cp's OEIS Frontend

Same as Pisot sequences E(1, 4), L(1, 4), P(1, 4), T(1, 4). Essentially same as Pisot sequences E(4, 16), L(4, 16), P(4, 16), T(4, 16). See A008776 for definitions of Pisot sequences.

The convolution square root of this sequence is A000984, the central binomial coefficients: C(2n,n). - T. D. Noe, Jun 11 2002

With P(n) being the number of integer partitions of n, p(i) as the number of parts of the i-th partition of n, d(i) as the number of different parts of the i-th partition of n, m(i, j) the multiplicity of the j-th part of the i-th partition of n, one has a(n) = Sum_{i = 1..P(n)} p(i)!/(Product_{j = 1..d(i)} m(i, j)!) * 2^(n-1). - Thomas Wieder, May 18 2005

Sums of rows of the triangle in A122366. - Reinhard Zumkeller, Aug 30 2006

Hankel transform of A076035. - Philippe Deléham, Feb 28 2009

Equals the Catalan sequence: (1, 1, 2, 5, 14, ...), convolved with A032443: (1, 3, 11, 42, ...). - Gary W. Adamson, May 15 2009

Sum of coefficients of expansion of (1 + x + x^2 + x^3)^n.

a(n) is number of compositions of natural numbers into n parts less than 4. For example, a(2) = 16 since there are 16 compositions of natural numbers into 2 parts less than 4.

The compositions of n in which each natural number is colored by one of p different colors are called p-colored compositions of n. For n >= 1, a(n) equals the number of 4-colored compositions of n such that no adjacent parts have the same color. - Milan Janjic, Nov 17 2011

Squares in A002984. - Reinhard Zumkeller, Dec 28 2011

Row sums of Pascal's triangle using the rule that going left increases the value by a factor of k = 3. For example, the first three rows are {1}, {3, 1}, and {9, 6, 1}. Using this rule gives row sums as (k+1)^n. - Jon Perry, Oct 11 2012

First differences of A002450. - Omar E. Pol, Feb 20 2013

Sum of all peak heights in Dyck paths of semilength n+1. - David Scambler, Apr 22 2013

Powers of 4 exceed powers of 2 by A020522 which is the m-th oblong number A002378(m), m being the n-th Mersenne number A000225(n); hence, we may write, a(n) = A000079(n) + A002378(A000225(n)). - Lekraj Beedassy, Jan 17 2014

a(n) is equal to 1 plus the sum for 0 < k < 2^n of the numerators and denominators of the reduced fractions k/2^n. - J. M. Bergot, Jul 13 2015

Binomial transform of A000244. - Tony Foster III, Oct 01 2016

From Ilya Gutkovskiy, Oct 01 2016: (Start)

Number of nodes at level n regular 4-ary tree.

Partial sums of A002001. (End)

Satisfies Benford's law [Berger-Hill, 2011]. - N. J. A. Sloane, Feb 08 2017

Also the number of connected dominating sets in the (n+1)-barbell graph. - Eric W. Weisstein, Jun 29 2017

Side length of the cells at level n in a pyramid scheme where a square grid is decomposed into overlapping 2 X 2 blocks (cf. Kropatsch, 1985). - Felix Fröhlich, Jul 04 2019

a(n-1) is the number of 3-compositions of n; see Hopkins & Ouvry reference. - Brian Hopkins, Aug 15 2020

a(n) is also the number of different lines determined by pair of vertices in an n-dimensional hypercube. The number of these lines modulo being parallel is in A003462. - Ola Veshta (olaveshta(AT)my-deja.com), Feb 15 2001

Let G_n be the elementary Abelian group G_n = (C_2)^n for n >= 1: A006516 is the number of times the number -1 appears in the character table of G_n and A007582 is the number of times the number 1. Together the two sequences cover all the values in the table, i.e., A006516(n) + A007582(n) = 2^(2n). - Ahmed Fares (ahmedfares(AT)my-deja.com), Jun 01 2001

a(n) is the number of n-letter words formed using four distinct letters, one of which appears an odd number of times. - Lekraj Beedassy, Jul 22 2003 [See, e.g., the Balakrishnan reference, problems 2.67 and 2.68, p. 69. - Wolfdieter Lang, Jul 16 2017]

Number of 0's making up the central triangle in a Pascal's triangle mod 2 gasket. - Lekraj Beedassy, May 14 2004

m-th triangular number, where m is the n-th Mersenne number, i.e., a(n)=A000217(A000225(n)). - Lekraj Beedassy, May 25 2004

