f(a/b) = (a + 3*b)/(a + b). Taking a = b = 1 to start with and carrying out this mapping repeatedly on each new (reduced) rational number gives the following sequence 1/1, 2/1, 5/3, 7/4, 19/11, ... converging to 3^(1/2). Sequence contains the denominators. The same mapping for N, i.e., f(a/b) = (a + Nb)/(a + b) gives fractions converging to N^(1/2). - Amarnath Murthy, Mar 22 2003

Sqrt(3) = 2/2 + 2/3 + 2/(3*11) + 2/(11*41) + 2/(41*153) + 2/(153*571), ...; the sum of the first 6 terms of this series = 1.7320490367..., while sqrt(3) = 1.7320508075... - Gary W. Adamson, Dec 15 2007

From Clark Kimberling, Aug 27 2008: (Start)

Related convergents (numerator/denominator):

lower principal convergents: A001834/A001835

upper principal convergents: A001075/A001353

intermediate convergents: A005320/A001075

principal and intermediate convergents: A143642/A140827

lower principal and intermediate convergents: A143643/A005246. (End)

Row sums of triangle A152063 = (1, 3, 4, 11, ...). - Gary W. Adamson, Nov 26 2008

From Alois P. Heinz, Apr 13 2011: (Start)

Also number of domino tilings of the 3 X (n-1) rectangle with upper left corner removed iff n is even. For n=4 the 4 domino tilings of the 3 X 3 rectangle with upper left corner removed are:

. ._. . ._. . ._. . ._.

.|__| .|__| .| | | .|___|

| |_| | | | | | ||| |_| |

||__| |||_| ||__| |_|_| (End)

This is the sequence of Lehmer numbers u_n(sqrt(R),Q) with the parameters R = 2 and Q = -1. It is a strong divisibility sequence, that is, gcd(a(n),a(m)) = a(gcd(n,m)) for all natural numbers n and m. - Peter Bala, Apr 18 2014

2^(-floor(n/2))*(1 + sqrt(3))^n = A002531(n) + a(n)*sqrt(3); integers in the real quadratic number field Q(sqrt(3)). - Wolfdieter Lang, Feb 11 2018

Let T(n) = 2^(n mod 2), U(n) = a(n), V(n) = A002531(n), x(n) = V(n)/U(n). Then T(n*m) * U(n+m) = U(n)*V(m) + U(m)*V(n), T(n*m) * V(n+m) = 3*U(n)*U(m) + V(m)*V(n), x(n+m) = (3 + x(n)*x(m))/(x(n) + x(m)). - Michael Somos, Nov 29 2022

Inverse binomial transform of A057083. Binomial transform of A061347. The sums of consecutive pairs of elements give A084103. - Paul Barry, May 15 2003

Hexaperiodic sequence identical to its third differences. - Paul Curtz, Dec 13 2007

a(n+1) is the Hankel transform of A001700(n+1)-A001700(n). - Paul Barry, Apr 21 2009

Non-simple continued fraction expansion of 1 = 1+1/(2+1/(1+1/(-1+...))). - R. J. Mathar, Mar 08 2012

Pisano period lengths: 1, 3, 2, 6, 6, 6, 6, 6, 6, 6, 6, 6, 6, 6, 6, 6, 6, 6, 6, 6, ... - R. J. Mathar, Aug 10 2012

Alternating row sums of Riordan triangle A111125. - Wolfdieter Lang, Oct 18 2012

Periodic sequences of this type can be also calculated by a(n) = c + floor(q/(p^m-1)*p^n) mod p, where c is a constant, q is the number representing the periodic digit pattern and m is the period length. c, p and q can be calculated as follows: Let D be the array representing the number pattern to be repeated, m = size of D, max = maximum value of elements in D, min = minimum value of elements in D. Than c := min, p := max - min + 1 and q := p^m*sum_{i=1..m} (D(i)-min)/p^i. Example: D = (1, 2, 1, -1, -2, -1), c = -2, m = 6, p = 5 and q = 12276 for this sequence. - Hieronymus Fischer, Jan 04 2013

T(n,k) is the number of leaves at level k+1 in all ordered trees with n+1 edges. - Emeric Deutsch, Jan 15 2005

