A003215 -id:A003215 - cp's OEIS frontend

A056108 Fourth spoke of a hexagonal spiral.

1, 5, 15, 31, 53, 81, 115, 155, 201, 253, 311, 375, 445, 521, 603, 691, 785, 885, 991, 1103, 1221, 1345, 1475, 1611, 1753, 1901, 2055, 2215, 2381, 2553, 2731, 2915, 3105, 3301, 3503, 3711, 3925, 4145, 4371, 4603, 4841, 5085, 5335, 5591, 5853, 6121, 6395

Offset: 0

Henry Bottomley, Jun 09 2000

a(n) = sum of (n+1)-th row terms of triangle A134234. - Gary W. Adamson, Oct 14 2007
If Y is a 4-subset of an n-set X then, for n >= 4, a(n-4) is the number of 4-subsets of X having at least two elements in common with Y. - Milan Janjic, Dec 08 2007
Equals binomial transform of [1, 4, 6, 0, 0, 0, ...] - Gary W. Adamson, Apr 30 2008
From A.K. Devaraj, Sep 18 2009: (Start)
Let f(x) be a polynomial in x. Then f(x + n*f(x)) is congruent to 0 (mod f(x)); here n belongs to N.
There is nothing interesting in the quotients f(x + n*f(x))/f(x) when x belongs to Z.
However, when x is irrational these quotients consist of two parts, a) rational integers and b) integer multiples of x.
The present sequence is the integer part when the polynomial is x^2 + x + 1 and x = sqrt(2),
f(x+n*f(x))/f(x) = a(n) + A005563(n)*sqrt(2).
Equals triangle A128229 as an infinite lower triangular matrix * A016777 as a vector, where A016777(n) = (3*n+1). (End)
Numbers of the form ((-h^2+h+1)^2+(h^2-h+1)^2+(h^2+h-1)^2)/(h^2+h+1) for h=n+1. - Bruno Berselli, Mar 13 2013

G. C. Greubel, Table of n, a(n) for n = 0..5000
Henry Bottomley, Illustration of initial terms
G. Nebe and N. J. A. Sloane, Home page for hexagonal (or triangular) lattice A2
Luis Manuel Rivera, Integer sequences and k-commuting permutations, arXiv preprint arXiv:1406.3081 [math.CO], 2014-2015.
Index entries for linear recurrences with constant coefficients, signature (3,-3,1).

Cf. A134234, A000217.
Cf. A005563, A016777, A128229.
Other spokes: A003215, A056105, A056106, A056107, A056109.
Other spirals: A054552.

[3*n^2+n+1: n in [0..50]]; // Bruno Berselli, Mar 13 2013

Table[3 n^2 + n + 1, {n, 0, 50}] (* Bruno Berselli, Mar 13 2013 *)
LinearRecurrence[{3,-3,1},{1,5,15},50] (* Harvey P. Dale, Dec 26 2023 *)

a(n)=3*n^2+n+1 \\ Charles R Greathouse IV, Jun 17 2017

a(n) = 3*n^2 + n + 1.
a(n) = a(n-1) + 6*n - 2 = 2*a(n-1) - a(n-2) + 6
a(n) = 3*a(n-1) - 3*a(n-2) + a(n-3).
a(n) = A056105(n) + 3*n = A056106(n) + 2*n = A056107(n) + n = A056109(n) - n = A003215(n) - 2*n.
a(n) = A096777(3n+1) . - Reinhard Zumkeller, Dec 29 2007
a(n) = 6*n+a(n-1)-2 with n>0, a(0)=1. - Vincenzo Librandi, Aug 07 2010
G.f.: (1+2*x+3*x^2)/(1-3*x+3*x^2-x^3). - Colin Barker, Jan 04 2012
a(-n) = A056106(n). - Bruno Berselli, Mar 13 2013
E.g.f.: (3*x^2 + 4*x + 1)*exp(x). - G. C. Greubel, Jul 19 2017

A069129 Centered 16-gonal numbers.

Original entry on oeis.org

1, 17, 49, 97, 161, 241, 337, 449, 577, 721, 881, 1057, 1249, 1457, 1681, 1921, 2177, 2449, 2737, 3041, 3361, 3697, 4049, 4417, 4801, 5201, 5617, 6049, 6497, 6961, 7441, 7937, 8449, 8977, 9521, 10081, 10657, 11249, 11857, 12481, 13121, 13777, 14449, 15137, 15841

Offset: 1

Terrel Trotter, Jr., Apr 07 2002

Also, sequence found by reading the line from 1, in the direction 1, 17, ..., in the square spiral whose vertices are the triangular numbers A000217. Opposite numbers to the members of A139098 in the same spiral. - Omar E. Pol, Apr 26 2008
The subsequence of primes begins: 17, 97, 241, 337, 449, 577, 881, 1249, 3041, 3361, 3697, 4049, 4801, 6961, 7937, 9521,  10657, 13121, 14449. See A184899: n such that the n-th centered 12-gonal number is prime. Indices of prime star numbers. - Jonathan Vos Post, Feb 27 2011
Binomial transform of [1, 16, 16, 0, 0, 0, ...] and Narayana transform (A001263) of [1, 16, 0, 0, 0, ...]. - Gary W. Adamson, Jul 28 2011
Centered hexadecagonal numbers or centered hexakaidecagonal numbers. - Omar E. Pol, Oct 03 2011
a(n) = m(n,n) for an array constructed by using the terms in A016813 as the antidiagonals; the first few antidiagonals are 1; 5,9; 13,17,21; 25,29,33,37. - J. M. Bergot, Jul 05 2013
[The first five rows begin: 1,9,21,37,57; 5,17,33,53,77; 13,29,49,73,101; 25,45,69,97,129; 41,65,93,125,161.]

			a(5) = 161 because 8*5^2 - 8*5 + 1 = 200 - 40 + 1 = 161.

