A062728 -id:A062728 - cp's OEIS frontend

A014105 Second hexagonal numbers: a(n) = n(2n + 1).

0, 3, 10, 21, 36, 55, 78, 105, 136, 171, 210, 253, 300, 351, 406, 465, 528, 595, 666, 741, 820, 903, 990, 1081, 1176, 1275, 1378, 1485, 1596, 1711, 1830, 1953, 2080, 2211, 2346, 2485, 2628, 2775, 2926, 3081, 3240, 3403, 3570, 3741, 3916, 4095, 4278

Offset: 0

N. J. A. Sloane, Jun 14 1998

Note that when starting from a(n)^2, equality holds between series of first n+1 and next n consecutive squares: a(n)^2 + (a(n) + 1)^2 + ... + (a(n) + n)^2 = (a(n) + n + 1)^2 + (a(n) + n + 2)^2 + ... + (a(n) + 2*n)^2; e.g., 10^2 + 11^2 + 12^2 = 13^2 + 14^2. - Henry Bottomley, Jan 22 2001; with typos fixed by Zak Seidov, Sep 10 2015
a(n) = sum of second set of n consecutive even numbers - sum of the first set of n consecutive odd numbers: a(1) = 4-1, a(3) = (8+10+12) - (1+3+5) = 21. - Amarnath Murthy, Nov 07 2002
Partial sums of odd numbers 3 mod 4, that is, 3, 3+7, 3+7+11, ... See A001107. - Jon Perry, Dec 18 2004
If Y is a fixed 3-subset of a (2n+1)-set X then a(n) is the number of (2n-1)-subsets of X intersecting Y. - Milan Janjic, Oct 28 2007
More generally (see the first comment), for n > 0, let b(n,k) = a(n) + k*(4*n + 1). Then b(n,k)^2 + (b(n,k) + 1)^2 + ... + (b(n,k) + n)^2 = (b(n,k) + n + 1 + 2*k)^2 + ... + (b(n,k) + 2*n + 2*k)^2 + k^2; e.g., if n = 3 and k = 2, then b(n,k) = 47 and 47^2 + ... + 50^2 = 55^2 + ... + 57^2 + 2^2. - Charlie Marion, Jan 01 2011
Sequence found by reading the line from 0, in the direction 0, 10, ..., and the line from 3, in the direction 3, 21, ..., in the square spiral whose vertices are the triangular numbers A000217. - Omar E. Pol, Nov 09 2011
a(n) is the number of positions of a domino in a pyramidal board with base 2n+1. - César Eliud Lozada, Sep 26 2012
Differences of row sums of two consecutive rows of triangle A120070, i.e., first differences of A016061. - J. M. Bergot, Jun 14 2013 [In other words, the partial sums of this sequence give A016061. - Leo Tavares, Nov 23 2021]
a(n)*Pi is the total length of half circle spiral after n rotations. See illustration in links. - Kival Ngaokrajang, Nov 05 2013
For corresponding sums in first comment by Henry Bottomley, see A059255. - Zak Seidov, Sep 10 2015
a(n) also gives the dimension of the simple Lie algebras B_n (n >= 2) and C_n (n >= 3). - Wolfdieter Lang, Oct 21 2015
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 unsigned A130757, analogous to A132440 for the Pascal matrix. - Tom Copeland, Dec 13 2015
Partial sums of squares with alternating signs, ending in an even term: a(n) = 0^2 - 1^2 +- ... + (2*n)^2, cf. Example & Formula from Berselli, 2013. - M. F. Hasler, Jul 03 2018
Also numbers k with the property that in the symmetric representation of sigma(k) the smallest Dyck path has a central peak and the largest Dyck path has a central valley, n > 0. (Cf. A237593.) - Omar E. Pol, Aug 28 2018
a(n) is the area of a triangle with vertices at (0,0), (2*n+1, 2*n), and ((2*n+1)^2, 4*n^2). - Art Baker, Dec 12 2018
This sequence is the largest subsequence of A000217 such that gcd(a(n), 2*n) = a(n) mod (2*n) = n, n > 0 up to a given value of n. It is the interleave of A033585 (a(n) is even) and A033567 (a(n) is odd). - Torlach Rush, Sep 09 2019
A generalization of Hasler's Comment (Jul 03 2018) follows. Let P(k,n) be the n-th k-gonal number. Then for k > 1, partial sums of {P(k,n)} with alternating signs, ending in an even term, = n*((k-2)*n + 1). - Charlie Marion, Mar 02 2021
Let U_n(H) = {A in M_n(H): A*A^H = I_n} be the group of n X n unitary matrices over the quaternions (A^H is the conjugate transpose of A. Note that over the quaternions we still have A*A^H = I_n <=> A^H*A = I_n by mapping A and A^H to (2n) X (2n) complex matrices), then a(n) is the dimension of its Lie algebra u_n(H) = {A in M_n(H): A + A^H = 0} as a real vector space. A basis is given by {(E_{st}-E_{ts}), i*(E_{st}+E_{ts}), j*(E_{st}+E_{ts}), k*(E_{st}+E_{ts}): 1 <= s < t <= n} U {i*E_{tt}, j*E_{tt}, k*E_{tt}: t = 1..n}, where E_{st} is the matrix with all entries zero except that its (st)-entry is 1. - Jianing Song, Apr 05 2021

			For n=6, a(6) = 0^2 - 1^2 + 2^2 - 3^2 + 4^2 - 5^2 + 6^2 - 7^2 + 8^2 - 9^2 + 10^2 - 11^2 + 12^2 = 78. - _Bruno Berselli_, Aug 29 2013

Louis Comtet, Advanced Combinatorics, Reidel, 1974, pp. 77-78. (In the integral formula on p. 77 a left bracket is missing for the cosine argument.)

