cp's OEIS Frontend

Hankel transform is (-1)^n.

Row sums of the Riordan array ((1-2*x)/(1+x+x^2), x/(1+x+x^2))^(-1). - Paul Barry, Nov 06 2008

Binary representation of the Kempner-Mahler number Sum_{k>=0} 1/2^(2^k) = A007404.

Also a(n) = A(n) mod 2 where A is any of A001700, A005573, A007854, A026641, A049027, A064063, A064088, A064090, A064092, A064325, A064327, A064329, A064331, A064613, A076026, A105523, A123273, A126694, A126930, A126931, A126982, A126983, A126987, A127016, A127053, A127358, A127360, A127361, A127363. - Philippe Deléham, May 26 2007

a(n) = (product of digits of n; n in binary notation) mod 2. This sequence is a transformation of the Thue-Morse sequence (A010060), since there exists a function f such that f(sum of digits of n) = (product of digits of n). - Ctibor O. Zizka, Feb 12 2008

a(n-1), n >= 1, the characteristic sequence for powers of 2, A000079, is the unique solution of the following formal product and formal power series identity: Product_{j>=1} (1 + a(j-1)*x^j) = 1 + Sum_{k>=1} x^k = 1/(1-x). The product is therefore Product_{l>=1} (1 + x^(2^l)). Proof. Compare coefficients of x^n and use the binary representation of n. Uniqueness follows from the recurrence relation given for the general case under A147542. - Wolfdieter Lang, Mar 05 2009

a(n) is also the number of orbits of length n for the map x -> 1-cx^2 on [-1,1] at the Feigenbaum critical value c=1.401155... . - Thomas Ward, Apr 08 2009

A054525 (Mobius transform) * A001511 = A036987 = A047999^(-1) * A001511 = the inverse of Sierpiński's gasket * the ruler sequence. - Gary W. Adamson, Oct 26 2009 [Of course this is only vaguely correct depending on how the fuzzy indexing in these formulas is made concrete. - R. J. Mathar, Jun 20 2014]

Characteristic function of A000225. - Reinhard Zumkeller, Mar 06 2012

Also parity of the Catalan numbers A000108. - Omar E. Pol, Jan 17 2012

For n >= 2, also the largest exponent k >= 0 such that n^k in binary notation does not contain both 0 and 1. Unlike for the decimal version of this sequence, A062518, where the terms are only conjectural, for this sequence the values of a(n) can be proved to be the characteristic function of A000225, as follows: n^k will contain both 0 and 1 unless n^k = 2^r-1 for some r. But this is a special case of Catalan's equation x^p = y^q-1, which was proved by Preda Mihăilescu to have no nontrivial solution except 2^3 = 3^2 - 1. - Christopher J. Smyth, Aug 22 2014

Image, under the coding a,b -> 1; c -> 0, of the fixed point, starting with a, of the morphism a -> ab, b -> cb, c -> cc. - Jeffrey Shallit, May 14 2016

Number of nonisomorphic Boolean algebras of order n+1. - Jianing Song, Jan 23 2020

Equivalently, a triangle read by rows, where the rows are obtained by sorting the elements of the rows of Pascal's triangle (A007318) into descending order. - Philippe Deléham, May 21 2005

Equivalently, as a triangle read by rows, this is T(n,k)=binomial(n,floor((n-k)/2)); column k then has e.g.f. Bessel_I(k,2x)+Bessel_I(k+1,2x). - Paul Barry, Feb 28 2006

Antidiagonal sums are A037952(n+1) = C(n+1,[n/2]). Matrix inverse is the row reversal of triangle A066170. Eigensequence is A125094(n) = Sum_{k=0..n-1} A125093(n-1,k)*A125094(k). - Paul D. Hanna, Nov 20 2006

Riordan array (1/(1-x-x^2*c(x^2)),x*c(x^2)); where c(x)=g.f.for Catalan numbers A000108. - Philippe Deléham, Mar 17 2007