Vincenzo Librandi, Table of n, a(n) for n = 1..1000
Omar E. Pol, Determinacion geometrica de los numeros primos y perfectos.
Leo Tavares, Illustration: Octagonal Stars.
Eric Weisstein's World of Mathematics, Centered Polygonal Numbers.
R. Yin, J. Mu, and T. Komatsu, The p-Frobenius Number for the Triple of the Generalized Star Numbers, Preprints 2024, 2024072280. See p. 2.
Index entries for sequences related to centered polygonal numbers.
Index entries for linear recurrences with constant coefficients, signature (3,-3,1).

Cf. A005448, A001844, A005891, A003215, A069099, A000217, A035008, A139098.

Bisection of A077221.

[8*n^2-8*n+1: n in [0..50]]; // Vincenzo Librandi, Feb 05 2013

FoldList[#1 + #2 &, 1, 16 Range@ 45] (* Robert G. Wilson v, Feb 02 2011 *)
Rest[CoefficientList[Series[-x(1+14x+x^2)/(x-1)^3,{x,0,50}],x]]  (* Harvey P. Dale, Apr 22 2011 *)

a(n)=8*n^2-8*n+1 \\ Charles R Greathouse IV, Sep 24 2015

a(n) = 8*n^2 - 8*n + 1.
a(n) = A035008(n-1) + 1. - Omar E. Pol, Apr 26 2008
a(n) = 16*n + a(n-1) - 16 with n > 1, a(1)=1. - Vincenzo Librandi, Aug 08 2010
G.f.: -x*(1+14*x+x^2) / (x-1)^3. - R. J. Mathar, Feb 04 2011
E.g.f.: (8*x^2 + 1)*exp(x). - G. C. Greubel, Jul 18 2017
a(n) = A056220(2n-1). - Bruce J. Nicholson, Aug 31 2017
Sum_{n>=1} 1/a(n) = Pi * tan(Pi/(2*sqrt(2))) / (4*sqrt(2)). - Vaclav Kotesovec, Jul 23 2019
From Amiram Eldar, Jun 21 2020: (Start)
Sum_{n>=1} a(n)/n! = 9*e - 1.
Sum_{n>=1} (-1)^n * a(n)/n! = 9/e - 1. (End)
Product_{n>=2} (a(n) - 1) / (a(n) + 1) = Pi/4. - Dimitris Valianatos, Jun 27 2020
a(n) = A016754(n-1) + 8*A000217(n-1). - Leo Tavares, Jul 19 2021

A002407 Cuban primes: primes which are the difference of two consecutive cubes.

Original entry on oeis.org

7, 19, 37, 61, 127, 271, 331, 397, 547, 631, 919, 1657, 1801, 1951, 2269, 2437, 2791, 3169, 3571, 4219, 4447, 5167, 5419, 6211, 7057, 7351, 8269, 9241, 10267, 11719, 12097, 13267, 13669, 16651, 19441, 19927, 22447, 23497, 24571, 25117, 26227, 27361, 33391

Offset: 1

N. J. A. Sloane

Primes of the form p = (x^3 - y^3)/(x - y) where x=y+1. See A007645 for generalization. I first saw the name "cuban prime" in Cunningham (1923). Values of x are in A002504 and y are in A111251. - N. J. A. Sloane, Jan 29 2013
Prime hex numbers (cf. A003215).
Equivalently, primes of the form p=1+3k(k+1) (and then k=floor(sqrt(p/3))). Also: primes p such that n^2(p+n) is a cube for some n>0. - M. F. Hasler, Nov 28 2007
Primes p such that 4p = 1+3s^2 for some integer s (A121259). - Michael Somos, Sep 15 2005
This sequence is believed to be infinite. - N. J. A. Sloane, May 07 2020

			a(1) = 7 = 1+3k(k+1) (with k=1) is the smallest prime of this form.
a(10^5) = 1792617147127 since this is the 100000th prime of this form.

Allan Joseph Champneys Cunningham, On quasi-Mersennian numbers, Mess. Math., 41 (1912), 119-146.
Allan Joseph Champneys Cunningham, Binomial Factorisations, Vols. 1-9, Hodgson, London, 1923-1929; see Vol. 1, pp. 245-259.
J.-M. De Koninck & A. Mercier, 1001 Problèmes en Théorie Classique des Nombres, Problem 241 pp. 39; 179, Ellipses Paris 2004.
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).