Number of walks of length 2n+1 between two nodes at distance 3 in the cycle graph C_8. - Herbert Kociemba, Jul 02 2004

The sequence of fractions a(n+1)/(n+1) is the 3rd binomial transform of (1, 0, 1/3, 0, 1/5, 0, 1/7, ...). - Paul Barry, Aug 05 2005

Number of monic irreducible polynomials of degree 2 in GF(2^n)[x]. - Max Alekseyev, Jan 23 2006

(A007582(n))^2 + a(n)^2 = A007582(2n). E.g., A007582(3) = 36, a(3) = 28; A007582(6) = 2080. 36^2 + 28^2 = 2080. - Gary W. Adamson, Jun 17 2006

The sequence 6*a(n), n>=1, gives the number of edges of the Hanoi graph H_4^{n} with 4 pegs and n>=1 discs. - Daniele Parisse, Jul 28 2006

8*a(n) is the total border length of the 4*n masks used when making an order n regular DNA chip, using the bidimensional Gray code suggested by Pevzner in the book "Computational Molecular Biology." - Bruno Petazzoni (bruno(AT)enix.org), Apr 05 2007

If we start with 1 in binary and at each step we prepend 1 and append 0, we construct this sequence: 1 110 11100 1111000 etc.; see A109241(n-1). - Artur Jasinski, Nov 26 2007

Let P(A) be the power set of an n-element set A. Then a(n) = the number of pairs of elements {x,y} of P(A) for which x does not equal y. - Ross La Haye, Jan 02 2008

Wieder calls these "conjoint usual 2-combinations." The set of "conjoint strict k-combinations" is the subset of conjoint usual k-combinations where the empty set and the set itself are excluded from possible selection. These numbers C(2^n - 2,k), which for k = 2 (i.e., {x,y} of the power set of a set) give {1, 0, 1, 15, 91, 435, 1891, 7875, 32131, 129795, 521731, ...}. - Ross La Haye, Jan 15 2008

If n is a member of A000043 then a(n) is also a perfect number (A000396). - Omar E. Pol, Aug 30 2008

a(n) is also the number whose binary representation is A109241(n-1), for n>0. - Omar E. Pol, Aug 31 2008

From Daniel Forgues, Nov 10 2009: (Start)

If we define a spoof-perfect number as:

A spoof-perfect number is a number that would be perfect if some (one or more) of its odd composite factors were wrongly assumed to be prime, i.e., taken as a spoof prime.

And if we define a "strong" spoof-perfect number as:

A "strong" spoof-perfect number is a spoof-perfect number where sigma(n) does not reveal the compositeness of the odd composite factors of n which are wrongly assumed to be prime, i.e., taken as a spoof prime.

The odd composite factors of n which are wrongly assumed to be prime then have to be obtained additively in sigma(n) and not multiplicatively.

Then:

If 2^n-1 is odd composite but taken as a spoof prime then 2^(n-1)*(2^n - 1) is an even spoof perfect number (and moreover "strong" spoof-perfect).

For example:

a(8) = 2^(8-1)*(2^8 - 1) = 128*255 = 32640 (where 255 (with factors 3*5*17) is taken as a spoof prime);

sigma(a(8)) = (2^8 - 1)*(255 + 1) = 255*256 = 2*(128*255) = 2*32640 = 2n is spoof-perfect (and also "strong" spoof-perfect since 255 is obtained additively);

a(11) = 2^(11-1)*(2^11 - 1) = 1024*2047 = 2096128 (where 2047 (with factors 23*89) is taken as a spoof prime);

sigma(a(11)) = (2^11 - 1)*(2047 + 1) = 2047*2048 = 2*(1024*2047) = 2*2096128 = 2n is spoof-perfect (and also "strong" spoof-perfect since 2047 is obtained additively).

I did a Google search and didn't find anything about the distinction between "strong" versus "weak" spoof-perfect numbers. Maybe some other terminology is used.

An example of an even "weak" spoof-perfect number would be:

n = 90 = 2*5*9 (where 9 (with factors 3^2) is taken as a spoof prime);

sigma(n) = (1+2)*(1+5)*(1+9) = 3*(2*3)*(2*5) = 2*(2*5*(3^2)) = 2*90 = 2n is spoof-perfect (but is not "strong" spoof-perfect since 9 is obtained multiplicatively as 3^2 and is thus revealed composite).

Euler proved:

If 2^k - 1 is a prime number, then 2^(k-1)*(2^k - 1) is a perfect number and every even perfect number has this form.

