A005449 -id:A005449 - cp's OEIS frontend

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

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

A151542 Generalized pentagonal numbers: a(n) = 12n + 3n*(n-1)/2.

Original entry on oeis.org

0, 12, 27, 45, 66, 90, 117, 147, 180, 216, 255, 297, 342, 390, 441, 495, 552, 612, 675, 741, 810, 882, 957, 1035, 1116, 1200, 1287, 1377, 1470, 1566, 1665, 1767, 1872, 1980, 2091, 2205, 2322, 2442, 2565, 2691, 2820, 2952, 3087, 3225, 3366, 3510, 3657, 3807, 3960

Offset: 0

N. J. A. Sloane, May 15 2009

G. C. Greubel, Table of n, a(n) for n = 0..1000
Sela Fried, Counting r X s rectangles in nondecreasing and Smirnov words, arXiv:2406.18923 [math.CO], 2024. See p. 5.
Leo Tavares, Illustration: Star Triangles
Index entries for linear recurrences with constant coefficients, signature (3,-3,1).

The generalized pentagonal numbers b*n + 3*n*(n-1)/2, for b = 1 through 12, form sequences A000326, A005449, A045943, A115067, A140090, A140091, A059845, A140672, A140673, A140674, A140675, A151542.

Cf. A003154, A060544.

s=0;lst={};Do[AppendTo[lst,s+=n],{n,12,6!,3}];lst (* Vladimir Joseph Stephan Orlovsky, Mar 05 2010 *)
LinearRecurrence[{3,-3,1}, {0,12,27}, 50] (* or *) With[{nn = 50}, CoefficientList[Series[(3/2)*(8*x + x^2)*Exp[x], {x, 0, nn}], x] Range[0, nn]!] (* G. C. Greubel, May 26 2017 *)

x='x+O('x^50); concat([0], Vec(serlaplace((3/2)*(8*x + x^2)*exp(x)))) \\ G. C. Greubel, May 26 2017

a(n)=(3*n^2+21*n)/2 \\ Charles R Greathouse IV, Jun 16 2017

a(n) = a(n-1) + 3*n + 9 (with a(0)=0). - Vincenzo Librandi, Nov 26 2010
G.f.: 3*x*(4 - 3*x)/(1 - x)^3. - Arkadiusz Wesolowski, Dec 24 2011
From G. C. Greubel, May 26 2017: (Start)
a(n) = 3*n*(n+7)/2.
a(n) = 3*a(n-1) - 3*a(n-2) + a(n-3).
E.g.f.: (3/2)*(8*x + x^2)*exp(x). (End)
From Amiram Eldar, Feb 25 2022: (Start)
Sum_{n>=1} 1/a(n) = 121/490.
Sum_{n>=1} (-1)^(n+1)/a(n) = 4*log(2)/21 - 319/4410. (End)
a(n) = A003154(n+1) - A060544(n). - Leo Tavares, Mar 26 2022

A165355 a(n) = 3n + 1 if n is even, or a(n) = (3n + 1)/2 if n is odd.

Original entry on oeis.org

1, 2, 7, 5, 13, 8, 19, 11, 25, 14, 31, 17, 37, 20, 43, 23, 49, 26, 55, 29, 61, 32, 67, 35, 73, 38, 79, 41, 85, 44, 91, 47, 97, 50, 103, 53, 109, 56, 115, 59, 121, 62, 127, 65, 133, 68, 139, 71, 145, 74, 151, 77, 157, 80, 163, 83, 169, 86, 175, 89, 181, 92, 187, 95, 193, 98

