A005891 -id:A005891 - cp's OEIS frontend

A092185 a(n) = (5/6)n^3+(5/2)n^2+(8/3)*n.

0, 6, 22, 53, 104, 180, 286, 427, 608, 834, 1110, 1441, 1832, 2288, 2814, 3415, 4096, 4862, 5718, 6669, 7720, 8876, 10142, 11523, 13024, 14650, 16406, 18297, 20328, 22504, 24830, 27311, 29952, 32758, 35734, 38885, 42216, 45732, 49438, 53339, 57440, 61746, 66262

Offset: 0

N. J. A. Sloane, Apr 02 2004

Harvey P. Dale, Table of n, a(n) for n = 0..1000
Index entries for linear recurrences with constant coefficients, signature (4,-6,4,-1).

Partial sums of A005891.

LinearRecurrence[{4,-6,4,-1},{0,6,22,53},50] (* Harvey P. Dale, May 27 2012 *)

a(n)=(5/6)*n^3+(5/2)*n^2+(8/3)*n \\ Charles R Greathouse IV, Oct 18 2022

a(n)= +4*a(n-1) -6*a(n-2) +4*a(n-3) -a(n-4). G.f.: x*(6-2*x+x^2)/(x-1)^4. - R. J. Mathar, Jun 21 2010

A121990 Expansion of x(1+9x+2x^2)/((1-x)(1-3*x+x^2)).

Original entry on oeis.org

1, 13, 50, 149, 409, 1090, 2873, 7541, 19762, 51757, 135521, 354818, 928945, 2432029, 6367154, 16669445, 43641193, 114254146, 299121257, 783109637, 2050207666, 5367513373, 14052332465, 36789484034, 96316119649, 252158874925

Offset: 1

Roger L. Bagula, Sep 10 2006

nonn

G. C. Greubel, Table of n, a(n) for n = 1..1000
Index entries for linear recurrences with constant coefficients, signature (4,-4,1).

Cf. A000045, A003215, A005891.

F:=Fibonacci;; List([1..30], n-> 12*F(2*n-1) +F(2*n-3) -12 ); # G. C. Greubel, Nov 21 2019

F:= Fibonacci; [12*F(2*n-1) +F(2*n-3) -12: n in [1..30]]; // G. C. Greubel, Nov 21 2019

with(combinat); seq(12*fibonacci(2*n-1) +fibonacci(2*n-3) -12, n=1..30); # G. C. Greubel, Nov 21 2019

LinearRecurrence[{4,-4,1}, {1,13,50}, 30] (* G. C. Greubel, Sep 14 2017 *)
With[{F=Fibonacci}, Table[12*(F[2*n-1]-1) + F[2*n-3], {n,30}]] (* G. C. Greubel, Nov 21 2019 *)

x='x+O('x^30); Vec(x*(1+9*x+2*x^2)/((1-x)*(x^2-3*x+1))) \\ G. C. Greubel, Sep 14 2017

vector(30, n, 12*fibonacci(2*n-1) +fibonacci(2*n-3) -12) \\ G. C. Greubel, Nov 21 2019

f=fibonacci; [12*f(2*n-1) + f(2*n-3) -12 for n in (1..30)] # G. C. Greubel, Nov 21 2019

a(n) = 3*a(n - 1) - a(n - 2) + 12.
a(n) = (1/10)*(-120 + (65 - 11*sqrt(5))*((1/2)*(3 - sqrt(5)))^n + ((1/2)*(3 + sqrt(5)))^n*(65 + 11*sqrt(5))).
From R. J. Mathar, Apr 04 2009: (Start)
a(n) = 4*a(n-1) - 4*a(n-2) + a(n-3).
G.f.: x*(1+9*x+2*x^2)/((1-x)*(1-3*x+x^2)). (End)
a(n) = 12*Fibonacci(2*n-1) + Fibonacci(2*n-3) - 12. - G. C. Greubel, Nov 21 2019

Edited and new name based on g.f. by G. C. Greubel and Joerg Arndt, Sep 14 2017

A121991 a(n) = 3*a(n-1) - a(n-2) - a(n-3) + 12.

Original entry on oeis.org

0, 1, 13, 50, 148, 393, 993, 2450, 5976, 14497, 35077, 84770, 204748, 494409, 1193721, 2882018, 6957936, 16798081, 40554301, 97906898, 236368324, 570643785, 1377656145, 3325956338, 8029569096, 19385094817

Offset: 0

Roger L. Bagula, Sep 10 2006

G. C. Greubel, Table of n, a(n) for n = 0..1000
Index entries for linear recurrences with constant coefficients, signature (4, -4, 0, 1).