David A. Corneth, Table of n, a(n) for n = 1..10000 (first 1000 terms from T. D. Noe)
A. J. C. Cunningham, On quasi-Mersennian numbers, Mess. Math., 41 (1912), 119-146. [Annotated scan of page 144 only]
A. J. C. Cunningham, Binomial Factorisations, Vols. 1-9, Hodgson, London, 1923-1929. [Annotated scans of a few pages from Volumes 1 and 2]
R. K. Guy, Letter to N. J. A. Sloane, 1987.
G. L. Honaker, Jr., Prime curio for 127.
Michael Penn, Nearly cubic primes., YouTube video, 2023.
Project Euler, Problem 131: Prime cube partnership.
Eric Weisstein's World of Mathematics, Cuban Prime
Wikipedia, Cuban prime.

Cf. A000217, A002504, A003215, A111251, A113478, A145203.

Cf. A002648 (with x=y+2), A003627, A007645, A201477, A334520.

[a: n in [0..100] | IsPrime(a) where a is (3*n^2+3*n+1)]; // Vincenzo Librandi, Jan 20 2020

lst={}; Do[If[PrimeQ[p=(n+1)^3-n^3], AppendTo[lst, p]], {n, 10^2}]; lst (* Vladimir Joseph Stephan Orlovsky, Aug 21 2008 *)
Select[Table[3x^2+3x+1,{x,100}],PrimeQ] (* or *) Select[Last[#]- First[#]&/@ Partition[Range[100]^3,2,1],PrimeQ] (* Harvey P. Dale, Mar 10 2012 *)
Select[Differences[Range[100]^3],PrimeQ] (* Harvey P. Dale, Jan 19 2020 *)

{a(n)= local(m, c); if(n<1, 0, c=0; m=1; while( cMichael Somos, Sep 15 2005 */

A002407(n,k=1)=until(isprime(3*k*k+++1) && !n--,);3*k*k--+1
list_A2407(Nmax)=for(k=1,sqrt(Nmax/3),isprime(t=3*k*(k+1)+1) && print1(t",")) \\ M. F. Hasler, Nov 28 2007

from sympy import isprime
def aupto(limit):
    alst, k, d = [], 1, 7
    while d <= limit:
        if isprime(d): alst.append(d)
        k += 1; d = 1+3*k*(k+1)
    return alst
print(aupto(34000)) # Michael S. Branicky, Jul 19 2021

a(n) = 6*A000217(A111251(n)) + 1. - Christopher Hohl, Jul 01 2019
From Rémi Guillaume, Nov 07 2023: (Start)
a(n) = A003215(A111251(n)).
a(n) = (3*(2*A002504(n) - 1)^2 + 1)/4.
a(n) = (3*A121259(n)^2 + 1)/4.
a(n) = prime(A145203(n)). (End)

More terms from James Sellers, Aug 08 2000

Entry revised by N. J. A. Sloane, Jan 29 2013

A147875 Second heptagonal numbers: a(n) = n(5n+3)/2.

Original entry on oeis.org

0, 4, 13, 27, 46, 70, 99, 133, 172, 216, 265, 319, 378, 442, 511, 585, 664, 748, 837, 931, 1030, 1134, 1243, 1357, 1476, 1600, 1729, 1863, 2002, 2146, 2295, 2449, 2608, 2772, 2941, 3115, 3294, 3478, 3667, 3861, 4060, 4264, 4473, 4687, 4906, 5130, 5359, 5593

Offset: 0

Vladimir Joseph Stephan Orlovsky, Nov 16 2008

Zero followed by partial sums of A016897.
Apparently = every 2nd term of A111710 and A085787.
Bisection of A085787. Sequence found by reading the line from 0, in the direction 0, 13, ... and the line from 4, in the direction 4, 27, ..., in the square spiral whose vertices are the generalized heptagonal numbers A085787. - Omar E. Pol, Jul 18 2012
Numbers of the form m^2 + k*m*(m+1)/2: in this case is k=3. See also A254963. - Bruno Berselli, Feb 11 2015

			G.f. = 4*x + 13*x^2 + 27*x^3 + 46*x^4 + 70*x^5 + 99*x^6 + 133*x^7 + ... - _Michael Somos_, Jan 25 2019

Harvey P. Dale, Table of n, a(n) for n = 0..1000
Leo Tavares, Illustration: Sliced Hexagons
Index entries for linear recurrences with constant coefficients, signature (3,-3,1).

Cf. A016897, A111710, A000217, A085787, A224419 (positions of squares).
Second n-gonal numbers: A005449, A014105, A045944, A179986, A033954, A062728, A135705.
Cf. A000566.
Cf. A003215, A000096, A000290.

List([0..50], n-> n*(5*n+3)/2); # G. C. Greubel, Jul 04 2019

[n*(5*n+3)/2: n in [0..50]]; // G. C. Greubel, Jul 04 2019

Table[(n(5n+3))/2, {n, 0, 50}] (* or *) LinearRecurrence[{3, -3, 1}, {0, 4, 13}, 50] (* Harvey P. Dale, May 15 2013 *)