Offset: 0

Paul Curtz, Sep 16 2009

Second trisection of A026741.
A111329(n+1) = A000041(a(n)). - Reinhard Zumkeller, Nov 19 2009
We observe that this sequence is a particular case of sequence for which there exists q: a(n+3) = (a(n+2)*a(n+1)+q)/a(n) for every n >= n0. Here q=-9 and n0=0. - Richard Choulet, Mar 01 2010
The entries are also encountered via the bilinear transform approximation to the natural log (unit circle). Specifically, evaluating 2(z-1)/(z+1) at z = 2, 3, 4, ..., A165355 entries stem from the pair (sums) seen 2 ahead of each new successive prime. For clarity, the evaluation output is 2, 3, 1, 1, 6, 5, 4, 3, 10, 7, 3, 2, 14, 9, 8, 5, 18, 11, ..., where (1+1), (4+3), (3+2), (8+5), ... generate the A165355 entries (after the first). As an aside, the same mechanism links A165355 to A140777. - Bill McEachen, Jan 08 2015
As a follow-up to the previous comment, it appears that the numerators and denominators of 2(z-1)/(z+1) are respectively given by A145979 and A060819, but with different offsets. - Michel Marcus, Jan 14 2015
Odd parts of the terms give A067745. E.g.: 1, 2/2, 7, 5, 13, 8/8 .... - Joe Slater, Nov 30 2016

G. C. Greubel, Table of n, a(n) for n = 0..1000
D. H. Lehmer, Continued fractions containing arithmetic progressions, Scripta Mathematica, 29 (1973): 17-24. [Annotated copy of offprint]
Index entries for linear recurrences with constant coefficients, signature (0,2,0,-1).

Cf. A165351, A165367, A067745, A007814.

f[n_] := If[ OddQ@ n, (3n +1)/2, (3n +1)]; Array[f, 66, 0] (* Robert G. Wilson v, Jan 26 2015 *)
f[n_] := (3 (-1)^(2n) + (-1)^(1 + n)) (-2 + 3n)/4; Array[f, 66] (* or *)
CoefficientList[ Series[(x^3 + 5x^2 + 2x + 1)/(x^2 - 1)^2, {x, 0, 65}], x] (* or *)
LinearRecurrence[{0, 2, 0, -1}, {1, 2, 7, 5}, 66] (* Robert G. Wilson v, Apr 13 2017 *)

a(n)=n+=2*n+1; if(n%2,n,n/2) \\ Charles R Greathouse IV, Jan 13 2015

a(n) = A026741(3*n+1).
a(n)*A026741(n) = A005449(n).
a(n)*A022998(n+1) = A000567(n+1).
a(n) = A026741(n+1) + A022998(n).
a(2n) = A016921(n). a(2n+1) = A016789(n).
a(2n+1)*A026741(2n) = A045944(n).
G.f.: (1+2*x+5*x^2+x^3)/((x-1)^2 * (1+x)^2). - R. J. Mathar, Sep 26 2009
a(n) = (3+9*n)/4 + (-1)^n*(1+3*n)/4. - R. J. Mathar, Sep 26 2009
a(n) = 2*(3n+1)/(4-((2n+2) mod 4)). - Bill McEachen, Jan 09 2015
If a(2n-1) = x then a(2n) = 2x+3. - Robert G. Wilson v, Jan 26 2015
Let the reduced Collatz procedure be defined as Cr(n) = (3*n+1)/2. For odd n, a(n) = Cr(n). For even n, a(n) = Cr(4*n+1)/2. - Joe Slater, Nov 29 2016
a(n) = A067745(n+1) * 2^A007814((3n+1)/2). - Joe Slater, Nov 30 2016
a(n) = 2*a(n-2) - a(n-4). - G. C. Greubel, Apr 13 2017

All comments changed to formulas by R. J. Mathar, Sep 26 2009
New name from Charles R Greathouse IV, Jan 13 2015
Name corrected by Joe Slater, Nov 29 2016

A261718 Number A(n,k) of partitions of n where each part i is marked with a word of length i over a k-ary alphabet whose letters appear in alphabetical order; square array A(n,k), n>=0, k>=0, read by antidiagonals.

