cp's OEIS Frontend

For n > 0, a(n) is the degree (n-1) "numbral" power of 5 (see A048888 for the definition of numbral arithmetic). Example: a(3) = 21, since the numbral square of 5 is 5(*)5 = 101(*)101(base 2) = 101 OR 10100 = 10101(base 2) = 21, where the OR is taken bitwise. - John W. Layman, Dec 18 2001

a(n) is composite for all n > 2 and has factors x, (3*x + 2*(-1)^n) where x belongs to A001045. In binary the terms greater than 0 are 1, 101, 10101, 1010101, etc. - John McNamara, Jan 16 2002

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

The Collatz-function iteration started at a(n), for n >= 1, will end at 1 after 2*n+1 steps. - Labos Elemer, Sep 30 2002 [corrected by Wolfdieter Lang, Aug 16 2021]

Second binomial transform of A001045. - Paul Barry, Mar 28 2003

All members of sequence are also generalized octagonal numbers (A001082). - Matthew Vandermast, Apr 10 2003

Also sum of squares of divisors of 2^(n-1): a(n) = A001157(A000079(n-1)), for n > 0. - Paul Barry, Apr 11 2003

Binomial transform of A000244 (with leading zero). - Paul Barry, Apr 11 2003

Number of walks of length 2n between two vertices at distance 2 in the cycle graph C_6. For n = 2 we have for example 5 walks of length 4 from vertex A to C: ABABC, ABCBC, ABCDC, AFABC and AFEDC. - Herbert Kociemba, May 31 2004

Also number of walks of length 2n + 1 between two vertices at distance 3 in the cycle graph C_12. - Herbert Kociemba, Jul 05 2004

a(n+1) is the number of steps that are made when generating all n-step random walks that begin in a given point P on a two-dimensional square lattice. To make one step means to mark one vertex on the lattice (compare A080674). - Pawel P. Mazur (Pawel.Mazur(AT)pwr.wroc.pl), Mar 13 2005

a(n+1) is the sum of square divisors of 4^n. - Paul Barry, Oct 13 2005

a(n+1) is the decimal number generated by the binary bits in the n-th generation of the Rule 250 elementary cellular automaton. - Eric W. Weisstein, Apr 08 2006

a(k) = [M^k]2,1, where M is the 3 X 3 matrix defined as follows: M = [1, 1, 1; 1, 3, 1; 1, 1, 1]. - _Simone Severini, Jun 11 2006

a(n-1) / a(n) = percentage of wasted storage if a single image is stored as a pyramid with a each subsequent higher resolution layer containing four times as many pixels as the previous layer. n is the number of layers. - Victor Brodsky (victorbrodsky(AT)gmail.com), Jun 15 2006

k is in the sequence if and only if C(4k + 1, k) (A052203) is odd. - Paul Barry, Mar 26 2007

This sequence also gives the number of distinct 3-colorings of the odd cycle C(2*n - 1). - Keith Briggs, Jun 19 2007

All numbers of the form m*4^m + (4^m-1)/3 have the property that they are sums of two squares and also their indices are the sum of two squares. This follows from the identity m*4^m + (4^m-1)/3 = 4(4(..4(4m + 1) + 1) + 1) + 1 ..) + 1. - Artur Jasinski, Nov 12 2007

For n > 0, terms are the numbers that, in base 4, are repunits: 1_4, 11_4, 111_4, 1111_4, etc. - Artur Jasinski, Sep 30 2008

Let A be the Hessenberg matrix of order n, defined by: A[1, j] = 1, A[i, i] := 5, (i > 1), A[i, i - 1] = -1, and A[i, j] = 0 otherwise. Then, for n >= 1, a(n) = charpoly(A,1). - Milan Janjic, Jan 27 2010

This is the sequence A(0, 1; 3, 4; 2) = A(0, 1; 4, 0; 1) of the family of sequences [a, b : c, d : k] considered by G. Detlefs, and treated as A(a, b; c, d; k) in the W. Lang link given below. - Wolfdieter Lang, Oct 18 2010

6*a(n) + 1 is every second Mersenne number greater than or equal to M3, hence all Mersenne primes greater than M2 must be a 6*a(n) + 1 of this sequence. - Roderick MacPhee, Nov 01 2010

Smallest number having alternating bit sum n. Cf. A065359.

