

A000384


Hexagonal numbers: a(n) = n*(2*n1).
(Formerly M4108 N1705)


410



0, 1, 6, 15, 28, 45, 66, 91, 120, 153, 190, 231, 276, 325, 378, 435, 496, 561, 630, 703, 780, 861, 946, 1035, 1128, 1225, 1326, 1431, 1540, 1653, 1770, 1891, 2016, 2145, 2278, 2415, 2556, 2701, 2850, 3003, 3160, 3321, 3486, 3655, 3828, 4005, 4186, 4371, 4560
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OFFSET

0,3


COMMENTS

Number of edges in the join of two complete graphs, each of order n, K_n * K_n.  Roberto E. Martinez II, Jan 07 2002
The power series expansion of the entropy function H(x) = (1+x)log(1+x) + (1x)log(1x) has 1/a_i as the coefficient of x^(2i) (the odd terms being zero).  Tommaso Toffoli (tt(AT)bu.edu), May 06 2002
Sequence also gives the greatest semiperimeter of primitive Pythagorean triangles having inradius n1. Such a triangle has consecutive longer sides, with short leg 2n1, hypotenuse a(n)  (n1) = A001844(n), and area (n1)*a(n) = 6*A000330(n1).  Lekraj Beedassy, Apr 23 2003
More generally, if p1 and p2 are two arbitrarily chosen distinct primes then a(n) is the number of divisors of (p1^2*p2)^(n1) or equivalently of any member of A054753^(n1).  Ant King, Aug 29 2011
Number of standard tableaux of shape (2n1,1,1) (n>=1).  Emeric Deutsch, May 30 2004
It is well known that for n>0, A014105(n) [0,3,10,21,...] is the first of 2n+1 consecutive integers such that the sum of the squares of the first n+1 such integers is equal to the sum of the squares of the last n; e.g., 10^2 + 11^2 + 12^2 = 13^2 + 14^2.
Less well known is that for n>1, a(n) [0,1,6,15,28,...] is the first of 2n consecutive integers such that sum of the squares of the first n such integers is equal to the sum of the squares of the last n1 plus n^2; e.g., 15^2 + 16^2 + 17^2 = 19^2 + 20^2 + 3^2.  Charlie Marion, Dec 16 2006
Sequence found by reading the line from 0, in the direction 0, 6, ... and the line from 1, in the direction 1, 15, ..., in the square spiral whose vertices are the generalized hexagonal numbers A000217.  Omar E. Pol, Jan 09 2009
Let Hex(n)=hexagonal number, T(n)=triangular number, then Hex(n)=T(n)+3*T(n1).  Vincenzo Librandi, Nov 10 2010
For n>=1, 1/a(n) = Sum_{k=0..2*n1} ((1)^(k+1)*binomial(2*n1,k)*binomial(2*n1+k,k)*H(k)/(k+1)) with H(k) harmonic number of order k.
The number of possible distinct colorings of any 2 colors chosen from n colors of a square divided into quadrants.  Paul Cleary, Dec 21 2010
For n>0, a(n1) is the number of triples (w,x,y) with all terms in {0,...,n} and max(wx,xy) = wy.  Clark Kimberling, Jun 12 2012
a(n) is the number of positions of one domino in an even pyramidal board with base 2n.  César Eliud Lozada, Sep 26 2012
Let a triangle have T(0,0) = 0 and T(r,c) = r^2  c^2. The sum of the differences of the terms in row(n) and row(n1) is a(n).  J. M. Bergot, Jun 17 2013
With T_(i+1,i)=a(i+1) and all other elements of the lower triangular matrix T zero, T is the infinitesimal generator for A176230, analogous to A132440 for the Pascal matrix.  Tom Copeland, Dec 11 2013
