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A003215 Hex (or centered hexagonal) numbers: 3*n*(n+1)+1 (crystal ball sequence for hexagonal lattice).
(Formerly M4362)
209
1, 7, 19, 37, 61, 91, 127, 169, 217, 271, 331, 397, 469, 547, 631, 721, 817, 919, 1027, 1141, 1261, 1387, 1519, 1657, 1801, 1951, 2107, 2269, 2437, 2611, 2791, 2977, 3169, 3367, 3571, 3781, 3997, 4219, 4447, 4681, 4921, 5167, 5419, 5677, 5941, 6211, 6487 (list; graph; refs; listen; history; text; internal format)
OFFSET

0,2

COMMENTS

The hexagonal lattice is the familiar 2-dimensional lattice in which each point has 6 neighbors. This is sometimes called the triangular lattice.

Crystal ball sequence for A_2 lattice. - Michael Somos, Jun 03 2012

Sixth spoke of hexagonal spiral (cf. A056105-A056109).

Number of ordered triples (a,b,c), -n <= a,b,c <= n, such that a+b+c=0. - Benoit Cloitre, Jun 14 2003

Also the number of partitions of 6n into at most 3 parts, A001399(6n). - R. K. Guy, Oct 20 2003

Also, a(n) is the number of partitions of 6(n+1) into exactly 3 distinct parts. - William J. Keith, Jul 01 2004

Number of dots in a centered hexagonal figure with n+1 dots on each side.

Values of second Bessel polynomial y_2(n) (see A001498).

First differences of cubes (A000578). - Cecilia Rossiter (cecilia(AT)noticingnumbers.net), Dec 15 2004

Final digits of Hex numbers (hex(n) mod 10) are periodic with palindromic period of length 5 {1, 7, 9, 7, 1}. Last two digits of Hex numbers (hex(n) mod 100) are periodic with palindromic period of length 100. - Alexander Adamchuk, Aug 11 2006

All divisors of a(n) are congruent to 1, modulo 6. Proof: If p is an odd prime different from 3 then 3n^2 + 3n + 1 = 0 (mod p) implies 9(2n + 1)^2 = -3 (mod p), whence p = 1 (mod 6). - Nick Hobson, Nov 13 2006

For n>=1, a(n) = side of Outer Napoleon Triangle whose reference triangle is a right triangle with legs (3a(n))^(1/2) and 3n(a(n))^(1/2). - Tom Schicker (tschicke(AT)email.smith.edu), Apr 25 2007

Number of triples (a,b,c) where 0<=(a,b)<=n and c=n (at least once the term n). E.g., for n = 1: (0,0,1),0,1,0),(1,0,0),(0,1,1),(1,0,1),(1,1,0),(1,1,1), then a(1)=7. - Philippe Lallouet (philip.lallouet(AT)wanadoo.fr), Aug 20 2007

Equals the triangular numbers convolved with [1, 4, 1, 0, 0, 0, ...]. - Gary W. Adamson and Alexander R. Povolotsky, May 29 2009

From Terry Stickels, Dec 07 2009: (Start)

Also the maximum number of viewable cubes from any one static point while viewing a cube stack of identical cubes of varying magnitude.

For example, viewing a 2 X 2 X 2 stack will yield 7 maximum viewable cubes.

If the stack is 3 X 3 X 3, the maximum number of viewable cubes from any one static position is 19, and so on.

The number of cubes in the stack must always be the same number for width, length, height (at true regular cubic stack) and the maximum number of visible cubes can always be found by taking any cubic number and subtracting the number of the cube that is one less.

Examples: 125 - 64 = 61, 64 - 27 = 37, 27 - 8 = 19. (End)

The sequence of digital roots of the a(n) is period 3: repeat [1,7,1]. - Ant King, Jun 17 2012

The average of the first n (n>0) centered hexagonal numbers is the n-th square. - Philippe Deléham, Feb 04 2013

A002024 is the following array A read along antidiagonals:

  1,  2,  3,  4,  5,  6, ...