Cf. A000129, A003215, A005891.

RecurrenceTable[{a[n] == 3*a[n - 1] - a[n - 2] - a[n - 3] + 12, a[0] == 0, a[1] == 1, a[2] == 13}, a, {n,0,50}] (* or *) LinearRecurrence[{4,-4,0,1}, {0,1,13,50}, 50] (* G. C. Greubel, Sep 14 2017 *)

x='x+O('x^50); concat([0], Vec(-x(1+9x+2x^2)/((1-x)^2*(x^2+2x-1))) ) \\ G. C. Greubel, Sep 14 2017

a(n) = ((11 - 7*sqrt(2))*(1 - sqrt(2))^n + (1 + sqrt(2))^n*(11 + 7*sqrt(2)) - 24*n - 22)/4.
O.g.f.: -x(1+9x+2x^2)/((1-x)^2*(x^2+2x-1)). - R. J. Mathar, Aug 22 2008
a(n) = -6(n+1)+(1+11*A000129(n+1)+3*A000129(n))/2. - R. J. Mathar, Aug 22 2008
E.g.f.: (1/2)*(11*cosh(sqrt(2)*x) + 7*sqrt(2)*sinh(sqrt(2)*x) - (12*x + 11))*exp(x). - G. C. Greubel, Sep 14 2017
a(n) = 4*a(n-1) - 4*a(n-2) + a(n-4). - Wesley Ivan Hurt, May 04 2024

Edited by N. J. A. Sloane, Aug 24 2008, Dec 30 2008

A141534 Derived from the centered polygonal numbers: start with the first triangular number, then the sum of the first square number and the second triangular number, then the sum of first pentagonal number, the second square number and the third triangular number, and so on and so on...

Original entry on oeis.org

1, 4, 11, 26, 55, 105, 184, 301, 466, 690, 985, 1364, 1841, 2431, 3150, 4015, 5044, 6256, 7671, 9310, 11195, 13349, 15796, 18561, 21670, 25150, 29029, 33336, 38101, 43355, 49130, 55459, 62376, 69916, 78115, 87010, 96639, 107041, 118256, 130325

Offset: 1

Dan Graybill (clopen(AT)comcast.net), Aug 12 2008

Consider the array of triangular, square and centered polygonal numbers (irregular variant of A086272 and A086273):
 3   6  10  15  21  28  36  45  55  A000217
 4   9  16  25  36  49  64  81 100  A000290
 6  16  31  51  76 106 141 181 226  A005891
 7  19  37  61  91 127 169 217 271  A003215
 8  22  43  71 106 148 197 253 316  A069099
 9  25  49  81 121 169 225 289 361  A016754
10  28  55  91 136 190 253 325 406  A060544
11  31  61 101 151 211 281 361 451  A062786
12  34  67 111 166 232 309 397 496  A069125
13  37  73 121 181 253 337 433 541  A003154
14  40  79 131 196 274 365 469 586  A069126
15  43  85 141 211 295 393 505 631  A069127
etc. The sequence contains the antidiagonal sums of this array. - R. J. Mathar, Jun 05 2011
For comparison, the antidiagonal sums of A086270 are essentially A006522 starting at the 4th term. - R. J. Mathar, Sep 20 2008

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

Cf. A000217.

a(n) = (n-1)*(n^3+11*n^2-38*n+120)/24, n>1. - R. J. Mathar, Sep 12 2008

G.f.: x*(1-x+x^2+x^3-x^5)/(1-x)^5. - Alexander R. Povolotsky, Jun 06 2011

A193249 Snub dodecahedron with faces of centered polygons.

Original entry on oeis.org

1, 153, 755, 2107, 4509, 8261, 13663, 21015, 30617, 42769, 57771, 75923, 97525, 122877, 152279, 186031, 224433, 267785, 316387, 370539, 430541, 496693, 569295, 648647, 735049, 828801, 930203, 1039555, 1157157, 1283309, 1418311, 1562463, 1716065, 1879417

Offset: 1

Craig Ferguson, Jul 19 2011

The sequence starts with a central dot and expands outward with (n-1) centered polygonal pyramids producing a snub dodecahedron. Each iteration requires the addition of (n-2) edges and (n-1) vertices to complete the centered polygon of each face. [centered triangles (A005448) and centered pentagons (A005891)]

Bruno Berselli, Table of n, a(n) for n = 1..1000
Index entries for linear recurrences with constant coefficients, signature (4,-6,4,-1).