The following seems to be true (is there a proof?):

If 2^k - 1 is an odd composite number taken as a spoof prime, then 2^(k-1)*(2^k - 1) is a "strong" spoof-perfect number and every even "strong" spoof-perfect number has this form?

There is only one known odd spoof-perfect number (found by Rene Descartes) but it is a "weak" spoof-perfect number (cf. 'Descartes numbers' and 'Unsolved problems in number theory' links below). (End)

a(n+1) = A173787(2*n+1,n); cf. A020522, A059153. - Reinhard Zumkeller, Feb 28 2010

Also, row sums of triangle A139251. - Omar E. Pol, May 25 2010

Starting with "1" = (1, 1, 2, 4, 8, ...) convolved with A002450: (1, 5, 21, 85, 341, ...); and (1, 3, 7, 15, 31, ...) convolved with A002001: (1, 3, 12, 48, 192, ...). - Gary W. Adamson, Oct 26 2010

a(n) is also the number of toothpicks in the corner toothpick structure of A153006 after 2^n - 1 stages. - Omar E. Pol, Nov 20 2010

The number of n-dimensional odd theta functions of half-integral characteristic. (Gunning, p.22) - Michael Somos, Jan 03 2014

a(n) = A000217((2^n)-1) = 2^(2n-1) - 2^(n-1) is the nearest triangular number below 2^(2n-1); cf. A007582, A233327. - Antti Karttunen, Feb 26 2014

a(n) is the sum of all the remainders when all the odd numbers < 2^n are divided by each of the powers 2,4,8,...,2^n. - J. M. Bergot, May 07 2014

Let b(m,k) = number of ways to form a sequence of m selections, without replacement, from a circular array of m labeled cells, such that the first selection of a cell whose adjacent cells have already been selected (a "first connect") occurs on the k-th selection. b(m,k) is defined for m >=3, and for 3 <= k <= m. Then b(m,k)/2m ignores rotations and reflection. Let m=n+2, then a(n) = b(m,m-1)/2m. Reiterated, a(n) is the (m-1)th column of the triangle b(m,k)/2m, whose initial rows are (1), (1 2), (2 6 4), (6 18 28 8), (24 72 128 120 16), (120 360 672 840 496 32), (720 2160 4128 5760 5312 2016 64); see A249796. Note also that b(m,3)/2m = n!, and b(m,m)/2m = 2^n. Proofs are easy. - Tony Bartoletti, Oct 30 2014

Beginning at a(1) = 1, this sequence is the sum of the first 2^(n-1) numbers of the form 4*k + 1 = A016813(k). For example, a(4) = 120 = 1 + 5 + 9 + 13 + 17 + 21 + 25 + 29. - J. M. Bergot, Dec 07 2014

a(n) is the number of edges in the (2^n - 1)-dimensional simplex. - Dimitri Boscainos, Oct 05 2015

a(n) is the number of linear elements in a complete plane graph in 2^n points. - Dimitri Boscainos, Oct 05 2015

a(n) is the number of linear elements in a complete parallelotope graph in n dimensions. - Dimitri Boscainos, Oct 05 2015

a(n) is the number of lattices L in Z^n such that the quotient group Z^n / L is C_4. - Álvar Ibeas, Nov 26 2015

a(n) gives the quadratic coefficient of the polynomial ((x + 1)^(2^n) + (x - 1)^(2^n))/2, cf. A201461. - Martin Renner, Jan 14 2017

Let f(x)=x+2*sqrt(x) and g(x)=x-2*sqrt(x). Then f(4^n*x)=b(n)*f(x)+a(n)*g(x) and g(4^n*x)=a(n)*f(x)+b(n)*g(x), where b is A007582. - Luc Rousseau, Dec 06 2018

For n>=1, a(n) is the covering radius of the first order Reed-Muller code RM(1,2n). - Christof Beierle, Dec 22 2021

a(n) =

This sequence is the normalized length per iteration of the space-filling Peano-Hilbert curve. The curve remains in a square, but its length increases without bound. The length of the curve, after n iterations in a unit square, is a(n)*2^(-n) where a(n) = 4*a(n-1)+3. This is the sequence of a(n) values. a(n)*(2^(-n)*2^(-n)) tends to 1, the area of the square where the curve is generated, as n increases. The ratio between the number of segments of the curve at the n-th iteration (A015521) and a(n) tends to 4/5 as n increases. - Giorgio Balzarotti, Mar 16 2006

Numbers whose base-4 representation is 333....3. - Zerinvary Lajos, Feb 03 2007

From Eric Desbiaux, Jun 28 2009: (Start)

It appears that for a given area, a square n^2 can be divided into n^2+1 other squares.