a(n)=n*(5*n+3)/2 \\ Charles R Greathouse IV, Sep 24 2015

[n*(5*n+3)/2 for n in (0..50)] # G. C. Greubel, Jul 04 2019

G.f.: x*(4+x)/(1-x)^3.
a(n) = Sum_{k=0..n-1} A016897(k).
a(n) - a(n-1) = 5*n -1. - Vincenzo Librandi, Nov 26 2010
G.f.: U(0) where U(k) = 1 + 2*(2*k+3)/(k + 2 - x*(k+2)^2*(k+3)/(x*(k+2)*(k+3) + (2*k+2)*(2*k+3)/U(k+1))); (continued fraction, 3-step). - Sergei N. Gladkovskii, Nov 14 2012
E.g.f.: U(0) where U(k) = 1 + 2*(2*k+3)/(k + 2 - 2*x*(k+2)^2*(k+3)/(2*x*(k+2)*(k+3) + (2*k+2)^2*(2*k+3)/U(k+1))); (continued fraction, 3rd kind, 3-step). - Sergei N. Gladkovskii, Nov 14 2012
a(n) = A130520(5n+3). - Philippe Deléham, Mar 26 2013
a(n) = A131242(10n+7)/2. - Philippe Deléham, Mar 27 2013
a(n) = 3*a(n-1) - 3*a(n-2) + a(n-3); a(0)=0, a(1)=4, a(2)=13. - Harvey P. Dale, May 15 2013
Sum_{n>=1} 1/a(n) = 10/9 + sqrt(1 - 2/sqrt(5))*Pi/3 - 5*log(5)/6 + sqrt(5)*log((1 + sqrt(5))/2)/3 = 0.4688420784500060750083432... . - Vaclav Kotesovec, Apr 27 2016
a(n) = A000217(n) + A000217(2*n). - Bruno Berselli, Jul 01 2016
From Ilya Gutkovskiy, Jul 01 2016: (Start)
E.g.f.: x*(8 + 5*x)*exp(x)/2.
Dirichlet g.f.: (5*zeta(s-2) + 3*zeta(s-1))/2. (End)
a(n) = A000566(-n) for all n in Z. - Michael Somos, Jan 25 2019
From Leo Tavares, Feb 14 2022: (Start)
a(n) = A003215(n) - A000217(n+1). See Sliced Hexagons illustration in links.
a(n) = A000096(n) + 2*A000290(n). (End)

Edited by Klaus Brockhaus and R. J. Mathar, Nov 20 2008

New name from Bruno Berselli, Jan 13 2011

A049451 Twice second pentagonal numbers.

Original entry on oeis.org

0, 4, 14, 30, 52, 80, 114, 154, 200, 252, 310, 374, 444, 520, 602, 690, 784, 884, 990, 1102, 1220, 1344, 1474, 1610, 1752, 1900, 2054, 2214, 2380, 2552, 2730, 2914, 3104, 3300, 3502, 3710, 3924, 4144, 4370, 4602, 4840, 5084, 5334, 5590, 5852, 6120, 6394, 6674, 6960, 7252, 7550, 7854

Offset: 0

Joe Keane (jgk(AT)jgk.org)

From Floor van Lamoen, Jul 21 2001: (Start)
Write 1,2,3,4,... in a hexagonal spiral around 0, then a(n) is the sequence found by reading the line from 0 in the direction 0,4,... . The spiral begins:
                            .
                          52
                          . \
            33--32--31--30  51
            /           . \   \
          34  16--15--14  29  50
          /   /       . \   \   \
        35  17   5---4  13  28  49
        /   /   /   . \   \   \   \
      36  18   6   0   3  12  27  48
      /   /   /   /   /   /   /   /
    37  19   7   1---2  11  26  47
      \   \   \         /   /   /
      38  20   8---9--10  25  46
        \   \             /   /
        39  21--22--23--24  45
          \                 /
          40--41--42--43--44
(End)
Number of edges in the join of the complete bipartite graph of order 2n and the cycle graph of order n, K_n,n * C_n. - Roberto E. Martinez II, Jan 07 2002
The average of the first n elements starting from a(1) is equal to (n+1)^2. - Mario Catalani (mario.catalani(AT)unito.it), Apr 10 2003
If Y is a 4-subset of an n-set X then, for n >= 4, a(n-4) is the number of (n-4)-subsets of X having either one element or two elements in common with Y. - Milan Janjic, Dec 28 2007
With offset 1: the maximum possible sum of numbers in an N x N standard Minesweeper grid. - Dmitry Kamenetsky, Dec 14 2008
a(n) = A001399(6*n-2), number of partitions of 6*n-2 into parts < 4. For example a(2)=14 where the partitions of 6*2-2=10 into parts < 4 are [1,1,1,1,1,1,1,1,1,1], [1,1,1,1,1,1,1,1,2], [1,1,1,1,1,1,1,3], [1,1,1,1,1,1,2,2], [1,1,1,1,1,2,3], [1,1,1,1,2,2,2], [1,1,1,1,3,3], [1,1,1,2,2,3], [1,1,2,2,2,2], [1,1,2,3,3], [1,2,2,2,3], [2,2,2,2,2], [1,3,3,3], [2,2,3,3]. - Adi Dani, Jun 07 2011
A003056 is the following array A read by antidiagonals:
  0,  1,  2,  3,  4,  5, ...
  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, ...
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)*Pi is the total length of 3 points circle center spiral after n rotations. The spiral length at each rotation (L(n)) is A016957. The spiral length ratio rounded down [floor(L(n)/L(1))] is A001651. See illustration in links. - Kival Ngaokrajang, Dec 27 2013
Partial sums give A114364. - Leo Tavares, Feb 25 2022
For n >= 1, the continued fraction expansion of sqrt(27*a(n)) is [9n+1; {2, 2n-1, 1, 4, 1, 2n-1, 2, 18n+2}]. - Magus K. Chu, Oct 13 2022

