A003154 -id:A003154 - cp's OEIS frontend

A081266 Staggered diagonal of triangular spiral in A051682.

0, 6, 21, 45, 78, 120, 171, 231, 300, 378, 465, 561, 666, 780, 903, 1035, 1176, 1326, 1485, 1653, 1830, 2016, 2211, 2415, 2628, 2850, 3081, 3321, 3570, 3828, 4095, 4371, 4656, 4950, 5253, 5565, 5886, 6216, 6555, 6903, 7260, 7626, 8001, 8385, 8778, 9180

Offset: 0

Paul Barry, Mar 15 2003

Staggered diagonal of triangular spiral in A051682, between (0,4,17) spoke and (0,7,23) spoke.
Binomial transform of (0, 6, 9, 0, 0, 0, ...).
If Y is a fixed 3-subset of a (3n+1)-set X then a(n) is the number of (3n-1)-subsets of X intersecting Y. - Milan Janjic, Oct 28 2007
Partial sums give A085788. - Leo Tavares, Nov 23 2023

			a(1)=9*1+0-3=6, a(2)=9*2+6-3=21, a(3)=9*3+21-3=45.
For n=3, a(3) = -0^2+1^2-2^2+3^2-4^2+5^2-6^2+7^2-8^2+9^2 = 45.

Muniru A Asiru, Table of n, a(n) for n = 0..10000
Tomislav Došlić and Luka Podrug, Sweet division problems: from chocolate bars to honeycomb strips and back, arXiv:2304.12121 [math.CO], 2023.
Milan Janjic, Two Enumerative Functions
Milan Janjic and B. Petkovic, A Counting Function, arXiv:1301.4550 [math.CO], 2013.
Amelia Carolina Sparavigna, The groupoid of the Triangular Numbers and the generation of related integer sequences, Politecnico di Torino, Italy (2019).
Leo Tavares, Star illustration
Index entries for linear recurrences with constant coefficients, signature (3,-3,1).

Cf. A000217, A000290, A005449, A014105, A022266, A033585, A062725, A144312, A144314, A218470.

Cf. A003154, A133694.

List([0..50],n->Binomial(3*n+1,2)); # Muniru A Asiru, Feb 28 2019

seq(binomial(3*n+1,2), n=0..45); # Zerinvary Lajos, Jan 21 2007

LinearRecurrence[{3,-3,1},{0,6,21},50] (* Harvey P. Dale, Aug 29 2015 *)

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

a(n) = 6*C(n,1) + 9*C(n,2).
a(n) = 3*n*(3*n+1)/2.
G.f.: (6*x+3*x^2)/(1-x)^3.
a(n) = A000217(3*n); a(2*n) = A144314(n). - Reinhard Zumkeller, Sep 17 2008
a(n) = 3*A005449(n). - R. J. Mathar, Mar 27 2009
a(n) = 9*n+a(n-1)-3 for n>0, a(0)=0. - Vincenzo Librandi, Aug 08 2010
a(n) = A218470(9n+5). - Philippe Deléham, Mar 27 2013
a(n) = Sum_{k=0..3n} (-1)^(n+k)*k^2. - Bruno Berselli, Aug 29 2013
E.g.f.: 3*exp(x)*x*(4 + 3*x)/2. - Stefano Spezia, Jun 06 2021
From Amiram Eldar, Aug 11 2022: (Start)
Sum_{n>=1} 1/a(n) = 2 - Pi/(3*sqrt(3)) - log(3).
Sum_{n>=1} (-1)^(n+1)/a(n) = 2*Pi/(3*sqrt(3)) + 4*log(2)/3 - 2. (End)
From Leo Tavares, Nov 23 2023: (Start)
a(n) = 3*A000217(n) + 3*A000290(n).
a(n) = A003154(n+1) - A133694(n+1). (End)

A069190 Centered 24-gonal numbers.