Reinhard Zumkeller, Table of n, a(n) for n = 0..10000
Matthew Cho, Anton Dochtermann, Ryota Inagaki, Suho Oh, Dylan Snustad, and Bailee Zacovic, Chip-firing and critical groups of signed graphs, arXiv:2306.09315 [math.CO], 2023. See p. 22.
Robert FERREOL, Illustration: triangular numbers of even order
Guo-Niu Han, Enumeration of Standard Puzzles, 2011. [Cached copy]
Guo-Niu Han, Enumeration of Standard Puzzles, arXiv:2006.14070 [math.CO], 2020.
Milan Janjic, Two Enumerative Functions, University of Banja Luka (Bosnia and Herzegovina, 2017).
Ângela Mestre and José Agapito, Square Matrices Generated by Sequences of Riordan Arrays, J. Int. Seq., Vol. 22 (2019), Article 19.8.4.
Kival Ngaokrajang, Illustration of half circle spiral.
Markus Scheuer, show that; strange sum yields triangular numbers, Mathematics StackExchange.
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).
Amelia Carolina Sparavigna, The groupoid of the Triangular Numbers and the generation of related integer sequences, Politecnico di Torino, Italy (2019).
Leo Tavares, Illustration: Squared Hexagons.
Index entries for two-way infinite sequences
Index entries for linear recurrences with constant coefficients, signature (3,-3,1).

Cf. A000217, A000290, A000384, A002378, A002943, A100040, A100041, A081266, A144312.
Cf. A033567, A033585, A059255, A130757, A132440, A027907.
Second column of array A094416.
Equals A033586(n) divided by 4.
See Comments of A132124.
Second n-gonal numbers: A005449, A147875, A045944, A179986, A033954, A062728, A135705.
Row sums in triangle A253580.
Cf. A016061, A003215, A000290, A188859.

List([0..50],n->n*(2*n+1)); # Muniru A Asiru, Oct 31 2018

a014105 n = n * (2 * n + 1)
a014105_list = scanl (+) 0 a004767_list  -- Reinhard Zumkeller, Oct 03 2012

[ n*(2*n+1) : n in [0..50] ]; // Wesley Ivan Hurt, Jun 14 2014

seq(binomial(2*n+1,2), n=0..46); # Zerinvary Lajos, Jan 21 2007

Table[n*(2*n+1), {n,0,100}] (* Vladimir Joseph Stephan Orlovsky, Nov 16 2008 *)
LinearRecurrence[{3,-3,1},{0,3,10},50] (* Harvey P. Dale, Feb 10 2015 *)
CoefficientList[Series[x*(3 + x)/(1 - x)^3,{x, 0, 50}], x] (* Stefano Spezia, Sep 02 2018 *)

PARI
```
a(n)=n*(2*n+1)
```

[n*(2*n+1) for n in range(50)] # G. C. Greubel, Dec 16 2018

a(n) = 3*Sum_{k=1..n} tan^2(k*Pi/(2*(n + 1))). - Ignacio Larrosa Cañestro, Apr 17 2001
a(n)^2 = n*(a(n) + 1 + a(n) + 2 + ... + a(n) + 2*n); e.g., 10^2 = 2*(11 + 12 + 13 + 14). - Charlie Marion, Jun 15 2003
From N. J. A. Sloane, Sep 13 2003: (Start)
G.f.: x*(3 + x)/(1 - x)^3.
E.g.f.: exp(x)*(3*x + 2*x^2).
a(n) = A000217(2*n) = A000384(-n). (End)
a(n) = A084849(n) - 1; A100035(a(n) + 1) = 1. - Reinhard Zumkeller, Oct 31 2004
a(n) = A126890(n, k) + A126890(n, n-k), 0 <= k <= n. - Reinhard Zumkeller, Dec 30 2006
a(2*n) = A033585(n); a(3*n) = A144314(n). - Reinhard Zumkeller, Sep 17 2008
a(n) = a(n-1) + 4*n - 1 (with a(0) = 0). - Vincenzo Librandi, Dec 24 2010
a(n) = Sum_{k=0.2*n} (-1)^k*k^2. - Bruno Berselli, Aug 29 2013
a(n) = A242342(2*n + 1). - Reinhard Zumkeller, May 11 2014
a(n) = Sum_{k=0..2} C(n-2+k, n-2) * C(n+2-k, n), for n > 1. - J. M. Bergot, Jun 14 2014
a(n) = floor(Sum_{j=(n^2 + 1)..((n+1)^2 - 1)} sqrt(j)). Fractional portion of each sum converges to 1/6 as n -> infinity. See A247112 for a similar summation sequence on j^(3/2) and references to other such sequences. - Richard R. Forberg, Dec 02 2014
a(n) = 3*a(n-1) - 3*a(n-2) + a(n-3) for n >= 3, with a(0) = 0, a(1) = 3, and a(2) = 10. - Harvey P. Dale, Feb 10 2015
Sum_{n >= 1} 1/a(n) = 2*(1 - log(2)) = 0.61370563888010938... (A188859). - Vaclav Kotesovec, Apr 27 2016
From Wolfdieter Lang, Apr 27 2018: (Start)
a(n) = trinomial(2*n, 2) = trinomial(2*n, 2*(2*n-1)), for n >= 1, with the trinomial irregular triangle A027907; i.e., trinomial(n,k) = A027907(n,k).
a(n) = (1/Pi) * Integral_{x=0..2} (1/sqrt(4 - x^2)) * (x^2 - 1)^(2*n) * R(4*(n-1), x), for n >= 0, with the R polynomial coefficients given in A127672, and R(-m, x) = R(m, x). [See Comtet, p. 77, the integral formula for q = 3, n -> 2*n, k = 2, rewritten with x = 2*cos(phi).] (End)
a(n) = A002943(n)/2. - Ralf Steiner, Jul 23 2019
a(n) = A000290(n) + A002378(n). - Torlach Rush, Nov 02 2020
a(n) = A003215(n) - A000290(n+1). See Squared Hexagons illustration. Leo Tavares, Nov 23 2021
Sum_{n>=1} (-1)^(n+1)/a(n) = Pi/2 + log(2) - 2. - Amiram Eldar, Nov 28 2021