=50*ROW()^3-75*ROW()^2+27*ROW()-1 fill down to desired size.

[50*n^3-75*n^2+27*n-1: n in [1..34]];  // Bruno Berselli, Jul 22 2011

A193249:=n->(2*n-1)*(25*n^2-25*n+1); seq(A193249(n), n=1..50); # Wesley Ivan Hurt, Apr 30 2014

Table[(2 n - 1) (25 n^2 - 25 n + 1), {n, 50}] (* Wesley Ivan Hurt, Apr 30 2014 *)

for(n=1, 34, print1(50*n^3-75*n^2+27*n-1", "));  \\ Bruno Berselli, Jul 21 2011

a(n) = 50*n^3-75*n^2+27*n-1 = (2*n-1)*(25*n^2-25*n+1).

G.f.: x*(1+x)*(1+148*x+x^2)/(1-x)^4. - Bruno Berselli, Jul 22 2011

A193251 Small rhombicosidodecahedron with faces of centered polygons.

Original entry on oeis.org

1, 123, 605, 1687, 3609, 6611, 10933, 16815, 24497, 34219, 46221, 60743, 78025, 98307, 121829, 148831, 179553, 214235, 253117, 296439, 344441, 397363, 455445, 518927, 588049, 663051, 744173, 831655, 925737, 1026659, 1134661, 1249983, 1372865, 1503547, 1642269

Offset: 1

Craig Ferguson, Jul 19 2011

The sequence starts with a central dot and expands outward with (n-1) centered polygonal pyramids producing a small rhombicosidodecahedron. Each iteration requires the addition of (n-2) edges and (n-1) vertices to complete the centered polygon of each face. [centered triangles (A005448), centered squares (A001844) and centered pentagons (A005891)]

Vincenzo Librandi, Table of n, a(n) for n = 1..10000
OEIS Wiki, (Centered_polygons) pyramidal numbers.
Wikipedia, Tetrahedral number.
Wikipedia, Triangular number.
Wikipedia, Centered polygonal number.
Index entries for linear recurrences with constant coefficients, signature (4,-6,4,-1).

Cf. A001844, A005448, A005891, A060747, A195317.

[40*n^3-60*n^2+22*n-1: n in [1..50]]; // Vincenzo Librandi, Aug 30 2011

A193251[n_] := (2*n - 1)*(20*(n - 1)*n + 1); Array[A193251, 50] (* or *)
LinearRecurrence[{4, -6, 4, -1}, {1, 123, 605, 1687}, 50] (* Paolo Xausa, Aug 25 2025 *)

a(n)=40*n^3-60*n^2+22*n-1 \\ Charles R Greathouse IV, Oct 19 2022

a(n) = 40*n^3 - 60*n^2 + 22*n - 1.
G.f.: x*(1+x)*(x^2 + 118*x + 1)/(x-1)^4. - R. J. Mathar, Aug 26 2011
From Elmo R. Oliveira, Aug 22 2025: (Start)
E.g.f.: 1 + exp(x)*(-1 + 2*x + 60*x^2 + 40*x^3).
a(n) = 4*a(n-1) - 6*a(n-2) + 4*a(n-3) - a(n-4) for n > 4.
a(n) = A060747(n)*A195317(n). (End)

A216318 Number of peaks in all Dyck n-paths after changing each valley to a peak by the transform DU -> UD.

Original entry on oeis.org

0, 1, 2, 8, 31, 119, 456, 1749, 6721, 25883, 99892, 386308, 1496782, 5809478, 22584160, 87922215, 342741285, 1337698515, 5226732060, 20442936360, 80031775890, 313585934610, 1229695855440, 4825705232010, 18950613058026, 74467158658974, 292797216620776, 1151895428382104

Offset: 0

David Scambler, Sep 03 2012

nonn

			The 5 Dyck 3-paths after changing DU to UD become two copies of UUUDDD with one peak each and three copies of UUDUDD with two peaks each giving a(3)=8.

Vincenzo Librandi, Table of n, a(n) for n = 0..1000

Cf. A001700, A003516, A005891.