For n = 1, 2, ..., the last digit of a(n) is 1, 5, 1, 5, ... . - Washington Bomfim, Jan 21 2011

Rule 50 elementary cellular automaton generates this sequence. This sequence also appears in the second column of array in A173588. - Paul Muljadi, Jan 27 2011

Sequence found by reading the line from 0, in the direction 0, 5, ... and the line from 1, in the direction 1, 21, ..., in the square spiral whose edges are the Jacobsthal numbers A001045 and whose vertices are the numbers A000975. These parallel lines are two semi-diagonals in the spiral. - Omar E. Pol, Sep 10 2011

a(n), n >= 1, is also the inverse of 3, denoted by 3^(-1), Modd(2^(2*n - 1)). For Modd n see a comment on A203571. E.g., a(2) = 5, 3 * 5 = 15 == 1 (Modd 8), because floor(15/8) = 1 is odd and -15 == 1 (mod 8). For n = 1 note that 3 * 1 = 3 == 1 (Modd 2) because floor(3/2) = 1 and -3 == 1 (mod 2). The inverse of 3 taken Modd 2^(2*n) coincides with 3^(-1) (mod 2^(2*n)) given in A007583(n), n >= 1. - Wolfdieter Lang, Mar 12 2012

If an AVL tree has a leaf at depth n, then the tree can contain no more than a(n+1) nodes total. - Mike Rosulek, Nov 20 2012

Also, this is the Lucas sequence V(5, 4). - Bruno Berselli, Jan 10 2013

Also, for n > 0, a(n) is an odd number whose Collatz trajectory contains no odd number other than n and 1. - Jayanta Basu, Mar 24 2013

Sum_{n >= 1} 1/a(n) converges to (3*(log(4/3) - QPolyGamma[0, 1, 1/4]))/log(4) = 1.263293058100271... = A321873. - K. G. Stier, Jun 23 2014

Consider n spheres in R^n: the i-th one (i=1, ..., n) has radius r(i) = 2^(1-i) and the coordinates of its center are (0, 0, ..., 0, r(i), 0, ..., 0) where r(i) is in position i. The coordinates of the intersection point in the positive orthant of these spheres are (2/a(n), 4/a(n), 8/a(n), 16/a(n), ...). For example in R^2, circles centered at (1, 0) and (0, 1/2), and with radii 1 and 1/2, meet at (2/5, 4/5). - Jean M. Morales, May 19 2015

From Peter Bala, Oct 11 2015: (Start)

a(n) gives the values of m such that binomial(4*m + 1,m) is odd. Cf. A003714, A048716, A263132.

2*a(n) = A020988(n) gives the values of m such that binomial(4*m + 2, m) is odd.

4*a(n) = A080674(n) gives the values of m such that binomial(4*m + 4, m) is odd. (End)

Collatz Conjecture Corollary: Except for powers of 2, the Collatz iteration of any positive integer must eventually reach a(n) and hence terminate at 1. - Gregory L. Simay, May 09 2016

Number of active (ON, black) cells at stage 2^n - 1 of the two-dimensional cellular automaton defined by "Rule 598", based on the 5-celled von Neumann neighborhood. - Robert Price, May 16 2016

From Luca Mariot and Enrico Formenti, Sep 26 2016: (Start)

a(n) is also the number of coprime pairs of polynomials (f, g) over GF(2) where both f and g have degree n + 1 and nonzero constant term.

a(n) is also the number of pairs of one-dimensional binary cellular automata with linear and bipermutive local rule of neighborhood size n+1 giving rise to orthogonal Latin squares of order 2^m, where m is a multiple of n. (End)

Except for 0, 1 and 5, all terms are Brazilian repunits numbers in base 4, and so belong to A125134. For n >= 3, all these terms are composite because a(n) = {(2^n-1) * (2^n + 1)}/3 and either (2^n - 1) or (2^n + 1) is a multiple of 3. - Bernard Schott, Apr 29 2017

