A115067 -id:A115067 - cp's OEIS frontend

A077414 a(n) = n(n - 1)(n + 2)/2.

0, 4, 15, 36, 70, 120, 189, 280, 396, 540, 715, 924, 1170, 1456, 1785, 2160, 2584, 3060, 3591, 4180, 4830, 5544, 6325, 7176, 8100, 9100, 10179, 11340, 12586, 13920, 15345, 16864, 18480, 20196, 22015, 23940, 25974, 28120, 30381, 32760, 35260

Offset: 1

Wolfdieter Lang, Nov 29 2002

Number of independent components of a certain 3-tensor in n-space.
a(n) is the number of independent components of a 3-tensor t(a,b,c) which satisfies t(a,b,c) = t(b,a,c) and Sum_{a=1..n} t(a,a,c) = 0 for all c, with a,b,c range 1..n. (3-tensor in n-dimensional space which is symmetric and traceless in one pair of its indices.)
Row 2 of the convolution array A213761.  - Clark Kimberling, Jul 04 2012
Also, the number of ways to place two dominoes horizontally in the same row on an (n+2) X (n+2) chessboard. - Ralf Stephan, Jun 09 2014
Also, the sum of all the numbers in a completely filled n X n tic-tac-toe board with n-1 players using the numbers 0, 1, 2,... n-2. See "Sums of Square Tic Tac Toe Boards that end in a Draw" in links for proof. - Tanner Robeson, Aug 23 2020
a(n) is the number of degrees of freedom in a tetrahedral cell for a Raviart-Thomas finite element space of order n. - Matthew Scroggs, Jan 02 2021

			For n=6, a(6) = 1*(3*5+1)+2*(3*4+1)+3*(3*3+1)+4*(3*2+1)+5*(3*1+1) = 120. - _Bruno Berselli_, Feb 13 2014
G.f. = 4*x^2 + 15*x^3 + 36*x^4 + 70*x^5 + 120*x^6 + 189*x^7 + 280*x^8 + ...

Vincenzo Librandi, Table of n, a(n) for n = 1..1000
DefElement, Raviart-Thomas
Sela Fried, Counting r X s rectangles in nondecreasing and Smirnov words, arXiv:2406.18923 [math.CO], 2024. See p. 9.
Tanner Robeson, Sums of Square Tic Tac Toe Boards that end in a Draw.
Index entries for linear recurrences with constant coefficients, signature (4,-6,4,-1).
Index to sequences related to polygonal numbers.

Cf. A000096, A005564, A057145, A115067 (first differences), A213761.

Cf. similar sequences of the type m*(m+1)*(m+k)/2 listed in A267370.

[n*(n-1)*(n+2)/2: n in [1..30]]; // G. C. Greubel, Jan 18 2018

A077414:=n->n*(n-1)*(n+2)/2: seq(A077414(n), n=1..60); # Wesley Ivan Hurt, Apr 09 2017

Table[(n (n - 1) (n + 2))/2, {n, 50}] (* or *) LinearRecurrence[{4, -6, 4, -1}, {0, 4, 15, 36}, 50] (* Harvey P. Dale, Jun 04 2012 *)
CoefficientList[Series[x (4 - x)/(1 - x)^4, {x, 0, 40}], x] (* Vincenzo Librandi, Feb 14 2014 *)

a(n)=n*(n-1)*(n+2)/2 \\ Charles R Greathouse IV, Oct 07 2015

concat(0, Vec(x^2*(4-x)/(1-x)^4 + O(x^200))) \\ Altug Alkan, Jan 15 2016

a(n) = n * ( binomial(n+1, 2)-1 ).
G.f.: x^2*(4-x)/(1-x)^4.
a(n) = n*Sum_{j=2..n} j. - Zerinvary Lajos, Sep 12 2006
a(1)=0, a(2)=4, a(3)=15, a(4)=36; for n>4, a(n) = 4*a(n-1)-6*a(n-2)+4*a(n-3)-a(n-4). - Harvey P. Dale, Jun 04 2012
a(n) = Sum_{i=1..n-1} i*(3*(n-i)+1). - Bruno Berselli, Feb 13 2014
a(-n) = -A005564(n). - Michael Somos, Jun 09 2014
a(n) = A057145(n,n+2). - R. J. Mathar, Jul 28 2016
a(n) = t(n,t(n,1)) + n, where t(n,k) = n*(n+1)/2 + k*n and t(n,1) = A000096(n). - Bruno Berselli, Feb 28 2017
a(n) = n^3/2 + n^2/2 - n. - Tanner Robeson, Aug 23 2020
Sum_{n>=2} 1/a(n) = 7/18. - Amiram Eldar, Oct 07 2020
Sum_{n>=2} (-1)^n/a(n) = 4*log(2)/3 - 13/18. - Amiram Eldar, Feb 22 2022
E.g.f.: exp(x)*x^2*(4 + x)/2. - Stefano Spezia, Jan 03 2023

A302909 Determinant of n X n matrix whose main diagonal consists of the first n 5-gonal numbers and all other elements are 1's.