Link added and minor errors corrected by Johannes W. Meijer, Feb 04 2010

A045944 Rhombic matchstick numbers: a(n) = n(3n+2).

Original entry on oeis.org

0, 5, 16, 33, 56, 85, 120, 161, 208, 261, 320, 385, 456, 533, 616, 705, 800, 901, 1008, 1121, 1240, 1365, 1496, 1633, 1776, 1925, 2080, 2241, 2408, 2581, 2760, 2945, 3136, 3333, 3536, 3745, 3960, 4181, 4408, 4641, 4880, 5125, 5376, 5633, 5896, 6165, 6440

Offset: 0

R. K. Guy

From Floor van Lamoen, Jul 21 2001: (Start)
Write 1,2,3,4,... in a hexagonal spiral around 0, then a(n) is the n-th term of the sequence found by reading the line from 0 in the direction 0,5,.... The spiral begins:
                 .
                 85--84--83--82--81--80
                   .                   \
                   56--55--54--53--52  79
                   / .               \   \
                 57  33--32--31--30  51  78
                 /   / .           \   \   \
               58  34  16--15--14  29  50  77
               /   /   / .       \   \   \   \
             59  35  17   5---4  13  28  49  76
             /   /   /   / .   \   \   \   \   \
           60  36  18   6   0   3  12  27  48  75
           /   /   /   /   /   /   /   /   /   /
         61  37  19   7   1---2  11  26  47  74
           \   \   \   \         /   /   /   /
           62  38  20   8---9--10  25  46  73
             \   \   \             /   /   /
             63  39  21--22--23--24  45  72
               \   \                 /   /
               64  40--41--42--43--44  71
                 \                     /
                 65--66--67--68--69--70
(End)
Connection to triangular numbers: a(n) = 4*T_n + S_n where T_n is the n-th triangular number and S_n is the n-th square. - William A. Tedeschi, Sep 12 2010
Also, second octagonal numbers. - Bruno Berselli, Jan 13 2011
Sequence found by reading the line from 0, in the direction 0, 16, ... and the line from 5, in the direction 5, 33, ..., in the square spiral whose vertices are the generalized octagonal numbers A001082. - Omar E. Pol, Jul 18 2012
Let P denote the points from the n X n grid. A(n-1) also coincides with the minimum number of points Q needed to "block" P, that is, every line segment spanned by two points from P must contain one point from Q. - Manfred Scheucher, Aug 30 2018
Also the number of internal edges of an (n+1)*(n+1) "square" of hexagons; i.e., n+1 rows, each of n+1 edge-adjacent hexagons, stacked with minimal overhang. - Jon Hart, Sep 29 2019
For n >= 1, the continued fraction expansion of sqrt(27*a(n)) is [9n+2; {1, 2n-1, 1, 1, 1, 2n-1, 1, 18n+4}]. - Magus K. Chu, Oct 13 2022

Ivan Panchenko, Table of n, a(n) for n = 0..1000
Ghislain R. Franssens, On a Number Pyramid Related to the Binomial, Deleham, Eulerian, MacMahon and Stirling number triangles, Journal of Integer Sequences, Vol. 9 (2006), Article 06.4.1.
M. Janjic and B. Petkovic, A Counting Function, arXiv:1301.4550 [math.CO], 2013.
Leo Tavares, Illustration: Square Stars
Leo Tavares, Illustration: Split Stars
Index entries for linear recurrences with constant coefficients, signature (3,-3,1).

Bisection of A001859. See Comments of A135713.
Cf. A000217, A000567, A001082, A002378, A016969, A049450, A174709.
Cf. second n-gonal numbers: A005449, A014105, A147875, A179986, A033954, A062728, A135705.
Cf. numbers of the form n*(d*n+10-d)/2: A008587, A056000, A028347, A140090, A014106, A028895, A186029, A007742, A022267, A033429, A022268, A049452, A186030, A135703, A152734, A139273.
Cf. A000290, A022144, A005563, A001105.
Cf. A056109.
Cf. A003154.