Riordan array ((1-2x-sqrt(1-4x))/(2x^2),(1-2x-sqrt(1-4x))/(2x)). Inverse array is A053122. - Paul Barry, Mar 17 2005

T(n,k) is the number of walks of n steps, each in direction N, S, W, or E, starting at the origin, remaining in the upper half-plane and ending at height k (see the R. K. Guy reference, p. 5). Example: T(3,2)=6 because we have ENN, WNN, NEN, NWN, NNE and NNW. - Emeric Deutsch, Apr 15 2005

Triangle T(n,k), 0<=k<=n, read by rows given by T(0,0)=1, T(n,k)=0 if k<0 or if k>n, T(n,0) = 2*T(n-1,0) + T(n-1,1), T(n,k) = T(n-1,k-1) + 2*T(n-1,k) + T(n-1,k+1) for k>=1. - Philippe Deléham, Mar 30 2007

Number of (2n+1)-step walks from (0,0) to (2n+1,2k+1) and consisting of steps u=(1,1) and d=(1,-1) in which the path stays in the nonnegative quadrant. Examples: T(2,0)=5 because we have uuudd, uudud, uuddu, uduud, ududu; T(2,1)=4 because we have uuuud, uuudu, uuduu, uduuu; T(2,2)=1 because we have uuuuu. - Philippe Deléham, Apr 16 2007, Apr 18 2007

Triangle read by rows: T(n,k)=number of lattice paths from (0,0) to (n,k) that do not go below the line y=0 and consist of steps U=(1,1), D=(1,-1) and two types of steps H=(1,0); example: T(3,1)=14 because we have UDU, UUD, 4 HHU paths, 4 HUH paths and 4 UHH paths. - Philippe Deléham, Sep 25 2007

This triangle belongs to the family of triangles defined by T(0,0)=1, T(n,k)=0 if k<0 or if k>n, T(n,0) = x*T(n-1,0) + T(n-1,1), T(n,k) = T(n-1,k-1) + y*T(n-1,k) + T(n-1,k+1) for k>=1. Other triangles arise by choosing different values for (x,y): (0,0) -> A053121; (0,1) -> A089942; (0,2) -> A126093; (0,3) -> A126970; (1,0) -> A061554; (1,1) -> A064189; (1,2) -> A039599; (1,3) -> A110877; (1,4) -> A124576; (2,0) -> A126075; (2,1) -> A038622; (2,2) -> A039598; (2,3) -> A124733; (2,4) -> A124575; (3,0) -> A126953; (3,1) -> A126954; (3,2) -> A111418; (3,3) -> A091965; (3,4) -> A124574; (4,3) -> A126791; (4,4) -> A052179; (4,5) -> A126331; (5,5) -> A125906. - Philippe Deléham, Sep 25 2007

With offset [1,1] this is the (ordinary) convolution triangle a(n,m) with o.g.f. of column m given by (c(x)-1)^m, where c(x) is the o.g.f. for Catalan numbers A000108. See the Riordan comment by Paul Barry.

T(n, k) is also the number of order-preserving full transformations (of an n-chain) with exactly k fixed points. - Abdullahi Umar, Oct 02 2008

T(n,k)/2^(2n+1) = coefficients of the maximally flat lowpass digital differentiator of the order N=2n+3. - Pavel Holoborodko (pavel(AT)holoborodko.com), Dec 19 2008

