cp's OEIS Frontend

Row sums are:

{1, 2, 2, -4, -20, -8, 184, 464, -1648, -10720, 8224}

In real quantum mechanical 2 dimensional orthogonal partitions it would be:

p(x,y,n,m)=T(x,n)*H(y,m).

Here I have made x=y and n=m to get a new sort of polynomial with an odd number of vector coefficients.

The traditional Schoedinger wave mechanics solution of hydrogen is a partition of four (not two dimensions): wave_function=Bessel(r,n)*Legendre(theta,l)*Fourier(phi,m)*Spin(t,s).

The row sums are all zero.

The idea of roots of polynomials of the sort came from the realization that in Umbral calculus for the expansion function:

p(x,t)=Sum(P(xd,n)*t^n/n!,{n,0,Infinity}];

to actually work there has to be a convergent limit:

Limit[P(x,n)*t^n/n!,n->Infinity]=0;

The idea that a point gets "trapped" in complex dynamics is the iterative:

Pc[x,n]=x

So if we look at polynomials as iterative steps, at a fixed point

the roots would be important dynamically.

Row sums are all 2.

The rationale behind this interpolation is that Bessel functions have 1/2 values, so what about other orthogonal polynomials?

The integration shows that they are "mostly" orthogonal when three away from the diagonal.

TableForm[Table[Integrate[p[x, m]*p[x, n]/Sqrt[1 - x^2], {x, -1, 1}], {n, 0, 10}, {m, 0, 10}]]

These polynomials would also be related to two-dimensional Chladni-Chebyshev type standing waves as:

Chladni[x,y,n,m]=ChebyshevT[n, x] + ChebyshevT[m, y].

Second differences downwards columns of the Chebyshev triangle A053120.

Row sums are 2, 6, 16, 40, 96, 224, 512, 1152, 2560, 5632, 12288,..., A057711.

Row sums are constantly two.

Cf. A000204 for Lucas numbers beginning with 1.

Also the number of independent vertex sets and vertex covers for the cycle graph C_n for n >= 2. - Eric W. Weisstein, Jan 04 2014

Also the number of matchings in the n-cycle graph C_n for n >= 3. - Eric W. Weisstein, Oct 01 2017

Also the number of maximal independent vertex sets (and maximal vertex covers) for the n-helm graph for n >= 3. - Eric W. Weisstein, May 27 2017

Also the number of maximal independent vertex sets (and maximal vertex covers) for the n-sunlet graph for n >= 3. - Eric W. Weisstein, Aug 07 2017

This is also the Horadam sequence (2, 1, 1, 1). - Ross La Haye, Aug 18 2003

For distinct primes p, q, L(p) is congruent to 1 mod p, L(2p) is congruent to 3 mod p and L(pq) is congruent 1 + q(L(q) - 1) mod p. Also, L(m) divides F(2km) and L((2k + 1)m), k, m >= 0.

a(n) = Sum_{k=0..ceiling((n - 1)/2)} P(3; n - 1 - k, k), n >= 1, with a(0) = 2. These are the sums over the SW-NE diagonals in P(3; n, k), the (3, 1) Pascal triangle A093560. Observation by Paul Barry, Apr 29 2004. Proof via recursion relations and comparison of inputs. Also SW-NE diagonal sums of the (1, 2) Pascal triangle A029635 (with T(0, 0) replaced by 2).

Suppose psi = log(phi) = A002390. We get the representation L(n) = 2*cosh(n*psi) if n is even; L(n) = 2*sinh(n*psi) if n is odd. There is a similar representation for Fibonacci numbers (A000045). Many Lucas formulas now easily follow from appropriate sinh- and cosh-formulas. For example: the identity cosh^2(x) - sinh^2(x) = 1 implies L(n)^2 - 5*F(n)^2 = 4*(-1)^n (setting x = n*psi). - Hieronymus Fischer, Apr 18 2007

From John Blythe Dobson, Oct 02 2007, Oct 11 2007: (Start)

The parity of L(n) follows easily from its definition, which shows that L(n) is even when n is a multiple of 3 and odd otherwise.

The first six multiplication formulas are:

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

L(3n) = L(n)^3 - 3*(-1)^n*L(n);

L(4n) = L(n)^4 - 4*(-1)^n*L(n)^2 + 2;

L(5n) = L(n)^5 - 5*(-1)^n*L(n)^3 + 5*L(n);

L(6n) = L(n)^6 - 6*(-1)^n*L(n)^4 + 9*L(n)^2 - 2*(-1)^n.

Generally, L(n) | L(mn) if and only if m is odd.

In the expansion of L(mn), where m represents the multiplier and n the index of a known value of L(n), the absolute values of the coefficients are the terms in the m-th row of the triangle A034807. When m = 1 and n = 1, L(n) = 1 and all the terms are positive and so the row sums of A034807 are simply the Lucas numbers. (End)

From John Blythe Dobson, Nov 15 2007: (Start)

The comments submitted by Miklos Kristof on Mar 19 2007 for the Fibonacci numbers (A000045) contain four important identities that have close analogs in the Lucas numbers:

For a >= b and odd b, L(a + b) + L(a - b) = 5*F(a)*F(b).

For a >= b and even b, L(a + b) + L(a - b) = L(a)*L(b).

For a >= b and odd b, L(a + b) - L(a - b) = L(a)*L(b).

For a >= b and even b, L(a + b) - L(a - b) = 5*F(a)*F(b).

A particularly interesting instance of the difference identity for even b is L(a + 30) - L(a - 30) = 5*F(a)*832040, since 5*832040 is divisible by 100, proving that the last two digits of Lucas numbers repeat in a cycle of length 60 (see A106291(100)). (End)

From John Blythe Dobson, Nov 15 2007: (Start)

The Lucas numbers satisfy remarkable difference equations, in some cases best expressed using Fibonacci numbers, of which representative examples are the following:

L(n) - L(n - 3) = 2*L(n - 2);

L(n) - L(n - 4) = 5*F(n - 2);

L(n) - L(n - 6) = 4*L(n - 3);

L(n) - L(n - 12) = 40*F(n - 6);

L(n) - L(n - 60) = 4160200*F(n - 30).

These formulas establish, respectively, that the Lucas numbers form a cyclic residue system of length 3 (mod 2), of length 4 (mod 5), of length 6 (mod 4), of length 12 (mod 40) and of length 60 (mod 4160200). The divisibility of the last modulus by 100 accounts for the fact that the last two digits of the Lucas numbers begin to repeat at L(60).