It's a rotation and zoom out of a Cartesian plan, which creates squares with side

= sqrt( (n^2) / (n^2+1) ) --> A010503|A010532|A010541... --> limit 1,

and diagonal sqrt(2*sqrt((n^2)/(n^2+1))) --> A010767|... --> limit A002193.

(End)

Also the total number of line segments after the n-th stage in the H tree, if 4^(n-1) H's are added at the n-th stage to the structure in which every "H" is formed by 3 line segments. A164346 (the first differences of this sequence) gives the number of line segments added at the n-th stage. - Omar E. Pol, Feb 16 2013

a(n) is the cumulative number of segment deletions in a Koch snowflake after (n+1) iterations. - Ivan N. Ianakiev, Nov 22 2013

Inverse binomial transform of A005057. - Wesley Ivan Hurt, Apr 04 2014

For n > 0, a(n) is one-third the partial sums of A002063(n-1). - J. M. Bergot, May 23 2014

Also the cyclomatic number of the n-Sierpinski tetrahedron graph. - Eric W. Weisstein, Sep 18 2017

a(n) is twice the area of the trapezoid created by the four points (2^n,2^(n+1)), (2^(n+1), 2^n), (2^(n+1), 2^(n+2)), (2^(n+2), 2^(n+1)). - J. M. Bergot, May 23 2014

These are squares that can be expressed as sum of exactly two distinct powers of two. For instance, a(4) = 9*4^4 = 2304 = 2^11 + 2^8 . It is conjectured that these are the only squares with this characteristic (tested on squares up to (10^7)^2). - Andres Cicuttin, Apr 23 2016

Conjecture is true. It is equivalent to prove that the Diophantine equation m^2 = 2^k*(1+2^h), where h>0, has solutions only when h=3. Dividing by 2^k we must obtain an odd square on the left, since 1+2^h is odd, so we can write (2*r+1)^2 = 1+2^h. Expanding, we have 4*r*(r+1) = 2^h, from which it follows that r must be equal to 1 and thus h=3, since r and r+1 must be powers of 2. - Giovanni Resta, Jul 27 2017

Binomial transform of A083232. Inverse binomial transform of A066443.

Numbers k such that, except for some first term, k^2 = [A000302]^3 + [A004171]^3 + [A002001]^3; e.g., 3072^2 = 64^3 + 128^3 + 192^3; 51539607552^2 = 4194304^3 + 8388608^3 + 12582912^3. - Vincenzo Librandi, Aug 08 2010

With the exception of the first term, these numbers cannot be written as the sum of two integer cubes but can be written as the sum of two positive rational cubes (i.e., 6*8^n = (17*2^n/21)^3 + (37*2^n/21)^3). - Arkadiusz Wesolowski, Aug 15 2013

a(n+1) is the number of unit square faces on the convex hull of a level n Menger sponge. This follows since it has six exterior faces, each of which is a Sierpinski carpet with 8^n squares. - Allan Bickle, Nov 28 2022

a(n) = a(3n)/a(0) = a(3n+1)/a(1). a(n) mod 2 = A039966(n). Row sums of A117939.

Observation: if this is written as a triangle (see example) then at least the first five row sums coincide with A002001. - Omar E. Pol, Nov 28 2011

Riordan array (1-x,1) read by rows; Riordan inverse is (1/(1-x),1). Columns have g.f. (1-x)x^k. Diagonal sums are A033999. Unsigned version in A097806.

Table T(n,k) read by antidiagonals. T(n,1) = 1, T(n,2) = -1, T(n,k) = 0, k > 2. - Boris Putievskiy, Jan 17 2013

Finite difference operator (pair difference): left multiplication by T of a sequence arranged as a column vector gives a running forward difference, a(k+1)-a(k), or first finite difference (modulo sign), of the elements of the sequence. T^n gives the n-th finite difference (mod sign). T is the inverse of the summation matrix A000012 (regarded as lower triangular matrices). - Tom Copeland, Mar 26 2014

Binomial transform of A000244 without initial 1.

Second binomial transform of A007283.

Third binomial transform of A010701.

Inverse binomial transform of A005053 without initial 1.

First differences of A024036. - Omar E. Pol, Feb 16 2013

Let f(k) be the sum of the smallest three positive divisors of k, g(k) be the sum of the largest two positive divisors of k, this sequence from a(2) onwards contains the numbers k for which g(k) is a positive integer power of f(k). - Yifan Xie, Jan 27 2024