cp's OEIS Frontend

Nonnegative X values of solutions to the equation X^3 + 4*X^2 = Y^2. The respective Y values are n*(n^2 - 4). - Mohamed Bouhamida, Nov 06 2007

Discriminants of binary forms x^2 + n*x*y + y^2 (for n > 1). - Artur Jasinski, Apr 28 2008

a(n)*a(n-1) + 4 = (a(n)-n)^2. This is the case d = 4 in the general (n^2-d)*((n-1)^2-d) + d = (n^2-n-d)^2. - Bruno Berselli, Dec 07 2011

Interleaving of A134582 and A078371. - Bruce J. Nicholson, Oct 14 2019

a(n) is the number of edges in (n+1) X (n+1) square grid with all horizontal, vertical and diagonal segments filled in. - Asher Auel, Jan 12 2000

In other words, the edge count of the (n+1) X (n+1) king graph. - Eric W. Weisstein, Jun 20 2017

Write 0,1,2,... in clockwise spiral; sequence gives numbers on one of 4 diagonals. (See Example section.)

The identity (4*n+1)^2 - (4*n^2+2*n)*(2)^2 = 1 can be written as A016813(n)^2 - a(n)*2^2 = 1. - Vincenzo Librandi, Jul 20 2010 - Nov 25 2012

Starting with "6" = binomial transform of [6, 14, 8, 0, 0, 0, ...]. - Gary W. Adamson, Aug 27 2010

The hyper-Wiener index of the crown graph G(n) (n>=3). The crown graph G(n) is the graph with vertex set {x(1), x(2), ..., x(n), y(1), y(2), ..., y(n)} and edge set {(x(i), y(j)): 1 <= i,j <= n, i != j} (= the complete bipartite graph K(n,n) with horizontal edges removed). The Hosoya-Wiener polynomial of G(n) is n(n-1)(t+t^2)+nt^3. - Emeric Deutsch, Aug 29 2013

Sum of the numbers from n to 3n. - Wesley Ivan Hurt, Oct 27 2014

Subsequence of A004613: all numbers in this sequence have all prime factors of the form 4k+1. E.g., 40001 = 13*17*181, 13 = 4*3 + 1, 17 = 4*4 + 1, 181 = 4*45 + 1. - Cino Hilliard, Aug 26 2006, corrected by Franklin T. Adams-Watters, Mar 22 2011

A000466(n), A008586(n) and a(n) are Pythagorean triples. - Zak Seidov, Jan 16 2007

Solutions x of the Mordell equation y^2 = x^3 - 3a^2 - 1 for a = 0, 1, 2, ... - Michel Lagneau, Feb 12 2010

Ulam's spiral (NW spoke). - Robert G. Wilson v, Oct 31 2011

For n >= 1, a(n) is numerator of radius r(n) of circle with sagitta = n and cord length = 1. The denominator is A008590(n). - Kival Ngaokrajang, Jun 13 2014

a(n)+6 is prime for n = 0..6 and for n = 15..20. - Altug Alkan, Sep 28 2015

In the following the expression [n odd] is 1 if n is odd, 0 otherwise.

(+) W_n(0) = E_n are the Euler (or secant) numbers A122045.

(+) W_n(1) = T_n are the signed tangent numbers, see A009006.

(+) W_{n-1}(1) n / (4^n - 2^n) = B_n gives for n > 1 the Bernoulli number A027641/A027642.

(+) W_n(-1) 2^{-n}(n+1) = G_n the Genocchi number A036968.

(+) W_n(1/2) 2^{n} are the signed generalized Euler (Springer) number, see A001586.

(+) | W_n([n odd]) | the number of alternating permutations A000111.

(+) | W_n([n odd]) / n! | for 0<=n the Euler zeta number A099612/A099617 (see Wikipedia on Bernoulli number). - Peter Luschny, Dec 29 2008

The diagonals in the full triangle (with zero coefficients) of the polynomials have the general form E(k)*binomial(n+k,k) (k>=0 fixed, n=0,1,...) where E(n) are the Euler numbers in the enumeration A122045. For k=2 we find the triangular numbers A000217 and for k=4 A154286. - Peter Luschny, Jan 06 2009

From Peter Bala, Jun 10 2009: (Start)

The Swiss-Knife polynomials W_n(x) may be expressed in terms of the Bernoulli polynomials B(n,x) as

... W_n(x) = 4^(n+1)/(2*n+2)*[B(n+1,(x+3)/4) - B(n+1,(x+1)/4)].

The Swiss-Knife polynomials are, apart from a multiplying factor, examples of generalized Bernoulli polynomials.