The divisibility properties of the Lucas numbers are very complex and still not fully understood, but several important criteria are established in Zhi-Hong Sun's 2003 survey of congruences for Fibonacci numbers. (End)

Sum_{n>0} a(n)/(n*2^n) = 2*log(2). - Jaume Oliver Lafont, Oct 11 2009

A010888(a(n)) = A030133(n). - Reinhard Zumkeller, Aug 20 2011

The powers of phi, the golden ratio, approach the values of the Lucas numbers, the odd powers from above and the even powers from below. - Geoffrey Caveney, Apr 18 2014

Inverse binomial transform is (-1)^n * a(n). - Michael Somos, Jun 03 2014

Lucas numbers are invariant to the following transformation for all values of the integers j and n, including negative values, thus: L(n) = (L(j+n) + (-1)^n * L(j-n))/L(j). The same transformation applied to all sequences of the form G(n+1) = m * G(n) + G(n-1) yields Lucas numbers for m = 1, except where G(j) = 0, regardless of initial values which may be nonintegers. The corresponding sequences for other values of m are: for m = 2, 2*A001333; for m = 3, A006497; for m = 4, 2*A001077; for m = 5, A087130; for m = 6, 2*A005667; for m = 7, A086902. The invariant ones all have G(0) = 2, G(1) = m. A related family of sequences is discussed at A059100. - Richard R. Forberg, Nov 23 2014

If x=a(n), y=a(n+1), z=a(n+2), then -x^2 - z*x - 3*y*x - y^2 + y*z + z^2 = 5*(-1)^(n+1). - Alexander Samokrutov, Jul 04 2015

A conjecture on the divisibility of infinite subsequences of Lucas numbers by prime(n)^m, m >= 1, is given in A266587, together with the prime "entry points". - Richard R. Forberg, Dec 31 2015

A trapezoid has three lengths of sides in order L(n-1), L(n+1), L(n-1). For increasing n a very close approximation to the maximum area will have the fourth side equal to 2*L(n). For a trapezoid with sides L(n-1), L(n-3), L(n-1), the fourth side will be L(n). - J. M. Bergot, Mar 17 2016

Satisfies Benford's law [Brown-Duncan, 1970; Berger-Hill, 2017]. - N. J. A. Sloane, Feb 08 2017

Lucas numbers L(n) and Fibonacci numbers F(n), being related by the formulas F(n) = (F(n-1) + L(n-1))/2 and L(n) = 2 F(n+1) - F(n), are a typical pair of "autosequences" (see the link to OEIS Wiki). - Jean-François Alcover, Jun 09 2017

For n >= 3, the Lucas number L(n) is the dimension of a commutative Hecke algebra of affine type A_n with independent parameters. See Theorem 1.4, Corollary 1.5, and the table on page 524 in the link "Hecke algebras with independent parameters". - Jia Huang, Jan 20 2019

From Klaus Purath, Apr 19 2019: (Start)

While all prime numbers appear as factors in the Fibonacci numbers, this is not the case with the Lucas numbers. For example, L(n) is never divisible by the following prime numbers < 150: 5, 13, 17, 37, 53, 61, 73, 89, 97, 109, 113, 137, 149 ... See A053028. Conjecture: Three properties can be determined for these prime numbers:

First observation: The prime factors > 3 occur in the Fibonacci numbers with an odd index.

Second observation: These are the prime numbers p congruent to 2, 3 (modulo 5), which occur both in Fibonacci(p+1) and in Fibonacci((p+1)/2) as prime factors, or the prime numbers p congruent to 1, 4 (modulo 5), which occur both in Fibonacci((p-1)/2) and in Fibonacci((p-1)/(2^k)) with k >= 2.

Third observation: The Pisano period lengths of these prime numbers, given in A001175, are always divisible by 4, but not by 8. In contrast, those of the prime factors of Lucas numbers are divisible either by 2, but not by 4, or by 8. (See also comment in A053028 by N. J. A. Sloane, Feb 21 2004). (End)

L(n) is the sum of 4*k consecutive terms of the Fibonacci sequence (A000045) divided by Fibonacci(2*k): (Sum_{i=0..4*k-1, k>=1} F(n+i))/F(2*k) = L(n+2*k+1). Sequences extended to negative indices, following the rule a(n-1) = a(n+1) - a(n). - Klaus Purath, Sep 15 2019

If one forms a sequence (A) of the Fibonacci type with the initial values A(0) = A022095(n) and A(1) = A000285(n), then A(n+1) = L(n+1)^2 always applies. - Klaus Purath, Sep 29 2019

From Kai Wang, Dec 18 2019: (Start)

L((2*m+1)k)/L(k) = Sum_{i=0..m-1} (-1)^(i*(k+1))*L((2*m-2*i)*k) + (-1)^(m*k).

Example: k=5, m=2, L(5)=11, L(10)=123, L(20)=15127, L(25)=167761. L(25)/L(5) = 15251, L(20) + L(10) + 1 = 15127 + 123 + 1 = 15251. (End)

From Peter Bala, Dec 23 2021: (Start)

The Gauss congruences a(n*p^k) == a(n*p^(k-1)) ( mod p^k ) hold for all prime p and positive integers n and k.

For a positive integer k, the sequence (a(n))n>=1 taken modulo k becomes a purely periodic sequence. For example, taken modulo 11, the sequence becomes [1, 3, 4, 7, 0, 7, 7, 3, 10, 2, 1, 3, 4, 7, 0, 7, 7, 3, 10, 2, ...], a periodic sequence with period 10. (End)

For any sequence with recurrence relation b(n) = b(n-1) + b(n-2), it can be shown that the recurrence relation for every k-th term is given by: b(n) = A000032(k) * b(n-k) + (-1)^(k+1) * b(n-2k), extending to negative indices as necessary. - Nick Hobson, Jan 19 2024

For n >= 3, L(n) is the number of (n-1)-digit numbers where all consecutive pairs of digits have a difference of at least 8. - Edwin Hermann, Apr 19 2025

Apart from initial term, same as A000079 (powers of 2).

Number of compositions (ordered partitions) of n. - Toby Bartels, Aug 27 2003

Number of ways of putting n unlabeled items into (any number of) labeled boxes where every box contains at least one item. Also "unimodal permutations of n items", i.e., those which rise then fall. (E.g., for three items: ABC, ACB, BCA and CBA are unimodal.) - Henry Bottomley, Jan 17 2001

Number of permutations in S_n avoiding the patterns 213 and 312. - Tuwani Albert Tshifhumulo, Apr 20 2001. More generally (see Simion and Schmidt), the number of permutations in S_n avoiding (i) the 123 and 132 patterns; (ii) the 123 and 213 patterns; (iii) the 132 and 213 patterns; (iv) the 132 and 231 patterns; (v) the 132 and 312 patterns; (vi) the 213 and 231 patterns; (vii) the 213 and 312 patterns; (viii) the 231 and 312 patterns; (ix) the 231 and 321 patterns; (x) the 312 and 321 patterns.

a(n+2) is the number of distinct Boolean functions of n variables under action of symmetric group.

Number of unlabeled (1+2)-free posets. - Detlef Pauly, May 25 2003

Image of the central binomial coefficients A000984 under the Riordan array ((1-x), x*(1-x)). - Paul Barry, Mar 18 2005

Binomial transform of (1, 0, 1, 0, 1, 0, 1, 0, 1, 0, ...); inverse binomial transform of A007051. - Philippe Deléham, Jul 04 2005

Also, number of rationals in [0, 1) whose binary expansions terminate after n bits. - Brad Chalfan, May 29 2006

Equals row sums of triangle A144157. - Gary W. Adamson, Sep 12 2008

Prepend A089067 with a 1, getting (1, 1, 3, 5, 13, 23, 51, ...) as polcoeff A(x); then (1, 1, 2, 4, 8, 16, ...) = A(x)/A(x^2). - Gary W. Adamson, Feb 18 2010

An elephant sequence, see A175655. For the central square four A[5] vectors, with decimal values 2, 8, 32 and 128, lead to this sequence. For the corner squares these vectors lead to the companion sequence A094373. - Johannes W. Meijer, Aug 15 2010

From Paul Curtz, Jul 20 2011: (Start)

Array T(m,n) = 2*T(m,n-1) + T(m-1,n):

1, 1, 2, 4, 8, 16, ... = a(n)

1, 3, 8, 20, 48, 112, ... = A001792,

1, 5, 18, 56, 160, 432, ... = A001793,

1, 7, 32, 120, 400, 1232, ... = A001794,

1, 9, 50, 220, 840, 2912, ... = A006974, followed with A006975, A006976, gives nonzero coefficients of Chebyshev polynomials of first kind A039991 =

1,

1, 0,

2, 0, -1,

4, 0, -3, 0,

8, 0, -8, 0, 1.

