cp's OEIS Frontend

Satisfies a recurrence of S_r type for r=36: 0, 1, 36 and a(n-1)*a(n+1)=(a(n)-1)^2. First observed by Colin Dickson in alt.math.recreational, Mar 07 2004. - Rainer Rosenthal, Mar 14 2004

For every n, a(n) is the first of three triangular numbers in geometric progression. The third number in the progression is a(n+1). The middle triangular number is sqrt(a(n)*a(n+1)). Chen and Fang prove that four distinct triangular numbers are never in geometric progression. - T. D. Noe, Apr 30 2007

The sum of any two terms is never equal to a Fermat number. - Arkadiusz Wesolowski, Feb 14 2012

Conjecture: No a(2^k), where k is a nonnegative integer, can be expressed as a sum of a positive square number and a positive triangular number. - Ivan N. Ianakiev, Sep 19 2012

For n=2k+1, A010888(a(n))=1 and for n=2k, k > 0, A010888(a(n))=9. - Ivan N. Ianakiev, Oct 12 2013

For n > 0, these are the triangular numbers which are the sum of two consecutive triangular numbers, for instance 36 = 15 + 21 and 1225 = 595 + 630. - Michel Marcus, Feb 18 2014

The sequence is the case P1 = 36, P2 = 68, Q = 1 of the 3-parameter family of 4th order linear divisibility sequences found by Williams and Guy. - Peter Bala, Apr 03 2014

For n=2k, k > 0, a(n) is divisible by 12 and is therefore abundant. I conjecture that for n=2k+1 a(n) is deficient [true for k up to 43 incl.]. - Ivan N. Ianakiev, Sep 30 2014

The conjecture is true for all k > 0 because: For n=2k+1, k > 0, a(n) is odd. If a(n) is a prime number, it is deficient; otherwise a(n) has one or two distinct prime factors and is therefore deficient again. So for n=2k+1, k > 0, a(n) is deficient. - Muniru A Asiru, Apr 13 2016

Numbers k for which A139275(k) is a perfect square. - Bruno Berselli, Jan 16 2018

Row sums are A011782. Inverse is A065547.

Triangle T(n,k), 0 <= k <= n, read by rows, given by [0, 1, -1, 0, 0, 0, 0, 0, ...] DELTA [1, 0, 1, 0, 0, 0, 0, 0, ...] where DELTA is the operator defined in A084938. - Philippe Deléham, Jul 29 2006

Sum of entries in column k is A001519(k+1) (the odd-indexed Fibonacci numbers). - Philippe Deléham, Dec 02 2008

Number of permutations of length n avoiding simultaneously the patterns 123 and 132 with k left-to-right minima. A left-to-right minimum in a permutation a(1)a(2)...a(n) is position i such that a(j) > a(i) for all j < i. - Tian Han, Nov 16 2023

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

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

Partial sums of Pell numbers A000129.

W(n){1,3;2,-1,1} = Sum_{i=1..n} W(i){1,2;2,-1,0}, where W(n){a,b; p,q,r} implies x(n) = p*x(n-1) - q*x(n-2) + r; x(0)=a, x(1)=b.

Number of 2 X (n+1) binary arrays with path of adjacent 1's from upper left to lower right corner. - R. H. Hardin, Mar 16 2002

Binomial transform of A029744. - Paul Barry, Apr 23 2004

Number of (s(0), s(1), ..., s(n+2)) such that 0 < s(i) < 4 and |s(i) - s(i-1)| <= 1 for i = 1,2,...,n+2, s(0) = 1, s(n+2) = 3. - Herbert Kociemba, Jun 16 2004

Equals row sums of triangle A153346. - Gary W. Adamson, Dec 24 2008

Equals the sum of the terms of the antidiagonals of A142978. - J. M. Bergot, Nov 13 2012

a(p-2) == 0 mod p where p is an odd prime, see A270342. - Altug Alkan, Mar 15 2016

