OFFSET
0,2
COMMENTS
Consider all Pythagorean triples (X,Y,Z=Y+1) ordered by increasing Z; sequence gives Y values. X values are 1, 3, 5, 7, 9, ... (A005408), Z values are A001844.
In the triple (X, Y, Z) we have X^2=Y+Z. Actually, the triple is given by {x, (x^2 -+ 1)/2}, where x runs over the odd numbers (A005408) and x^2 over the odd squares (A016754). - Lekraj Beedassy, Jun 11 2004
a(n) is the number of edges in n X n square grid with all horizontal and vertical segments filled in. - Asher Auel, Jan 12 2000 [Corrected by Felix Huber, Apr 09 2024]
a(n) is the only number satisfying an inequality related to zeta(2) and zeta(3): Sum_{i>a(n)+1} 1/i^2 < Sum_{i>n} 1/i^3 < Sum_{i>a(n)} 1/i^2. - Benoit Cloitre, Nov 02 2001
Number of right triangles made from vertices of a regular n-gon when n is even. - Sen-Peng Eu, Apr 05 2001
Number of ways to change two non-identical letters in the word aabbccdd..., where there are n type of letters. - Zerinvary Lajos, Feb 15 2005
a(n) is the number of (n-1)-dimensional sides of an (n+1)-dimensional hypercube (e.g., squares have 4 corners, cubes have 12 edges, etc.). - Freek van Walderveen (freek_is(AT)vanwal.nl), Nov 11 2005
From Nikolaos Diamantis (nikos7am(AT)yahoo.com), May 23 2006: (Start)
Consider a triangle, a pentagon, a heptagon, ..., a k-gon where k is odd. We label a triangle with n=1, a pentagon with n=2, ..., a k-gon with n = floor(k/2). Imagine a player standing at each vertex of the k-gon.
Initially there are 2 frisbees, one held by each of two neighboring players. Every time they throw the frisbee to one of their two nearest neighbors with equal probability. Then a(n) gives the average number of steps needed so that the frisbees meet.
I verified this by simulating the processes with a computer program. For example, a(2) = 12 because in a pentagon that's the expected number of trials we need to perform. That is an exercise in Concrete Mathematics and it can be done using generating functions. (End)
A diagonal of A059056. - Zerinvary Lajos, Jun 18 2007
If X_1,...,X_n is a partition of a 2n-set X into 2-blocks then a(n-1) is equal to the number of 2-subsets of X containing none of X_i, (i=1,...,n). - Milan Janjic, Jul 16 2007
X values of solutions to the equation 2*X^3 + X^2 = Y^2. To find Y values: b(n) = 2n(n+1)(2n+1). - Mohamed Bouhamida, Nov 06 2007
Number of (n+1)-permutations of 3 objects u,v,w, with repetition allowed, containing n-1 u's. Example: a(1)=4 because we have vv, vw, wv and ww; a(2)=12 because we can place u in each of the previous four 2-permutations either in front, or in the middle, or at the end. - Zerinvary Lajos, Dec 27 2007
Sequence found by reading the line from 0, in the direction 0, 4, ... and the same line from 0, in the direction 0, 12, ..., in the square spiral whose vertices are the triangular numbers A000217. - Omar E. Pol, May 03 2008
a(n) is also the least weight of self-conjugate partitions having n different even parts. - Augustine O. Munagi, Dec 18 2008
From Peter Luschny, Jul 12 2009: (Start)
The general formula for alternating sums of powers of even integers is in terms of the Swiss-Knife polynomials P(n,x) A153641 (P(n,1)-(-1)^k P(n,2k+1))/2. Here n=2, thus
a(k) = |(P(2,1) - (-1)^k*P(2,2k+1))/2|. (End)
The sum of squares of n+1 consecutive numbers between a(n)-n and a(n) inclusive equals the sum of squares of n consecutive numbers following a(n). For example, for n = 2, a(2) = 12, and the corresponding equation is 10^2 + 11^2 + 12^2 = 13^2 + 14^2. - Tanya Khovanova, Jul 20 2009
Number of roots in the root system of type D_{n+1} (for n>2). - Tom Edgar, Nov 05 2013
Draw n ellipses in the plane (n>0), any 2 meeting in 4 points; sequence gives number of intersections of these ellipses (cf. A051890, A001844); a(n) = A051890(n+1) - 2 = A001844(n) - 1. - Jaroslav Krizek, Dec 27 2013
a(n) appears also as the second member of the quartet [p0(n), a(n), p2(n), 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), p2(n) = A054000(n+1) and p3(n) = A139570(n). See a comment on A147973, also with a reference. - Wolfdieter Lang, Oct 15 2014
