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A003215 Hex (or centered hexagonal) numbers: 3*n*(n+1)+1 (crystal ball sequence for hexagonal lattice).

Original entry on oeis.org

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, 6769
Offset: 0

Views

Author

N. J. A. Sloane

Keywords

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 integer 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) is the 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), so 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) is 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 numbers (zero included), at most 3 of which are greater than 1. - Mauro Fiorentini, Jan 01 2018
Area enclosed by the segment of Archimedean spiral between n*Pi/2 and (n+1)*Pi/2 in Pi^3/48 units. - Carmine Suriano, Apr 10 2018
This sequence contains all numbers k such that 12*k - 3 is a square. - Klaus Purath, Oct 19 2021
The continued fraction expansion of sqrt(3*a(n)) is [3n+1; {1, 1, 2n, 1, 1, 6n+2}]. For n = 0, this collapses to [1; {1, 2}]. - Magus K. Chu, Sep 12 2022

Examples

			G.f. = 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)
From _Klaus Purath_, Dec 03 2021: (Start)
(1) a(19) is not a prime number, because besides a(19) = a(9) + P(29), a(19) = a(15) + P(20) = a(2) + P(33) is also true.
(2) a(25) is prime, because except for a(25) = a(12) + P(38) there is no other equation of this pattern. (End)

References

  • John H. Conway and Richard K. Guy, The Book of Numbers, New York: Springer-Verlag, 1996. See p. 81.
  • 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

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, A056220, A130298, A132111 (second diagonal), A158405, A215630, A239449, A243201.
Column k=3 of A080853, and column k=2 of A047969.
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).
Cf. A008292.
Cf. A005448, A140091, A001844, A000384, A060544, A045943.
Cf. A154105.

Programs

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) + A056220(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
a(n) = 3*Integral_{x=n..n+1} x^2 dx. - Carmine Suriano, Apr 10 2018
a(n) = A287326(A000124(n), 1). - Kolosov Petro, Oct 22 2018
From Amiram Eldar, Jun 20 2020: (Start)
Sum_{n>=0} a(n)/n! = 10*e.
Sum_{n>=0} (-1)^(n+1)*a(n)/n! = 2/e. (End)
G.f.: polylog(-3, x)*(1-x)/x. See the Simon Plouffe formula above, and the g.f. of the rows of A008292 by Vladeta Jovovic, Sep 02 2002. - Wolfdieter Lang, May 08 2021
a(n) = T(n-1)^2 - 2*T(n)^2 + T(n+1)^2, n >= 1, T = triangular number A000217. - Klaus Purath, Oct 11 2021
a(n) = 1 + 2*Sum_{j=n..2n} j. - Klaus Purath, Oct 19 2021
a(n) = A069099(n+1) - A000217(n). - Klaus Purath, Nov 03 2021
From Leo Tavares, Dec 03 2021: (Start)
a(n) = A005448(n) + A140091(n);
a(n) = A001844(n) + A002378(n);
a(n) = A005891(n) + A000217(n);
a(n) = A000290(n) + A000384(n+1);
a(n) = A060544(n-1) + 3*A000217(n);
a(n) = A060544(n-1) + A045943(n).
a(2*n+1) = A154105(n).
(End)

Extensions

Partially edited by Joerg Arndt, Mar 11 2010

Started in 1964 by Neil J. A. Sloane | Maintained by The OEIS Foundation Inc.

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