Original entry on oeis.org

1, 25, 73, 145, 241, 361, 505, 673, 865, 1081, 1321, 1585, 1873, 2185, 2521, 2881, 3265, 3673, 4105, 4561, 5041, 5545, 6073, 6625, 7201, 7801, 8425, 9073, 9745, 10441, 11161, 11905, 12673, 13465, 14281, 15121, 15985, 16873, 17785, 18721, 19681, 20665, 21673

Offset: 1

Terrel Trotter, Jr., Apr 10 2002

Sequence found by reading the line from 1, in the direction 1, 25, ..., in the square spiral whose vertices are the generalized octagonal numbers A001082. Semi-axis opposite to A135453 in the same spiral. - Omar E. Pol, Sep 16 2011

			a(5) = 241 because 12*5^2 - 12*5 + 1 = 300 - 60 + 1 = 241.

Ivan Panchenko, Table of n, a(n) for n = 1..1000
John Elias, Illustration: Odd Ordered Star Perimeters.
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. centered k-gonal numbers with k=3..25: A005448, A001844, A005891, A003215, A069099, A016754, A060544, A062786, A069125, A003154, A069126, A069127, A069128, A069129, A069130, A069131, A069132, A069133, A069178, A069173, A069174, A069190, A262221.

FoldList[#1 + #2 &, 1, 24 Range@ 45] (* Robert G. Wilson v *)
Table[12n^2-12n+1,{n,50}] (* or *) LinearRecurrence[{3,-3,1},{1,25,73},50] (* Harvey P. Dale, Jul 17 2011 *)

a(n)=12*n^2-12*n+1 \\ Charles R Greathouse IV, Sep 24 2015

a(n) = 12*n^2 - 12*n + 1.
a(n) = 24*n + a(n-1) - 24 with a(1)=1. - Vincenzo Librandi, Aug 08 2010
a(n) = 3*a(n-1) - 3*a(n-2) + a(n-3); a(1)=1, a(2)=25, a(3)=73. - Harvey P. Dale, Jul 17 2011
G.f.: x*(1+22*x+x^2)/(1-x)^3. - Harvey P. Dale, Jul 17 2011
Binomial transform of [1, 24, 24, 0, 0, 0, ...] and Narayana transform (cf. A001263) of [1, 24, 0, 0, 0, ...]. - Gary W. Adamson, Jul 26 2011
From Amiram Eldar, Jun 21 2020: (Start)
Sum_{n>=1} 1/a(n) = Pi*tan(Pi/sqrt(6))/(4*sqrt(6)).
Sum_{n>=1} a(n)/n! = 13*e - 1.
Sum_{n>=1} (-1)^n * a(n)/n! = 13/e - 1. (End)
E.g.f.: exp(x)*(1 + 12*x^2) - 1. - Stefano Spezia, May 31 2022

More terms from Harvey P. Dale, Jul 17 2011

A152749 a(n) = (n+1)(3n+1)/4 for n odd, a(n) = n(3n+2)/4 for n even.

Original entry on oeis.org

0, 2, 4, 10, 14, 24, 30, 44, 52, 70, 80, 102, 114, 140, 154, 184, 200, 234, 252, 290, 310, 352, 374, 420, 444, 494, 520, 574, 602, 660, 690, 752, 784, 850, 884, 954, 990, 1064, 1102, 1180, 1220, 1302, 1344, 1430, 1474, 1564, 1610, 1704, 1752, 1850, 1900, 2002