[n*(3*n+2) : n in [0..100]]; // Wesley Ivan Hurt, Sep 24 2017

Table[n*(3n+2), {n,0,60}] (* Harvey P. Dale, May 05 2011 *)
LinearRecurrence[{3,-3,1},{0,5,16},60] (* Harvey P. Dale, Jan 19 2016 *)
CoefficientList[Series[x*(5 + x)/(1 - x)^3,{x, 0, 60}], x] (* Stefano Spezia, Sep 01 2018 *)

a(n)=n*(3*n+2) \\ Charles R Greathouse IV, Nov 20 2012

O.g.f.: x*(5+x)/(1-x)^3. - R. J. Mathar, Jan 07 2008
a(n) = 3*a(n-1) - 3*a(n-2) + a(n-3), with a(0)=0, a(1)=5, a(2)=16. - Harvey P. Dale, May 06 2011
a(n) = a(n-1) + 6*n - 1 (with a(0)=0). - Vincenzo Librandi, Nov 18 2010
For n > 0, a(n)^3 + (a(n)+1)^3 + ... + (a(n)+n)^3 + 2*A000217(n)^2 = (a(n) + n + 1)^3 + ... + (a(n) + 2n)^3; see also A033954. - Charlie Marion, Dec 08 2007
a(n) = Sum_{i=0..n-1} A016969(i) for n > 0. - Bruno Berselli, Jan 13 2011
a(n) = A174709(6*n+4). - Philippe Deléham, Mar 26 2013
a(n) = A001082(2*n). - Michael Turniansky, Aug 24 2013
Sum_{n>=1} 1/a(n) = (9 + sqrt(3)*Pi - 9*log(3))/12 = 0.3794906245574721941... . - Vaclav Kotesovec, Apr 27 2016
a(n) = A002378(n) + A014105(n). - J. M. Bergot, Apr 24 2018
Sum_{n>=1} (-1)^(n+1)/a(n) = Pi/sqrt(12) - 3/4. - Amiram Eldar, Jul 03 2020
E.g.f.: exp(x)*x*(5 + 3*x). - Stefano Spezia, Jun 08 2021
From Leo Tavares, Oct 14 2021: (Start)
a(n) = A000290(n) + 4*A000217(n). See Square Stars illustration.
a(n) = A000567(n+2) - A022144(n+1)
a(n) = A005563(n) + A001105(n).
a(n) = A056109(n) - 1. (End)
From Leo Tavares, Oct 06 2022: (Start)
a(n) = A003154(n+1) - A000567(n+1). See Split Stars illustration.
a(n) = A014105(n) + 2*A000217(n). (End)

A033954 Second 10-gonal (or decagonal) numbers: n(4n+3).

Original entry on oeis.org

0, 7, 22, 45, 76, 115, 162, 217, 280, 351, 430, 517, 612, 715, 826, 945, 1072, 1207, 1350, 1501, 1660, 1827, 2002, 2185, 2376, 2575, 2782, 2997, 3220, 3451, 3690, 3937, 4192, 4455, 4726, 5005, 5292, 5587, 5890, 6201, 6520, 6847, 7182, 7525, 7876, 8235

Offset: 0

N. J. A. Sloane

Same as A033951 except start at 0. See example section.

Bisection of A074377. Also sequence found by reading the line from 0, in the direction 0, 22, ... and the line from 7, in the direction 7, 45, ..., in the square spiral whose vertices are the generalized 10-gonal numbers A074377. - Omar E. Pol, Jul 24 2012

			  36--37--38--39--40--41--42
   |                       |
  35  16--17--18--19--20  43
   |   |               |   |
  34  15   4---5---6  21  44
   |   |   |       |   |   |
  33  14   3   0===7==22==45==76=>
   |   |   |   |   |   |
  32  13   2---1   8  23
   |   |           |   |
  31  12--11--10---9  24
   |                   |
  30--29--28--27--26--25

S. M. Ellerstein, The square spiral, J. Recreational Mathematics 29 (#3, 1998) 188; 30 (#4, 1999-2000), 246-250.
R. L. Graham, D. E. Knuth and O. Patashnik, Concrete Mathematics. Addison-Wesley, Reading, MA, 2nd ed., 1994, p. 99.

Ivan Panchenko, Table of n, a(n) for n = 0..1000
Emilio Apricena, A version of the Ulam spiral.
Leo Tavares, Illustration: V numbers
Index entries for linear recurrences with constant coefficients, signature (3,-3,1).

Cf. A002620, A033951.
Sequences from spirals: A001107, A002939, A007742, A033951, A033952, A033953, A033954, A033989, A033990, A033991, A002943, A033996, A033988.
Sequences on the four axes of the square spiral: Starting at 0: A001107, A033991, A007742, A033954; starting at 1: A054552, A054556, A054567, A033951.
Sequences on the four diagonals of the square spiral: Starting at 0: A002939 = 2*A000384, A016742 = 4*A000290, A002943 = 2*A014105, A033996 = 8*A000217; starting at 1: A054554, A053755, A054569, A016754.
Sequences obtained by reading alternate terms on the X and Y axes and the two main diagonals of the square spiral: Starting at 0: A035608, A156859, A002378 = 2*A000217, A137932 = 4*A002620; starting at 1: A317186, A267682, A002061, A080335.
Second n-gonal numbers: A005449, A014105, A147875, A045944, A179986, this sequence, A062728, A135705.
Cf. A060544.