Original entry on oeis.org

1, 1, 0, 1, 1, 0, 1, 2, 2, 0, 1, 3, 7, 3, 0, 1, 4, 15, 18, 5, 0, 1, 5, 26, 55, 50, 7, 0, 1, 6, 40, 124, 216, 118, 11, 0, 1, 7, 57, 235, 631, 729, 301, 15, 0, 1, 8, 77, 398, 1470, 2780, 2621, 684, 22, 0, 1, 9, 100, 623, 2955, 8001, 12954, 8535, 1621, 30, 0

Offset: 0

Alois P. Heinz, Aug 29 2015

			A(3,2) = 18: 3aaa, 3aab, 3abb, 3bbb, 2aa1a, 2aa1b, 2ab1a, 2ab1b, 2bb1a, 2bb1b, 1a1a1a, 1a1a1b, 1a1b1a, 1a1b1b, 1b1a1a, 1b1a1b, 1b1b1a, 1b1b1b.
Square array A(n,k) begins:
  1,  1,   1,    1,     1,      1,      1,       1, ...
  0,  1,   2,    3,     4,      5,      6,       7, ...
  0,  2,   7,   15,    26,     40,     57,      77, ...
  0,  3,  18,   55,   124,    235,    398,     623, ...
  0,  5,  50,  216,   631,   1470,   2955,    5355, ...
  0,  7, 118,  729,  2780,   8001,  19158,   40299, ...
  0, 11, 301, 2621, 12954,  45865, 130453,  317905, ...
  0, 15, 684, 8535, 55196, 241870, 820554, 2323483, ...

Alois P. Heinz, Antidiagonals n = 0..140, flattened

Columns k=0-10 give: A000007, A000041, A074141, A261737, A261738, A261739, A261740, A261741, A261742, A261743, A261744.
Rows n=0-2 give: A000012, A001477, A005449.
Main diagonal gives A209668.
Cf. A144064, A261719, A261780.

b:= proc(n, i, k) option remember; `if`(n=0, 1, `if`(i<1, 0,
      b(n, i-1, k)+`if`(i>n, 0, b(n-i, i, k)*binomial(i+k-1, k-1))))
    end:
A:= (n, k)-> b(n, n, k):
seq(seq(A(n, d-n), n=0..d), d=0..12);

b[n_, i_, k_] := b[n, i, k] = If[n == 0, 1, If[i < 1, 0, b[n, i - 1, k] + If[i > n, 0, b[n - i, i, k]*Binomial[i + k - 1, k - 1]]]]; A[n_, k_] := b[n, n, k]; Table[Table[A[n, d - n], {n, 0, d}], {d, 0, 12}] // Flatten (* Jean-François Alcover, Feb 22 2016, after Alois P. Heinz *)

A(n,k) = Sum_{i=0..k} C(k,i) * A261719(n,k-i).

A033580 Four times second pentagonal numbers: a(n) = 2n(3*n+1).

Original entry on oeis.org

0, 8, 28, 60, 104, 160, 228, 308, 400, 504, 620, 748, 888, 1040, 1204, 1380, 1568, 1768, 1980, 2204, 2440, 2688, 2948, 3220, 3504, 3800, 4108, 4428, 4760, 5104, 5460, 5828, 6208, 6600, 7004, 7420, 7848, 8288, 8740, 9204, 9680, 10168, 10668, 11180, 11704, 12240

Offset: 0

N. J. A. Sloane

Subsequence of A062717: A010052(6*a(n)+1) = 1. - Reinhard Zumkeller, Feb 21 2011
Sequence found by reading the line from 0, in the direction 0, 8,..., in the square spiral whose vertices are the generalized pentagonal numbers A001318. Opposite numbers to the members of A139267 in the same spiral - Omar E. Pol, Sep 09 2011
a(n) is the number of edges of the octagonal network O(n,n); O(m,n) is defined by Fig. 1 of the Siddiqui et al. reference. - Emeric Deutsch May 13 2018
The partial sums of this sequence give A035006. - Leo Tavares, Oct 03 2021

Ivan Panchenko, Table of n, a(n) for n = 0..1000
M. K. Siddiqui, M. Naeem, N. A. Rahman, and M. Imran, Computing topological indices of certain networks, J. of Optoelectronics and Advanced Materials, 18, No. 9-10 (2016), pp. 884-892.
Leo Tavares, Illustration: Crossed Stars
Leo Tavares, Illustration: Four Quarter Star Crosses
Leo Tavares, Illustration: Triangulated Star Crosses
Leo Tavares, Illustration: Oblong Star Crosses
Leo Tavares, Illustration: Crossed Diamond Stars
Index entries for linear recurrences with constant coefficients, signature (3,-3,1).