Also, the lexicographically earliest sequence of positive integers such that for n > 3, {sqrt(2)*a(n)} is located strictly between {sqrt(2)*a(n-1)} and {sqrt(2)*a(n-2)} where {} denotes the fractional part. - Ivan Neretin, May 02 2017

a(n+1) is the number of weak orderings on {1,...,n} that are weakly single-peaked w.r.t. the total ordering 1 < ... < n. - J. Devillet, Oct 06 2017

Let x_n be the sequence 1,3,7,17,41,99,239,... (this sequence or A001333) and let y_n = 1,2,5,12,29,70,169,... (A000129). Then {+- x_n +- y_n*sqrt(2) } are the units in the ring of algebraic integers Z[ sqrt(2) ].

Consider a string of n red, blue and green beads (with start and end points distinct and not interchangeable). If one pairing is disallowed, so that a red bead cannot immediately follow a blue bead or vice versa, how many different strings exist of any given length? Answer is a(n). E.g., a(3)=17 because there are 17 strings of length 3: RRR, RRG, RGR, RGG, RGB, GRR, GRG, GGR, GGG, GGB, GBG, GBB, BGR, BGG, BGB, BBG, BBB - Wayne VanWeerthuizen, May 02 2004

The number of Khalimsky-continuous functions with one fixed endpoint. - Shiva Samieinia (shiva(AT)math.su.se), Oct 08 2007

The sequence (-1)^C(n+1,2)*a(n) with g.f. (1-3x-x^2-x^3)/(1+6x^2+x^4) is the Hankel transform of the signed central binomial coefficients (-1)^C(n+1,2)*A001405(n). - Paul Barry, Jun 24 2008

An elephant sequence, see A175655. For the central square six A[5] vectors, with decimal values between 21 and 336, lead to this sequence. For the corner squares these vectors lead to the companion sequence A000129 (without the leading 0). - Johannes W. Meijer, Aug 15 2010

Sequence is related to rhombus substitution tilings showing 8-fold rotational symmetry (see A001333). - L. Edson Jeffery, Apr 04 2011

Number of length-n strings of 3 letters {0,1,2} with no two adjacent nonzero letters identical. The general case (strings of L letters) is the sequence with g.f. (1+x)/(1-(L-1)*x-x^2). - Joerg Arndt, Oct 11 2012

Row sums of A035607, when seen as a triangle read by rows. - Reinhard Zumkeller, Jul 20 2013

Interpretation via Cartier-Foata traces. Consider the trace monoid on Sigma = {a, b, c} in which only a and b commute. Let h(t) be the CF height (number of cliques) and |t| the length (number of letters). Then a(n) counts traces t with h(t) = n and |t| = n. In CF normal form these are exactly sequences of n singleton cliques {a},{b},{c} with the rule: whenever two consecutive cliques lie in {{a},{b}} they must be equal (so {a}{b} and {b}{a} are forbidden, while {a}{a}, {b}{b}, and any occurrence of {c} are allowed). Equivalently, no direct change a<->b without an intervening {c}. Examples: valid {a}{a}{c}{b}{b}; invalid {a}{b}{a}. This CF trace model is equivalent to Wayne VanWeerthuizen's bead model given above, where {a} = red, {b} = blue, {c} = green. In both cases, a red bead cannot immediately follow a blue bead, and a blue bead cannot immediately follow a red bead, without a green bead in between. - Constantinos Kourouzides, Aug 13 2025

b(0)=0, c(0)=1, b(i+1)=b(i)+c(i), c(i+1)=b(i+1)+b(i); then a(i) (the number in the sequence) is 2b(i)^2 if i is even, c(i)^2 if i is odd and b(n)=A000129(n) and c(n)=A001333(n). - Darin Stephenson (stephenson(AT)cs.hope.edu) and Alan Koch

For n > 1 gives solutions to A007913(2x) = A007913(x+1). - Benoit Cloitre, Apr 07 2002

If (X,X+1,Z) is a Pythagorean triple, then Z-X-1 and Z+X are in the sequence.