a(n) is the number of length 2n binary sequences that have exactly two 1's. a(2) = 6 because we have: {0,0,1,1}, {0,1,0,1}, {0,1,1,0}, {1,0,0,1}, {1,0,1,0}, {1,1,0,0}. The ordinary generating function with interpolated zeros is: (x^2 + 3*x^4)/(1x^2)^3.  Geoffrey Critzer, Jan 02 2014
For n > 0, a(n) is the largest integer k such that k^2 + n^2 is a multiple of k + n. More generally, for m > 0 and n > 0, the largest integer k such that k^(2*m) + n^(2*m) is a multiple of k + n is given by k = 2*n^(2*m)  n.  Derek Orr, Sep 04 2014
Binomial transform of (0, 1, 4, 0, 0, 0, ...) and second partial sum of (0, 1, 4, 4, 4, ...).  Gary W. Adamson, Oct 05 2015
a(n) also gives the dimension of the simple Lie algebras D_n, for n >= 4.  Wolfdieter Lang, Oct 21 2015
For n > 0, a(n) equals the number of compositions of n+11 into n parts avoiding parts 2, 3, 4.  Milan Janjic, Jan 07 2016
Also the number of minimum dominating sets and maximal irredundant sets in the ncocktail party graph.  Eric W. Weisstein, Jun 29 and Aug 17 2017
As Beedassy's formula shows, this Hexagonal number sequence is the odd bisection of the Triangle number sequence. Both of these sequences are figurative number sequences. For A000384, a(n) can be found by multiplying its triangle number by its hexagonal number. For example let's use the number 153. 153 is said to be the 17th triangle number but is also said to be the 9th hexagonal number. Triangle(17) Hexagonal(9). 17*9=153. Because the Hexagonal number sequence is a subset of the Triangle number sequence, the Hexagonal number sequence will always have both a triangle number and a hexagonal number. n* (2*n1) because (2*n1) renders the triangle number.  Bruce J. Nicholson, Nov 05 2017
Also numbers k with the property that in the symmetric representation of sigma(k) the smallest Dyck path has a central valley and the largest Dyck path has a central peak, n >= 1. Thus all hexagonal numbers > 0 have middle divisors. (Cf. A237593.)  Omar E. Pol, Aug 28 2018
Consider all Pythagorean triples (X, Y, Z=Y+1) ordered by increasing Z: a(n+1) gives the semiperimeter of related triangles; A005408, A046092 and A001844 give the X, Y and Z values.  Ralf Steiner, Feb 25 2020
It appears that these are the numbers k with the property that the smallest subpart in the symmetric representation of sigma(k) is 1.  Omar E. Pol, Aug 28 2021
The nth hexagonal number equals the sum of the n consecutive integers with the same parity starting at 2*n1; for example, 1, 2+4, 3+5+7, 4+6+8+10, etc. In general, the nth 2kgonal number is the sum of the n consecutive integers with the same parity starting at (k2)*n  (k3). When k = 1 and 2, this result generates the positive integers, A000027, and the squares, A000290, respectively.  Charlie Marion, Mar 02 2022
Conjecture: For n>0, min{k such that there exist subsets A,B of {0,1,2,...,a(n)} such that A=B=k and A+B={0,1,2,...,2*a(n)}} = 2*n.  Michael Chu, Mar 09 2022