Cf. A033579, A049451, A045945, A186423.
Cf. sequences listed in A254963.
Cf. A003154, A016813.
Cf. A035006.
Cf. A005449, A046092, A033996, A002378, A016742, A000290, A016754, A001105.

List([0..50], n-> 2*n*(3*n+1)); # G. C. Greubel, Oct 09 2019

[2*n*(3*n+1): n in [0..50]]; // G. C. Greubel, Oct 09 2019

seq(2*n*(3*n+1), n=0..50); # G. C. Greubel, Oct 09 2019

4*Binomial[3*Range[50]-2, 2]/3 (* G. C. Greubel, Oct 09 2019 *)

a(n)=2*n*(3*n+1) \\ Charles R Greathouse IV, Sep 28 2015

[2*n*(3*n+1) for n in (0..50)] # G. C. Greubel, Oct 09 2019

a(n) = a(n-1) +12*n -4 (with a(0)=0). - Vincenzo Librandi, Aug 05 2010
G.f.: 4*x*(2+x)/(1-x)^3. - Colin Barker, Feb 13 2012
a(-n) = A033579(n). - Michael Somos, Jun 09 2014
E.g.f.: 2*x*(4 + 3*x)*exp(x). - G. C. Greubel, Oct 09 2019
From Amiram Eldar, Jan 14 2021: (Start)
Sum_{n>=1} 1/a(n) = 3/2 - Pi/(4*sqrt(3)) - 3*log(3)/4.
Sum_{n>=1} (-1)^(n+1)/a(n) =  -3/2 + Pi/(2*sqrt(3)) + log(2). (End)
From Leo Tavares, Oct 12 2021: (Start)
a(n) = A003154(n+1) - A016813(n). See Crossed Stars illustration.
a(n) = 4*A005449(n). See Four Quarter Star Crosses illustration.
a(n) = 2*A049451(n).
a(n) = A046092(n-1) + A033996(n). See Triangulated Star Crosses illustration.
a(n) = 4*A000217(n-1) + 8*A000217(n).
a(n) = 4*A000217(n-1) + 4*A002378. See Oblong Star Crosses illustration.
a(n) = A016754(n) + 4*A000217(n). See Crossed Diamond Stars illustration.
a(n) = 2*A001105(n) + 4*A000217(n).
a(n) = A016742(n) + A046092(n).
a(n) = 4*A000290(n) + 4*A000217(n). (End)

A140673 a(n) = 3n(n + 5)/2.

Original entry on oeis.org

0, 9, 21, 36, 54, 75, 99, 126, 156, 189, 225, 264, 306, 351, 399, 450, 504, 561, 621, 684, 750, 819, 891, 966, 1044, 1125, 1209, 1296, 1386, 1479, 1575, 1674, 1776, 1881, 1989, 2100, 2214, 2331, 2451, 2574, 2700, 2829, 2961, 3096

Offset: 0

Omar E. Pol, May 22 2008

a(n) equals the number of vertices of the A256666(n)-th graph (see Illustration of initial terms in A256666 Links). - Ivan N. Ianakiev, Apr 20 2015

G. C. Greubel, Table of n, a(n) for n = 0..5000
Sela Fried, Counting r X s rectangles in nondecreasing and Smirnov words, arXiv:2406.18923 [math.CO], 2024. See p. 5.
Index entries for linear recurrences with constant coefficients, signature (3,-3,1).

Cf. A055998.

The generalized pentagonal numbers b*n+3*n*(n-1)/2, for b = 1 through 12, form sequences A000326, A005449, A045943, A115067, A140090, A140091, A059845, A140672, A140673, A140674, A140675, A151542.