For n >= 2, a(n) gives exactly the positive integers m such that 1,2,...,m has a perfect median. The sequence of associated perfect medians is A001109. Let a_1,...,a_m be an (ordered) sequence of real numbers, then a term a_k is a perfect median if Sum_{j=1..k-1} a_j = Sum_{j=k+1..m} a_j. See Puzzle 1 in MSRI Emissary, Fall 2005. - Asher Auel, Jan 12 2006

This is the r=8 member of the r-family of sequences S_r(n) defined in A092184 where more information can be found.

Also, 1^3 + 2^3 + 3^3 + ... + a(n)^3 = k(n)^4 where k(n) is A001109. - Anton Vrba (antonvrba(AT)yahoo.com), Nov 18 2006

If T_x = y^2 is a triangular number which is also a square, the least number which is both triangular and square and greater than T_x is T_(3*x + 4*y + 1) = (2*x + 3*y + 1)^2 (W. Sierpiński 1961). - Richard Choulet, Apr 28 2009

If (a,b) is a solution of the Diophantine equation 0 + 1 + 2 + ... + x = y^2, then a or (a+1) is a perfect square. If (a,b) is a solution of the Diophantine equation 0 + 1 + 2 + ... + x = y^2, then a or a/8 is a perfect square. If (a,b) and (c,d) are two consecutive solutions of the Diophantine equation 0 + 1 + 2 + ... + x = y^2 with a < c, then a+b = c-d and ((d+b)^2, d^2-b^2) is a solution, too. If (a,b), (c,d) and (e,f) are three consecutive solutions of the Diophantine equation 0 + 1 + 2 + ... + x = y^2 with a < c < e, then (8*d^2, d*(f-b)) is a solution, too. - Mohamed Bouhamida, Aug 29 2009

If (p,q) and (r,s) are two consecutive solutions of the Diophantine equation 0 + 1 + 2 + ... + x = y^2 with p < r, then r = 3p + 4q + 1 and s = 2p + 3q + 1. - Mohamed Bouhamida, Sep 02 2009

Also numbers k such that (ceiling(sqrt(k*(k+1)/2)))^2 - k*(k+1)/2 = 0. - Ctibor O. Zizka, Nov 10 2009

From Lekraj Beedassy, Mar 04 2011: (Start)

Let x=a(n) be the index of the associated triangular number T_x=1+2+3+...+x and y=A001109(n) be the base of the associated perfect square S_y=y^2. Now using the identity S_y = T_y + T_{y-1}, the defining T_x = S_y may be rewritten as T_y = T_x - T_{y-1}, or 1+2+3+...+y = y+(y+1)+...+x. This solves the Strand Magazine House Number problem mentioned in A001109 in references from Poo-Sung Park and John C. Butcher. In a variant of the problem, solving the equation 1+3+5+...+(2*x+1) = (2*x+1)+(2*x+3)+...+(2*y-1) implies S_(x+1) = S_y - S_x, i.e., with (x,x+1,y) forming a Pythagorean triple, the solutions are given by pairs of x=A001652(n), y=A001653(n). (End)

If P = 8*n +- 1 is a prime, then P divides a((P-1)/2); e.g., 7 divides a(3) and 41 divides a(20). Also, if P = 8*n +- 3 is prime, then 4*P divides (a((P-1)/2) + a((P+1)/2) + 3). - Kenneth J Ramsey, Mar 05 2012

Starting at a(2), a(n) gives all the dimensions of Euclidean k-space in which the ratio of outer to inner Soddy hyperspheres' radii for k+1 identical kissing hyperspheres is rational. The formula for this ratio is (1+3k+2*sqrt(2k*(k+1)))/(k-1) where k is the dimension. So for a(3) = 49, the ratio is 6 in the 49th dimension. See comment for A010502. - Frank M Jackson, Feb 09 2013

Conjecture: For n>1 a(n) is the index of the first occurrence of -n in sequence A123737. - Vaclav Kotesovec, Jun 02 2015