  2,  3,  4,  5,  6,  7, ...

  3,  4,  5,  6,  7,  8, ...

  4,  5,  6,  7,  8,  9, ...

  5,  6,  7,  8,  9, 10, ...

  6,  7,  8,  9, 10, 11, ...

and a(n) is the hook sum Sum_{k=0..n} A(n,k) + Sum_{r=0..n-1} A(r,n). - R. J. Mathar, Jun 30 2013

a(n) = the sum of the terms in the n+1 X n+1 matrices minus those in n X n matrices in an array formed by considering A158405 an array (the beginning terms in each row are 1,3,5,7,9,11,...). - J. M. Bergot, Jul 05 2013

The formula also equals the product of the three distinct combinations of two consecutive numbers: n^2, (n+1)^2, and n*(n+1). - J. M. Bergot, Mar 28 2014

The sides of any triangle ABC are divided into 2n + 1 equal segments by 2n points: A_1, A_2, ..., A_2n in side a, and also on the sides b and c cyclically. If A'B'C' is the triangle delimited by AA_n, BB_n and CC_n cevians, we have (ABC)/(A'B'C') = a(n) (see Java applet link). - Ignacio Larrosa Cañestro, Jan 02 2015

a(n) is the maximal number of parts into which (n+1) triangles can intersect one another. - Ivan N. Ianakiev, Feb 18 2015

((2^m-1)n)^t mod a(n) = ((2^m-1)(n+1))^t mod a(n) = ((2^m-1)(2n+1))^t mod a(n), where m any positive integer, and t = 0(mod 6). - Alzhekeyev Ascar M, Oct 07 2016

((2^m-1)n)^t mod a(n) = ((2^m-1)(n+1))^t mod a(n) = a(n) - (((2^m-1)(2n+1))^t mod a(n)), where m any positive integer, and t = 3(mod 6). - Alzhekeyev Ascar M, Oct 07 2016

(3n+1)^(a(n)-1) mod a(n) = (3n+2)^(a(n)-1) mod a(n) = 1. If a(n) not prime, then always strong pseudoprime. - Alzhekeyev Ascar M, Oct 07 2016

Every positive integer is the sum of 8 Hex number (zero included), at most 3 of which greater than 1. - Mauro Fiorentini, Jan 01 2018

REFERENCES

M. Gardner, Time Travel and Other Mathematical Bewilderments. Freeman, NY, 1988, p. 18.

N. J. A. Sloane and Simon Plouffe, The Encyclopedia of Integer Sequences, Academic Press, 1995 (includes this sequence).

LINKS

T. D. Noe, Table of n, a(n) for n = 0..1000

G. L. Alexanderson, John E. Wetzel, Dissections of a tetrahedron, J. Combinatorial Theory Ser. B 11 (1971), 58--66. MR0303412 (46 #2549). See p. 58.

B. T. Bennett and R. B. Potts, Arrays and brooks, J. Austral. Math. Soc., 7 (1967), 23-31 (see p. 30).

B. T. Bennett and R. B. Potts, Arrays and brooks, J. Austral. Math. Soc., 7 (1967), 23-31. [Annotated scanned copy]

Aran Bingham, Commutative n-ary Arithmetic, University of New Orleans Theses and Dissertations, Paper 1959, 2015.

H. Bottomley, Illustration of initial terms

J. H. Conway and N. J. A. Sloane, Low-Dimensional Lattices VII: Coordination Sequences, Proc. Royal Soc. London, A453 (1997), 2369-2389 (pdf).

M. Gardner & N. J. A. Sloane, Correspondence, 1973-74

R. K. Guy, Letter to N. J. A. Sloane, 1987

R. K. Guy, The strong law of small numbers. Amer. Math. Monthly 95 (1988), no. 8, 697-712.

R. K. Guy, The strong law of small numbers. Amer. Math. Monthly 95 (1988), no. 8, 697-712. [Annotated scanned copy]

G. S. Kazandzidis, On a Conjecture of Moessner and a General Problem, Bull. Soc. Math. Grece, Nouvelle Serie - vol. 2, fasc. 1-2, pp. 23-30. (1961)

Ignacio Larrosa Cañestro, Hexágono y estrella determinados por tres pares de cevianas simétricas, (java applet).