Table[Sum[i + n - 3, {i, 6, n}], {n, 5, 52}] (* Zerinvary Lajos, Jul 11 2009 *)
Table[3 n (n + 5)/2, {n, 0, 50}] (* Bruno Berselli, Sep 05 2018 *)
LinearRecurrence[{3,-3,1},{0,9,21},50] (* Harvey P. Dale, Jul 20 2023 *)

concat(0, Vec(3*x*(3 - 2*x)/(1 - x)^3 + O(x^100))) \\ Michel Marcus, Apr 20 2015

a(n) = 3*n*(n+5)/2; \\ Altug Alkan, Sep 05 2018

a(n) = A055998(n)*3 = (3*n^2 + 15*n)/2 = n*(3*n + 15)/2.
a(n) = 3*n + a(n-1) + 6 for n>0, a(0)=0. - Vincenzo Librandi, Aug 03 2010
G.f.: 3*x*(3 - 2*x)/(1 - x)^3. - Arkadiusz Wesolowski, Dec 24 2011
E.g.f.: (1/2)*(3*x^2 + 18*x)*exp(x). - G. C. Greubel, Jul 17 2017
From Amiram Eldar, Feb 25 2022: (Start)
Sum_{n>=1} 1/a(n) = 137/450.
Sum_{n>=1} (-1)^(n+1)/a(n) = 4*log(2)/15 - 47/450. (End)

A140674 a(n) = n(3n + 17)/2.

Original entry on oeis.org

0, 10, 23, 39, 58, 80, 105, 133, 164, 198, 235, 275, 318, 364, 413, 465, 520, 578, 639, 703, 770, 840, 913, 989, 1068, 1150, 1235, 1323, 1414, 1508, 1605, 1705, 1808, 1914, 2023, 2135, 2250, 2368, 2489, 2613, 2740, 2870, 3003, 3139

Offset: 0

Omar E. Pol, May 22 2008

G. C. Greubel, Table of n, a(n) for n = 0..5000
Sela Fried, Counting r X s rectangles in nondecreasing and Smirnov words, arXiv:2406.18923 [math.CO], 2024. See p. 5.
Index entries for linear recurrences with constant coefficients, signature (3,-3,1).

The generalized pentagonal numbers b*n+3*n*(n-1)/2, for b = 1 through 12, form sequences A000326, A005449, A045943, A115067, A140090, A140091, A059845, A140672, A140673, A140674, A140675, A151542.

[n*(3*n + 17)/2: n in [0..70]]; // Wesley Ivan Hurt, Apr 21 2021

Table[(3*n^2 + 17*n)/2, {n, 0, 50}] (* G. C. Greubel, Jul 17 2017 *)

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

a(n) = (3*n^2 + 17*n)/2.
a(n) = 7*n + 3*A000217(n). - Reinhard Zumkeller, May 28 2008
a(n) = 3*n + a(n-1) + 7 (with a(0)=0). - Vincenzo Librandi, Aug 03 2010
G.f.: x*(10 - 7*x)/(1 - x)^3. - Arkadiusz Wesolowski, Dec 24 2011
E.g.f.: (1/2)*(3*x^2 + 20*x)*exp(x). - G. C. Greubel, Jul 17 2017

A101986 Maximum sum of products of successive pairs in a permutation of order n+1.

Original entry on oeis.org

0, 2, 9, 23, 46, 80, 127, 189, 268, 366, 485, 627, 794, 988, 1211, 1465, 1752, 2074, 2433, 2831, 3270, 3752, 4279, 4853, 5476, 6150, 6877, 7659, 8498, 9396, 10355, 11377, 12464, 13618, 14841, 16135, 17502, 18944, 20463, 22061, 23740, 25502

Offset: 0

Eugene McDonnell (eemcd(AT)mac.com), Jan 29 2005

1 3 5 4 2 is the 11th permutation, in lexical order. of order 5. Its reverse 2 4 5 3 1 is the 41st. The earliest permutation of order 6 is the 41st, 1 3 5 6 4 2. This pattern continues as far as I have looked, so its reversal 2 4 6 5 3 1 is the 191st and the earliest permutation of order 7 is the 191st, et cetera.
Comments from Dmitry Kamenetsky, Dec 15 2006: (Start)
This sequence is related to A026035, except here we take the maximum sum of products of successive pairs. Here is a method for generating such permutations. Start with two lists, the first has numbers 1 to n, while the second is empty.
Repeat the following operations until the first list is empty: 1. Move the smallest number of the first list to the leftmost available position in the second list. The move operation removes the original number from the first list. 2. Move the smallest number of the first list to the rightmost available position in the second list. For example when n=8, the permutation is 1, 3, 5, 7, 8, 6, 4, 2. (End)
Convolution of odd numbers and integers greater than 1. - Reinhard Zumkeller, Mar 30 2012
For n>0, a(n) is row 2 of the convolution array A213751. - Clark Kimberling, Jun 20 2012

			The permutations of order 5 with maximum sum of products is 1 3 5 4 2 and its reverse, since (1*3)+(3*5)+(5*4)+(4*2) is 46. All others are empirically less than 46. So a(4) = 46.