			From _Dmitry Kamenetsky_, Dec 14 2008, with slight rewording by Raymond Martineau (mart0258(AT)yahoo.com), Dec 16 2008: (Start)
For an N x N Minesweeper grid the highest sum of numbers is (N-1)(3*N-2). This is achieved by filling every second row with mines (shown as 'X'). For example, when N=5 the best grids are:
.
  X X X X X
  4 6 6 6 4
  X X X X X
  4 6 6 6 4
  X X X X X
.
  and
.
  2 3 3 3 2
  X X X X X
  4 6 6 6 4
  X X X X X
  2 3 3 3 2
.
each giving a total of 52. (End)

L. B. W. Jolley, Summation of Series, Dover Publications, 1961, p. 12.

Ivan Panchenko, Table of n, a(n) for n = 0..1000
Raghavendra N. Bhat, Cristian Cobeli, and Alexandru Zaharescu, A lozenge triangulation of the plane with integers, arXiv:2403.10500 [math.NT], 2024.
Kival Ngaokrajang, Illustration of 3 points circle center spiral.
Leo Tavares, Illustration: Double Hexagonal Trapezoids.
Index entries for linear recurrences with constant coefficients, signature (3,-3,1).

Cf. A000567, A001105, A002378, A005449, A033580, A049450.
Similar sequences are listed in A316466.
Cf. A003215, A005408, A114364.

List([0..55], n-> n*(3*n+1)); # G. C. Greubel, Sep 01 2019

a049451 n = n * (3 * n + 1)  -- Reinhard Zumkeller, Jul 07 2012

[n*(3*n+1): n in [0..55]]; // G. C. Greubel, Sep 01 2019

Table[n(3n+1), {n,0,55}] (* or *) CoefficientList[Series[2x(2+x)/(1-x)^3, {x,0,55}], x] (* Michael De Vlieger, Apr 05 2017 *)

a(n)=n*(3*n+1) \\ Charles R Greathouse IV, Sep 24 2015

[n*(3*n+1) for n in range(60)] # Gennady Eremin, Feb 27 2022

[n*(3*n+1) for n in (0..55)] # G. C. Greubel, Sep 01 2019

a(n) = n*(3*n+1).
G.f.: 2*x*(2+x)/(1-x)^3.
Sum_{i=1..n} a(i) = A045991(n+1). - Gary W. Adamson, Dec 20 2006
a(n) = 2*A005449(n). - Omar E. Pol, Dec 18 2008
a(n) = a(n-1) + 6*n -2, n > 0. - Vincenzo Librandi, Aug 06 2010
a(n) = A100104(n+1) - A100104(n). - Reinhard Zumkeller, Jul 07 2012
a(n) = 3*a(n-1) - 3*a(n-2) + a(n-3) with a(0) = 0, a(1) = 4, a(2) = 14. - Philippe Deléham, Mar 26 2013
a(n) = A174709(6*n+3). - Philippe Deléham, Mar 26 2013
a(n) = (24/(n+2)!)*Sum_{j=0..n} (-1)^(n-j)*binomial(n,j)*j^(n+2). - Bruno Berselli, Jun 04 2013 - after the similar formula of Vladimir Kruchinin in A002411
a(n) = A002061(n+1) + A056220(n). - Bruce J. Nicholson, Sep 21 2017
a(n) = Sum_{i = 2..5} P(i,n), where P(i,m) = m*((i-2)*m-(i-4))/2. - Bruno Berselli, Jul 04 2018
E.g.f.: x*(4 + 3*x)*exp(x). - G. C. Greubel, Sep 01 2019
a(n) = A003215(n) - A005408(n). - Leo Tavares, Feb 25 2022
From Amiram Eldar, Feb 27 2022: (Start)
Sum_{n>=1} 1/a(n) = 3 - Pi/(2*sqrt(3)) - 3*log(3)/2.
Sum_{n>=1} (-1)^(n+1)/a(n) = Pi/sqrt(3) + 2*log(2) - 3. (End)
a(n) = A001105(n) + A002378(n). - Torlach Rush, Jul 11 2022

A056106 Second spoke of a hexagonal spiral.

Original entry on oeis.org

1, 3, 11, 25, 45, 71, 103, 141, 185, 235, 291, 353, 421, 495, 575, 661, 753, 851, 955, 1065, 1181, 1303, 1431, 1565, 1705, 1851, 2003, 2161, 2325, 2495, 2671, 2853, 3041, 3235, 3435, 3641, 3853, 4071, 4295, 4525, 4761, 5003, 5251, 5505, 5765, 6031, 6303

Offset: 0

Henry Bottomley, Jun 09 2000

First differences of A027444. - J. M. Bergot, Jun 04 2012
Numbers of the form ((h^2+h+1)^2+(-h^2+h+1)^2+(h^2+h-1)^2)/(h^2-h+1) for h=n-1. - Bruno Berselli, Mar 13 2013
For n > 0: 2*a(n) = A058331(n) + A001105(n) + A001844(n-1) = A251599(3*n-2) + A251599(3*n-1) + A251599(3*n). - Reinhard Zumkeller, Dec 13 2014
For all n >= 6, a(n+1) expressed in base n is "353". - Mathew Englander, Jan 06 2021

Vincenzo Librandi, Table of n, a(n) for n = 0..10000
Henry Bottomley, Spokes of a hexagonal spiral (illustration of initial terms).
G. Nebe and N. J. A. Sloane, Home page for hexagonal (or triangular) lattice A2.
Index entries for linear recurrences with constant coefficients, signature (3,-3,1).