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)=T(n-1,0)+T(n-1,1), T(n,k)=T(n-1,k-1)+T(n-1,k+1) for k>=1. - Philippe Deléham, Mar 27 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

T(n,k) is the number of paths from (0,k) to some (n,m) which never dip below y=0, touch y=0 at least once and are made up only of the steps (1,1) and (1,-1). This can be proved using the recurrence supplied by Deléham. - Gerald McGarvey, Oct 15 2008

Triangle read by rows = partial sums of A053121 terms starting from the right. - Gary W. Adamson, Oct 24 2008

As a subset of the "family of triangles" (Deleham comment of Sep 25 2007), beginning with a signed variant of A061554, M = (-1,0) = (1; -1, 1; 2, -1, 1; -3, 3, -1, 1; ...) successive binomial transforms of M yield (0,1) - A089942; (1,2) - A039599; (2,3) - A124733; (3,4) - A124574; (4,5) - A126331; ... such that the binomial transform of the triangle generated from (n,n+1) = the triangle generated from (n+1,n+2). Similarly, another subset beginning with A053121 - (0,0), and taking successive binomial transforms yields (1,1) - A064189; (2,2) - A039598; (3,3) - A091965, ... By rows, the triangle generated from (n,n) can be obtained by taking pairwise sums from the (n-1,n) triangle starting from the right. For example, row 2 of (1,2) - A039599 = (2, 3, 1); and taking pairwise sums from the right we obtain (5, 4, 1) = row 2 of (2,2) - A039598. - Gary W. Adamson, Aug 04 2011

The triangle by rows (n) with alternating signs (+-+...) from the top as a set of simultaneous equations solves for diagonal lengths of odd N (N = 2n+1) regular polygons. The constants in each case are powers of c = 2*cos(2*Pi/N). By way of example, the first 3 rows relate to the heptagon and the simultaneous equations are (1,0,0) = 1; (-1,1,0) = c = 1.24697...; and (2,-1,1) = c^2. The answers are 1, 2.24697..., and 1.801...; the 3 distinct diagonal lengths of the heptagon with edge = 1. - Gary W. Adamson, Sep 07 2011

Riordan array (c(x^2)/(1-2xc(x^2)),xc(x^2)) where c(x)=g.f. of Catalan numbers A000108. - Philippe Deléham, Mar 18 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

a(n) = # Dyck (n+1)-paths all of whose components are symmetric. A strict Dyck path is one with exactly one return to ground level (necessarily at the end). Every nonempty Dyck path is expressible uniquely as a concatenation of one or more strict Dyck paths, called its components. - David Callan, Mar 02 2005

a(n) = # 2-Motzkin paths (i.e., Motzkin paths with blue and red level steps) with no level steps at positive height. Example: a(2)=5 because, denoting U=(1,1), D=(1,-1), B=blue (1,0), R=red (1,0), we have BB, BR, RB, RR, and UD. - Emeric Deutsch, Jun 07 2011

Inverse Chebyshev transform of the second kind applied to 2^n. This maps g(x) -> c(x^2)g(xc(x^2)). - Paul Barry, Sep 14 2005

Hankel transform of this sequence gives A000012 = [1,1,1,1,1,1,1,...]. - Philippe Deléham, Oct 24 2007

Inverse binomial transform of A059738. - Philippe Deléham, Nov 24 2009

By rows, equals partial sums of A053121 reversed rows. Example: Row 4 of A053121 = (2, 0, 3, 0, 1) -> (1, 0, 3, 0, 2) -> (1, 1, 4, 4, 6). - Gary W. Adamson, Dec 28 2008, edited by Michel Marcus, Sep 22 2015

The triangular view of A061554 is:

1;

1, 1;

2, 1, 1;

3, 3, 1, 1;

6, 4, 4, 1, 1;

The matrix square of this (infinite) triangle is this triangle here.