Offset: 0

Vincenzo Librandi, Dec 31 2009

Interleaving of A049450 and A049451 (for n > 0).
Also, integer values of k*(k+1)/3. - Charles R Greathouse IV, Dec 11 2010
The nonzero coefficients of the expansion of f(a) = Product_{k>=1} (1-a^(2k)), see A194159, occur at the terms of the sequence given above, i.e., f(a) = 1 - a^2 - a^4 + a^10 + a^14 - a^24 - a^30 + a^44 + a^52 - a^70 - a^80 + ... = Sum_{n>=0} (-1)^binomial(n+1,2)*a^A152749(n). - Johannes W. Meijer, Aug 21 2011
Partial sums of A109043. - Reinhard Zumkeller, Mar 31 2012
Nonnegative k such that 12*k+1 is a square. - Vicente Izquierdo Gomez, Jul 22 2013
Equivalently, numbers of the form h*(3*h+1), where h = 0, -1, 1, -2, 2, -3, 3, -4, 4, ... (see also the fifth comment of A062717). - Bruno Berselli, Feb 02 2017
For n > 0, a(n-1) is the sum of the largest parts of the partitions of 2n into two even parts. - Wesley Ivan Hurt, Dec 19 2017
The sequence terms occur as exponents in the expansion of Sum_{n >= 0} q^(n*(n+1)/2) * Product_{k >= n+1} 1 - q^k = 1 - q^2 - q^4 + q^10 + q^14 - q^24 - q^30 + + - -  .... - Peter Bala, Dec 15 2024
Sequence terms occur as exponents in the expansions of Sum_{n >= 0} q^(n*(2*n+1)) * Product_{k >= 2*n+2} 1 - q^k  = Sum_{n >= 0} q^(n*(2*n-1)) * Product_{k >= 2*n+1} 1 - q^k = 1 - q^2 - q^4 + q^10 + q^14 - q^24 - q^30 + + - - .... - Peter Bala, Jun 23 2025

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).

Cf. A049450 (n*(3*n-1)), A049451 (n*(3*n+1)), A153383 (12n+1 is not prime).

a152749 n = a152749_list !! (n-1)
a152749_list = scanl1 (+) a109043_list
-- Reinhard Zumkeller, Mar 31 2012

[IsOdd(n) select (n+1)*(3*n+1)/4 else n*(3*n+2)/4: n in [0..52]];

f:=func; [0] cat [f(n*m): m in [-1,1], n in [1..30]]; // Bruno Berselli, Nov 13 2012

A152749 := proc(n): if type(n,even) then n*(3*n+2)/4  else (n+1)*(3*n+1)/4 fi: end: seq(A152749(n), n=0..51); # Johannes W. Meijer, Aug 21 2011

Table[If[OddQ[n],(n+1)*(3*n+1)/4,n*(3*n+2)/4],{n,0,60}] (* Vladimir Joseph Stephan Orlovsky, Feb 03 2012 *)
LinearRecurrence[{1,2,-2,-1,1}, {0, 2, 4, 10, 14}, 50] (* Vincenzo Librandi, Feb 22 2012 *)
Select[Range[1,1000], IntegerQ[Sqrt[12#+1]]&] (* Vicente Izquierdo Gomez, Jul 22 2013 *)

From R. J. Mathar, Jan 03-06 2009: (Start)
G.f.: 2*x*(1+x+x^2)/((1+x)^2*(1-x)^3).
a(n) = a(n-1) + 2*a(n-2) - 2*a(n-3) - a(n-4) + a(n-5) = A003154(n+1)/8 - (-1)^n*A005408(n)/8.
a(n) = 2*A001318(n) = ((6*n^2+6*n+1) - (2*n+1)*(-1)^n)/8. (End)
From Amiram Eldar, Mar 15 2022: (Start)
Sum_{n>=1} 1/a(n) = 3 - Pi/sqrt(3).
Sum_{n>=1} (-1)^(n+1)/a(n) = 3*(log(3)-1). (End)

Edited, typo corrected and extended by Klaus Brockhaus, Jan 02 2009

Leading term a(0)=0 added by Johannes W. Meijer, Aug 21 2011

A287326 Triangle read by rows: T(n, k) = 6k(n-k) + 1; n >= 0, 0 <= k <= n.