a(n) appears also as the third and fourth member of the quartet [p0(n), p0(n), a(n), a(n)] of the square of [n, n, n+1, n+1] in the Clifford algebra Cl_2 for n >= 0. p0(n) = A001105(n). - Wolfdieter Lang, Oct 16 2014
Consider two equal rectangles composed of unit squares. Then surround the 1st rectangle with 1-unit-wide layers to build larger rectangles, and surround the 2nd rectangle just to hide the previous layers. If r(n) and h(n) are the number of unit squares needed for n layers in the 1st case and the 2nd case, then for all rectangles, we have a(n) = r(n) - h(n) for n>=1. - Michel Marcus, Sep 28 2015
When greater than 4, a(n) is the perimeter of a Pythagorean triangle with an even short leg 2*n. - Agola Kisira Odero, Apr 26 2016
Also the number of minimum connected dominating sets in the (n+1)-cocktail party graph. - Eric W. Weisstein, Jun 29 2017
Consider a circular cake from which wedges of equal center angle c are cut out in clockwise succession and turned around so that the bottom comes to the top. This goes on until the cake shows its initial surface again. An interesting case occurs if 360°/c is not an integer. Then, with n = floor(360°/c), the number of wedges which have to be cut out and turned equals a(n). (For the number of cutting line segments see A005408.) - According to Peter Winkler's book "Mathematical Mind-Benders", which presents the problem and its solution (see Winkler, pp. 111, 115) the problem seems to be of French origin but little is known about its history. - Manfred Boergens, Apr 05 2022
a(n-3) is the maximum irregularity over all maximal 2-degenerate graphs with n vertices. The extremal graphs are 2-stars (K_2 joined to n-2 independent vertices). (The irregularity of a graph is the sum of the differences between the degrees over all edges of the graph.) - Allan Bickle, May 29 2023
Number of ways of placing a domino on a (n+1)X(n+1) board of squares. - R. J. Mathar, Apr 24 2024
REFERENCES
Tom M. Apostol, Introduction to Analytic Number Theory, Springer-Verlag, 1976, page 3.
Albert H. Beiler, Recreations in the Theory of Numbers. New York: Dover, p. 125, 1964.
Ronald L. Graham, D. E. Knuth and Oren Patashnik, Concrete Mathematics, Reading, Massachusetts: Addison-Wesley, 1994.
Peter Winkler, Mathematical Mind-Benders, Wellesley, Massachusetts: A K Peters, 2007.
LINKS
Vincenzo Librandi, Table of n, a(n) for n = 0..10000
Allan Bickle and Zhongyuan Che, Irregularities of Maximal k-degenerate Graphs, Discrete Applied Math. 331 (2023) 70-87.
Allan Bickle, A Survey of Maximal k-degenerate Graphs and k-Trees, Theory and Applications of Graphs 0 1 (2024) Article 5.
H. J. Brothers, Pascal's Prism: Supplementary Material.
Hsien-Kuei Hwang, Svante Janson, and Tsung-Hsi Tsai, Exact and asymptotic solutions of the recurrence f(n) = f(floor(n/2)) + f(ceiling(n/2)) + g(n): theory and applications, preprint, 2016.
Hsien-Kuei Hwang, Svante Janson, and Tsung-Hsi Tsai, Exact and Asymptotic Solutions of a Divide-and-Conquer Recurrence Dividing at Half: Theory and Applications, ACM Transactions on Algorithms, Vol. 13, No. 4 (2017), Article #47.
Milan Janjic, Two Enumerative Functions
Ron Knott, Pythagorean Triples and Online Calculators
Tanya Khovanova, A Miracle Equation.
Augustine O. Munagi, Pairing conjugate partitions by residue classes, Discrete Math., 308 (2008), 2492-2501. [From Augustine O. Munagi, Dec 18 2008]
Enrique Navarrete and Daniel Orellana, Finding Prime Numbers as Fixed Points of Sequences, arXiv:1907.10023 [math.NT], 2019.
Amelia Carolina Sparavigna, The groupoids of Mersenne, Fermat, Cullen, Woodall and other Numbers and their representations by means of integer sequences, Politecnico di Torino, Italy (2019), [math.NT].
Amelia Carolina Sparavigna, Some Groupoids and their Representations by Means of Integer Sequences, International Journal of Sciences (2019) Vol. 8, No. 10.
Rusliansyah D. Suprijanto, Observation on Sums of Powers of Integers Divisible by Four, Applied Mathematical Sciences, Vol. 8, 2014, no. 45, 2219 - 2226.
Leo Tavares, Illustration: Diamond Rows
Herman Tulleken, Polyominoes 2.2: How they fit together, (2019).