Original entry on oeis.org

1, 4, 44, 924, 31416, 1570800, 108385200, 9863053200, 1144114171200, 164752440652800, 28831677114240000, 6025820516876160000, 1482351847151535360000, 423952628285339112960000, 139480414705876568163840000, 52305155514703713061440000000

Offset: 1

Muniru A Asiru, Apr 15 2018

nonn

			The 7 X 7 matrix (as below) has determinant 108385200.
  1  1  1  1  1  1  1
  1  5  1  1  1  1  1
  1  1 12  1  1  1  1
  1  1  1 22  1  1  1
  1  1  1  1 35  1  1
  1  1  1  1  1 51  1
  1  1  1  1  1  1 70

Muniru A Asiru, Table of n, a(n) for n = 1..100

Cf. A000326, A115067.

Cf. Determinant of n X n matrix whose main diagonal consists of the first n k-gonal numbers and all other elements are 1's: A000142 (k=2), A067550 (k=3), A010791 (k=4, with offset 1), this sequence (k=5), A302910 (k=6), A302911 (k=7), A302912 (k=8), A302913 (k=9), A302914 (k=10).

d:=(i,j)->`if`(i<>j,1,i*(3*i-1)/2):
seq(LinearAlgebra[Determinant](Matrix(n,d)),n=1..17);

Table[FullSimplify[Gamma[n] * Gamma[n + 5/3] * 3^(n + 1) / (5 * Gamma[2/3] * 2^n)], {n, 1, 15}] (* Vaclav Kotesovec, Apr 16 2018 *)
Module[{nn=20,pn5},pn5=PolygonalNumber[5,Range[nn]];Table[Det[DiagonalMatrix[Take[pn5,n]]/.(0->1)],{n,nn}]] (* Harvey P. Dale, Feb 07 2025 *)

a(n) = matdet(matrix(n, n, i, j, if (i!=j, 1, i*(3*i-1)/2))); \\ Michel Marcus, Apr 16 2018

first(n) = my(res = vector(n)); res[1] = 1; for(i = 1, n - 1, res[i + 1] = res[i] * i*(3*i + 5)/2); res \\ David A. Corneth, Apr 16 2018

From Vaclav Kotesovec, Apr 16 2018: (Start)
a(n) = Gamma(n) * Gamma(n + 5/3) * 3^(n + 1) / (5 * Gamma(2/3) * 2^n).
a(n) ~ Gamma(1/3) * 3^(n + 3/2) * n^(2*n + 2/3) / (5 * 2^n * exp(2*n)).
(End)
a(n + 1) = A115067(n + 1) * a(n) = a(n) * n*(3*n + 5)/2. - David A. Corneth, Apr 16 2018

A194455 a(n) = 2^n + 3n + 1.

Original entry on oeis.org

2, 6, 11, 18, 29, 48, 83, 150, 281, 540, 1055, 2082, 4133, 8232, 16427, 32814, 65585, 131124, 262199, 524346, 1048637, 2097216, 4194371, 8388678, 16777289, 33554508, 67108943, 134217810, 268435541, 536871000, 1073741915, 2147483742, 4294967393, 8589934692, 17179869287

Offset: 0

Bruno Berselli, Sep 01 2011

Inverse binomial transform of this sequence: 2,4,1,1 (1 continued).

Bruno Berselli, Table of n, a(n) for n = 0..1000
Index entries for linear recurrences with constant coefficients, signature (4,-5,2).

Cf. A000051, A005126, A176691; A120845.
Cf. A062709 (first differences), A000079 (second and successive differences).
Cf. A146529 (differences between alternate terms, for n>2).
Cf. A086653, A115067.