Original entry on oeis.org

1, 1, 1, 1, 7, 1, 1, 13, 13, 1, 1, 19, 25, 19, 1, 1, 25, 37, 37, 25, 1, 1, 31, 49, 55, 49, 31, 1, 1, 37, 61, 73, 73, 61, 37, 1, 1, 43, 73, 91, 97, 91, 73, 43, 1, 1, 49, 85, 109, 121, 121, 109, 85, 49, 1, 1, 55, 97, 127, 145, 151, 145, 127, 97, 55, 1, 1, 61, 109, 145, 169, 181, 181, 169, 145, 109, 61, 1

Offset: 0

Kolosov Petro, Aug 31 2017

From Kolosov Petro, Apr 12 2020: (Start)
Let A(m, r) = A302971(m, r) / A304042(m, r).
Let L(m, n, k) = Sum_{r=0..m} A(m, r) * k^r * (n - k)^r.
Then T(n, k) = L(1, n, k), i.e T(n, k) is partial case of L(m, n, k) for m = 1.
T(n, k) is symmetric: T(n, k) = T(n, n-k). (End)

			Triangle begins:
  ----------------------------------------
  k=    0   1   2   3   4   5   6   7   8
  ----------------------------------------
  n=0:  1;
  n=1:  1,  1;
  n=2:  1,  7,  1;
  n=3:  1, 13, 13,  1;
  n=4:  1, 19, 25, 19,  1;
  n=5:  1, 25, 37, 37, 25,  1;
  n=6:  1, 31, 49, 55, 49, 31,  1;
  n=7:  1, 37, 61, 73, 73, 61, 37,  1;
  n=8:  1, 43, 73, 91, 97, 91, 73, 43,  1;

Georg Fischer, Table of n, a(n) for n = 0..495 [rows 0..10 and 12..30 from Kolosov Petro]
Petro Kolosov, On the link between binomial theorem and discrete convolution, arXiv:1603.02468 [math.NT], 2016-2025.
Petro Kolosov, Polynomial identity involving binomial theorem and Faulhaber's formula, 2023.
Petro Kolosov, History and overview of the polynomial P_b^m(x), 2024.

Columns k=0..6 give A000012, A016921, A017533, A161705, A103214, A128470, A158065.
Column sums k=0..4 give A000027, A000567, A051866, A051872, A255185.
Row sums give A001093.
Various cases of L(m, n, k): This sequence (m=1), A300656(m=2), A300785(m=3). See comments for L(m, n, k).
Differences of cubes n^3 are T(A000124(n), 1).
Cf. A000124, A000578, A003154, A003215, A007318, A008458, A016969.
Cf. A038593, A055012, A068601, A077028, A094053, A166873, A227776.
Cf. A294317, A302971, A304042.

Flat(List([0..11],n->List([0..n],k->6*k*(n-k)+1))); # Muniru A Asiru, Oct 09 2018

/* As triangle */ [[6*k*(n-k) + 1: k in [0..n]]: n in [0.. 15]]; // Vincenzo Librandi, Oct 26 2018

T := (n, k) -> 6*k*(n-k) + 1:
seq(seq(T(n, k), k=0..n), n=0..11); # Muniru A Asiru, Oct 09 2018

T[n_, k_] := 6 k (n - k) + 1; Column[Table[T[n, k], {n, 0, 10}, {k, 0, n}], Center] (* Kolosov Petro, Jun 02 2019 *)

t(n, k) = 6*k*(n-k)+1
trianglerows(n) = for(x=0, n-1, for(y=0, x, print1(t(x, y), ", ")); print(""))
/* Print initial 9 rows of triangle as follows */
trianglerows(9) \\ Felix Fröhlich, Jan 09 2018

def A287326(n,k): return 6*k*(n-k) + 1
flatten([[A287326(n,k) for k in range(n+1)] for n in range(13)]) # G. C. Greubel, Sep 25 2024