First differences of A053698, A027444, and A188947.
Cf. A113524 (semiprime terms), A002061.
Cf. A001105, A001844, A058331, A251599.
Other spokes: A003215, A056105, A056107, A056108, A056109.
Other spirals: A054552.

a056106 n = n * (3 * n - 1) + 1  -- Reinhard Zumkeller, Dec 13 2014

I:=[1,3]; [n le 2 select I[n] else 2*Self(n-1)-Self(n-2)+6: n in [1..50]]; // Vincenzo Librandi, Nov 14 2011

Table[3*n^2 - n + 1, {n,0,50}] (* G. C. Greubel, Jul 19 2017 *)

PARI
```
a(n) = 3*n^2-n+1;
```

a(n) = 3*n^2 - n + 1.
a(n) = a(n-1) + 6*n - 4 = 2*a(n-1) - a(n-2) + 6.
a(n) = 3*a(n-1) - 3*a(n-2) + a(n-3).
E.g.f.: (1+2*x+3*x^2)*exp(x). - Paul Barry, Mar 13 2003
a(n) = A096777(3*n) for n>0. - Reinhard Zumkeller, Dec 29 2007
G.f.: (1+5*x^2)/(1-3*x+3*x^2-x^3). - Colin Barker, Jan 04 2012
a(n) = n*A002061(n+1) - (n-1)*A002061(n). - Bruno Berselli, Jan 15 2013
a(-n) = A056108(n). - Bruno Berselli, Mar 13 2013

A014209 a(n) = n^2 + 3*n - 1.

Original entry on oeis.org

-1, 3, 9, 17, 27, 39, 53, 69, 87, 107, 129, 153, 179, 207, 237, 269, 303, 339, 377, 417, 459, 503, 549, 597, 647, 699, 753, 809, 867, 927, 989, 1053, 1119, 1187, 1257, 1329, 1403, 1479, 1557, 1637, 1719, 1803, 1889, 1977, 2067, 2159, 2253, 2349, 2447, 2547, 2649

Offset: 0

N. J. A. Sloane

Difference between n-th centered hexagonal number and (2*n)^2. - Alonso del Arte, Jul 06 2004
Given the roots to n^2 + 3*n - 1, a = -3.302775..., b = 0.302775...; then a(n) = (n + 3 + a)*(n + 3 + b). Example: a(3) = 17 = (6 - 3.302...)*(6 + 0.302775). - Gary W. Adamson, Jul 29 2009
a(n-1) = n*(n+1) - 3, with a(-1) = -3, gives the values for a*c of indefinite binary quadratic forms [a, b, c] of discriminant D = 13 for b = 2*n + 1. In general D = b^2 - 4*a*c > 0 and the form [a, b, c] is a*x^2 + b*x*y + c*y^2. - Wolfdieter Lang, Aug 15 2013
Numbers m >= -1 such that 4*m + 13 is a square. - Bruce J. Nicholson, Jul 17 2017

Vincenzo Librandi, Table of n, a(n) for n = 0..1000
Wikipedia, Centered Hexagonal Numbers.
Index entries for linear recurrences with constant coefficients, signature (3,-3,1).

Cf. A002522, A003215, A176271.

Table[n^2+3*n-1, {n, 0, 100}] (* Vladimir Joseph Stephan Orlovsky, Oct 08 2009 *)
CoefficientList[Series[(- 1 + 6 x - 3 x^2)/(1 - x)^3, {x, 0, 60}], x] (* Vincenzo Librandi, Oct 15 2013 *)

a(n)=n^2+3*n-1 \\ Charles R Greathouse IV, Sep 24 2015

For n > 0: a(n) = A176271(n+1,n). - Reinhard Zumkeller, Apr 13 2010
a(n) = a(n-1) + 2*n + 2, with n > 0, a(0)=-1. - Vincenzo Librandi, Nov 20 2010
From Colin Barker, Feb 12 2012: (Start)
a(n) = 3*a(n-1) - 3*a(n-2) + a(n-3).
G.f.: (-1+6*x-3*x^2)/(1-x)^3. (End)
Sum_{n>=0} 1/a(n) = 1/3 + tan(sqrt(13)*Pi/2)*Pi/sqrt(13). - Amiram Eldar, Jan 08 2023
E.g.f.: exp(x)*(-1 + 4*x + x^2). - Elmo R. Oliveira, Oct 31 2024

A022521 a(n) = (n+1)^5 - n^5.