For n=2*k, k>0, a(n) is divisible by 8 (deficient), so since all proper divisors of deficient numbers are deficient, then a(n) is deficient. For n=2*k+1, k>0, a(n) is odd. If a(n) is a prime number, it is deficient; otherwise a(n) has one or two distinct prime factors and is therefore deficient again. sigma(a(5)) = 1723 < 3362 = 2*a(5). In either case, a(n) is deficient. - Muniru A Asiru, Apr 14 2016

The squares of NSW numbers (A008843) interleaved with twice squares from A084703, where A008843(n) = A002315(n)^2 and A084703(n) = A001542(n)^2. Conjecture: Also numbers n such that sigma(n) = A000203(n) and sigma(n-th triangular number) = A074285(n) are both odd numbers. - Jaroslav Krizek, Aug 05 2016

For n > 0, numbers for which the number of odd divisors of both n and of n + 1 is odd. - Gionata Neri, Apr 30 2018

a(n) will be solutions to some (A000217(k) + A000217(k+1))/2. - Art Baker, Jul 16 2019

For n >= 2, a(n) is the base for which A058331(A001109(n)) is a length-3 repunit. Example: for n=2, A001109(2)=6 and A058331(6)=73 and 73 in base a(2)=8 is 111. See Grantham and Graves. - Michel Marcus, Sep 11 2020

Row sums of triangle A054458.

a(n) = term (1,1) in M^n, M = the 3 X 3 matrix [1,1,1; 1,1,1; 2,2,1]. - Gary W. Adamson, Mar 12 2009

Equals the INVERT transform of A001333: (1, 3, 7, 17, 41, 99, ...). - Gary W. Adamson, Aug 14 2010

a(n) is the number of one sided n-step walks taking steps from {(0,1), (-1,0), (1,0), (1,1)}. - Shanzhen Gao, May 13 2011

Number of quaternary words of length n on {0,1,2,3} containing no subwords 03 or 30. - Philippe Deléham, Apr 27 2012

Pisano period lengths: 1, 1, 4, 1, 24, 4, 48, 1, 12, 24, 30, 4, 12, 48, 24, 2, 272, 12, 18, 24, ... - R. J. Mathar, Aug 10 2012

a(n) = A007481(2*n+1) - A007481(2*n) = A007481(2*(n+1)) - A007481(2*n+1). - Reinhard Zumkeller, Oct 25 2015

Number of length-n words on a,b,c,d avoiding aa and ab. For n >= 1, the number of such words ending with a or the number of those ending with b is A007482(n-1), and the number of those ending with c or the number of those ending with d is a(n-1). - Jianing Song, Jun 01 2022

Normally sequences like this are not included, since with the alternating 0's deleted it is already in the database.

Inverse binomial transform of A001333. - Paul Barry, Feb 25 2003

"Sloping binary representation" of powers of 2 (A000079), slope=-1 (see A037095 and A102370). - Philippe Deléham, Jan 04 2008

0,1,0,2,0,4,0,8,0,16,... is the inverse binomial transform of A000129 (Pell numbers). - Philippe Deléham, Oct 28 2008

Number of maximal self-avoiding walks from the NW to SW corners of a 3 X n grid.

Row sums of the triangle in A204293. - Reinhard Zumkeller, Jan 14 2012

Pisano period lengths: 1, 1, 4, 1, 8, 4, 6, 1, 12, 8, 20, 4, 24, 6, 8, 1, 16, 12, 36, 8, ... . - R. J. Mathar, Aug 10 2012

This sequence occurs in the length L(n) = sqrt(2)^n of Lévy's C-curve at the n-th iteration step. Therefore, L(n) is the Q(sqrt(2)) integer a(n) + a(n-1)*sqrt(2), with a(-1) = 0. For a variant of this C-curve see A251732 and A251733. - Wolfdieter Lang, Dec 08 2014

a(n) counts walks (closed) on the graph G(1-vertex,2-loop,2-loop). Equivalently the middle entry (2,2) of A^n where the adjacency matrix of digraph is A=(0,1,0;1,0,1;0,1,0). - David Neil McGrath, Dec 19 2014

a(n-2) is the number of compositions of n into even parts. For example, there are 4 compositions of 6 into even parts: (6), (222), (42), and (24). - David Neil McGrath, Dec 19 2014