T(n, k) = 6*k*(n-k) + 1.
G.f. of column k: n^k*(1+(6*k-1)*n)/(1-n)^2.
G.f.: (1 - x - x*y + 7*x^2*y)/((1 - x)^2*(1 - x*y)^2). - Stefano Spezia, Oct 09 2018 [Adapted by Stefano Spezia, Sep 25 2024]
From Kolosov Petro, Jun 05 2019: (Start)
T(n, k) = 1/2 * T(A294317(n, k), k) + 1/2.
T(n+1, k) = 2*T(n, k) - T(n-1, k), for n >= k.
T(n, k) = 6*A077028(n, k) - 5.
T(2n, n) = A227776(n).
T(2n+1, n) = A003154(n+1).
T(2n+3, n) = A166873(n+1).
Sum_{k=0..n-1} T(n, k) = Sum_{k=1..n} T(n, k) = A000578(n).
Sum_{k=1..n-1} T(n, k) = A068601(n).
(n+1)^3 - n^3 = T(A000124(n), 1). (End)
Sum_{k=0..n} (-1)^k*T(n, k) = (-1/2)*(1 + (-1)^n)*A016969(floor(n/2) - 1). - G. C. Greubel, Sep 25 2024

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

A140811 a(n) = 6*n^2 - 1.

Original entry on oeis.org

-1, 5, 23, 53, 95, 149, 215, 293, 383, 485, 599, 725, 863, 1013, 1175, 1349, 1535, 1733, 1943, 2165, 2399, 2645, 2903, 3173, 3455, 3749, 4055, 4373, 4703, 5045, 5399, 5765, 6143, 6533, 6935, 7349, 7775, 8213, 8663, 9125, 9599, 10085, 10583, 11093, 11615

Offset: 0

Paul Curtz, Jul 16 2008

Also: The numerators in the j=2 column of the array a(i,j) defined in A140825, where the columns j=0 and j=1 are represented by A000012 and A005408. This could be extended to column j=3: 1, -1, 9, 55, 161, ... The common feature of these sequences derived from a(i,j) is that their j-th differences are constant sequences defined by A091137(j).
a(n) is the set of all k such that 6*k + 6 is a perfect square. - Gary Detlefs, Mar 04 2010
The identity (6*n^2 - 1)^2 - (9*n^2 - 3)*(2*n)^2 = 1 can be written as a(n+1)^2 - A157872(n)*A005843(n+1)^2 = 1. - Vincenzo Librandi, Feb 05 2012
Apart from first term, sequence found by reading the line from 5, in the direction 5, 23, ..., in the square spiral whose vertices are the generalized pentagonal numbers A001318. - Omar E. Pol, Jul 18 2012
From Paul Curtz, Sep 17 2018: (Start)
Terms from center to right in the following spiral:
.
                    65--63--61--59
                    /             \
                  67  31--29--27  57
                  /   /         \   \
                69  33   9---7  25  55
                /   /   /     \   \   \
              71  35  11  -1===5==23==53==>
                  /   /   /   /   /   /
                37  13   1---3  21  51
                  \   \         /   /
                  39  15--17--19  49
                    \             /
                    41--43--45--47 (End)

P. Curtz, Intégration numérique des systèmes différentiels à conditions initiales, Note 12, Centre de Calcul Scientifique de l'Armement, Arcueil, 1969, 132 pages, pp. 28-36. CCSA, then CELAR. Now DGA Maitrise de l'Information 35131 Bruz.

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

Cf. A005843, A157872, A060747, A103115(n+1), A141417 (array).
Cf. A000012, A001318, A005408, A091137, A140825.
Cf. A003154, A032528, A033581, A016777, A017593.

[6*n^2 - 1: n in [0..50]]; // Vincenzo Librandi, Jun 02 2011

LinearRecurrence[{3,-3,1},{-1,5,23},40] (* Vincenzo Librandi, Feb 05 2012 *)
CoefficientList[Series[(1-8*x-5*x^2)/(x-1)^3 , {x, 0, 40}], x] (* Stefano Spezia, Sep 17 2018 *)