T(m,n) third vertical: 2*n^2, n positive (A001105).

Fourth vertical appears in Janet table even rows, last vertical (A168342 array, A138509, rank 3, 13, = A166911)). (End)

A131577(n) and differences are:

0, 1, 2, 4, 8, 16,

1, 1, 2, 4, 8, 16, = a(n),

0, 1, 2, 4, 8, 16,

1, 1, 2, 4, 8, 16.

Number of 2-color necklaces of length 2n equal to their complemented reversal. For length 2n+1, the number is 0. - David W. Wilson, Jan 01 2012

Edges and also central terms of triangle A198069: a(0) = A198069(0,0) and for n > 0: a(n) = A198069(n,0) = A198069(n,2^n) = A198069(n,2^(n-1)). - Reinhard Zumkeller, May 26 2013

These could be called the composition numbers (see the second comment) since the equivalent sequence for partitions is A000041, the partition numbers. - Omar E. Pol, Aug 28 2013

Number of self conjugate integer partitions with exactly n parts for n>=1. - David Christopher, Aug 18 2014

The sequence is the INVERT transform of (1, 0, 1, 1, 2, 3, 5, 8, 13, 21, 34, ...). - Gary W. Adamson, Jul 16 2015

Number of threshold graphs on n nodes [Hougardy]. - Falk Hüffner, Dec 03 2015

Number of ternary words of length n in which binary subwords appear in the form 10...0. - Milan Janjic, Jan 25 2017

a(n) is the number of words of length n over an alphabet of two letters, of which one letter appears an even number of times (the empty word of length 0 is included). See the analogous odd number case in A131577, and the Balakrishnan reference in A006516 (the 4-letter odd case), pp. 68-69, problems 2.66, 2.67 and 2.68. - Wolfdieter Lang, Jul 17 2017

Number of D-equivalence classes of Łukasiewicz paths. Łukasiewicz paths are D-equivalent iff the positions of pattern D are identical in these paths. - Sergey Kirgizov, Apr 08 2018

Number of color patterns (set partitions) for an oriented row of length n using two or fewer colors (subsets). Two color patterns are equivalent if we permute the colors. For a(4)=8, the 4 achiral patterns are AAAA, AABB, ABAB, and ABBA; the 4 chiral patterns are the 2 pairs AAAB-ABBB and AABA-ABAA. - Robert A. Russell, Oct 30 2018

The determinant of the symmetric n X n matrix M defined by M(i,j) = (-1)^max(i,j) for 1 <= i,j <= n is equal to a(n) * (-1)^(n*(n+1)/2). - Bernard Schott, Dec 29 2018

For n>=1, a(n) is the number of permutations of length n whose cyclic representations can be written in such a way that when the cycle parentheses are removed what remains is 1 through n in natural order. For example, a(4)=8 since there are exactly 8 permutations of this form, namely, (1 2 3 4), (1)(2 3 4), (1 2)(3 4), (1 2 3)(4), (1)(2)(3 4), (1)(2 3)(4), (1 2)(3)(4), and (1)(2)(3)(4). Our result follows readily by conditioning on k, the number of parentheses pairs of the form ")(" in the cyclic representation. Since there are C(n-1,k) ways to insert these in the cyclic representation and since k runs from 0 to n-1, we obtain a(n) = Sum_{k=0..n-1} C(n-1,k) = 2^(n-1). - Dennis P. Walsh, May 23 2020

Maximum number of preimages that a permutation of length n + 1 can have under the consecutive-231-avoiding stack-sorting map. - Colin Defant, Aug 28 2020

a(n) is the number of occurrences of the empty set {} in the von Neumann ordinals from 0 up to n. Each ordinal k is defined as the set of all smaller ordinals: 0 = {}, 1 = {0}, 2 = {0,1}, etc. Since {} is the foundational element of all ordinals, the total number of times it appears grows as powers of 2. - Kyle Wyonch, Mar 30 2025

Sometimes also called lambda numbers.

Also denominators of continued fraction convergents to sqrt(2): 1, 3/2, 7/5, 17/12, 41/29, 99/70, 239/169, 577/408, 1393/985, 3363/2378, 8119/5741, 19601/13860, 47321/33461, 114243/80782, ... = A001333/A000129.

Number of lattice paths from (0,0) to the line x=n-1 consisting of U=(1,1), D=(1,-1) and H=(2,0) steps (i.e., left factors of Grand Schroeder paths); for example, a(3)=5, counting the paths H, UD, UU, DU and DD. - Emeric Deutsch, Oct 27 2002

a(2*n) with b(2*n) := A001333(2*n), n >= 1, give all (positive integer) solutions to Pell equation b^2 - 2*a^2 = +1 (see Emerson reference). a(2*n+1) with b(2*n+1) := A001333(2*n+1), n >= 0, give all (positive integer) solutions to Pell equation b^2 - 2*a^2 = -1.

Bisection: a(2*n+1) = T(2*n+1, sqrt(2))/sqrt(2) = A001653(n), n >= 0 and a(2*n) = 2*S(n-1,6) = 2*A001109(n), n >= 0, with T(n,x), resp. S(n,x), Chebyshev's polynomials of the first, resp. second kind. S(-1,x)=0. See A053120, resp. A049310. - Wolfdieter Lang, Jan 10 2003

Consider the mapping f(a/b) = (a + 2b)/(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, 3/2, 7/5, 17/12, 41/29, ... converging to 2^(1/2). Sequence contains the denominators. - Amarnath Murthy, Mar 22 2003

This is also the Horadam sequence (0,1,1,2). Limit_{n->oo} a(n)/a(n-1) = sqrt(2) + 1 = A014176. - Ross La Haye, Aug 18 2003

Number of 132-avoiding two-stack sortable permutations.

From Herbert Kociemba, Jun 02 2004: (Start)

For n > 0, the number of (s(0), s(1), ..., s(n)) such that 0 < s(i) < 4 and |s(i) - s(i-1)| <= 1 for i = 1,2,...,n, s(0) = 2, s(n) = 3.