List([0..50], n-> n*(4*n+3)) # G. C. Greubel, May 24 2019

[n*(4*n+3): n in [0..50]]; // G. C. Greubel, May 24 2019

Table[n(4n+3),{n,0,50}] (* or *) LinearRecurrence[{3,-3,1},{0,7,22},50] (* Harvey P. Dale, May 06 2018 *)

PARI
```
a(n)=4*n^2+3*n
```

[n*(4*n+3) for n in (0..50)] # G. C. Greubel, May 24 2019

a(n) = A001107(-n) = A074377(2*n).
G.f.: x*(7+x)/(1-x)^3. - Michael Somos, Mar 03 2003
a(n) = a(n-1) + 8*n - 1 with a(0)=0. - Vincenzo Librandi, Jul 20 2010
For n>0, Sum_{j=0..n} (a(n) + j)^4 + (4*A000217(n))^3 = Sum_{j=n+1..2n} (a(n) + j)^4; see also A045944. - Charlie Marion, Dec 08 2007, edited by Michel Marcus, Mar 14 2014
a(n) = 3*a(n-1) - 3*a(n-2) + a(n-3) with a(0) = 0, a(1) = 7, a(2) = 22. - Philippe Deléham, Mar 26 2013
a(n) = A118729(8n+6). - Philippe Deléham, Mar 26 2013
a(n) = A002943(n) + n = A007742(n) + 2n = A016742(n) + 3n = A033991(n) + 4n = A002939(n) + 5n = A001107(n) + 6n = A033996(n) - n. - Philippe Deléham, Mar 26 2013
Sum_{n>=1} 1/a(n) = 4/9 + Pi/6 - log(2) = 0.2748960394827980081... . - Vaclav Kotesovec, Apr 27 2016
E.g.f.: exp(x)*x*(7 + 4*x). - Stefano Spezia, Jun 08 2021
Sum_{n>=1} (-1)^(n+1)/a(n) = Pi/(3*sqrt(2)) + log(2)/3 - 4/9 - sqrt(2)*arcsinh(1)/3. - Amiram Eldar, Nov 28 2021
For n>0, (a(n)^2 + n)/(a(n) + n) = (4*n + 1)^2/4, a ratio of two squares. - Rick L. Shepherd, Feb 23 2022
a(n) = A060544(n+1) - A000217(n+1). - Leo Tavares, Mar 31 2022

A195160 Generalized 11-gonal (or hendecagonal) numbers: m(9m - 7)/2 with m = 0, 1, -1, 2, -2, 3, -3, ...

Original entry on oeis.org

0, 1, 8, 11, 25, 30, 51, 58, 86, 95, 130, 141, 183, 196, 245, 260, 316, 333, 396, 415, 485, 506, 583, 606, 690, 715, 806, 833, 931, 960, 1065, 1096, 1208, 1241, 1360, 1395, 1521, 1558, 1691, 1730, 1870, 1911, 2058, 2101, 2255, 2300, 2461, 2508, 2676

Offset: 0

Omar E. Pol, Sep 10 2011

Exponents of q in the expansion of Product_{n >= 1} (1 - q^(9*n))*(1 + q^(9*n-1))*(1 + q^(9*n-8)) = 1 + q + q^8 + q^11 + q^25 + q^30 + .... - Peter Bala, Nov 21 2024

Vincenzo Librandi, Table of n, a(n) for n = 0..1000
Index entries for linear recurrences with constant coefficients, signature (1,2,-2,-1,1).

Partial sums of A195159.
Column 7 of A195152.
Cf. A316672.
Sequences of generalized k-gonal numbers: A001318 (k=5), A000217 (k=6), A085787 (k=7), A001082 (k=8), A118277 (k=9), A074377 (k=10), this sequence (k=11), A195162 (k=12), A195313 (k=13), A195818 (k=14), A277082 (k=15), A274978 (k=16), A303305 (k=17), A274979 (k=18), A303813 (k=19), A218864 (k=20), A303298 (k=21), A303299 (k=22), A303303 (k=23), A303814 (k=24), A303304 (k=25), A316724 (k=26), A316725 (k=27), A303812 (k=28), A303815 (k=29), A316729 (k=30).

I:=[0, 1, 8, 11, 25]; [n le 5 select I[n] else Self(n-1)+2*Self(n-2)-2*Self(n-3)-Self(n-4)+Self(n-5): n in [1..50]]; // Vincenzo Librandi, Apr 09 2013

CoefficientList[Series[x (1 + 7 x + x^2)/((1 + x)^2 (1 - x)^3), {x, 0, 60}], x] (* Vincenzo Librandi, Apr 09 2013 *)

a(n)=(18*n*(n+1)+5*(2*n+1)*(-1)^n-5)/16 \\ Charles R Greathouse IV, Sep 24 2015

From Bruno Berselli, Sep 14 2011: (Start)
G.f.: x*(1+7*x+x^2)/((1+x)^2*(1-x)^3).
a(n) = (18*n*(n+1)+5*(2*n+1)*(-1)^n-5)/16.
a(2n) = A062728(n), a(2n-1) = A051682(n). (End)
Sum_{n>=1} 1/a(n) = 18/49 + 2*Pi*cot(2*Pi/9)/7. - Vaclav Kotesovec, Oct 05 2016

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

A179986 Second 9-gonal (or nonagonal) numbers: a(n) = n(7n+5)/2.

Original entry on oeis.org

0, 6, 19, 39, 66, 100, 141, 189, 244, 306, 375, 451, 534, 624, 721, 825, 936, 1054, 1179, 1311, 1450, 1596, 1749, 1909, 2076, 2250, 2431, 2619, 2814, 3016, 3225, 3441, 3664, 3894, 4131, 4375, 4626, 4884, 5149, 5421, 5700, 5986, 6279, 6579, 6886

Offset: 0

Bruno Berselli, Jan 13 2011

This sequence is a bisection of A118277 (even part).
Sequence found by reading the line from 0, in the direction 0, 19... and the line from 6, in the direction 6, 39,..., in the square spiral whose vertices are the generalized 9-gonal numbers A118277. - Omar E. Pol, Jul 24 2012
The early part of this sequence is a strikingly close approximation to the early part of A100752. - Peter Munn, Nov 14 2019

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

Cf. second k-gonal numbers: A005449 (k=5), A014105 (k=6), A147875 (k=7), A045944 (k=8), this sequence (k=9), A033954 (k=10), A062728 (k=11), A135705 (k=12).