Reinhard Zumkeller, Table of n, a(n) for n = 0..1000
Leroy Quet, Max/Min Sums From Permutations - a message on Jan 28 2005 to the sequence list, asking for someone to provide more values and to submit these to OEIS.
Index entries for linear recurrences with constant coefficients, signature (4,-6,4,-1).

Pairwise sums of A005581.

Cf. A000290, A000330, A008865, A026035, A160805.

a101986 n = sum $ zipWith (*) [1,3..] (reverse [2..n+1])
-- Reinhard Zumkeller, Mar 30 2012

0 1 9 2 & p. % 6 & p. (A) NB. the polynomial P such that P(n) is a(n).
NB. where 0 1 9 2 are the coefficients in ascending order of the numerator of a rational polynomial and 6 is the (constant) coefficient of its denominator. J's primitive function p. produces a polynomial with these coefficients. Division is indicated by % . Thus the J expression (A) is equivalent to the formula above.

a:=n->add((n+j^2),j=1..n): seq(a(n),n=0..41); # Zerinvary Lajos, Jul 27 2006

Table[(n + 9 n^2 + 2 n^3)/6, {n, 0, 41}] (* Robert G. Wilson v, Feb 04 2005 *)

a(n)=n*(2*n^2+9*n+1)/6 \\ Charles R Greathouse IV, Jan 17 2012

a(n) = n*(2*n^2 + 9*n + 1)/6.
a(n+1) = a(n) + A008865(n+2); a(n) = A160805(n) - 4. [Reinhard Zumkeller, May 26 2009]
G.f.: x*(1+x)*(2-x)/(1-x)^4. - L. Edson Jeffery, Jan 17 2012
a(n) = 4*a(n-1)-6*a(n-2)+4*a(n-3)-a(n-4) for n>3, a(0)=0, a(1)=2, a(2)=9, a(3)=23. - L. Edson Jeffery, Jan 17 2012
a(n) = A000330(n) + A005449(n) - A000217(n). - Richard R. Forberg, Aug 07 2013
a(n) = 1 + sum( A008865(i), i=1..n+1 ). [Bruno Berselli, Jan 13 2015]
a(n) = A000290(n) + A000330(n). - J. M. Bergot, Apr 26 2018

Edited by Bruno Berselli, Jan 13 2015

Name edited by Alois P. Heinz, Feb 02 2019

A225470 Triangle read by rows, s_3(n, k) where s_m(n, k) are the Stirling-Frobenius cycle numbers of order m; n >= 0, k >= 0.

Original entry on oeis.org

1, 2, 1, 10, 7, 1, 80, 66, 15, 1, 880, 806, 231, 26, 1, 12320, 12164, 4040, 595, 40, 1, 209440, 219108, 80844, 14155, 1275, 57, 1, 4188800, 4591600, 1835988, 363944, 39655, 2415, 77, 1, 96342400, 109795600, 46819324, 10206700, 1276009, 95200, 4186, 100, 1

Offset: 0

Peter Luschny, May 16 2013

The Stirling-Frobenius subset numbers S_{m}(n,k), for m >= 1 fixed, regarded as an infinite lower triangular matrix, can be inverted by Sum_{k} S_{m}(n,k)*s_{m}(k,j)*(-1)^(n-k) = [j = n]. The inverse numbers s_{m}(k,j), which are unsigned, are the Stirling-Frobenius cycle numbers. For m = 1 this gives the classical Stirling cycle numbers A132393. The Stirling-Frobenius subset numbers are defined in A225468.