Magma
```
[2^n+3*n+1: n in [0..31]];
```

Table[2^n + 3 n + 1, {n, 0, 40}] (* Vincenzo Librandi, Mar 26 2013 *)
LinearRecurrence[{4,-5,2},{2,6,11},40] (* Harvey P. Dale, Oct 01 2014 *)

PARI
```
for(n=0, 31, print1(2^n+3*n+1", "));
```

G.f.: (2 - 2*x - 3*x^2)/((1 - 2*x)*(1 - x)^2).
a(n) = A086653(n) - 1  for n > 0.
Sum_{i=0..n} a(i) = A115067(n+1) + 2^(n+1).
a(n) = 3*a(n-1) - 2*a(n-2) - 3  for n > 1.
a(n)^2 = 2^(n+1)*(a(n-1) + 3) + (3*n + 1)^2  for n > 2.
E.g.f.: exp(x)*(1 + exp(x) + 3*x). - Stefano Spezia, May 06 2023

A216172 Number of all possible tetrahedra of any size, having reverse orientation to the original regular tetrahedron, formed when intersecting the latter by planes parallel to its sides and dividing its edges into n equal parts.

Original entry on oeis.org

0, 0, 1, 4, 10, 21, 39, 66, 105, 159, 231, 325, 445, 595, 780, 1005, 1275, 1596, 1974, 2415, 2926, 3514, 4186, 4950, 5814, 6786, 7875, 9090, 10440, 11935, 13585, 15400, 17391, 19569, 21945, 24531, 27339, 30381, 33670, 37219, 41041, 45150, 49560, 54285, 59340

Offset: 1

V.J. Pohjola, Sep 03 2012

The number of all possible tetrahedra of any size, having the same orientation as the original regular tetrahedron is given by A000332(n+3).

Create a sequence wherein the sum of three consecutive numbers is a triangular number: 0,0,0,1,2,3,5,7...; then find the partial sums of this sequence: 0,0,0,1,3,6,11,18...; then take the partial sums of this sequence: 0,0,0,1,4,10,21,39,66... and after dropping the first two zeros, you get this sequence. - J. M. Bergot, Apr 14 2016

			For n=9 the numbers of the reversely oriented tetrahedra, starting from the smallest size, are A000292(7)=84, A000292(4)=20, and A000292(1)=1, the sum being a(9)=105.

Vincenzo Librandi, Table of n, a(n) for n = 1..1000
Index entries for linear recurrences with constant coefficients, signature (4,-6,5,-5,6,-4,1).

Cf. A000292, A000332, A216173, A216175.

I:=[0, 0, 1, 4, 10, 21, 39]; [n le 7 select I[n] else 4*Self(n-1)-6*Self(n-2)+5*Self(n-3)-5*Self(n-4)+6*Self(n-5)-4*Self(n-6)+Self(n-7): n in [1..50]]; // Vincenzo Librandi, Sep 12 2012

nnn = 100; Tev[n_] := (n - 2) (n - 1) n/6; Table[Sum[Tev[n - nn], {nn, 0, n - 1, 3}], {n, nnn}]
Table[(1/72) (-6 n - 5 n^2 + 2 n^3 + n^4 + 4 - 4 (-1)^Mod[n, 3]), {n, 50}]
CoefficientList[Series[x^2 / ((1 - x)^5*(1 + x + x^2)), {x, 0, 50}], x] (* Vincenzo Librandi, Sep 12 2012 *)
LinearRecurrence[{4,-6,5,-5,6,-4,1},{0,0,1,4,10,21,39},50] (* Harvey P. Dale, Feb 18 2018 *)

a(n)=(n^4+2*n^3-5*n^2-6*n+4-4*(-1)^(n%3))/72 \\ Charles R Greathouse IV, Sep 12 2012

a(n) = (1/72)*(-6*n -5*n^2 +2*n^3 +n^4 +4 -4*(-1)^(n mod 3)).
G.f.: x^3/((1-x)^5*(1+x+x^2)). - Bruno Berselli, Sep 11 2012
a(3*n-1) = A000217(A115067(n)); a(3*n) = A000217(A095794(n)); a(3*n+1) = A000217(A143208(n+2)) + A000217(n). - J. M. Bergot, Apr 14 2016
E.g.f.: (1/216)*(8 - 24*x + 24*x^2 + 24*x^3 + 3*x^4)*exp(x) - (1/27)*(cos(sqrt(3)*x/2) - sqrt(3)*sin(sqrt(3)*x/2))*exp(-x/2). - Ilya Gutkovskiy, Apr 14 2016

A027625 Numerator of n(n+5)/((n+2)(n+3)).