REFERENCES

Albert H. Beiler, Recreations in the Theory of Numbers, Dover, NY, 1964, p. 189.
Louis Comtet, Advanced Combinatorics, Reidel, 1974, pp. 7778. (In the integral formula on p. 77 a left bracket is missing for the cosine argument.)
E. Deza and M. M. Deza, Figurate numbers, World Scientific Publishing (2012), page 6.
L. E. Dickson, History of the Theory of Numbers. Carnegie Institute Public. 256, Washington, DC, Vol. 1, 1919; Vol. 2, 1920; Vol. 3, 1923, see vol. 2, p. 2.
N. J. A. Sloane, A Handbook of Integer Sequences, Academic Press, 1973 (includes this sequence).
N. J. A. Sloane and Simon Plouffe, The Encyclopedia of Integer Sequences, Academic Press, 1995 (includes this sequence).


LINKS

Jonathan M. Borwein, Dirk Nuyens, Armin Straub and James Wan, Random Walk Integrals, The Ramanujan Journal, October 2011, 26:109. DOI: 10.1007/s111390119325y.
Michel Waldschmidt, Continued fractions, Ecole de recherche CIMPAOujda, Théorie des Nombres et ses Applications, 18  29 mai 2015: Oujda (Maroc).


FORMULA

G.f.: x*(1+3*x)/(1x)^3.  Simon Plouffe in his 1992 dissertation, dropping the initial zero
a(n) = right term of M^n * [1,0,0], where M = the 3 X 3 matrix [1,0,0; 1,1,0; 1,4,1]. Example: a(5) = 45 since M^5 *[1,0,0] = [1,5,45].  Gary W. Adamson, Dec 24 2006
Starting with offset 1, = binomial transform of [1, 5, 4, 0, 0, 0, ...]. Also, A004736 * [1, 4, 4, 4, ...].  Gary W. Adamson, Oct 25 2007
a(n)^2 + (a(n)+1)^2 + ... + (a(n)+n1)^2 = (a(n)+n+1)^2 + ... + (a(n)+2n1)^2 + n^2; e.g., 6^2 + 7^2 = 9^2 + 2^2; 28^2 + 29^2 + 30^2 + 31^2 = 33^2 + 34^2 + 35^2 + 4^2.  Charlie Marion, Nov 10 2007
a(n) = binomial(n+1,2) + 3*binomial(n,2).
a(n) = 3*a(n1)  3*a(n2) + a(n3), a(0)=0, a(1)=1, a(2)=6.  Jaume Oliver Lafont, Dec 02 2008
a(n) = 2*a(n1)  a(n2) + 4.  Ant King, Aug 26 2011
a(n+1) = trinomial(2*n+1, 2) = trinomial(2*n+1, 4*n), for n >= 0, with the trinomial irregular triangle A027907. a(n+1) = (n+1)*(2*n+1) = (1/Pi)*Integral_{x=0..2} (1/sqrt(4  x^2))*(x^2  1)^(2*n+1)*R(4*n2, x) with the R polynomial coefficients given in A127672. [Comtet, p. 77, the integral formula for q=3, n > 2*n+1, k = 2, rewritten with x = 2*cos(phi)].  Wolfdieter Lang, Apr 19 2018
Product_{n>=2} (1  1/a(n)) = 2/3.  Amiram Eldar, Jan 21 2021
a(n) = floor(Sum_{k=(n1)^2..n^2} sqrt(k)), for n >= 1.  Amrit Awasthi, Jun 13 2021


MAPLE



MATHEMATICA

LinearRecurrence[{3, 3, 1}, {0, 1, 6}, 50] (* Harvey P. Dale, Sep 10 2015 *)
Join[{0}, Accumulate[Range[1, 312, 4]]] (* Harvey P. Dale, Mar 26 2016 *)
(* For Mathematica 10.4+ *) Table[PolygonalNumber[RegularPolygon[6], n], {n, 0, 48}] (* Arkadiusz Wesolowski, Aug 27 2016 *)
CoefficientList[Series[x*(1 + 3*x)/(1  x)^3 , {x, 0, 100}], x] (* Stefano Spezia, Sep 02 2018 *)


PROG

(PARI) a(n)=n*(2*n1)
(PARI) a(n) = binomial(2*n, 2) \\ Altug Alkan, Oct 06 2015
(Haskell)
a000384 n = n * (2 * n  1)
a000384_list = scanl (+) 0 a016813_list
(Python 3) # Intended to compute the initial segment of the sequence, not isolated terms.
def aList():
x, y = 1, 1
yield 0
while True:
yield x
x, y = x + y + 4, y + 4


CROSSREFS

a(n)= A093561(n+1, 2), (4, 1)Pascal column.
Cf. A002939 (twice a(n): sums of Pythagorean triples (X, Y, Z=Y+1).


KEYWORD

nonn,easy,nice


AUTHOR



EXTENSIONS



STATUS

approved