Original entry on oeis.org

1, 31, 211, 781, 2101, 4651, 9031, 15961, 26281, 40951, 61051, 87781, 122461, 166531, 221551, 289201, 371281, 469711, 586531, 723901, 884101, 1069531, 1282711, 1526281, 1803001, 2115751, 2467531

Offset: 0

N. J. A. Sloane

Last digit of a(n) is always 1. Last two digits of a(n) (i.e., a(n) mod 100) are repeated periodically with palindromic part of period 20 {1,31,11,81,1,51,31,61,81,51,51,81,61,31,51,1,81,11,31,1}. Last three digits of a(n) (i.e., a(n) mod 1000) are repeated periodically with palindromic part of period 200. - Alexander Adamchuk, Aug 11 2006
In Conway and Guy, these numbers are called nexus numbers of order 5. - M. F. Hasler, Jan 27 2013
Numbers that can be arranged in a triangular-antitegmatic icosachoron (the 4D version of "rhombic dodecahedal numbers" (A005917)). - Steven Lu, Mar 28 2023

John H. Conway and Richard K. Guy, The Book of Numbers, Copernicus Press, NY, 1996, p. 54.

Vincenzo Librandi, Table of n, a(n) for n = 0..10000
Polytope Wiki, Triangular-antitegmatic Icosachoron
Index entries for linear recurrences with constant coefficients, signature (5,-10,10,-5,1).

First differences of A000584.
Column k=4 of array A047969.
Cf. A003215, A005917, A006322, A008292, A019916, A019952, A022522.

[(n+1)^5-n^5: n in [0..30]]; // Vincenzo Librandi, Nov 23 2011

Table[(n+1)^5-n^5, {n,0,30}] (* Vincenzo Librandi, Nov 23 2011 *)

a(n)=(n+1)^5-n^5 \\ Charles R Greathouse IV, Sep 24 2015

a(n) = A003215(n) + 24 * A006322(n). - Xavier Acloque, Oct 11 2003
G.f.: (-1-x^4-26*x^3-66*x^2-26*x)/(x-1)^5. - Maksym Voznyy (voznyy(AT)mail.ru), Aug 11 2009
G.f.: polylog(-5, x)*(1-x)/x. See the g.f. of the rows of A008292 by Vladeta Jovovic, Sep 02 2002. - Wolfdieter Lang, May 10 2021
Sum_{n>=0} 1/a(n) = c1*tanh(c2/2) - c2*tanh(c1/2), where c1 = tan(3*Pi/10)*Pi and c2 = tan(Pi/10)*Pi. - Amiram Eldar, Jan 27 2022

A101321 Table T(n,m) = 1 + nm(m+1)/2 read by antidiagonals: centered polygonal numbers.

Original entry on oeis.org

1, 1, 1, 1, 2, 1, 1, 4, 3, 1, 1, 7, 7, 4, 1, 1, 11, 13, 10, 5, 1, 1, 16, 21, 19, 13, 6, 1, 1, 22, 31, 31, 25, 16, 7, 1, 1, 29, 43, 46, 41, 31, 19, 8, 1, 1, 37, 57, 64, 61, 51, 37, 22, 9, 1, 1, 46, 73, 85, 85, 76, 61, 43, 25, 10, 1, 1, 56, 91, 109, 113, 106, 91, 71, 49, 28, 11, 1, 1, 67

Offset: 0

Eugene McDonnell (eemcd(AT)mac.com), Dec 24 2004

Row n gives the centered figurate numbers of the n-gon.

Antidiagonal sums are in A101338.

			The upper left corner of the infinite array T is
|0| 1   1   1   1   1   1   1   1   1   1 ... A000012
|1| 1   2   4   7  11  16  22  29  37  46 ... A000124
|2| 1   3   7  13  21  31  43  57  73  91 ... A002061
|3| 1   4  10  19  31  46  64  85 109 136 ... A005448
|4| 1   5  13  25  41  61  85 113 145 181 ... A001844
|5| 1   6  16  31  51  76 106 141 181 226 ... A005891
|6| 1   7  19  37  61  91 127 169 217 271 ... A003215
|7| 1   8  22  43  71 106 148 197 253 316 ... A069099
|8| 1   9  25  49  81 121 169 225 289 361 ... A016754
|9| 1  10  28  55  91 136 190 253 325 406 ... A060544

Cf. A000217, A001263, A101338.

A101321 := proc(n,k)
        n*k*(k+1)/2+1 ;
end proc: # R. J. Mathar, Jul 28 2016

T[n_, m_] := 1 + n m (m + 1)/2;
Table[T[n - m, m], {n, 0, 12}, {m, n, 0, -1}] // Flatten (* Jean-François Alcover, Mar 23 2020 *)

T(n,m) = 1 + n*m*(m+1)/2 \\ Charles R Greathouse IV, Jul 28 2016

T(n,2) = A016777(n). T(n,3) = A016921(n). T(n,4) = A017281(n).
T(10,m) = A062786(m+1).
T(11,m) = A069125(m+1).
T(12,m) = A003154(m+1).
T(13,m) = A069126(m+1).
T(14,m) = A069127(m+1).
T(15,m) = A069128(m+1).
T(16,m) = A069129(m+1).
T(17,m) = A069130(m+1).
T(18,m) = A069131(m+1).
T(19,m) = A069132(m+1).
T(20,m) = A069133(m+1).
T(n+1,m) = T(n,m) + m*(m+1)/2. - Gary W. Adamson and Michel Marcus, Oct 13 2015

Edited by R. J. Mathar, Oct 21 2009

A001921 a(n) = 14*a(n-1) - a(n-2) + 6 for n>1, a(0)=0, a(1)=7.