G. Nebe and N. J. A. Sloane, Home page for hexagonal (or triangular) lattice A2

Simon Plouffe, Approximations de séries génératrices et quelques conjectures, Dissertation, Université du Québec à Montréal, 1992.

Simon Plouffe, 1031 Generating Functions and Conjectures, Université du Québec à Montréal, 1992.

B. K. Teo and N. J. A. Sloane, Magic numbers in polygonal and polyhedral clusters, Inorgan. Chem. 24 (1985), 4545-4558.

Eric Weisstein's World of Mathematics, Hex Number

Eric Weisstein's World of Mathematics, Nexus Number

Eric Weisstein's World of Mathematics, Outer Napoleon Triangle.

Index entries for sequences related to centered polygonal numbers

Index entries for crystal ball sequences

Index entries for sequences related to A2 = hexagonal = triangular lattice

Index entries for linear recurrences with constant coefficients, signature (3,-3,1).

FORMULA

a(n) = 3*n*(n+1) + 1, n >= 0 (see the name).

a(n) = (n+1)^3 - n^3 = a(-1-n).

G.f.: (1 + 4*x + x^2) / (1 - x)^3. - Simon Plouffe in his 1992 dissertation

a(n) = 6*A000217(n) + 1.

a(n) = a(n-1) + 6*n = 2a(n-1) - a(n-2) + 6 = 3*a(n-1) - 3*a(n-2) + a(n-3) = A056105(n) + 5n = A056106(n) + 4*n = A056107(n) + 3*n = A056108(n) + 2*n = A056108(n) + n.

n-th partial arithmetic mean is n^2. - Amarnath Murthy, May 27 2003

a(n) = 1 + Sum_{j=0..n} (6*j). E.g., a(2)=19 because 1+ 6*0 + 6*1 + 6*2 = 19. - Xavier Acloque, Oct 06 2003

The sum of the first n hexagonal numbers is n^3. That is, Sum_{n>=1} (3*n*(n-1) + 1) = n^3. - Edward Weed (eweed(AT)gdrs.com), Oct 23 2003

a(n) = right term in M^n * [1 1 1], where M = the 3 X 3 matrix [1 0 0 / 2 1 0 / 3 3 1]. M^n * [1 1 1] = [1 2n+1 a(n)]. E.g., a(4) = 61, right term in M^4 * [1 1 1], since M^4 * [1 1 1] = [1 9 61] = [1 2n+1 a(4)]. - Gary W. Adamson, Dec 22 2004

Row sums of triangle A130298. - Gary W. Adamson, Jun 07 2007

a(n) = 3*n^2 + 3*n + 1. Proof: 1) If n occurs once, it may be in 3 positions; for the two other ones, n terms are independently possible, then we have 3*n^2 different triples. 2) If the term n occurs twice, the third one may be placed in 3 positions and have n possible values, then we have 3*n more different triples. 3) The term n may occurs 3 times in one way only that gives the formula. - Philippe Lallouet (philip.lallouet(AT)wanadoo.fr), Aug 20 2007

Binomial transform of [1, 6, 6, 0, 0, 0, ...]; Narayana transform (A001263) of [1, 6, 0, 0, 0, ...]. - Gary W. Adamson, Dec 29 2007

a(n) = (n-1)*A000166(n) + (n-2)*A000166(n-1) = (n-1)floor(n!*e^(-1)+1) + (n-2)*floor((n-1)!*e^(-1)+1) (with offset 0). - Gary Detlefs, Dec 06 2009

a(n) = A028896(n) + 1. - Omar E. Pol, Oct 03 2011

a(n) = integral( (sin((n+1/2)x)/sin(x/2))^3, x=0..Pi)/Pi. - Yalcin Aktar, Dec 03 2011