Given the 3 X 3 matrix A = [2, 1, 1; 1, 2, 1; 1, 1, 2] and the 3 X 3 unit matrix I_3, A^n = a(n)(A - I_3) + I_3. - Nicolas Patrois, Jul 05 2017

The binary expansion of a(n) (n >= 1) consists of n 1's alternating with n - 1 0's. Example: a(4) = 85 = 1010101_2. - Emeric Deutsch, Aug 30 2017

a(n) (n >= 1) is the viabin number of the integer partition [n, n - 1, n - 2, ..., 2, 1] (for the definition of viabin number see comment in A290253). Example: a(4) = 85 = 1010101_2; consequently, the southeast border of the Ferrers board of the corresponding integer partition is ENENENEN, where E = (1, 0), N = (0, 1); this leads to the integer partition [4, 3, 2, 1]. - Emeric Deutsch, Aug 30 2017

Numbers whose binary and Gray-code representations are both palindromes (i.e., intersection of A006995 and A281379). - Amiram Eldar, May 17 2021

Starting with n = 1 the sequence satisfies {a(n) mod 6} = repeat{1, 5, 3}. - Wolfdieter Lang, Jan 14 2022

Terms >= 5 are those q for which the multiplicative order of 2 mod q is floor(log_2(q)) + 2 (and which is 1 more than the smallest possible order for any q). - Tim Seuré, Mar 09 2024

The order of 2 modulo a(n) is 2*n for n >= 2. - Joerg Arndt, Mar 09 2024

Riordan array (c(x)/sqrt(1-4*x),x*c(x)^2) where c(x) is g.f. of A000108. Unsigned version of A113187. Diagonal sums are A014301(n+1).

Triangle T(n,k),0<=k<=n, read by rows defined by :T(0,0)=1, T(n,k)=0 if k<0 or if k>n, T(n,0)=3*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 22 2007

Reversal of A122366. - Philippe Deléham, Mar 22 2007

Column k has e.g.f. exp(2x)(Bessel_I(k,2x)+Bessel_I(k+1,2x)). - Paul Barry, Jun 06 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

Diagonal sums are A014301(n+1). - Paul Barry, Mar 08 2011

This triangle T(n,k) appears in the expansion of odd powers of Fibonacci numbers F=A000045 in terms of F-numbers with multiples of odd numbers as indices. See the Ozeki reference, p. 108, Lemma 2. The formula is: F_l^(2*n+1) = sum(T(n,k)*(-1)^((n-k)*(l+1))* F_{(2*k+1)*l}, k=0..n)/5^n, n >= 0, l >= 0. - Wolfdieter Lang, Aug 24 2012

Central terms give A052203. - Reinhard Zumkeller, Mar 14 2014

This triangle appears in the expansion of (4*x)^n in terms of the polynomials Todd(n, x):= T(2*n+1, sqrt(x))/sqrt(x) = sum(A084930(n,m)*x^m), n >= 0. This follows from the inversion of the lower triangular Riordan matrix built from A084930 and comparing the g.f. of the row polynomials. - Wolfdieter Lang, Aug 05 2014

From Wolfdieter Lang, Aug 15 2014: (Start)

This triangle is the inverse of the signed Riordan triangle (-1)^(n-m)*A111125(n,m).

This triangle T(n,k) appears in the expansion of x^n in terms of the polynomials todd(k, x):= T(2*k+1, sqrt(x)/2)/(sqrt(x)/2) = S(k, x-2) - S(k-1, x-2) with the row polynomials T and S for the triangles A053120 and A049310, respectively: x^n = sum(T(n,k)*todd(k, x), k=0..n). Compare this with the preceding comment.

The A- and Z-sequences for this Riordan triangle are [1, 2, 1, repeated 0] and [3, 1, repeated 0]. For A- and Z-sequences for Riordan triangles see the W. Lang link under A006232. This corresponds to the recurrences given in the Philippe Deléham, Mar 22 2007 comment above. (End)

From Gus Wiseman, Sep 28 2022: (Start)

Also the number of integer compositions of 4n with alternating sum 2n, where the alternating sum of a sequence (y_1,...,y_k) is Sum_i (-1)^(i-1) y_i. These compositions are ranked by A348614. The a(12) = 21 compositions are:

(6,2) (1,2,5) (1,1,5,1) (1,1,1,1,4)

(2,2,4) (2,1,4,1) (1,1,2,1,3)

(3,2,3) (3,1,3,1) (1,1,3,1,2)

(4,2,2) (4,1,2,1) (1,1,4,1,1)

(5,2,1) (5,1,1,1) (2,1,1,1,3)

(2,1,2,1,2)

(2,1,3,1,1)

(3,1,1,1,2)

(3,1,2,1,1)

(4,1,1,1,1)

The following pertain to this interpretation:

- The case of partitions is A000712, reverse A006330.