Original entry on oeis.org

0, 7, 104, 1455, 20272, 282359, 3932760, 54776287, 762935264, 10626317415, 148005508552, 2061450802319, 28712305723920, 399910829332567, 5570039304932024, 77580639439715775, 1080558912851088832, 15050244140475527879, 209622859053806301480

Offset: 0

N. J. A. Sloane

(a(n)+1)^3 - a(n)^3 is a square (that of A001570(n)).
The ratio A001570(n)/a(n) tends to sqrt(3) = 1.73205... as n increases. - Pierre CAMI, Apr 21 2005
Define a(1)=0 a(2)=7 such that 3*(a(1)^2) + 3*a(1) + 1 = j(1)^2 = 1^2 and 3*(a(2)^2) + 3*a(2) + 1 = j(2)^2 = 13^2. Then a(n) = a(n-2) + 8*sqrt(3*(a(n-1)^2) + 3*a(n-1) + 1). Another definition : a(n) such that 3*(a(n)^2) + 3*a(n) + 1 = j(n)^2. - Pierre CAMI, Mar 30 2005
a(n) = A001353(n)*A001075(n+1). For n>0, the triple {a(n), a(n)+1=A001922(n), A001570(n)} forms a near-isosceles triangle with angle 2*Pi/3 bounded by the consecutive sides. - Lekraj Beedassy, Jul 21 2006
Numbers n such that A003215(n) is a square, cf. A006051. - Joerg Arndt, Jan 02 2017

			G.f. = 7*x + 104*x^2 + 1455*x^3 + 20272*x^4 + 282359*x^5 + 3932760*x^6 + ... - _Michael Somos_, Aug 17 2018

J. D. E. Konhauser et al., Which Way Did the Bicycle Go?, MAA 1996, p. 104.
E.-A. Majol, Note #2228, L'Intermédiaire des Mathématiciens, 9 (1902), pp. 183-185. - N. J. A. Sloane, Mar 03 2022
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).

T. D. Noe, Table of n, a(n) for n = 0..200
J. Brenner and E. P. Starke, Problem E702, Amer. Math. Monthly, 53 (1946), 465.
Simon Plouffe, Approximations de séries génératrices et quelques conjectures, Dissertation, Université du Québec à Montréal, 1992; arXiv:0911.4975 [math.NT], 2009.
Simon Plouffe, 1031 Generating Functions, Appendix to Thesis, Montreal, 1992
Eric Weisstein's World of Mathematics, Hex Number
Index entries for linear recurrences with constant coefficients, signature (15,-15,1).

Cf. A001570, A001922, A006051.

Cf. numbers m such that k*A000217(m)+1 is a square: A006451 for k=1; A233450 for k=3; A001652 for k=4; A129556 for k=5; this sequence for k=6. - Bruno Berselli, Dec 16 2013

[Round(-1/2 - (1/6)*Sqrt(3)*(7-4*Sqrt(3))^n + (1/6)*Sqrt(3)*(7+4*Sqrt(3))^n + (1/4)*(7+4*Sqrt(3))^n + (1/4)*(7-4*Sqrt(3))^n): n in [0..50]]; // G. C. Greubel, Nov 04 2017

A001921:=z*(-7+z)/(z-1)/(z**2-14*z+1); # Conjectured by Simon Plouffe in his 1992 dissertation.

t = {0, 7}; Do[AppendTo[t, 14*t[[-1]] - t[[-2]] + 6], {20}]; t (* T. D. Noe, Aug 17 2012 *)
LinearRecurrence[{15, -15, 1}, {0, 7, 104}, 19] (* Michael De Vlieger, Jan 02 2017 *)
a[ n_] := -1/2 + (ChebyshevT[n + 1, 7] - ChebyshevT[n, 7]) / 12; (* Michael Somos, Aug 17 2018 *)

concat(0, Vec(x*(x-7)/((x-1)*(x^2-14*x+1)) + O(x^100))) \\ Colin Barker, Jan 06 2015

{a(n) = -1/2 + (polchebyshev(n + 1, 1, 7) - polchebyshev(n, 1, 7)) / 12}; /* Michael Somos, Aug 17 2018 */

G.f.: x*(-7 + x)/(x - 1)/(x^2 - 14*x + 1) (see Simon Plouffe in Maple section).
a(n) = (A028230(n+1)-1)/2. - R. J. Mathar, Mar 19 2009
a(n) = 15*a(n-1) - 15*a(n-2) + a(n-3). - Colin Barker, Jan 06 2015
a(n) = -1 - a(-1-n) for all n in Z. - Michael Somos, Aug 17 2018

More terms from James Sellers, Jul 04 2000

cp's OEIS Frontend

A056108 Fourth spoke of a hexagonal spiral.

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A069129 Centered 16-gonal numbers.

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A002407 Cuban primes: primes which are the difference of two consecutive cubes.

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A147875 Second heptagonal numbers: a(n) = n*(5*n+3)/2.

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A049451 Twice second pentagonal numbers.

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A056106 Second spoke of a hexagonal spiral.

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A147875 Second heptagonal numbers: a(n) = n(5n+3)/2.

A101321 Table T(n,m) = 1 + nm(m+1)/2 read by antidiagonals: centered polygonal numbers.