Number of (s(0), s(1), ..., s(n)) such that 0 < s(i) < 4 and |s(i) - s(i-1)| <= 1 for i = 1,2,...,n, s(0) = 1, s(n) = 2. (End)

Counts walks of length n from a vertex of a triangle to another vertex to which a loop has been added. - Mario Catalani (mario.catalani(AT)unito.it), Jul 23 2004

Apart from initial terms, Pisot sequence P(2,5). See A008776 for definition of Pisot sequences. - David W. Wilson

Sums of antidiagonals of A038207 [Pascal's triangle squared]. - Ross La Haye, Oct 28 2004

The Pell primality test is "If N is an odd prime, then P(N)-Kronecker(2,N) is divisible by N". "Most" composite numbers fail this test, so it makes a useful pseudoprimality test. The odd composite numbers which are Pell pseudoprimes (i.e., that pass the above test) are in A099011. - Jack Brennen, Nov 13 2004

a(n) = sum of n-th row of triangle in A008288 = A094706(n) + A000079(n). - Reinhard Zumkeller, Dec 03 2004

Pell trapezoids (cf. A084158); for n > 0, A001109(n) = (a(n-1) + a(n+1))*a(n)/2; e.g., 1189 = (12+70)*29/2. - Charlie Marion, Apr 01 2006

(0!a(1), 1!a(2), 2!a(3), 3!a(4), ...) and (1,-2,-2,0,0,0,...) form a reciprocal pair under the list partition transform and associated operations described in A133314. - Tom Copeland, Oct 29 2007

Let C = (sqrt(2)+1) = 2.414213562..., then for n > 1, C^n = a(n)*(1/C) + a(n+1). Example: C^3 = 14.0710678... = 5*(0.414213562...) + 12. Let X = the 2 X 2 matrix [0, 1; 1, 2]; then X^n * [1, 0] = [a(n-1), a(n); a(n), a(n+1)]. a(n) = numerator of n-th convergent to (sqrt(2)-1) = 0.414213562... = [2, 2, 2, ...], the convergents being [1/2, 2/5, 5/12, ...]. - Gary W. Adamson, Dec 21 2007

A = sqrt(2) = 2/2 + 2/5 + 2/(5*29) + 2/(29*169) + 2/(169*985) + ...; B = ((5/2) - sqrt(2)) = 2/2 + 2/(2*12) + 2/(12*70) + 2/(70*408) + 2/(408*2378) + ...; A+B = 5/2. C = 1/2 = 2/(1*5) + 2/(2*12) + 2/(5*29) + 2/(12*70) + 2/(29*169) + ... - Gary W. Adamson, Mar 16 2008

From Clark Kimberling, Aug 27 2008: (Start)

Related convergents (numerator/denominator):

lower principal convergents: A002315/A001653

upper principal convergents: A001541/A001542

intermediate convergents: A052542/A001333

lower intermediate convergents: A005319/A001541

upper intermediate convergents: A075870/A002315

principal and intermediate convergents: A143607/A002965

lower principal and intermediate convergents: A143608/A079496

upper principal and intermediate convergents: A143609/A084068. (End)

Equals row sums of triangle A143808 starting with offset 1. - Gary W. Adamson, Sep 01 2008

Binomial transform of the sequence:= 0,1,0,2,0,4,0,8,0,16,..., powers of 2 alternating with zeros. - Philippe Deléham, Oct 28 2008

a(n) is also the sum of the n-th row of the triangle formed by starting with the top two rows of Pascal's triangle and then each next row has a 1 at both ends and the interior values are the sum of the three numbers in the triangle above that position. - Patrick Costello (pat.costello(AT)eku.edu), Dec 07 2008

Starting with offset 1 = eigensequence of triangle A135387 (an infinite lower triangular matrix with (2,2,2,...) in the main diagonal and (1,1,1,...) in the subdiagonal). - Gary W. Adamson, Dec 29 2008

Starting with offset 1 = row sums of triangle A153345. - Gary W. Adamson, Dec 24 2008

From Charlie Marion, Jan 07 2009: (Start)

In general, denominators, a(k,n) and numerators, b(k,n), of continued fraction convergents to sqrt((k+1)/k) may be found as follows:

let a(k,0) = 1, a(k,1) = 2k; for n > 0, a(k,2n) = 2*a(k,2n-1) + a(k,2n-2)

and a(k,2n+1) = (2k)*a(k,2n) + a(k,2n-1);

let b(k,0) = 1, b(k,1) = 2k+1; for n > 0, b(k,2n) = 2*b(k,2n-1) + b(k,2n-2)

and b(k,2n+1) = (2k)*b(k,2n) + b(k,2n-1).

For example, the convergents to sqrt(2/1) start 1/1, 3/2, 7/5, 17/12, 41/29.

In general, if a(k,n) and b(k,n) are the denominators and numerators, respectively, of continued fraction convergents to sqrt((k+1)/k) as defined above, then

k*a(k,2n)^2 - a(k,2n-1)*a(k,2n+1) = k = k*a(k,2n-2)*a(k,2n) - a(k,2n-1)^2 and

b(k,2n-1)*b(k,2n+1) - k*b(k,2n)^2 = k+1 = b(k,2n-1)^2 - k*b(k,2n-2)*b(k,2n);

for example, if k=1 and n=3, then a(1,n) = a(n+1) and

1*a(1,6)^2 - a(1,5)*a(1,7) = 1*169^2 - 70*408 = 1;

1*a(1,4)*a(1,6) - a(1,5)^2 = 1*29*169 - 70^2 = 1;

b(1,5)*b(1,7) - 1*b(1,6)^2 = 99*577 - 1*239^2 = 2;

b(1,5)^2 - 1*b(1,4)*b(1,6) = 99^2 - 1*41*239 = 2.

Cf. A001333, A142238, A142239, A153313, A153314, A153315, A153316, A153317, A153318.

(End)

Starting with offset 1 = row sums of triangle A155002, equivalent to the statement that the Fibonacci sequence convolved with the Pell sequence prefaced with a "1": (1, 1, 2, 5, 12, 29, ...) = (1, 2, 5, 12, 29, ...). - Gary W. Adamson, Jan 18 2009

It appears that P(p) == 8^((p-1)/2) (mod p), p = prime; analogous to [Schroeder, p. 90]: Fp == 5^((p-1)/2) (mod p). Example: Given P(11) = 5741, == 8^5 (mod 11). Given P(17) = 11336689, == 8^8 (mod 17) since 17 divides (8^8 - P(17)). - Gary W. Adamson, Feb 21 2009

Equals eigensequence of triangle A154325. - Gary W. Adamson, Feb 12 2009

Another combinatorial interpretation of a(n-1) arises from a simple tiling scenario. Namely, a(n-1) gives the number of ways of tiling a 1 X n rectangle with indistinguishable 1 X 2 rectangles and 1 X 1 squares that come in two varieties, say, A and B. For example, with C representing the 1 X 2 rectangle, we obtain a(4)=12 from AAA, AAB, ABA, BAA, ABB, BAB, BBA, BBB, AC, BC, CA and CB. - Martin Griffiths, Apr 25 2009

a(n+1) = 2*a(n) + a(n-1), a(1)=1, a(2)=2 was used by Theon from Smyrna. - Sture Sjöstedt, May 29 2009

The n-th Pell number counts the perfect matchings of the edge-labeled graph C_2 x P_(n-1), or equivalently, the number of domino tilings of a 2 X (n-1) cylindrical grid. - Sarah-Marie Belcastro, Jul 04 2009

As a fraction: 1/79 = 0.0126582278481... or 1/9799 = 0.000102051229...(1/119 and 1/10199 for sequence in reverse). - Mark Dols, May 18 2010

Limit_{n->oo} (a(n)/a(n-1) - a(n-1)/a(n)) tends to 2.0. Example: a(7)/a(6) - a(6)/a(7) = 169/70 - 70/169 = 2.0000845... - Gary W. Adamson, Jul 16 2010