a(n)=6*n^2-1 \\ Charles R Greathouse IV, Jun 01 2011

a(n) = 2*a(n-1) - a(n-2) + 12.
First differences: a(n+1) - a(n) = A017593(n).
Second differences: A071593(n+1) - A071593(n) = 12.
G.f.: (1-8*x-5*x^2)/(x-1)^3. - Jaume Oliver Lafont, Aug 30 2009
From Vincenzo Librandi, Feb 05 2012: (Start)
a(n) = a(n-1) + 12*n - 6.
a(n) = 3*a(n-1) - 3*a(n-2) + a(n-3). (End)
a(n) = A033581(n) - 1. - Omar E. Pol, Jul 18 2012
a(n) = A032528(2*n) - 1. - Adriano Caroli, Jul 21 2013
For n > 0, a(n) = floor(3/(cosh(1/n) - 1)) = floor(1/(n*sinh(1/n) - 1)); for similar formulas for cosine and sine, see A033581. - Clark Kimberling, Oct 19 2014, corrected by M. F. Hasler, Oct 21 2014
a(-n) = a(n). - Paul Curtz, Sep 17 2018
From Amiram Eldar, Feb 04 2021: (Start)
Sum_{n>=1} 1/a(n) = (1 - (Pi/sqrt(6))*cot(Pi/sqrt(6)))/2.
Sum_{n>=1} (-1)^(n+1)/a(n) = ((Pi/sqrt(6))*csc(Pi/sqrt(6)) - 1)/2.
Product_{n>=1} (1 + 1/a(n)) = (Pi/sqrt(6))*csc(Pi/sqrt(6)).
Product_{n>=1} (1 - 1/a(n)) = csc(Pi/sqrt(6))*sin(Pi/sqrt(3))/sqrt(2). (End)
a(n) = A003154(n+1) - 2*A016777(n). - Leo Tavares, May 13 2022
E.g.f.: exp(x)*(6*x^2 + 6*x - 1). - Elmo R. Oliveira, Jan 16 2025

Edited and extended by R. J. Mathar, Aug 06 2008

Better description Ray Chandler, Feb 03 2009

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

A005914 Number of points on surface of hexagonal prism: 12*n^2 + 2 for n > 0 (coordination sequence for W(2)).

Original entry on oeis.org

1, 14, 50, 110, 194, 302, 434, 590, 770, 974, 1202, 1454, 1730, 2030, 2354, 2702, 3074, 3470, 3890, 4334, 4802, 5294, 5810, 6350, 6914, 7502, 8114, 8750, 9410, 10094, 10802, 11534, 12290, 13070, 13874, 14702, 15554, 16430, 17330, 18254, 19202, 20174, 21170

Offset: 0

N. J. A. Sloane and Ralf W. Grosse-Kunstleve

For n >= 1, a(n) is equal to the number of functions f:{1,2,3,4}->{1,2,...,n,n+1} such that Im(f) contains 2 fixed elements. - Aleksandar M. Janjic and Milan Janjic, Feb 24 2007
Equals binomial transform of [1, 13, 23, 1, -1, 1, -1, 1, ...]. - Gary W. Adamson, Apr 22 2008
First bisection of A005918. After 1, all terms are in A000408 (see Formula section). - Bruno Berselli, Feb 07 2012
Also sequence found by reading the segment (1, 14) together with the line from 14, in the direction 14, 50, ..., in the square spiral whose vertices are the generalized octagonal numbers A001082. - Omar E. Pol, Nov 02 2012
Unique sequence such that for all n > 0, n*a(1) + (n-1)*a(2) + (n-3)*a(3) + ... + 2*a(2) + a(1) = n^4. - Warren Breslow, Dec 12 2014

Gmelin Handbook of Inorganic and Organometallic Chemistry, 8th Ed., 1994, TYPIX search code (229) cI2.
N. J. A. Sloane and Simon Plouffe, The Encyclopedia of Integer Sequences, Academic Press, 1995 (includes this sequence).