The signed triangle S(n,k) := (-1)^(n-k)*T(n,k) provides the transformation matrix between f(n,l) := L(2*l)*5^n*F(2*l)^(2*n+1) (F=Fibonacci numbers A000045, L=Lucas numbers A000032) and F(4*l*(k+1)), k = 0, ..., n, for each l>=0: f(n,l) = Sum_{k=0..n} S(n,k)*F(4*l*(k+1)), n>=0, l>=0. Proof: the o.g.f. of the l.h.s., G(l;x) := Sum_{n>=0} f(n,l)*x^n = F(4*l)/(1 - 5*F(2*l)^2*x) is shown to match the o.g.f. of the r.h.s.: after an interchange of the n- and k-summation, the Riordan property of S = (C(x)/x,C(x)) (compare with the above comments by Paul Barry), with C(x) := 1 - c(-x), with the o.g.f. c(x) of A000108 (Catalan numbers), is used, to obtain, after an index shift, first Sum_{k>=0} F(4*l*(k))*GS(k;x), with the o.g.f of column k of triangle S which is GS(k;x) := Sum_{n>=k} S(n,k)*x^n = C(x)^(k+1)/x. The result is GF(l;C(x))/x with the o.g.f. GF(l,x) := Sum_{k>=0} F(4*l*k)*x^k = x*F(4*l)/(1-L(4*l)*x+x^2) (see a comment on A049670, and A028412). If one uses then the identity L(4*n) - 5*F(2*n)^2 = 2 (in Koshy's book [reference under A065563] this is No. 15, p. 88, attributed to Lucas, 1876), the proof that one recovers the o.g.f. of the l.h.s. from above boils down to a trivial identity on the Catalan o.g.f., namely 1/c^2(-x) = 1 + 2*x - (x*c(-x))^2. - Wolfdieter Lang, Aug 27 2012

O.g.f. for row polynomials R(x) := Sum_{k=0..n} a(n,k)*x^k:

((1+x) - C(z))/(x - (1+x)^2*z) with C the o.g.f. of A000108 (Catalan numbers). From Riordan ((C(x)-1)/x,C(x)-1), compare with a Paul Barry comment above. This coincides with the o.g.f. given by Emeric Deutsch in the formula section. - Wolfdieter Lang, Nov 13 2012

The A-sequence for this Riordan triangle is [1,2,1] and the Z-sequence is [2,1]. See a W. Lang link under A006232 with details and references. - Wolfdieter Lang, Nov 13 2012

From Wolfdieter Lang, Sep 20 2013: (Start)

T(n, k) = A053121(2*n+1, 2*k+1). T(n, k) appears in the formula for the (2*n+1)-th power of the algebraic number rho(N) := 2*cos(Pi/N) = R(N, 2) in terms of the even-indexed diagonal/side length ratios R(N, 2*(k+1)) = S(2*k+1, rho(N)) in the regular N-gon inscribed in the unit circle (length unit 1). S(n, x) are Chebyshev's S polynomials (see A049310): rho(N)^(2*n+1) = Sum_{k=0..n} T(n, k)*R(N, 2*(k+1)), n >= 0, identical in N >= 1. For a proof see the Sep 21 2013 comment under A053121. Note that this is the unreduced version if R(N, j) with j > delta(N), the degree of the algebraic number rho(N) (see A055034), appears. For the even powers of rho(n) see A039599. (End)

The tridiagonal Toeplitz production matrix P in the Example section corresponds to the unsigned Cartan matrix for the simple Lie algebra A_n as n tends to infinity (cf. Damianou ref. in A053122). - Tom Copeland, Dec 11 2015 (revised Dec 28 2015)

T(n,k) is the number of pairs of non-intersecting walks of n steps, each in direction N or E, starting at the origin, and such that the end points of the two paths are separated by a horizontal distance of k. See Shapiro 1976. - Peter Bala, Apr 12 2017

Also the convolution triangle of the Catalan numbers A000108. - Peter Luschny, Oct 07 2022

Define the sequence T(a_0,a_1) by a_{n+2} is the greatest integer such that a_{n+2}/a_{n+1}= 0 . A004187 (with initial 0 omitted) is T(1,7).

This is a divisibility sequence.

For n>=2, a(n) equals the permanent of the (n-1) X (n-1) tridiagonal matrix with 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

a(n) and b(n) := A056854(n) are the proper and improper nonnegative solutions of the Pell equation b(n)^2 - 5*(3*a(n))^2 = +4. see the cross-reference to A056854 below. - Wolfdieter Lang, Jun 26 2013

For n>=1, a(n) equals the number of 01-avoiding words of length n-1 on alphabet {0,1,2,3,4,5,6}. - Milan Janjic, Jan 25 2015

The digital root is A253298, which shares its digital root with A253368. - Peter M. Chema, Jul 04 2016

Lim_{n->oo} a(n+1)/a(n) = 2 + 3*phi = 1+ A090550 = 6.854101... - Wolfdieter Lang, Nov 16 2023