CoefficientList[Series[(16*x*(1+Sqrt[1-4*x]+(5+3*Sqrt[1-4*x]-2*x)*(-1+x) x))/((1+Sqrt[1-4*x])^5*Sqrt[1-4*x]),{x,0,27}],x]

a(n):=if n<2 then n else binomial(2*n-2,n-1)*(5*(n-1)^2+5*(n-1)+2)/(2*n*(n+1)); /* Vladimir Kruchinin, Oct 30 2020 */

x='x+O('x^50); concat([0], Vec((16*x*(1+sqrt(1-4*x)-(5+3*sqrt(1-4*x)-2*x)*(1-x)*x)) / ((1+sqrt(1-4*x))^5*sqrt(1-4*x)))) \\ G. C. Greubel, Apr 01 2017

a(0)=0, a(1)=1, a(n>=2) = A001700(n-1) - Sum_{k=0..n-3} A001700(k) + Sum_{k=0..n-2} A003516(k) - 1.
G.f.: (16*x*(1+sqrt(1-4*x)+(5+3*sqrt(1-4*x)-2*x) * (-1+x)*x)) / ((1+sqrt(1-4*x))^5 * sqrt(1-4*x)).
a(n) ~ 5*2^(2*n-3)/sqrt(Pi*n). - Vaclav Kotesovec, Mar 21 2014
a(n) = C(2*n-2,n-1)*(5*(n-1)^2+5*(n-1)+2)/(2*n*(n+1)), n>1, a(0)=0, a(1)=1. - Vladimir Kruchinin, Oct 30 2020

A244911 Table read by antidiagonals: T(n,k) = n*k + T(n-1,k) for n >=1, T(0,k) = 1.

Original entry on oeis.org

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

Offset: 0

Kival Ngaokrajang, Jul 07 2014

T(n,k) is the total number of boxes, when we start with 1 center box (n = 0) then expand 1 box on k-arms for each n iteration. See illustration in links.
It seems that column C(k) = centered k-gonal numbers, and row R(n) = A000217(n)*k + 1.
The triangle under the main diagonal is A121722.
Column N (CN) is the Narayana transform (A001263) of (1, N, 0, 0, 0, ...). Example: C2 (1, 3, 7, 13, ...) is the Narayana transform of (1, 2, 0, 0, 0, ...). - Gary W. Adamson, Oct 01 2015

			Table begins:
       C0  C1  C2  C3  C4  C5
  n/k  0   1   2   3   4   5   ...
R0 0   1   1   1   1   1   1   ...
R1 1   1   2   3   4   5   6   ...
R2 2   1   4   7   10  13  16  ...
R3 3   1   7   13  19  25  31  ...
R4 4   1   11  21  31  41  51  ...
R5 5   1   16  31  46  61  76  ...
R6 6   1   22  43  64  85  106 ...
R7 7   1   29  57  85  113 141 ...
R8 8   1   37  73  109 145 181 ...
R9 9   1   46  91  136 181 226 ...
  ...  ... ... ... ... ... ... ...
C1 = A000124, C2 = A002061, C3 = A005448, C4 = A001844, C5 = A005891, C6 = A003215, C7 = A069099, C8 = A016754, C9 = A060544, C10 = A062786, C11 = A069125, C12  =  A003154.
R1 = A000027, R2 = A016777, R3 = A016921, R4 = A017281, R5 = 15*k + 1, R6 = A215146, R7 = A161714.

Kival Ngaokrajang, Illustration for n = 0..3, k = 1..4

Cf. A000217, A121722, A000124, A002061, A005448, A001844, A005891, A003215, A069099, A016754, A060544, A062786, A069125, A003154, A000027, A016777, A016921, A017281, A215146, A161714.

T(n,k) = n*k + T(n-1,k) for n >=1, T(0,k) = 1.

A276917 Numbers obtained by alternatively adding centered pentagonal layers of 5(2^n-1) and 5(3^n-1) elements.

Original entry on oeis.org

1, 6, 16, 31, 71, 106, 236, 311, 711, 866, 2076, 2391, 6031, 6666, 17596, 18871, 51671, 54226, 152636, 157751, 452991, 463226, 1348956, 1369431, 4026631, 4067586, 12039196, 12121111, 36035951, 36199786, 107944316, 108271991, 323505591, 324160946, 969861756

Offset: 0

Daniel Poveda Parrilla, Dec 29 2016

a(0), a(1), a(2) and a(3) are the first four centered pentagonal numbers, as they match the same pattern. From a(4) onwards all terms are a different kind of centered pentagonal numbers, as the number of elements in subsequent layers doesn't increase uniformly.
a(13) is the first palindromic number in the sequence. a(19) is the second one.
First prime terms are a(3), a(4), a(7), a(31), a(100) and a(115).

Daniel Poveda Parrilla, Table of n, a(n) for n = 0..1000
Daniel Poveda Parrilla, Illustration of initial terms
Index entries for linear recurrences with constant coefficients, signature (2,4,-10,-1,12,-6).