- Allowing any alternating sum gives A013777 (compositions of 4n).

- A011782 counts compositions of n.

- A034871 counts compositions of 2n with alternating sum 2k.

- A097805 counts compositions by alternating (or reverse-alternating) sum.

- A103919 counts partitions by sum and alternating sum (reverse: A344612).

- A345197 counts compositions by length and alternating sum.

(End)

In general, binomial(k*n,n)/k = binomial(k*n-1,n-1).

Sequences in the OEIS related to this identity are:

. C(2n,n) = A000984, C(2n,n)/2 = A001700;

. C(3n,n) = A005809, C(3n,n)/3 = A025174;

. C(4n,n) = A005810, C(4n,n)/4 = a(n);

. C(5n,n) = A001449, C(5n,n)/5 = A163456;

. C(6n,n) = A004355, C(6n,n)/6 is not in the OEIS.

Conjecture: a(n) == 1 (mod n^3) iff n is an odd prime.

It is known that a(p) == 1(mod p^3) for prime p >= 3. See Mestrovic, Section 3. - Peter Bala, Oct 09 2015

Sum of n-th row = A000302(n) = 4^n.

Central terms give A052203.

Reversal of A111418. - Philippe Deléham, Mar 22 2007

Coefficient triangle for the expansion of one half of odd powers of 2*x in terms of Chebyshev's T-polynomials: ((2*x)^(2*n+1))/2 = Sum_{k=0..n} a(n,k)*T(2*(n-k)+1,x) with Chebyshev's T-polynomials. See A053120. - Wolfdieter Lang, Mar 07 2007

The signed triangle T(n,k)*(-1)^(n-k) appears in the formula ((2*sin(phi))^(2*n+1))/2 = Sum_{k=0..n} ((-1)^(n-k))*a(n,k)*sin((2*(n-k)+1)*phi). - Wolfdieter Lang, Mar 07 2007

The signed triangle T(n,k)*(-1)^(n-k) appears therefore in the formula (4-x^2)^n = Sum_{k=0..n} ((-1)^(n-k))*a(n,k)*S(2*(n-k),x) with Chebyshev's S-polynomials. See A049310 for S(n,x). - Wolfdieter Lang, Mar 07 2007

From Wolfdieter Lang, Sep 18 2012: (Start)

The triangle T(n,k) appears also in the formula F(2*l+1)^(2*n+1) = (1/5^n)*Sum_{k=0..n} T(n,k)*F((2*(n-k)+1)*(2*l+1)), l >= 0, n >= 0, with F=A000045 (Fibonacci).

The signed triangle Ts(n,k):=T(n,k)*(-1)^k appears also in the formula

F(2*l)^(2*n+1) = (1/5^n)*Sum_{k=0..n} Ts(n,k)*F((2(n-k)+1)*2*l), l >= 0, n >= 0, with F=A000045 (Fibonacci).

This is Lemma 2 of the K. Ozeki reference, p. 108, written for odd and even indices separately.

(End)

a(n) is the number of paths from (0,0) to (5n,n) taking north and east steps while avoiding exactly 2 consecutive north steps. - Shanzhen Gao, Apr 15 2010

a(n) is the number of paths from (0,0) to (6n,n) taking north and east steps and avoiding north^{=2}. - Shanzhen Gao, Apr 15 2010

Row sums are A058622(n+1). Diagonal sums are A001791(n+1), with interpolated zeros. Inverse is A117179.

De-aerated and rows reversed, this matrix apparently becomes A014462. The nonzero antidiagonals are embedded in several entries and apparently contain partial sums of previous nonzero antidiagonals. - Tom Copeland, May 30 2017