Eric Weisstein's World of Mathematics, Aztec Diamond.
Eric Weisstein's World of Mathematics, Cocktail Party Graph.
Eric Weisstein's World of Mathematics, Connected Dominating Set.
Eric Weisstein's World of Mathematics, Gear Graph.
Eric Weisstein's World of Mathematics, Hamiltonian Path.
Eric Weisstein's World of Mathematics, Pythagorean Triple.
Index entries for linear recurrences with constant coefficients, signature (3,-3,1).
FORMULA
a(n) = A100345(n+1, n-1) for n>0.
a(n) = C(2n, 2) - n = 4*C(n, 2). - Zerinvary Lajos, Feb 15 2005
From Lekraj Beedassy, Jun 04 2006: (Start)
a(n) - a(n-1)=4*n.
Let k=a(n). Then a(n+1) = k + 2*(1 + sqrt(2k + 1)). (End)
O.g.f.:4*x/(1-x)^3; e.g.f.: exp(x)*(2*x^2+4*x). - Geoffrey Critzer, May 17 2009
From Stephen Crowley, Jul 26 2009: (Start)
a(n) = 1/int(-(x*n+x-1)*(step((-1+x*n)/n)-1)*n*step((x*n+x-1)/(n+1)),x=0..1) where step(x)=piecewise(x<0,0,0<=x,1) is the Heaviside step function.
Sum_{n>=1} 1/a(n) = 1/2. (End)
a(n) = 3*a(n-1) - 3*a(n-2) + a(n-3); a(0)=0, a(1)=4, a(2)=12. - Harvey P. Dale, Jul 25 2011
For n > 0, a(n) = 1/(Integral_{x=0..Pi/2} (sin(x))^(2*n-1)*(cos(x))^3). - Francesco Daddi, Aug 02 2011
a(n) = A001844(n) - 1. - Omar E. Pol, Oct 03 2011
(a(n) - A000217(k))^2 = A000217(2n-k)*A000217(2n+1+k) - (A002378(n) - A000217(k)), for all k. See also A001105. - Charlie Marion, May 09 2013
From Ivan N. Ianakiev, Aug 30 2013: (Start)
a(n)*(2m+1)^2 + a(m) = a(n*(2m+1)+m), for any nonnegative integers n and m.
t(k)*a(n) + t(k-1)*a(n+1) = a((n+1)*(t(k)-t(k-1)-1)), where k>=2, n>=1, t(k)=A000217(k). (End)
a(n) = A245300(n,n). - Reinhard Zumkeller, Jul 17 2014
Sum_{n>=1} (-1)^(n+1)/a(n) = log(2) - 1/2 = A187832. - Ilya Gutkovskiy, Mar 16 2017
a(n) = lcm(2*n,2*n+2). - Enrique Navarrete, Aug 30 2017
a(n)*a(n+k) + k^2 = m^2 (a perfect square), n >= 1, k >= 0. - Ezhilarasu Velayutham, May 13 2019
From Amiram Eldar, Jan 29 2021: (Start)
Product_{n>=1} (1 + 1/a(n)) = cosh(Pi/2)/(Pi/2).
Product_{n>=1} (1 - 1/a(n)) = -2*cos(sqrt(3)*Pi/2)/Pi. (End)
EXAMPLE
a(7)=112 because 112 = 2*7*(7+1).
The first few triples are (1,0,1), (3,4,5), (5,12,13), (7,24,25), ...
The first such partitions, corresponding to a(n)=1,2,3,4, are 2+2, 4+4+2+2, 6+6+4+4+2+2, 8+8+6+6+4+4+2+2. - Augustine O. Munagi, Dec 18 2008
MATHEMATICA
Table[2 n (n + 1), {n, 0, 50}] (* Stefan Steinerberger, Apr 03 2006 *)
LinearRecurrence[{3, -3, 1}, {0, 4, 12}, 50] (* Harvey P. Dale, Jul 25 2011 *)
4*Binomial[Range[50], 2] (* Harvey P. Dale, Jul 25 2011 *)
PROG
(PARI) a(n)=binomial(n+1, 2)<<2 \\ Charles R Greathouse IV, Jun 10 2011
(Magma) [2*n*(n+1): n in [0..50]]; // Vincenzo Librandi, Oct 04 2011
(Maxima) A046092(n):=2*n*(n+1)$
makelist(A046092(n), n, 0, 30); /* Martin Ettl, Nov 08 2012 */
(Haskell)
a046092 = (* 2) . a002378 -- Reinhard Zumkeller, Dec 15 2013
CROSSREFS
KEYWORD
nonn,easy,nice
AUTHOR
STATUS
approved