Let X be the Dirichlet character modulus 4 defined by X(4*n+1) = 1, X(4*n+3) = -1 and X(2*n) = 0. The generalized Bernoulli polynomials B(X;n,x), n = 1,2,..., associated with the character X are defined by means of the generating function

... t*exp(x*t)*(exp(t)-exp(3*t))/(exp(4*t)-1) = sum {n = 1..inf} B(X;n,x)*t^n/n!.

The first few values are B(X;1,x) = -1/2, B(X;2,x) = -x, B(X,3,x) = -3/2*(x^2-1) and B(X;4,x) = -2*(x^3-3*x).

In general, W_n(x) = -2/(n+1)*B(X;n+1,x).

For the theory of generalized Bernoulli polynomials associated to a periodic arithmetical function see [Cohen, Section 9.4].

The generalized Bernoulli polynomials may be used to evaluate twisted sums of k-th powers. For the present case the result is

sum{n = 0..4*N-1} X(n)*n^k = 1^k - 3^k + 5^k - 7^k + ... - (4*N-1)^k

= [B(X;k+1,4*N) - B(X;k+1,0)]/(k+1) = [W_k(0) - W_k(4*N)]/2.

For the proof apply [Cohen, Corollary 9.4.17 with m = 4 and x = 0].

The generalized Bernoulli polynomials and the Swiss-Knife polynomials are also related to infinite sums of powers through their Fourier series - see the formula section below. For a table of the coefficients of generalized Bernoulli polynomials attached to a Dirichlet character modulus 8 see A151751.

(End)

The Swiss-Knife polynomials provide a general formula for alternating sums of powers similar to the formula which are provided by the Bernoulli polynomials for non-alternating sums of powers (see the Luschny link). Sequences covered by this formula include A001057, A062393, A062392, A011934, A144129, A077221, A137501, A046092. - Peter Luschny, Jul 12 2009

The greatest common divisor of the nonzero coefficients of the decapitated Swiss-Knife polynomials is exp(Lambda(n)), where Lambda(n) is the von Mangoldt function for odd primes, symbolically:

gcd(coeffs(SKP_{n}(x) - x^n)) = A155457(n) (n>1). - Peter Luschny, Dec 16 2009

Another version is at A119879. - Philippe Deléham, Oct 26 2013

Triangles in rhombic matchstick arrangement of side n.

Maximum accumulated number of electrons at energy level n. - Scott A. Brown, Feb 28 2000

Let M_n denote the n X n matrix M_n(i,j)=i^2+j^2; then the characteristic polynomial of M_n is x^n - a(n)x^(n-1) - .... - Michael Somos, Nov 14 2002

Convolution of odds (A005408) and evens (A005843). - Graeme McRae, Jun 06 2006

a(n) is the number of non-monotonic functions with domain {0,1,2} and codomain {0,1,...,n}. - Dennis P. Walsh, Apr 25 2011

For any odd number 2n+1, find Sum_{aJ. M. Bergot, Jul 16 2011

a(n) gives the number of (n+1) X (n+1) symmetric (0,1)-matrices containing three ones (see [Cameron]). - L. Edson Jeffery, Feb 18 2012

a(n) is the number of 4-tuples (w,x,y,z) with all terms in {0,...,n} and |w - x| < y. - Clark Kimberling, Jun 02 2012

Partial sums of A001105. - Omar E. Pol, Jan 12 2013

Total number of square diagonals (of any size) in an n X n square grid. - Wesley Ivan Hurt, Mar 24 2015

Number of diagonal attacks of two queens on (n+1) X (n+1) chessboard. - Antal Pinter, Sep 20 2015

a(n) is the minimum value obtainable by partitioning either the set {x in the natural numbers | 1 <= x <= 2n} or the set {x in the natural numbers | 0 <= x <= 2n+1} into pairs, taking the product of all such pairs, and taking the sum of all such products. - Thomas Anton, Oct 21 2020

a(n) is the irregularity of the n-th power of a path of length at least 3*n. (The irregularity of a graph is the sum of the differences between the degrees over all edges of the graph.) - Allan Bickle, Jun 16 2023

a(n) is the maximum possible total number of inversions in all rows and all columns of a Latin square of order n+1. - Ivaylo Kortezov, Jun 28 2025

Draw n ellipses in the plane (n > 0); sequence gives maximum number of regions into which the plane is divided (cf. A014206, A386480).