Also the number of alternately constant compositions of n + 2, ranked by A351010. The alternately strict version gives A000213. The unordered version is A035363, ranked by A000290, strict A035457. - Gus Wiseman, Feb 19 2022

a(n) counts degree n fixed points of GF(2)[x]'s automorphisms. Proof: given a field k, k[x]'s automorphisms are determined by k's automorphisms and invertible affine maps x -> ax + b. GF(2) is rigid and has only one unit so its only nontrivial automorphism is x -> x + 1. For n = 0 we have 1 fixed point, the constant polynomial 1. (Taking the convention that 0 is not a degree 0 polynomial.) For n = 1 we have 0 fixed points as x -> x + 1 -> x are the only degree 1 polynomials. Note that if f(x) is a fixed point, then f(x) + 1 is also a fixed point. Given f(x) a degree n fixed point, we can assume WLOG x | f(x). Applying the automorphism, we then have x + 1 | f(x). Now note that f(x) / (x^2 + x) must be a fixed point, so any fixed point of degree n must either be of the form g(x) * (x^2 + x) or g(x) * (x^2 + x) + 1 for a unique degree n - 2 fixed point g(x). Therefore we have the recurrence relation a(n) = 2 * a(n - 2) as desired. - Keith J. Bauer, Mar 19 2024

Form the digraph with matrix A = [0,1,1,1; 1,0,1,1; 1,1,0,1; 1,0,1,1]. Then the sequence 0,1,1,5,... or (3^(n-1)-(-1)^n)/2+0^n/3 with g.f. x(1-x)/(1-2x-3x^2) corresponds to the (1,2) term of A^n. - Paul Barry, Oct 02 2004

3*a(n+1) + a(n) = 4*A060925(n); a(n+1) = A015518(n) + A060925(n); a(n+1) - 6*A015518(n) = (-1)^n. - Creighton Dement, Nov 15 2004

The sequence corresponds to the (1,1) term of the matrix [1,2;2,1]^n. - Simone Severini, Dec 04 2004

The same sequence may be obtained by the following process. Starting a priori with the fraction 1/1, the numerators of fractions built according to the rule: add top and bottom to get the new bottom, add top and 4 times the bottom to get the new top. The limit of the sequence of fractions is 2. - Cino Hilliard, Sep 25 2005

a(n)^2 + (2*A015518(n))^2 = a(2n). E.g., a(3) = 13, 2*A015518(3) = 14, A046717(6) = 365. 13^2 + 14^2 = 365. - Gary W. Adamson, Jun 17 2006

Equals INVERTi transform of A104934: (1, 2, 8, 28, 100, 356, 1268, ...). - Gary W. Adamson, Jul 21 2010

a(n) is the number of compositions of n when there are 1 type of 1 and 4 types of other natural numbers. - Milan Janjic, Aug 13 2010

An elephant sequence, see A175655. For the central square just one A[5] vector, with decimal value 341, leads to this sequence (without the first leading 1). For the corner squares this vector leads to the companion sequence A015518 (without the leading 0). - Johannes W. Meijer, Aug 15 2010

Pisano period lengths: 1, 1, 2, 1, 4, 2, 6, 4, 2, 4, 10, 2, 6, 6, 4, 8, 16, 2, 18, 4, ... - R. J. Mathar, Aug 10 2012

a(n) is the number of words of length n over a ternary alphabet whose position in the lexicographic order is a multiple of two. - Alois P. Heinz, Apr 13 2022

a(n) is the sum, for k=0..3, of the number of walks of length n between two vertices at distance k of the cube graph. - Miquel A. Fiol, Mar 09 2024