Original entry on oeis.org

0, 1, 7, 4, 6, 25, 11, 14, 52, 21, 25, 88, 34, 39, 133, 50, 56, 187, 69, 76, 250, 91, 99, 322, 116, 125, 403, 144, 154, 493, 175, 186, 592, 209, 221, 700, 246, 259, 817, 286, 300, 943, 329, 344, 1078, 375, 391, 1222

Offset: 0

N. J. A. Sloane

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

Cf. A027626 (denominator), A095794, A115067, A179436.

[Numerator(n*(n+5)/((n+2)*(n+3))): n in [0..50]]; // Vincenzo Librandi, Mar 04 2014

CoefficientList[Series[x*(1+7*x+4*x^2+3*x^3+4*x^4-x^5-x^6-2*x^7)/(1-x^3)^3, {x, 0, 50}], x] (* Vincenzo Librandi, Mar 04 2014 *)
Numerator[25*Binomial[Range[0, 50]/5 +1, 2]/3] (* G. C. Greubel, Aug 05 2022 *)

a(n) = numerator(n*(n+5)/6); \\ Altug Alkan, Apr 18 2018

[numerator(n*(n+5)/6) for n in (0..50)] # G. C. Greubel, Aug 05 2022

G.f.: x*(1 + 7*x + 4*x^2 + 3*x^3 + 4*x^4 - x^5 - x^6 - 2*x^7)/(1 - x^3)^3.
a(n) = numerator of n*(n+5)/6. - Altug Alkan, Apr 18 2018
From Peter Bala, Aug 06 2022: (Start)
a(n) is quasi-polynomial in n:
  a(3*n) = (1/2)*n*(3*n+5) = A115067(n+1).
  a(3*n+1) = (1/2)*(n+2)*(3*n+1) = A095794(n+1).
  a(3*n+2) = (1/2)*(3*n+2)*(3*n+7) = A179436(n). (End)
Sum_{n>=1} 1/a(n) = 4*Pi/(15*sqrt(3)) + 87/50. - Amiram Eldar, Aug 11 2022

A223544 T(n, k) = n*k - 1.

Original entry on oeis.org

0, 1, 3, 2, 5, 8, 3, 7, 11, 15, 4, 9, 14, 19, 24, 5, 11, 17, 23, 29, 35, 6, 13, 20, 27, 34, 41, 48, 7, 15, 23, 31, 39, 47, 55, 63, 8, 17, 26, 35, 44, 53, 62, 71, 80, 9, 19, 29, 39, 49, 59, 69, 79, 89, 99, 10, 21, 32, 43, 54, 65, 76, 87, 98, 109, 120, 11, 23, 35, 47, 59, 71, 83, 95, 107, 119, 131, 143

Offset: 1

Richard R. Forberg, Jul 19 2013

Previous name was: Triangle T(n,k), 0 < k <= n, T(n,1) = n - 1, T(n,k) = T(n,k-1) + n; read by rows.
This simple triangle arose analyzing f(x) = x/(n + e^(c/x)), for n <> 0. f(x) converges towards a rational number for large values of x, if x is rational. T(n+1,k)/(n+1)^2 equals the fractional portion of f(x) if x is large and restricted to the positive integers, c = 1 and n>=1, whereby the value of the fractional portion changes on a cycle with period n+1 (as k goes from 1 to n+1) for each n in the denominator of f(x). Other, somewhat similar triangles (or repeating fractional patterns) arise with other rational values of n or c, or other rational increments of x (even if a large irrational initial value of x is used).
Let S(n) = row sums = Sum(k>=1, T(n,k)), then:
S(n) = A077414(n);  S(n)/(n+2) = A000217(n); S(n)/n = A000096(n);
Let Sq(n) = sum of squares of row elements = Sum(k>=1, T(n,k)^2), then:
   Sq(n)/n^2 - 1/n = A058373(n)
Let D(n) = diagonal sums = Sum(k>=1, T(n-k+1, k)) then:
   D(2n) = A131423(n); D(2n-1) = 2/3*n^3 + 1/2*n^2 - 7/6*n;
   D(2n) - D(2n-1) = A000217(n); D(2n+1) - D(2n) = A115067(n);
   D(2n+2) - D(2n)= A056220(n+1);  D(2n+1) - D(2n -1) = A014106(n).
Equals A144204 with the first column of negative ones removed. - Georg Fischer, Jul 26 2023

			Triangle begins as:
0;
1,  3;
2,  5,  8;
3,  7, 11, 15;
4,  9  14, 19, 24;
5, 11, 17, 23, 29, 35;
6, 13, 20, 27, 34, 41, 48;
7, 15, 23, 31, 39, 47, 55, 63;
8, 17, 26, 35, 44, 53, 62, 71, 80;

Boris Putievskiy, Rows n = 1..140 of triangle, flattened

Cf. A075362, A144204.