Numbers k such that 2*k^2 +- 1 is a square. - Vincenzo Librandi, Jul 18 2010

Starting (1, 2, 5, ...) = INVERTi transform of A006190: (1, 3, 10, 33, 109, ...). - Gary W. Adamson, Aug 06 2010

[u,v] = [a(n), a(n-1)] generates all Pythagorean triples [u^2-v^2, 2uv, u^2+v^2] whose legs differ by 1. - James R. Buddenhagen, Aug 14 2010

An elephant sequence, see A175654. For the corner squares six A[5] vectors, with decimal values between 21 and 336, lead to this sequence (without the leading 0). For the central square these vectors lead to the companion sequence A078057. - Johannes W. Meijer, Aug 15 2010

Let the 2 X 2 square matrix A=[2, 1; 1, 0] then a(n) = the (1,1) element of A^(n-1). - Carmine Suriano, Jan 14 2011

Define a t-circle to be a first-quadrant circle tangent to the x- and y-axes. Such a circle has coordinates equal to its radius. Let C(0) be the t-circle with radius 1. Then for n > 0, define C(n) to be the next larger t-circle which is tangent to C(n - 1). C(n) has radius A001333(2n) + a(2n)*sqrt(2) and each of the coordinates of its point of intersection with C(n + 1) is a(2n + 1) + (A001333(2n + 1)*sqrt(2))/2. See similar Comments for A001109 and A001653, Sep 14 2005. - Charlie Marion, Jan 18 2012

A001333 and A000129 give the diagonal numbers described by Theon from Smyrna. - Sture Sjöstedt, Oct 20 2012

Pell numbers could also be called "silver Fibonacci numbers", since, for n >= 1, F(n+1) = ceiling(phi*F(n)), if n is even and F(n+1) = floor(phi*F(n)), if n is odd, where phi is the golden ratio, while a(n+1) = ceiling(delta*a(n)), if n is even and a(n+1) = floor(delta*a(n)), if n is odd, where delta = delta_S = 1+sqrt(2) is the silver ratio. - Vladimir Shevelev, Feb 22 2013

a(n) is the number of compositions (ordered partitions) of n-1 into two sorts of 1's and one sort of 2's. Example: the a(3)=5 compositions of 3-1=2 are 1+1, 1+1', 1'+1, 1'+1', and 2. - Bob Selcoe, Jun 21 2013

Between every two consecutive squares of a 1 X n array there is a flap that can be folded over one of the two squares. Two flaps can be lowered over the same square in 2 ways, depending on which one is on top. The n-th Pell number counts the ways n-1 flaps can be lowered. For example, a sideway representation for the case n = 3 squares and 2 flaps is \\., .//, \./, ./., .\., where . is an empty square. - Jean M. Morales, Sep 18 2013

Define a(-n) to be a(n) for n odd and -a(n) for n even. Then a(n) = A005319(k)*(a(n-2k+1) - a(n-2k)) + a(n-4k) = A075870(k)*(a(n-2k+2) - a(n-2k+1)) - a(n-4k+2). - Charlie Marion, Nov 26 2013

An alternative formulation of the combinatorial tiling interpretation listed above: Except for n=0, a(n-1) is the number of ways of partial tiling a 1 X n board with 1 X 1 squares and 1 X 2 dominoes. - Matthew Lehman, Dec 25 2013

Define a(-n) to be a(n) for n odd and -a(n) for n even. Then a(n) = A077444(k)*a(n-2k+1) + a(n-4k+2). This formula generalizes the formula used to define this sequence. - Charlie Marion, Jan 30 2014

a(n-1) is the top left entry of the n-th power of any of the 3 X 3 matrices [0, 1, 1; 1, 1, 1; 0, 1, 1], [0, 1, 1; 0, 1, 1; 1, 1, 1], [0, 1, 0; 1, 1, 1; 1, 1, 1] or [0, 0, 1; 1, 1, 1; 1, 1, 1]. - R. J. Mathar, Feb 03 2014

a(n+1) counts closed walks on K2 containing two loops on the other vertex. Equivalently the (1,1) entry of A^(n+1) where the adjacency matrix of digraph is A=(0,1;1,2). - David Neil McGrath, Oct 28 2014

For n >= 1, a(n) equals the number of ternary words of length n-1 avoiding runs of zeros of odd lengths. - Milan Janjic, Jan 28 2015

This is a divisibility sequence (i.e., if n|m then a(n)|a(m)). - Tom Edgar, Jan 28 2015

A strong divisibility sequence, that is, gcd(a(n), a(m)) = a(gcd(n, m)) for all positive integers n and m. - Michael Somos, Jan 03 2017

a(n) is the number of compositions (ordered partitions) of n-1 into two kinds of parts, n and n', when the order of the 1 does not matter, or equivalently, when the order of the 1' does not matter. Example: When the order of the 1 does not matter, the a(3)=5 compositions of 3-1=2 are 1+1, 1+1'=1+1, 1'+1', 2 and 2'. (Contrast with entry from Bob Selcoe dated Jun 21 2013.) - Gregory L. Simay, Sep 07 2017

Number of weak orderings R on {1,...,n} that are weakly single-peaked w.r.t. the total ordering 1 < ... < n and for which {1,...,n} has exactly one minimal element for the weak ordering R. - J. Devillet, Sep 28 2017

Also the number of matchings in the (n-1)-centipede graph. - Eric W. Weisstein, Sep 30 2017

Let A(r,n) be the total number of ordered arrangements of an n+r tiling of r red squares and white tiles of total length n, where the individual tile lengths can range from 1 to n. A(r,0) corresponds to a tiling of r red squares only, and so A(r,0)=1. Let A_1(r,n) = Sum_{j=0..n} A(r,j) and let A_s(r,n) = Sum_{j=0..n} A_(s-1)(r,j). Then A_0(1,n) + A_2(3,n-4) + A_4(5,n-8) + ... + A_(2j) (2j+1, n-4j) = a(n) without the initial 0. - Gregory L. Simay, May 25 2018

(1, 2, 5, 12, 29, ...) is the fourth INVERT transform of (1, -2, 5, -12, 29, ...), as shown in A073133. - Gary W. Adamson, Jul 17 2019

Number of 2-compositions of n restricted to odd parts (and allowed zeros); see Hopkins & Ouvry reference. - Brian Hopkins, Aug 17 2020

Also called the 2-metallonacci sequence; the g.f. 1/(1-k*x-x^2) gives the k-metallonacci sequence. - Michael A. Allen, Jan 23 2023

Named by Lucas (1878) after the English mathematician John Pell (1611-1685). - Amiram Eldar, Oct 02 2023

a(n) is the number of compositions of n when there are F(i) parts of size i, with i,n >= 1, F(n) the Fibonacci numbers, A000045(n) (see example below). - Enrique Navarrete, Dec 15 2023

The sequence contains exactly one square greater than 1, namely 4900 (according to Gardner). - Jud McCranie, Mar 19 2001, Mar 22 2007 [This is a result from Watson. - Charles R Greathouse IV, Jun 21 2013] [See A351830 for further related comments and references.]