Triangle T(n,k), read by rows, given by (2, 3, 5, 6, 8, 9, 11, 12, 14, 15, ... (A007494)) DELTA (1, 0, 1, 0, 1, 0, 1, 0, 1, 0, 1, 0, ...) where DELTA is the operator defined in A084938. - Philippe Deléham, May 14 2015

			Triangle starts:
  [n\k][    0,      1,     2,     3,    4,  5,  6]
  [0]       1,
  [1]       2,      1,
  [2]      10,      7,     1,
  [3]      80,     66,    15,     1,
  [4]     880,    806,   231,    26,    1,
  [5]   12320,  12164,  4040,   595,   40,  1,
  [6]  209440, 219108, 80844, 14155, 1275, 57,  1.
  ...
From _Wolfdieter Lang_, Aug 11 2017: (Start)
Recurrence (see Maple program): T(4, 2) = T(3, 1) + (3*4 - 1)*T(3, 2) = 66 + 11*15 = 231.
Boas-Buck type recurrence for column k = 2 and n = 4: T(4, 2) = (4!/2)*(3*(2 + 6*(5/12))*T(2, 2)/2! + 1*(2 + 6*(1/2))*T(3,2)/3!) = (4!/2)*(3*9/4 + 5*15/3!) = 231. (End)

Peter Bala, A 3 parameter family of generalized Stirling numbers
Peter Luschny, Generalized Eulerian polynomials
Peter Luschny, The Stirling-Frobenius numbers

Cf. A225468; A132393 (m=1), A028338 (m=2), A225471(m=4).
Column k=0..4 give A008544, A024395(n-1), A286722(n-2), A383221, A383222.
T(n, n) ~ A000012; T(n, n-1) ~ A005449; T(n, n-2) ~ A024391; T(n, n-3) ~ A024392.
row sums ~ A032031; alternating row sums ~ A007559.
Cf. A132393.

SF_C := proc(n, k, m) option remember;
if n = 0 and k = 0 then return(1) fi;
if k > n or  k < 0 then return(0) fi;
SF_C(n-1, k-1, m) + (m*n-1)*SF_C(n-1, k, m) end:
seq(print(seq(SF_C(n, k, 3), k = 0..n)), n = 0..8);

SFC[0, 0, ] = 1; SFC[n, k_, ] /; (k > n || k < 0) = 0; SFC[n, k_, m_] := SFC[n, k, m] = SFC[n-1, k-1, m] + (m*n-1)*SFC[n-1, k, m]; Table[SFC[n, k, 3], {n, 0, 8}, {k, 0, n}] // Flatten (* Jean-François Alcover, Jul 26 2013, after Maple *)

For a recurrence see the Maple program.
From Wolfdieter Lang, May 18 2017: (Start)
This is the Sheffer triangle (1/(1 - 3*x)^{-2/3}, -(1/3)*log(1-3*x)). See the P. Bala link where this is called exponential Riordan array, and the signed version is  denoted by s_{(3,0,2)}.
E.g.f. of row polynomials in the variable x (i.e., of the triangle): (1 - 3*z)^{-(2+x)/3}.
E.g.f. of column k: (1-3*x)^(-2/3)*(-(1/3)*log(1-3*x))^k/k!, k >= 0.
Recurrence for row polynomials R(n, x) = Sum_{k=0..n} T(n, k)*x^k: R(n, x) = (x+2)*R(n-1,x+3), with R(0, x) = 1.
R(n, x) = risefac(3,2;x,n) := Product_{j=0..(n-1)} (x + (2 + 3*j)). (See the P. Bala link, eq. (16) for the signed s_{3,0,2} row polynomials.)
T(n, k) = Sum_{j=0..(n-m)} binomial(n-j, k)* S1p(n, n-j)*2^(n-k-j)*3^j, with S1p(n, m) = A132393(n, m). (End)
Boas-Buck type recurrence for column sequence k: T(n, k) = (n!/(n - k)) * Sum_{p=k..n-1} 3^(n-1-p)*(2 + 3*k*beta(n-1-p))*T(p, k)/p!, for n > k >= 0, with input T(k, k) = 1, and beta(k) = A002208(k+1)/A002209(k+1), beginning {1/2, 5/12, 3/8, 251/720, ...}. See a comment and references in A286718. - Wolfdieter Lang, Aug 11 2017

A045945 Hexagonal matchstick numbers: a(n) = 3n(3*n+1).