Least k such that Z(k,2) <= Z(n,3) where Z(m,s) = Sum_{i>=m} 1/i^s = zeta(s) - Sum_{i=1..m-1} 1/i^s. - Benoit Cloitre, Nov 29 2002

For n > 2, third diagonal of A154685. - Vincenzo Librandi, Aug 06 2010

a(k) is also the Moore lower bound A198300(k,6) on the order A054760(k,6) of a (k,6)-cage. Equality is achieved if and only if there exists a finite projective plane of order k - 1. A sufficient condition for this is that k - 1 be a prime power. - Jason Kimberley, Oct 17 2011 and Jan 01 2013

From Jess Tauber, May 20 2013: (Start)

For neutron shell filling in spherical atomic nuclei, this sequence shows numerical differences between filled spin-split suborbitals sharing all quantum numbers except the principal quantum number n, and here all n's must differ by 1. Only a small handful of exceptions exist.

This sequence consists of summed pairs of every other doubled triangular number. It also can be created by taking differences between nuclear magic numbers from the harmonic oscillator (HO)(doubled tetrahedral) set and the spin-orbit (SO) set (2,6,14,28,50,82,126,184,...), with either set being larger. So SO-HO: 2-0=2, 6-0=6, 14-0=14, 28-2=26, 50-8=42, 82-20=62, 126-40=86, 184-70=114, and HO-SO: 2-0=2, 8-2=6, 20-6=14, 40-14=26, 70-28=42, 112-50=62, 168-82=86, 240-126=114. From the perspective of idealized HO periodic structure, with suborbitals in order from largest to smallest spin, alternating by parity, the HO-SO set is spaced two period analogs PLUS one suborbital, while the SO-HO set is spaced two period analogs MINUS one suborbital. (End)

The known values of f(k,6) and F(k,6) in Brown (1967), Table 1, closely match this sequence. - N. J. A. Sloane, Jul 09 2015

Numbers k such that 2*k - 3 is a square. - Bruno Berselli, Nov 08 2017

Numbers written 222 in number base B, including binary with 'digit' 2: 222(2)=14, 222(3)=26, ... - Ron Knott, Nov 14 2017

a(n) is the number of edges in (n+1) X (n+1) square grid with all horizontal, vertical and great diagonal segments filled in.

Nonnegative X values of integer solutions to the equation 2*X^3 + 4*X^2 = Y^2. To find Y values: b(n) = 2*n*(2*n^2 - 2). - Mohamed Bouhamida, Nov 06 2007

Second term of an arithmetic progression of 5 numbers with common difference 2n+1. The sum of squares of such 5 terms equals the sum of squares of 5 consecutive numbers starting a(n) + 2n + 1. - Carmine Suriano, Oct 16 2013

For m > 2, a(m-1) = 2*m*(m-2) is the number of Hamiltonian circuits on an m-gonal bipyramid with labeled vertices. - Stanislav Sykora, Jul 22 2014

a(n+1), n >= 0, appears also as the third member of the quartet [p0(n), p1(n), a(n+1), p3(n)] of the square of [n, n+1, n+2, n+3] in the Clifford algebra Cl_2 for n >= 0. p0(n) = -A147973(n+3), p1(n) = A046092(n) and p3(n) = A139570(n). See a comment on A147973, also with a reference. - Wolfdieter Lang, Oct 15 2014

From Bui Quang Tuan, Mar 31 2015: (Start)

For n >= 2, a(n) is the total sum of all numbers on the perimeter of a square consisting of n columns, each of which contains n numbers 1, 2, 3, ..., n.

Here is an example with n = 5:

1 1 1 1 1

2 2 2 2 2

3 3 3 3 3

4 4 4 4 4

5 5 5 5 5

where 1+1+1+1+1 + 2+2 + 3+3 + 4+4 + 5+5+5+5+5 = 48 = a(5).

(End)

Nonnegative k such that k/2+1 is a square. - Bruno Berselli, Apr 10 2018

Deleting the least significant digit yields the (n-1)-st triangular number: a(n) = 10*A000217(n-1) + 1. - Amarnath Murthy, Dec 11 2003

All divisors of a(n) are congruent to 1 or -1, modulo 10; that is, they end in the decimal digit 1 or 9. Proof: If p is an odd prime different from 5 then 5n^2 - 5n + 1 == 0 (mod p) implies 25(2n - 1)^2 == 5 (mod p), whence p == 1 or -1 (mod 10). - Nick Hobson, Nov 13 2006

Centered decagonal numbers. - Omar E. Pol, Oct 03 2011

The partial sums of this sequence give A004466. - Leo Tavares, Oct 04 2021

The continued fraction expansion of sqrt(5*a(n)) is [5n-3; {2, 2n-2, 2, 10n-6}]. For n=1, this collapses to [2; {4}]. - Magus K. Chu, Sep 12 2022

Numbers m such that 20*m + 5 is a square. Also values of the Fibonacci polynomial y^2 - x*y - x^2 for x = n and y = 3*n - 1. This is a subsequence of A089270. - Klaus Purath, Oct 30 2022

All terms can be written as a difference of two consecutive squares a(n) = A005891(n-1)^2 - A028895(n-1)^2, and they can be represented by the forms (x^2 + 2mxy + (m^2-1)y^2) and (3x^2 + (6m-2)xy + (3m^2-2m)y^2), both of discriminant 4. - Klaus Purath, Oct 17 2023

16 times the triangular numbers A000217.