Number of rhombi in an n X n rhombus. - Matti De Craene (Matti.DeCraene(AT)rug.ac.be), May 14 2000

Number of acute triangles made from the vertices of a regular n-polygon when n is odd (cf. A007290). - Sen-Peng Eu, Apr 05 2001

Gives number of squares with sides parallel to the axes formed from an n X n square. In a 1 X 1 square, one is formed. In a 2 X 2 square, five squares are formed. In a 3 X 3 square, 14 squares are formed and so on. - Kristie Smith (kristie10spud(AT)hotmail.com), Apr 16 2002; edited by Eric W. Weisstein, Mar 05 2025

a(n-1) = B_3(n)/3, where B_3(x) = x(x-1)(x-1/2) is the third Bernoulli polynomial. - Michael Somos, Mar 13 2004

Number of permutations avoiding 13-2 that contain the pattern 32-1 exactly once.

Since 3*r = (r+1) + r + (r-1) = T(r+1) - T(r-2), where T(r) = r-th triangular number r*(r+1)/2, we have 3*r^2 = r*(T(r+1) - T(r-2)) = f(r+1) - f(r-1) ... (i), where f(r) = (r-1)*T(r) = (r+1)*T(r-1). Summing over n, the right hand side of relation (i) telescopes to f(n+1) + f(n) = T(n)*((n+2) + (n-1)), whence the result Sum_{r=1..n} r^2 = n*(n+1)*(2*n+1)/6 immediately follows. - Lekraj Beedassy, Aug 06 2004

Also as a(n) = (1/6)*(2*n^3 + 3*n^2 + n), n > 0: structured trigonal diamond numbers (vertex structure 5) (cf. A006003 = alternate vertex; A000447 = structured diamonds; A100145 for more on structured numbers). - James A. Record (james.record(AT)gmail.com), Nov 07 2004

Number of triples of integers from {1, 2, ..., n} whose last component is greater than or equal to the others.

Kekulé numbers for certain benzenoids. - Emeric Deutsch, Jun 12 2005

Sum of the first n positive squares. - Cino Hilliard, Jun 18 2007

Maximal number of cubes of side 1 in a right pyramid with a square base of side n and height n. - Pasquale CUTOLO (p.cutolo(AT)inwind.it), Jul 09 2007

If a 2-set Y and an (n-2)-set Z are disjoint subsets of an n-set X then a(n-3) is the number of 4-subsets of X intersecting both Y and Z. - Milan Janjic, Sep 19 2007

We also have the identity 1 + (1+4) + (1+4+9) + ... + (1+4+9+16+ ... + n^2) = n(n+1)(n+2)(n+(n+1)+(n+2))/36; ... and in general the k-fold nested sum of squares can be expressed as n(n+1)...(n+k)(n+(n+1)+...+(n+k))/((k+2)!(k+1)/2). - Alexander R. Povolotsky, Nov 21 2007

The terms of this sequence are coefficients of the Engel expansion of the following converging sum: 1/(1^2) + (1/1^2)*(1/(1^2+2^2)) + (1/1^2)*(1/(1^2+2^2))*(1/(1^2+2^2+3^2)) + ... - Alexander R. Povolotsky, Dec 10 2007

Convolution of A000290 with A000012. - Sergio Falcon, Feb 05 2008

Hankel transform of binomial(2*n-3, n-1) is -a(n). - Paul Barry, Feb 12 2008

Starting (1, 5, 14, 30, ...) = binomial transform of [1, 4, 5, 2, 0, 0, 0, ...]. - Gary W. Adamson, Jun 13 2008

Starting (1,5,14,30,...) = second partial sums of binomial transform of [1,2,0,0,0,...]. a(n) = Sum_{i=0..n} binomial(n+2,i+2)*b(i), where b(i)=1,2,0,0,0,... - Borislav St. Borisov (b.st.borisov(AT)abv.bg), Mar 05 2009

Convolution of A001477 with A005408: a(n) = Sum_{k=0..n} (2*k+1)*(n-k). - Reinhard Zumkeller, Mar 07 2009

Sequence of the absolute values of the z^1 coefficients of the polynomials in the GF1 denominators of A156921. See A157702 for background information. - Johannes W. Meijer, Mar 07 2009

The sequence is related to A000217 by a(n) = n*A000217(n) - Sum_{i=0..n-1} A000217(i) and this is the case d = 1 in the identity n^2*(d*n-d+2)/2 - Sum_{i=0..n-1} i*(d*i-d+2)/2 = n*(n+1)(2*d*n-2*d+3)/6, or also the case d = 0 in n^2*(n+2*d+1)/2 - Sum_{i=0..n-1} i*(i+2*d+1)/2 = n*(n+1)*(2*n+3*d+1)/6. - Bruno Berselli, Apr 21 2010, Apr 03 2012

a(n)/n = k^2 (k = integer) for n = 337; a(337) = 12814425, a(n)/n = 38025, k = 195, i.e., the number k = 195 is the quadratic mean (root mean square) of the first 337 positive integers. There are other such numbers -- see A084231 and A084232. - Jaroslav Krizek, May 23 2010

Also the number of moves to solve the "alternate coins game": given 2n+1 coins (n+1 Black, n White) set alternately in a row (BWBW...BWB) translate (not rotate) a pair of adjacent coins at a time (1 B and 1 W) so that at the end the arrangement shall be BBBBB..BW...WWWWW (Blacks separated by Whites). Isolated coins cannot be moved. - Carmine Suriano, Sep 10 2010

From J. M. Bergot, Aug 23 2011: (Start)

Using four consecutive numbers n, n+1, n+2, and n+3 take all possible pairs (n, n+1), (n, n+2), (n, n+3), (n+1, n+2), (n+1, n+3), (n+2, n+3) to create unreduced Pythagorean triangles. The sum of all six areas is 60*a(n+1).

Using three consecutive odd numbers j, k, m, (j+k+m)^3 - (j^3 + k^3 + m^3) equals 576*a(n) = 24^2*a(n) where n = (j+1)/2. (End)

From Ant King, Oct 17 2012: (Start)

For n > 0, the digital roots of this sequence A010888(a(n)) form the purely periodic 27-cycle {1, 5, 5, 3, 1, 1, 5, 6, 6, 7, 2, 2, 9, 7, 7, 2, 3, 3, 4, 8, 8, 6, 4, 4, 8, 9, 9}.

For n > 0, the units' digits of this sequence A010879(a(n)) form the purely periodic 20-cycle {1, 5, 4, 0, 5, 1, 0, 4, 5, 5, 6, 0, 9, 5, 0, 6, 5, 9, 0, 0}. (End)

Length of the Pisano period of this sequence mod n, n>=1: 1, 4, 9, 8, 5, 36, 7, 16, 27, 20, 11, 72, 13, 28, 45, 32, 17, 108, 19, 40, ... . - R. J. Mathar, Oct 17 2012

Sum of entries of n X n square matrix with elements min(i,j). - Enrique Pérez Herrero, Jan 16 2013

The number of intersections of diagonals in the interior of regular n-gon for odd n > 1 divided by n is a square pyramidal number; that is, A006561(2*n+1)/(2*n+1) = A000330(n-1) = (1/6)*n*(n-1)*(2*n-1). - Martin Renner, Mar 06 2013

For n > 1, a(n)/(2n+1) = A024702(m), for n such that 2n+1 = prime, which results in 2n+1 = A000040(m). For example, for n = 8, 2n+1 = 17 = A000040(7), a(8) = 204, 204/17 = 12 = A024702(7). - Richard R. Forberg, Aug 20 2013