Cf. A001106, A022264, A022265, A024966, A055998, A100752, A174738, A186029, A218471.

[n*(7*n+5)/2: n in [0..50]]; // Bruno Berselli, Sep 23 2016

I:=[0, 6, 19]; [n le 3 select I[n] else 3*Self(n-1) -3*Self(n-2) +Self(n-3): n in [1..60]]; // Vincenzo Librandi, Oct 15 2012

f[n_] := n (7 n + 5)/2; f[Range[0, 60]] (* Vladimir Joseph Stephan Orlovsky, Feb 05 2011*)
LinearRecurrence[{3, -3, 1}, {0, 6, 19}, 60] (* or *) Array[(#(7# + 5))/2&, 60, 0] (* Harvey P. Dale, Aug 19 2011 *)
CoefficientList[Series[x (6 + x)/(1 - x)^3, {x, 0, 60}], x] (* Vincenzo Librandi, Oct 15 2012 *)

a(n)=n*(7*n+5)/2 \\ Charles R Greathouse IV, Sep 24 2015

G.f.: x*(6 + x)/(1 - x)^3.
a(n) = Sum_{i=0..(n-1)} A017053(i) for n>0.
a(-n) = A001106(n).
Sum_{i=0..n} (a(n)+i)^2 = ( Sum_{i=(n+1)..2*n} (a(n)+i)^2 ) + 21*A000217(n)^2 for n>0.
a(n) = a(n-1)+7*n-1 for n>0, with a(0)=0. - Vincenzo Librandi, Feb 05 2011
a(0)=0, a(1)=6, a(2)=19; for n>2, a(n) = 3*a(n-1)-3*a(n-2)+a(n-3). - Harvey P. Dale, Aug 19 2011
a(n) = A174738(7n+5). - Philippe Deléham, Mar 26 2013
a(n) = A001477(n) + 2*A000290(n) + 3*A000217(n). - J. M. Bergot, Apr 25 2014
a(n) = A055998(4*n) - A055998(3*n). - Bruno Berselli, Sep 23 2016
E.g.f.: (x/2)*(12 + 7*x)*exp(x). - G. C. Greubel, Aug 19 2017

A218470 Partial sums of floor(n/9).

Original entry on oeis.org

0, 0, 0, 0, 0, 0, 0, 0, 0, 1, 2, 3, 4, 5, 6, 7, 8, 9, 11, 13, 15, 17, 19, 21, 23, 25, 27, 30, 33, 36, 39, 42, 45, 48, 51, 54, 58, 62, 66, 70, 74, 78, 82, 86, 90, 95, 100, 105, 110, 115, 120, 125, 130, 135, 141, 147, 153, 159, 165, 171, 177, 183, 189, 196, 203, 210, 217, 224

Offset: 0

Philippe Deléham, Mar 26 2013

Apart from the initial zeros, the same as A008727.

			As square array:
..0....0....0....0....0....0....0....0....0....
..1....2....3....4....5....6....7....8....9....
.11...13...15...17...19...21...23...25...27....
.30...33...36...39...42...45...48...51...54....
.58...62...66...70...74...78...82...86...90....
.95..100..105..110..115..120..125..130..135....
141..147..153..159..165..171..177..183..189....
196..203..210..217..224..231..238..245..252....
...

G. C. Greubel, Table of n, a(n) for n = 0..1000
Index entries for linear recurrences with constant coefficients, signature (2,-1,0,0,0,0,0,0,1,-2,1).

Cf. similar sequences: A118729, A174109, A174738.

[&+[Floor(k/9): k in [0..n]]: n in [0..70]]; // Bruno Berselli, Mar 27 2013

Accumulate[Floor[Range[0, 100]/9]] (* Jean-François Alcover, Mar 27 2013 *)

for(n=0,50, print1(sum(k=0,n, floor(k/9)), ", ")) \\ G. C. Greubel, Dec 13 2016

a(n)=my(k=n\9); k*(9*k-7)/2 + k*(n-9*k) \\ Charles R Greathouse IV, Dec 13 2016

a(9n)   = A051682(n).
a(9n+1) = A062708(n).
a(9n+2) = A062741(n).
a(9n+3) = A022266(n).
a(9n+4) = A022267(n).
a(9n+5) = A081266(n).
a(9n+6) = A062725(n).
a(9n+7) = A062728(n).
a(9n+8) = A027468(n).
G.f.: x^9/((1-x)^2*(1-x^9)). - Bruno Berselli, Mar 27 2013

A135705 a(n) = 10binomial(n,2) + 9n.