Original entry on oeis.org

0, 12, 42, 90, 156, 240, 342, 462, 600, 756, 930, 1122, 1332, 1560, 1806, 2070, 2352, 2652, 2970, 3306, 3660, 4032, 4422, 4830, 5256, 5700, 6162, 6642, 7140, 7656, 8190, 8742, 9312, 9900, 10506, 11130, 11772, 12432, 13110, 13806, 14520, 15252, 16002, 16770, 17556

Offset: 0

R. K. Guy

This may also be construed as the number of line segments illustrating the isometric projection of a cube of side length n. Moreover, a(n) equals the number of rods making a cube of side length n+1 minus the number of rods making a cube of side length n. See the illustration in the links and formula below.

Ivan Panchenko, Table of n, a(n) for n = 0..1000
Peter M. Chema, Illustration of initial terms as the first difference of number of rods required to make a 3-D cube.
Craig Knecht, Number of positions a frame shifted H1 hexagon can occupy in a hexagon of order n.
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), [math.NT].
Index entries for linear recurrences with constant coefficients, signature (3,-3,1).

Cf. A000217, A000290, A005449, A033580, A059986.

The hexagon matchstick sequences are: Number of matchsticks: this sequence; size=1 triangles: A033581; larger triangles: A307253; total triangles: A045949. Analog for triangles: A045943; analog for stars: A045946. - John King, Apr 05 2019

a:= n-> 3*n*(3*n+1): seq(a(n), n=0..42); # Zerinvary Lajos, May 03 2007

f[n_]:=3*n*(3*n+1);f[Range[0,60]] (* Vladimir Joseph Stephan Orlovsky, Feb 05 2011 *)

a(n) = 3*n*(3*n+1) \\ Charles R Greathouse IV, Feb 27 2017

def a(n): return 3*n*(3*n+1) # Indranil Ghosh, Mar 26 2017

a(n) = a(n-1) + 6*(3*n-1) (with a(0)=0). - Vincenzo Librandi, Nov 18 2010
G.f.: 6*x*(2+x)/(1-x)^3. - Colin Barker, Feb 12 2012
a(n) = 6*A005449(n). - R. J. Mathar, Feb 13 2016
a(n) = A059986(n) - A059986(n-1). - Peter M. Chema, Feb 26 2017
a(n) = 6*(A000217(n) + A000290(n)). - Peter M. Chema, Mar 26 2017
From Amiram Eldar, Jan 14 2021: (Start)
Sum_{n>=1} 1/a(n) = 1 - Pi/(6*sqrt(3)) - log(3)/2.
Sum_{n>=1} (-1)^(n+1)/a(n) = -1 + Pi/(3*sqrt(3)) + 2*log(2)/3. (End)
From Elmo R. Oliveira, Dec 12 2024: (Start)
E.g.f.: 3*exp(x)*x*(4 + 3*x).
a(n) = 3*a(n-1) - 3*a(n-2) + a(n-3) for n >= 3. (End)

cp's OEIS Frontend

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

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A151542 Generalized pentagonal numbers: a(n) = 12*n + 3*n*(n-1)/2.

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A165355 a(n) = 3n + 1 if n is even, or a(n) = (3n + 1)/2 if n is odd.

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A261718 Number A(n,k) of partitions of n where each part i is marked with a word of length i over a k-ary alphabet whose letters appear in alphabetical order; square array A(n,k), n>=0, k>=0, read by antidiagonals.

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A033580 Four times second pentagonal numbers: a(n) = 2*n*(3*n+1).

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A140673 a(n) = 3*n*(n + 5)/2.

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A140674 a(n) = n*(3*n + 17)/2.

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

A151542 Generalized pentagonal numbers: a(n) = 12n + 3n*(n-1)/2.

A033580 Four times second pentagonal numbers: a(n) = 2n(3*n+1).

A140673 a(n) = 3n(n + 5)/2.

A140674 a(n) = n(3n + 17)/2.

A045945 Hexagonal matchstick numbers: a(n) = 3n(3*n+1).