A formula for the r-th successive summation of k^2, for k = 1 to n, is (2*n+r)*(n+r)!/((r+2)!*(n-1)!) (H. W. Gould). - Gary Detlefs, Jan 02 2014

The n-th square pyramidal number = the n-th triangular dipyramidal number (Johnson 12), which is the sum of the n-th + (n-1)-st tetrahedral numbers. E.g., the 3rd tetrahedral number is 10 = 1+3+6, the 2nd is 4 = 1+3. In triangular "dipyramidal form" these numbers can be written as 1+3+6+3+1 = 14. For "square pyramidal form", rebracket as 1+(1+3)+(3+6) = 14. - John F. Richardson, Mar 27 2014

Beukers and Top prove that no square pyramidal number > 1 equals a tetrahedral number A000292. - Jonathan Sondow, Jun 21 2014

Odd numbered entries are related to dissections of polygons through A100157. - Tom Copeland, Oct 05 2014

From Bui Quang Tuan, Apr 03 2015: (Start)

We construct a number triangle from the integers 1, 2, 3, ..., n as follows. The first column contains 2*n-1 integers 1. The second column contains 2*n-3 integers 2, ... The last column contains only one integer n. The sum of all the numbers in the triangle is a(n).

Here is an example with n = 5:

1

1 2

1 2 3

1 2 3 4

1 2 3 4 5

1 2 3 4

1 2 3

1 2

1

(End)

The Catalan number series A000108(n+3), offset 0, gives Hankel transform revealing the square pyramidal numbers starting at 5, A000330(n+2), offset 0 (empirical observation). - Tony Foster III, Sep 05 2016; see Dougherty et al. link p. 2. - Andrey Zabolotskiy, Oct 13 2016

Number of floating point additions in the factorization of an (n+1) X (n+1) real matrix by Gaussian elimination as e.g. implemented in LINPACK subroutines sgefa.f or dgefa.f. The number of multiplications is given by A007290. - Hugo Pfoertner, Mar 28 2018

The Jacobi polynomial P(n-1,-n+2,2,3) or equivalently the sum of dot products of vectors from the first n rows of Pascal's triangle (A007318) with the up-diagonal Chebyshev T coefficient vector (1,3,2,0,...) (A053120) or down-diagonal vector (1,-7,32,-120,400,...) (A001794). a(5) = 1 + (1,1).(1,3) + (1,2,1).(1,3,2) + (1,3,3,1).(1,3,2,0) + (1,4,6,4,1).(1,3,2,0,0) = (1 + (1,1).(1,-7) + (1,2,1).(1,-7,32) + (1,3,3,1).(1,-7,32,-120) + (1,4,6,4,1).(1,-7,32,-120,400))*(-1)^(n-1) = 55. - Richard Turk, Jul 03 2018

Coefficients in the terminating series identity 1 - 5*n/(n + 4) + 14*n*(n - 1)/((n + 4)*(n + 5)) - 30*n*(n - 1)*(n - 2)/((n + 4)*(n + 5)*(n + 6)) + ... = 0 for n = 1,2,3,.... Cf. A002415 and A108674. - Peter Bala, Feb 12 2019

n divides a(n) iff n == +- 1 (mod 6) (see A007310). (See De Koninck reference.) Examples: a(11) = 506 = 11 * 46, and a(13) = 819 = 13 * 63. - Bernard Schott, Jan 10 2020

For n > 0, a(n) is the number of ternary words of length n+2 having 3 letters equal to 2 and 0 only occurring as the last letter. For example, for n=2, the length 4 words are 2221,2212,2122,1222,2220. - Milan Janjic, Jan 28 2020

Conjecture: Every integer can be represented as a sum of three generalized square pyramidal numbers. A related conjecture is given in A336205 corresponding to pentagonal case. A stronger version of these conjectures is that every integer can be expressed as a sum of three generalized r-gonal pyramidal numbers for all r >= 3. In here "generalized" means negative indices are included. - Altug Alkan, Jul 30 2020

The natural number y is a term if and only if y = a(floor((3 * y)^(1/3))). - Robert Israel, Dec 04 2024

Also the number of directed bishop moves on an n X n chessboard, where two moves are considered the same if one can be obtained from the other by a rotation of the board. Reflections are ignored. Equivalently, number of directed bishop moves on an n X n chessboard, where two moves are considered the same if one can be obtained from the other by an axial reflection of the board (horizontal or vertical). Rotations and diagonal reflections are ignored. - Hilko Koning, Aug 22 2025

Number of n-step non-selfintersecting paths starting at (0,0) with steps of types (1,0), (-1,0) or (0,1) [Stanley].

Number of n steps one-sided prudent walks with east, west and north steps. - Shanzhen Gao, Apr 26 2011

Number of ternary strings of length n-1 with subwords (0,2) and (2,0) not allowed. - Olivier Gérard, Aug 28 2012

Number of symmetric 2n X 2 or (2n-1) X 2 crossword puzzle grids: all white squares are edge connected; at least 1 white square on every edge of grid; 180-degree rotational symmetry. - Erich Friedman

a(n+1) is the number of ways to put molecules on a 2 X n ladder lattice so that the molecules do not touch each other.

In other words, a(n+1) is the number of independent vertex sets and vertex covers in the n-ladder graph P_2 X P_n. - Eric W. Weisstein, Apr 04 2017

Number of (n-1) X 2 binary arrays with a path of adjacent 1's from top row to bottom row, see A359576. - R. H. Hardin, Mar 16 2002

a(2*n+1) with b(2*n+1) := A000129(2*n+1), n >= 0, give all (positive integer) solutions to Pell equation a^2 - 2*b^2 = -1.

a(2*n) with b(2*n) := A000129(2*n), n >= 1, give all (positive integer) solutions to Pell equation a^2 - 2*b^2 = +1 (see Emerson reference).

Bisection: a(2*n) = T(n,3) = A001541(n), n >= 0 and a(2*n+1) = S(2*n,2*sqrt(2)) = A002315(n), n >= 0, with T(n,x), resp. S(n,x), Chebyshev's polynomials of the first, resp. second kind. See A053120, resp. A049310.

Binomial transform of A077957. - Paul Barry, Feb 25 2003

For n > 0, the number of (s(0), s(1), ..., s(n)) such that 0 < s(i) < 4 and |s(i) - s(i-1)| <= 1 for i = 1,2,...,n, s(0) = 2, s(n) = 2. - Herbert Kociemba, Jun 02 2004

For n > 1, a(n) corresponds to the longer side of a near right-angled isosceles triangle, one of the equal sides being A000129(n). - Lekraj Beedassy, Aug 06 2004

Exponents of terms in the series F(x,1), where F is determined by the equation F(x,y) = xy + F(x^2*y,x). - Jonathan Sondow, Dec 18 2004

Number of n-words from the alphabet A={0,1,2} which two neighbors differ by at most 1. - Fung Cheok Yin (cheokyin_restart(AT)yahoo.com.hk), Aug 30 2006

Consider the mapping f(a/b) = (a + 2b)/(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, 3/2, 7/5, 17/12, 41/29, ... converging to 2^(1/2). Sequence contains the numerators. - Amarnath Murthy, Mar 22 2003 [Amended by Paul E. Black (paul.black(AT)nist.gov), Dec 18 2006]