Sum_{n>=0} 1/a(n) = Pi/sqrt(3)*tanh(Pi/(2*sqrt(3))) = 1.305284153013581... - Ant King, Jun 17 2012

a(n) = A000290(n) + A000217(2n+1). - Ivan N. Ianakiev, Sep 24 2013

a(n) = A002378(n+1) + A056620(n) = A005408(n) + 2*A005449(n) = 6*A000217(n) + 1. - Ivan N. Ianakiev, Sep 26 2013

a(n) = 6*A000124(n) - 5. - Ivan N. Ianakiev, Oct 13 2013

a(n) = A239426(n+1) / A239449(n+1) = A215630(2*n+1,n+1). - Reinhard Zumkeller, Mar 19 2014

a(n) = A243201(n) / A002061(n + 1). - Mathew Englander, Jun 03 2014

a(n) = A101321(6,n). - R. J. Mathar, Jul 28 2016

E.g.f.: (1 + 6*x + 3*x^2)*exp(x). - Ilya Gutkovskiy, Jul 28 2016

a(n) = (A001844(n) + A016754(n))/2. - Bruce J. Nicholson, Aug 06 2017

a(n) = A045943(2n+1). - Miquel Cerda, Jan 22 2018

EXAMPLE

1 + 7*x + 19*x^2 + 37*x^3 + 61*x^4 + 91*x^5 + 127*x^6 + 169*x^7 + 217*x^8 + ...

From Omar E. Pol, Aug 21 2011: (Start)

Illustration of initial terms:

.

.                                 o o o o

.                   o o o        o o o o o

.         o o      o o o o      o o o o o o

.   o    o o o    o o o o o    o o o o o o o

.         o o      o o o o      o o o o o o

.                   o o o        o o o o o

.                                 o o o o

.

.   1      7          19             37

.

(End)

MAPLE

A003215:=n->3*n*(n+1)+1; seq(A003215(n), n=0..100); # Wesley Ivan Hurt, Mar 28 2014

MATHEMATICA

FoldList[#1 + #2 &, 1, 6 Range@ 50] (* Robert G. Wilson v, Feb 02 2011 *)

LinearRecurrence[{3, -3, 1}, {1, 7, 19}, 47] (* Robert G. Wilson v, Jul 06 2013 *)

PROG

(PARI) {a(n) = 3*n*(n+1) + 1}

(Haskell)

a003215 n = 3 * n * (n + 1) + 1  -- Reinhard Zumkeller, Oct 22 2011

(Maxima) makelist(3*n*(n+1)+1, n, 0, 30); /* Martin Ettl, Nov 12 2012 */

(MAGMA) [3*n*(n+1)+1: n in [0..50]]; // G. C. Greubel, Nov 04 2017

CROSSREFS

Cf. A000124, A000166, A000217, A000290, A000578 (the cubes, or partial sums), A001263, A001498, A002061, A002378, A002407 (primes), A003514, A005408, A005449, A005891, A028896, A048766, A056105, A056106, A056107, A056108, A056109, A063496, A056620, A130298, A132111 (second diagonal), A158405, A215630, A239449, A243201.

Column k=3 of A080853.

See also A220083 for a list of numbers of the form n*P(s,n)-(n-1)*P(s,n-1), where P(s,n) is the n-th polygonal number with s sides.

Cf. A287326(A000124(n), 1).

Sequence in context: A177092 A023224 A113743 * A133323 A002407 A098484

Adjacent sequences:  A003212 A003213 A003214 * A003216 A003217 A003218

KEYWORD

nonn,easy,nice

AUTHOR

N. J. A. Sloane

EXTENSIONS

Partially edited by Joerg Arndt, Mar 11 2010

STATUS

approved

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Last modified June 16 00:19 EDT 2018. Contains 305468 sequences. (Running on oeis4.)