Original entry on oeis.org

0, 9, 28, 57, 96, 145, 204, 273, 352, 441, 540, 649, 768, 897, 1036, 1185, 1344, 1513, 1692, 1881, 2080, 2289, 2508, 2737, 2976, 3225, 3484, 3753, 4032, 4321, 4620, 4929, 5248, 5577, 5916, 6265, 6624, 6993, 7372, 7761, 8160, 8569, 8988, 9417, 9856, 10305, 10764

Offset: 0

N. J. A. Sloane, Mar 04 2008

Also, second 12-gonal (or dodecagonal) numbers. Identity for the numbers b(n)=n*(h*n+h-2)/2 (see Crossrefs): Sum_{i=0..n} (b(n)+i)^2 = (Sum_{i=n+1..2*n} (b(n)+i)^2) + h*(h-4)*A000217(n)^2 for n>0. - Bruno Berselli, Jan 15 2011
Sequence found by reading the line from 0, in the direction 0, 28, ..., and the line from 9, in the direction 9, 57, ..., in the square spiral whose vertices are the generalized 12-gonal numbers A195162. - Omar E. Pol, Jul 24 2012
Bisection of A195162. - Omar E. Pol, Aug 04 2012

Ivan Panchenko, Table of n, a(n) for n = 0..1000
L. Hogben, Choice and Chance by Cardpack and Chessboard, Vol. 1, Max Parrish and Co, London, 1950, p. 36.
Leo Tavares, Illustration: Diamond Clipped Stars
Index entries for linear recurrences with constant coefficients, signature (3,-3,1).

Second n-gonal numbers: A005449, A014105, A147875, A045944, A179986, A033954, A062728, this sequence.
Cf. A003154, A000290.
Cf. A195162.

List([0..50], n-> n*(5*n+4)); # G. C. Greubel, Jul 04 2019

[n*(5*n+4): n in [0..50]]; // G. C. Greubel, Jul 04 2019

LinearRecurrence[{3,-3,1}, {0,9,28}, 50] (* or *) Table[5*n^2 + 4*n, {n,0,50}] (* G. C. Greubel, Oct 29 2016 *)
Table[10 Binomial[n,2]+9n,{n,0,60}] (* Harvey P. Dale, Jun 14 2023 *)

a(n) = 10*binomial(n,2) + 9*n \\ Charles R Greathouse IV, Jun 11 2015

[n*(5*n+4) for n in (0..50)] # G. C. Greubel, Jul 04 2019

From R. J. Mathar, Mar 06 2008: (Start)
O.g.f.: x*(9+x)/(1-x)^3.
a(n) = n*(5*n+4). (End)
a(n) = a(n-1) + 10*n - 1 (with a(0)=0). - Vincenzo Librandi, Nov 24 2009
a(n) = Sum_{i=0..n-1} A017377(i) for n>0. - Bruno Berselli, Jan 15 2011
a(n) = A131242(10n+8). - Philippe Deléham, Mar 27 2013
Sum_{n>=1} 1/a(n) = 5/16 + sqrt(1 + 2/sqrt(5))*Pi/8 - 5*log(5)/16 - sqrt(5)*log((1 + sqrt(5))/2)/8 = 0.2155517745488486003038... . - Vaclav Kotesovec, Apr 27 2016
From G. C. Greubel, Oct 29 2016: (Start)
a(n) = 3*a(n-1) - 3*a(n-2) + a(n-3).
E.g.f.: x*(9 + 5*x)*exp(x). (End)
a(n) = A003154(n+1) - A000290(n+1). - Leo Tavares, Mar 29 2022

A125662 A convolution triangle of numbers based on A001906 (even-indexed Fibonacci numbers).

Original entry on oeis.org

1, 3, 1, 8, 6, 1, 21, 25, 9, 1, 55, 90, 51, 12, 1, 144, 300, 234, 86, 15, 1, 377, 954, 951, 480, 130, 18, 1, 987, 2939, 3573, 2305, 855, 183, 21, 1, 2584, 8850, 12707, 10008, 4740, 1386, 245, 24, 1, 6765, 26195, 43398, 40426, 23373, 8715, 2100, 316, 27, 1

Offset: 0

Philippe Deléham, Jan 28 2007

Subtriangle of the triangle given by [0,3,-1/3,1/3,0,0,0,0,0,...] DELTA [1,0,0,0,0,0,...] where DELTA is the operator defined in A084938. Unsigned version of A123965.
From Philippe Deléham, Feb 19 2012: (Start)
Riordan array (1/(1-3*x+x^2), x/(1-3*x+x^2)).
Equals A078812*A007318 as infinite lower triangular matrices.
Triangle of coefficients of Chebyshev's S(n,x+3) polynomials (exponents of x in increasing order). (End)
For 1 <= k <= n, T(n,k) equals the number of (n-1)-length  words over {0,1,2,3} containing k-1 letters equal 3 and avoiding 01. - Milan Janjic, Dec 20 2016

			Triangle begins:
   1;
   3,  1;
   8,  6,  1;
  21, 25,  9,  1;
  55, 90, 51, 12,  1;
  ...
Triangle [0,3,-1/3,1/3,0,0,0,...] DELTA [1,0,0,0,0,0,...] begins:
  1;
  0,  1;
  0,  3,  1;
  0,  8,  6,  1;
  0, 21, 25,  9,  1;
  0, 55, 90, 51, 12,  1;
  ...

Michael De Vlieger, Table of n, a(n) for n = 0..11475 (rows 0 <= n <= 150)
Milan Janjić, Words and Linear Recurrences, J. Int. Seq. 21 (2018), #18.1.4.