Also note: T(n+1,k) = T(n,k)+ k, and T(n,n) = n^2 - 1.
a(n) = A075362(n)-1; a(n)=i(t+1)-1, where i=n-t*(t+1)/2, t=floor((-1+sqrt(8*n-7))/2). - Boris Putievskiy, Jul 24 2013
T(n, k) = n*k - 1. - Georg Fischer, Jul 26 2023

Simpler name from Georg Fischer, Jul 26 2023

A236337 Expansion of (2 - x) / ((1 - x)^2 * (1 - x^3)) in powers of x.

Original entry on oeis.org

2, 3, 4, 7, 9, 11, 15, 18, 21, 26, 30, 34, 40, 45, 50, 57, 63, 69, 77, 84, 91, 100, 108, 116, 126, 135, 144, 155, 165, 175, 187, 198, 209, 222, 234, 246, 260, 273, 286, 301, 315, 329, 345, 360, 375, 392, 408, 424, 442, 459, 476, 495, 513, 531, 551, 570, 589

Offset: 0

Michael Somos, Jan 22 2014

The sequence is a quasi-polynomial sequence.

Given a sequence of Laurent polynomials defined by b(n) = (b(n-2)^2 - b(n-1)*b(n-3) * 2/x) / b(n-4), b(-4) = x, b(-3) = b(-2) = -b(-1) = 1. Then the denominator of b(n) is x^a(n).

			G.f. = 2 + 3*x + 4*x^2 + 7*x^3 + 9*x^4 + 11*x^5 + 15*x^6 + 18*x^7 + 21*x^8 + ...

G. C. Greubel, Table of n, a(n) for n = 0..2500
Paul Barry, Generalized Catalan recurrences, Riordan arrays, elliptic curves, and orthogonal polynomials, arXiv:1910.00875 [math.CO], 2019.
Cristian Cobeli, Aaditya Raghavan, and Alexandru Zaharescu, On the central ball in a translation invariant involutive field, arXiv:2408.01864 [math.NT], 2024. See p. 7.
Index entries for linear recurrences with constant coefficients, signature (2,-1,1,-2,1).

Cf. A236343. Trisections are A005449, A045943, A115067.

m:=60; R:=PowerSeriesRing(Integers(), m); Coefficients(R!((2-x)/((1-x)^2*(1-x^3)))); // G. C. Greubel, Aug 07 2018

CoefficientList[Series[(2-x)/((1-x)^2*(1-x^3)), {x, 0, 60}], x] (* Vaclav Kotesovec, Jan 22 2014 *)

{a(n) = ((n+1) * (n+6) + [6, 4, 0][n%3 + 1]) / 6};

{a(n) = if( n<0, polcoeff( x^4 * (-1 + 2*x) / ((1 - x)^2 * (1 - x^3)) + x * O(x^-n), -n), polcoeff( (2 - x) / ((1 - x)^2 * (1 - x^3)) + x * O(x^n), n))};

((2-x)/((1-x)^2*(1-x^3))).series(x, 60).coefficients(x, sparse=False) # G. C. Greubel, Feb 11 2019

0 = a(n)*(a(n+2) + a(n+3)) + a(n+1)*(-2*a(n+2) - a(n+3) + a(n+4))  + a(n+2)*(a(n+2) - 2*a(n+3) + a(n+4)) for all n in Z.
G.f.: (2 - x) / ((1 - x)^2 * (1 - x^3)).
Second difference is period 3 sequence [0, 2, -1, ...].
a(n) = 2*a(n-3) - a(n-6) + 3 = 2*a(n-1) - a(n-2) + a(n-3) - 2*a(n-4) + a(n-5).
a(-6 - n) = A236343(n).
a(3*n) = (1/2)*(n + 1)*(3*n + 4); a(3*n+1) = (1/2)*(n + 1)*(3*n + 6); a(3*n+2) = (1/2)*(n + 1)*(3*n + 8). - Peter Bala, Feb 11 2019

A345118 a(n) is the sum of the lengths of all the segments used to draw a square of side n representing a basketweave pattern where all the multiple strands are of unit width, the horizontal ones appearing as 1 X 3 rectangles, while the vertical ones as unit area squares.