Vincenzo Librandi, Table of n, a(n) for n = 0..10000
Ovidiu Bagdasar, On Some Functions Involving the lcm and gcd of Integer Tuples, Scientific Publications of the State University of Novi Pazar, Appl. Maths. Inform. and Mech., Vol. 6, No. 2 (2014), pp. 91-100.
Ralf W. Grosse-Kunstleve, Coordination Sequences and Encyclopedia of Integer Sequences, 1996.
Ralf W. Grosse-Kunstleve, G. O. Brunner and N. J. A. Sloane, Algebraic Description of Coordination Sequences and Exact Topological Densities for Zeolites, Acta Cryst., A52 (1996), pp. 879-889.
Milan Janjic, Enumerative Formulas for Some Functions on Finite Sets.
Claudio de J. Pita Ruiz V., Some Number Arrays Related to Pascal and Lucas Triangles, J. Int. Seq., Vol. 16 (2013), Article 13.5.7.
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.
Boon K. Teo and N. J. A. Sloane, Magic numbers in polygonal and polyhedral clusters, Inorgan. Chem., Vol. 24 (1985), pp. 4545-4558.
Index entries for linear recurrences with constant coefficients, signature (3,-3,1).

First differences of A005917.

Cf. A000408, A001082, A003154, A005918, A206399.

A005914:=-(z+1)*(z**2+10*z+1)/(z-1)**3; # Simon Plouffe in his 1992 dissertation.

Table[If[n == 0, 1, 12*n^2 + 2], {n, 0, 100}] (* Vladimir Joseph Stephan Orlovsky, Jun 19 2011 *)
Join[{1},LinearRecurrence[{3,-3,1},{14,50,110},50]] (* Harvey P. Dale, Oct 09 2012 *)

a(n)=12*n^2+2 \\ Charles R Greathouse IV, Jan 31 2012

G.f.: (1+x)*(1+10*x+x^2)/(1-x)^3. - Simon Plouffe (see MAPLE line)
a(n) = (2n-1)^2 + (2n)^2 + (2n+1)^2 for n > 0. - Bruno Berselli, Jan 30 2012
a(n) = 3*a(n-1) - 3*a(n-2) + a(n-3); a(0)=1, a(1)=14, a(2)=50, a(3)=110. - Harvey P. Dale, Oct 09 2012
E.g.f.: exp(x)*(12*x^2 + 12*x + 2) - 1. - Alois P. Heinz, Sep 10 2013
From Bruce J. Nicholson, Jan 19 2019: (Start)
Sum_{i=1..n} a(i) = A005917(n+1).
a(n) = A003154(n) + A003154(n+1). (End)
From Amiram Eldar, Jan 27 2022: (Start)
Sum_{n>=0} 1/a(n) = ((Pi/sqrt(6))*coth(Pi/sqrt(6)) + 3)/4.
Sum_{n>=0} (-1)^n/a(n) = ((Pi/sqrt(6))*cosech(Pi/sqrt(6)) + 3)/4. (End)

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)

A069125 a(n) = (11n^2 - 11n + 2)/2.

Original entry on oeis.org

1, 12, 34, 67, 111, 166, 232, 309, 397, 496, 606, 727, 859, 1002, 1156, 1321, 1497, 1684, 1882, 2091, 2311, 2542, 2784, 3037, 3301, 3576, 3862, 4159, 4467, 4786, 5116, 5457, 5809, 6172, 6546, 6931, 7327, 7734, 8152, 8581, 9021, 9472, 9934, 10407, 10891, 11386, 11892

Offset: 1

Terrel Trotter, Jr., Apr 07 2002

Centered hendecagonal (11-gonal) numbers. - Omar E. Pol, Oct 03 2011

Numbers of the form (2*m+1)^2 + k*m*(m+1)/2: in this case is k=3. See also A254963. - Bruno Berselli, Feb 11 2015

			a(5)=111 because 111 = (11*5^2 - 11*5 + 2)/2 = (275 - 55 + 2)/2 = 222/2.