Diagonal sums: A000244(powers of 3).
Row sums: A001353 (n+1).
Diagonals: A001906(n+1), A001871.
Cf. A000012, A008585, A062728.
Cf. Triangle of coefficients of Chebyshev's S(n,x+k) polynomials: A207824, A207823, A125662, A078812, A101950, A049310, A104562, A053122, A207815, A159764, A123967 for k = 5, 4, 3, 2, 1, 0, -1, -2, -3, -4, -5 respectively.

m:=12;
R:=PowerSeriesRing(Integers(), m+2);
A125662:= func< n,k | Abs( Coefficient(R!( Evaluate(ChebyshevU(n+1), (3-x)/2) ), k) ) >;
[A125662(n,k): k in [0..n], n in [0..m]]; // G. C. Greubel, Aug 20 2023

With[{n = 9}, DeleteCases[#, 0] & /@ CoefficientList[Series[1/(1 - 3 x + x^2 - y x), {x, 0, n}, {y, 0, n}], {x, y}]] // Flatten (* Michael De Vlieger, Apr 25 2018 *)
Table[Abs[CoefficientList[ChebyshevU[n,(x-3)/2], x]], {n,0,12}]//Flatten (* G. C. Greubel, Aug 20 2023 *)

def A125662(n,k): return abs( ( chebyshev_U(n, (3-x)/2) ).series(x, n+2).list()[k] )
flatten([[A125662(n,k) for k in range(n+1)] for n in range(13)]) # G. C. Greubel, Aug 20 2023

T(n,k) = T(n-1,k-1) + 3*T(n-1,k) - T(n-2,k); T(0,0)=1; T(n,k)=0 if k < 0 or k > n.
Sum_{k=0..n} T(n, k) = A001353(n+1).
Sum_{k=0..floor(n/2)} T(n-k, k) = A000244(n+1).
G.f.: 1/(1-3*x+x^2-y*x). - Philippe Deléham, Feb 19 2012
From G. C. Greubel, Aug 20 2023: (Start)
T(n, k) = abs( [x^k]( ChebyshevU(n, (3-x)/2) ) ).
Sum_{k=0..n} (-1)^k*T(n, k) = A000027(n+1).
Sum_{k=0..floor(n/2)} (-1)^k*T(n-k, k) = A000225(n). (End)

a(45) corrected and a(51) added by Philippe Deléham, Feb 19 2012

A211013 Second 13-gonal numbers: a(n) = n(11n+9)/2.

Original entry on oeis.org

0, 10, 31, 63, 106, 160, 225, 301, 388, 486, 595, 715, 846, 988, 1141, 1305, 1480, 1666, 1863, 2071, 2290, 2520, 2761, 3013, 3276, 3550, 3835, 4131, 4438, 4756, 5085, 5425, 5776, 6138, 6511, 6895, 7290, 7696, 8113, 8541, 8980, 9430, 9891, 10363

Offset: 0

Omar E. Pol, Aug 04 2012

Sequence found by reading the line from 0, in the direction 0, 31... and the line from 10, in the direction 10, 63,..., in the square spiral whose vertices are the generalized 13-gonal numbers A195313.

G. C. Greubel, Table of n, a(n) for n = 0..5000
Index entries for linear recurrences with constant coefficients, signature (3,-3,1).

Bisection of A195313.
Second k-gonal numbers (k=5..14): A005449, A014105, A147875, A045944, A179986, A033954, A062728, A135705, this sequence, A211014.
Cf. A051865.

List([0..50], n-> n*(11*n+9)/2); # G. C. Greubel, Jul 04 2019

[n*(11*n+9)/2: n in [0..50]]; // G. C. Greubel, Jul 04 2019

Table[n*(11*n+9)/2, {n,0,50}] (* G. C. Greubel, Jul 04 2019 *)

a(n)=n*(11*n+9)/2 \\ Charles R Greathouse IV, Jun 17 2017

[n*(11*n+9)/2 for n in (0..50)] # G. C. Greubel, Jul 04 2019

G.f.: x*(10+x)/(1-x)^3. - Philippe Deléham, Mar 27 2013
a(n) = 3*a(n-1) - 3*a(n-2) + a(n-3) with a(0) = 0, a(1) = 10, a(2) = 31. - Philippe Deléham, Mar 27 2013
a(n) = A051865(n) + 9n = A180223(n) + 8n = A022268(n) + 5n = A022269(n) + 4n = A152740(n) - n. - Philippe Deléham, Mar 27 2013
a(n) = A218530(11n+9). - Philippe Deléham, Mar 27 2013
E.g.f.: x*(20 + 11*x)*exp(x)/2. - G. C. Greubel, Jul 04 2019

cp's OEIS Frontend

A014105 Second hexagonal numbers: a(n) = n*(2*n + 1).

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A045944 Rhombic matchstick numbers: a(n) = n*(3*n+2).

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A033954 Second 10-gonal (or decagonal) numbers: n*(4*n+3).

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A195160 Generalized 11-gonal (or hendecagonal) numbers: m*(9*m - 7)/2 with m = 0, 1, -1, 2, -2, 3, -3, ...

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

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A179986 Second 9-gonal (or nonagonal) numbers: a(n) = n*(7*n+5)/2.

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A014105 Second hexagonal numbers: a(n) = n(2n + 1).

A045944 Rhombic matchstick numbers: a(n) = n(3n+2).

A033954 Second 10-gonal (or decagonal) numbers: n(4n+3).

A195160 Generalized 11-gonal (or hendecagonal) numbers: m(9m - 7)/2 with m = 0, 1, -1, 2, -2, 3, -3, ...

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

A179986 Second 9-gonal (or nonagonal) numbers: a(n) = n(7n+5)/2.

A135705 a(n) = 10binomial(n,2) + 9n.

A211013 Second 13-gonal numbers: a(n) = n(11n+9)/2.