Odd-indexed prime numerators are prime RMS numbers (A140480) and also NSW primes (A088165). - Ctibor O. Zizka, Aug 13 2008

The intermediate convergents to 2^(1/2) begin with 4/3, 10/7, 24/17, 58/41; essentially, numerators=A052542 and denominators here. - Clark Kimberling, Aug 26 2008

Equals right border of triangle A143966. Starting (1, 3, 7, ...) equals INVERT transform of (1, 2, 2, 2, ...) and row sums of triangle A143966. - Gary W. Adamson, Sep 06 2008

Inverse binomial transform of A006012; Hankel transform is := [1, 2, 0, 0, 0, 0, 0, 0, 0, ...]. - Philippe Deléham, Dec 04 2008

From Charlie Marion, Jan 07 2009: (Start)

In general, denominators, a(k,n) and numerators, b(k,n), of continued fraction convergents to sqrt((k+1)/k) may be found as follows:

let a(k,0) = 1, a(k,1) = 2k; for n>0, a(k,2n) = 2*a(k,2n-1) + a(k,2n-2) and a(k,2n+1) = (2k)*a(k,2n) + a(k,2n-1);

let b(k,0) = 1, b(k,1) = 2k+1; for n>0, b(k,2n) = 2*b(k,2n-1) + b(k,2n-2) and b(k,2n+1) = (2k)*b(k,2n) + b(k,2n-1).

For example, the convergents to sqrt(2/1) start 1/1, 3/2, 7/5, 17/12, 41/29.

In general, if a(k,n) and b(k,n) are the denominators and numerators, respectively, of continued fraction convergents to sqrt((k+1)/k) as defined above, then

k*a(k,2n)^2 - a(k,2n-1)*a(k,2n+1) = k = k*a(k,2n-2)*a(k,2n) - a(k,2n-1)^2 and

b(k,2n-1)*b(k,2n+1) - k*b(k,2n)^2 = k+1 = b(k,2n-1)^2 - k*b(k,2n-2)*b(k,2n);

for example, if k=1 and n=3, then b(1,n)=a(n+1) and

1*a(1,6)^2 - a(1,5)*a(1,7) = 1*169^2 - 70*408 = 1;

1*a(1,4)*a(1,6) - a(1,5)^2 = 1*29*169 - 70^2 = 1;

b(1,5)*b(1,7) - 1*b(1,6)^2 = 99*577 - 1*239^2 = 2;

b(1,5)^2 - 1*b(1,4)*b(1,6) = 99^2 - 1*41*239 = 2.

Cf. A000129, A142238-A142239, A153313-A153318.

(End)

This sequence occurs in the lower bound of the order of the set of equivalent resistances of n equal resistors combined in series and in parallel (A048211). - Sameen Ahmed Khan, Jun 28 2010

Let M = a triangle with the Fibonacci series in each column, but the leftmost column is shifted upwards one row. A001333 = lim_{n->infinity} M^n, the left-shifted vector considered as a sequence. - Gary W. Adamson, Jul 27 2010

a(n) is the number of compositions of n when there are 1 type of 1 and 2 types of other natural numbers. - Milan Janjic, Aug 13 2010

Equals the INVERTi transform of A055099. - Gary W. Adamson, Aug 14 2010

From L. Edson Jeffery, Apr 04 2011: (Start)

Let U be the unit-primitive matrix (see [Jeffery])

U = U_(8,2) = (0 0 1 0)

(0 1 0 1)

(1 0 2 0)

(0 2 0 1).

Then a(n) = (1/4)*Trace(U^n). (See also A084130, A006012.)

(End)

For n >= 1, row sums of triangle

m/k.|..0.....1.....2.....3.....4.....5.....6.....7

==================================================

.0..|..1

.1..|..1.....2

.2..|..1.....2.....4

.3..|..1.....4.....4.....8

.4..|..1.....4....12.....8....16

.5..|..1.....6....12....32....16....32

.6..|..1.....6....24....32....80....32....64

.7..|..1.....8....24....80....80...192....64...128

which is the triangle for numbers 2^k*C(m,k) with duplicated diagonals. - Vladimir Shevelev, Apr 12 2012

a(n) is also the number of ways to place k non-attacking wazirs on a 2 X n board, summed over all k >= 0 (a wazir is a leaper [0,1]). - Vaclav Kotesovec, May 08 2012

The sequences a(n) and b(n) := A000129(n) are entries of powers of the special case of the Brahmagupta Matrix - for details see Suryanarayan's paper. Further, as Suryanarayan remark, if we set A = 2*(a(n) + b(n))*b(n), B = a(n)*(a(n) + 2*b(n)), C = a(n)^2 + 2*a(n)*b(n) + 2*b(n)^2 we obtain integral solutions of the Pythagorean relation A^2 + B^2 = C^2, where A and B are consecutive integers. - Roman Witula, Jul 28 2012

Pisano period lengths: 1, 1, 8, 4, 12, 8, 6, 4, 24, 12, 24, 8, 28, 6, 24, 8, 16, 24, 40, 12, .... - R. J. Mathar, Aug 10 2012

This sequence and A000129 give the diagonal numbers described by Theon of Smyrna. - Sture Sjöstedt, Oct 20 2012

a(n) is the top left entry of the n-th power of any of the following six 3 X 3 binary matrices: [1, 1, 1; 1, 1, 1; 1, 0, 0] or [1, 1, 1; 1, 1, 0; 1, 1, 0] or [1, 1, 1; 1, 0, 1; 1, 1, 0] or [1, 1, 1; 1, 1, 0; 1, 0, 1] or [1, 1, 1; 1, 0, 1; 1, 0, 1] or [1, 1, 1; 1, 0, 0; 1, 1, 1]. - R. J. Mathar, Feb 03 2014

If p is prime, a(p) == 1 (mod p) (compare with similar comment for A000032). - Creighton Dement, Oct 11 2005, modified by Davide Colazingari, Jun 26 2016

a(n) = A000129(n) + A000129(n-1), where A000129(n) is the n-th Pell Number; e.g., a(6) = 99 = A000129(6) + A000129(5) = 70 + 29. Hence the sequence of fractions has the form 1 + A000129(n-1)/A000129(n), and the ratio A000129(n-1)/A000129(n)converges to sqrt(2) - 1. - Gregory L. Simay, Nov 30 2018

For n > 0, a(n+1) is the length of tau^n(1) where tau is the morphism: 1 -> 101, 0 -> 1. See Song and Wu. - Michel Marcus, Jul 21 2020

For n > 0, a(n) is the number of nonisomorphic quasitrivial semigroups with n elements, see Devillet, Marichal, Teheux. A292932 is the number of labeled quasitrivial semigroups. - Peter Jipsen, Mar 28 2021

a(n) is the permanent of the n X n tridiagonal matrix defined in A332602. - Stefano Spezia, Apr 12 2022

From Greg Dresden, May 08 2023: (Start)

For n >= 2, 4*a(n) is the number of ways to tile this T-shaped figure of length n-1 with two colors of squares and one color of domino; shown here is the figure of length 5 (corresponding to n=6), and it has 4*a(6) = 396 different tilings.

_

|| _

|||_|||

|_|

(End)

12*a(n) = number of walks of length n in the cyclic Kautz digraph CK(3,4). - Miquel A. Fiol, Feb 15 2024