Cf. A005891.

Table[5 (Sum[2^i, {i, 0, ((n + Mod[n, 2])/2)}] + Sum[3^j, {j, 0, ((n - Mod[n, 2])/2)}]) - 5 n - 9, {n, 0, 28}] (* or *)
CoefficientList[Series[(1 + 4 x - 15 x^3 + 6 x^4 - 6 x^5)/((-1 + x)^2 (1 - 5 x^2 + 6 x^4)), {x, 0, 28}], x] (* or *)
LinearRecurrence[{2, 4, -10, -1, 12, -6}, {1, 6, 16, 31, 71, 106}, 29]

Vec((1+4*x-15*x^3+6*x^4-6*x^5) / ((-1+x)^2*(1-5*x^2+6*x^4)) + O(x^40)) \\ Colin Barker, Dec 30 2016

a(n) = 5*(Sum_{i=0..((n+(n mod 2))/2)} 2^i + Sum_{j=0..((n-(n mod 2))/2)} 3^j) - 5*n - 9.
a(n) = a(n-1) + 5*((2+((n+1) mod 2))^((n+(n mod 2))/2) - 1) for n>0.
G.f.: (1+4*x-15*x^3+6*x^4-6*x^5)/((-1+x)^2*(1-5*x^2+6*x^4)).
From Colin Barker, Dec 30 2016: (Start)
a(n) = (-10*n + 5*3^(n/2+1) + 5*2^(n/2+2) - 33)/2 for n even.
a(n) = (-10*n + 5*3^(n/2+1/2) + 5*2^(n/2+5/2) - 33)/2 for n odd.
(End)

A279830 a(n) = the least integer that is centered polygonal in exactly n ways.

Original entry on oeis.org

4, 7, 37, 31, 91, 181, 211, 421, 631, 1891, 1261, 2521, 6931, 18481, 20791, 13861, 27721, 41581, 83161, 138601, 245701, 235621, 180181, 556921, 360361, 540541, 1670761, 1081081, 1413721, 2702701, 2162161, 6486481, 3063061, 8288281, 13430341, 6846841, 10270261, 6126121

Offset: 1

Daniel Sterman, Dec 20 2016

nonn

a(n) has exactly n representations as a centered r-gonal number P(r,m) = 1 + r*m*(m+1)/2, with m > 1, r > 0.
a(n) appears n+1 times in A101321, due to the second column containing every positive integer.
a(n)-1 is the first appearance of n+1 in A007862.

			a(4)=31, because 31 is a centered triangular number (A005448), a centered pentagonal number (A005891), a centered decagonal number (A062786), and a central polygonal number (A002061). No number less than 31 has 4 representations.

Lars Blomberg, Table of n, a(n) for n = 1..143

Cf. A007862 (see alternative definition: the number of ways to represent n+1 as a centered polygonal number).
Cf. A063778 (the equivalent for polygonal numbers).
Subset of A275340 (the list of nontrivial centered polygonal numbers).
Subset of A101321 (centered polygonal numbers read by antidiagonals).

f[n_] := Length@Select[Divisors[2 n - 2], IntegerQ@Sqrt[1 + 4 #] &] - 1;
Do[If[IntegerQ[A279830[f[i]]], , A279830[f[i]] = i], {i, 10000}];
A279830 /@ Range[13]
(* Davin Park, Dec 28 2016 *)

Corrected and extended by Davin Park, Dec 27 2016

cp's OEIS Frontend

A092185 a(n) = (5/6)*n^3+(5/2)*n^2+(8/3)*n.

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A121990 Expansion of x*(1+9*x+2*x^2)/((1-x)*(1-3*x+x^2)).

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A121991 a(n) = 3*a(n-1) - a(n-2) - a(n-3) + 12.

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A141534 Derived from the centered polygonal numbers: start with the first triangular number, then the sum of the first square number and the second triangular number, then the sum of first pentagonal number, the second square number and the third triangular number, and so on and so on...

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A193249 Snub dodecahedron with faces of centered polygons.

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A193251 Small rhombicosidodecahedron with faces of centered polygons.

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A216318 Number of peaks in all Dyck n-paths after changing each valley to a peak by the transform DU -> UD.

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A092185 a(n) = (5/6)n^3+(5/2)n^2+(8/3)*n.

A121990 Expansion of x(1+9x+2x^2)/((1-x)(1-3*x+x^2)).

A276917 Numbers obtained by alternatively adding centered pentagonal layers of 5(2^n-1) and 5(3^n-1) elements.