Centered 16-gonal numbers A069129, minus 1. Also, sequence found by reading the segment (0, 16) together with the line from 16, in the direction 16, 48, ..., in the square spiral whose vertices are the triangular numbers A000217. - Omar E. Pol, Apr 26 2008, Nov 20 2008

For n >= 1, number of permutations of n+1 objects selected from 5 objects v, w, x, y, z with repetition allowed, containing n-1 v's. Examples: at n=1, n-1=0 (i.e., zero v's), and a(1)=16 because we have ww, wx, wy, wz, xw, xx, xy, xz, yw, yx, yy, yz, zw, zx, zy, zz; at n=2, n-1=1 (i.e., one v), and there are 3 permutations corresponding to each one in the n=1 case (e.g., the single v can be inserted in any of three places in the 2-object permutation xy, yielding vxy, xvy, and xyv), so a(2) = 3*a(1) = 3*16 = 48; at n=3, n-1=2 (i.e., two v's), and a(3) = C(4,2)*a(1) = 6*16 = 96; etc. - Zerinvary Lajos, Aug 07 2008 (this needs clarification, Joerg Arndt, Feb 23 2014)

Sequence found by reading the line from 0, in the direction 0, 16, ... and the same line from 0, in the direction 0, 48, ..., in the square spiral whose vertices are the generalized 18-gonal numbers. - Omar E. Pol, Oct 03 2011

For n > 0, a(n) is the area of the triangle with vertices at ((n-1)^2, n^2), ((n+1)^2, (n+2)^2), and ((n+3)^2, (n+2)^2). - J. M. Bergot, May 22 2014

For n > 0, a(n) is the number of self-intersecting points in star polygon {4*(n+1)/(2*n+1)}. - Bui Quang Tuan, Mar 28 2015

Equivalently: integers k such that k$ / (k/2)! and k$ / (k/2+1)! are both squares when A000178 (k) = k$ = 1!*2!*...*k! is the superfactorial of k (see A348692 for further information). - Bernard Schott, Dec 02 2021

Schroeder, p. 330, states "For positive n, these winding numbers are precisely those whose continued fraction expansion is periodic and has period length 1".

Positive X values of solutions to the equation X^3 - 4*X^2 = Y^2. To find Y values: b(n) = n*(n^2 + 4). - Mohamed Bouhamida, Nov 06 2007

From Artur Jasinski, Oct 03 2008: (Start)

General formula for cotangent recurrences type:

a(n+1) = a(n)^3 + 3*a(n) and a(1)=k is

a(n) = floor(((k + sqrt(k^2 + 4))/2)^(3^(n-1))). (End)

Given sequences of the form S(n) = N*S(n-1) + S(n-2) starting (1, N, ...), and having convergents with discriminant (N^2 + 4), S(p) == (a(n))^((p-1)/2) mod p, for n>0, p = odd prime. Example: with N = 2 we have the Pell series (1, 2, 5, 12, 29, 70, 169, ...) with P(7) = 169. Then 169 == 8^3 mod 7, with a(2) = 8. Cf. Schroeder, "Number Theory in Science and Communication", p. 90, for N = 1: F(p) == 5^((p-1)/2) mod p. - Gary W. Adamson, Feb 23 2009

The only two real solutions of the form f(x) = A*x^p with positive p that satisfy f^(n)(x) = f^[-1](x), x >= 0, n >= 1, with f^(n) the n-th derivative and f^[-1] the compositional inverse of f, are obtained for p = p1(n) = (n + sqrt(a(n)))/2 and p = p2(n) = (n - sqrt(a(n)))/2, n >= 1, and A = A(n) = (fallfac(p,n))^(-p/(p+1)), for p = p1(n) and p = p2(n), respectively. Here fallfac(x, k) := product(x - j, j = 0..k-1), the falling factorials. See the T. Koshy reference, pp. 263-264 (there is also a solution for negative p if n is even; see the corresponding comment in A002522). - Wolfdieter Lang, Oct 21 2010, Oct 28 2010

(n + sqrt(a(n)))/2 = [n;n,n,...], with the regular continued fraction with period length 1. For a simple proof see, e.g., the Schroeder reference, pp. 330-331. See also the first comment above.