T. D. Noe, Table of n, a(n) for n = 1..1000
X. Acloque, Polynexus Numbers and other mathematical wonders. [broken link]
Leo Tavares, Illustration: Clipped Stars.
Eric Weisstein's World of Mathematics, Centered Polygonal Numbers.
Index entries for sequences related to centered polygonal numbers.
Index entries for linear recurrences with constant coefficients, signature (3,-3,1).

Cf. A001263, A001844, A003215, A005448, A005891, A069099.

Cf. A000217, A003154, A152740, A254963.

FoldList[#1 + #2 &, 1, 11 Range@ 45] (* Robert G. Wilson v *)
Table[(11n^2-11n+2)/2,{n,60}] (* or *) LinearRecurrence[{3,-3,1},{1,12,34},60] (* Harvey P. Dale, Jun 25 2011 *)

a(n)=(11*n^2-11*n+2)/2 \\ Charles R Greathouse IV, Sep 24 2015

a(n) = 1 + Sum_{j=0..n-1} (11*j). - Xavier Acloque, Oct 22 2003
Binomial transform of [1, 11, 11, 0, 0, 0, ...]; Narayana transform (A001263) of [1, 11, 0, 0, 0, ...]. - Gary W. Adamson, Dec 29 2007
a(n) = 11*n + a(n-1) - 11 with n > 1, a(1)=1. - Vincenzo Librandi, Aug 08 2010
G.f.: -x*(1+9*x+x^2)/(x-1)^3. - R. J. Mathar, Jun 05 2011
a(n) = 3*a(n-1) - 3*a(n-2) + a(n-3); a(0)=1, a(1)=12, a(2)=34. - Harvey P. Dale, Jun 25 2011
a(n) = A152740(n-1) + 1. - Omar E. Pol, Oct 03 2011
From Amiram Eldar, Jun 21 2020: (Start)
Sum_{n>=1} 1/a(n) = 2*Pi*tan(sqrt(3/11)*Pi/2)/sqrt(33).
Sum_{n>=1} a(n)/n! = 13*e/2 - 1.
Sum_{n>=1} (-1)^n * a(n)/n! = 13/(2*e) - 1. (End)
a(n) = A003154(n) - A000217(n-1). - Leo Tavares, Mar 29 2022
E.g.f.: exp(x)*(1 + 11*x^2/2) - 1. - Elmo R. Oliveira, Oct 18 2024

More terms from Harvey P. Dale, Jun 25 2011

Name rewritten by Bruno Berselli, Feb 11 2015

cp's OEIS Frontend

A081266 Staggered diagonal of triangular spiral in A051682.

Views

Author

Keywords

Comments

Examples

Links

Crossrefs

Programs

GAP

Maple

Mathematica

PARI

Formula

A069190 Centered 24-gonal numbers.

Views

Author

Keywords

Comments

Examples

Links

Crossrefs

Programs

Mathematica

PARI

Formula

Extensions

A152749 a(n) = (n+1)*(3*n+1)/4 for n odd, a(n) = n*(3*n+2)/4 for n even.

Views

Author

Keywords

Comments

Links

Crossrefs

Programs

Haskell

Magma

Magma

Maple

Mathematica

Formula

Extensions

A287326 Triangle read by rows: T(n, k) = 6*k*(n-k) + 1; n >= 0, 0 <= k <= n.

Views

Author

Keywords

Comments

Examples

Links

Crossrefs

Programs

GAP

Magma

Maple

Mathematica

PARI

SageMath

Formula

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

Views

Author

Keywords

Comments

Links

Crossrefs

Programs

GAP

Magma

Mathematica

PARI

Sage

Formula

A140811 a(n) = 6*n^2 - 1.

Views

Author

Keywords

Comments

References

Links

A152749 a(n) = (n+1)(3n+1)/4 for n odd, a(n) = n(3n+2)/4 for n even.

A287326 Triangle read by rows: T(n, k) = 6k(n-k) + 1; n >= 0, 0 <= k <= n.

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).

A069125 a(n) = (11n^2 - 11n + 2)/2.