Original entry on oeis.org

0, 4, 11, 20, 34, 50, 69, 92, 116, 144, 175, 208, 246, 286, 329, 376, 424, 476, 531, 588, 650, 714, 781, 852, 924, 1000, 1079, 1160, 1246, 1334, 1425, 1520, 1616, 1716, 1819, 1924, 2034, 2146, 2261, 2380, 2500, 2624, 2751, 2880, 3014, 3150, 3289, 3432, 3576, 3724

Offset: 0

Stefano Spezia, Jun 08 2021

			Illustrations for n = 1..8:
        _           _ _          _ _ _
       |_|         |_|_|        |_ _ _|
                   |_ _|        |_|_|_|
                                |_ _ _|
    a(1) = 4     a(2) = 11     a(3) = 20
     _ _ _ _     _ _ _ _ _    _ _ _ _ _ _
    |_ _|_|_|   |_ _|_|_ _|  |_|_|_ _ _|_|
    |_|_ _ _|   |_|_ _ _|_|  |_ _ _|_|_ _|
    |_ _|_|_|   |_ _|_|_ _|  |_|_|_ _ _|_|
    |_|_ _ _|   |_|_ _ _|_|  |_ _ _|_|_ _|
                |_ _|_|_ _|  |_|_|_ _ _|_|
                             |_ _ _|_|_ _|
    a(4) = 34    a(5) = 50     a(6) = 69
      _ _ _ _ _ _ _      _ _ _ _ _ _ _ _
     |_|_|_ _ _|_|_|    |_ _|_|_ _ _|_|_|
     |_ _ _|_|_ _ _|    |_|_ _ _|_|_ _ _|
     |_|_|_ _ _|_|_|    |_ _|_|_ _ _|_|_|
     |_ _ _|_|_ _ _|    |_|_ _ _|_|_ _ _|
     |_|_|_ _ _|_|_|    |_ _|_|_ _ _|_|_|
     |_ _ _|_|_ _ _|    |_|_ _ _|_|_ _ _|
     |_|_|_ _ _|_|_|    |_ _|_|_ _ _|_|_|
                        |_|_ _ _|_|_ _ _|
        a(7) = 92           a(8) = 116

Index entries for linear recurrences with constant coefficients, signature (3,-3,1,-1,3,-3,1).

Cf. A000035, A004773, A056594, A115067, A211014, A316316, A316317, A368052.

LinearRecurrence[{3,-3,1,-1,3,-3,1},{0,4,11,20,34,50,69},50]
a[ n_] := (3*n^2 + 5*n)/2 - (-1)^Floor[n/4]*Boole[Mod[n, 4] == 3]; (* Michael Somos, Jan 25 2024 *)

concat(0, Vec(x*(4 - x - x^2 + 3*x^3 + x^4)/((1 - x)^3*(1 + x^4)) + O(x^40))) \\ Felix Fröhlich, Jun 09 2021

{a(n) = (3*n^2 + 5*n)/2 - (-1)^(n\4)*(n%4==3)}; /* Michael Somos, Jan 25 2024 */

O.g.f.: x*(4 - x - x^2 + 3*x^3 + x^4)/((1 - x)^3*(1 + x^4)).
E.g.f.: (exp(x)*x*(8 + 3*x) + (-1)^(1/4)*(sinh((-1)^(1/4)*x) - sin((-1)^(1/4)*x)))/2.
a(n) = 3*a(n-1) - 3*a(n-2) + a(n-3) - a(n-4) + 3*a(n-5) - 3*a(n-6) + a(n-7) for n > 6.
a(n) = (n*(5 + 3*n) - (1 - (-1)^n)*sin((n-1)*Pi/4))/2.
a(n) = A211014(n/2) - A000035(n)*A056594((n-3)/2).
a(2*n) = A211014(n).
a(k) = A115067(k+1) for k not congruent to 3 mod 4 (A004773).
From Helmut Ruhland, Jan 29 2024: (Start)
For n > 1: a(n) - (2 * A368052(n+2) + A368052(n+3)) * 2 is periodic for n mod 8, i.e. a(n) = (2 * A368052(n+2) + A368052(n+3)) * 2 + f8(n) with
n mod 8 =   0   1   2   3   4   5   6   7
f8(n)   =   0   0  -3  -2  -2  -2   1   0   (End)

A363230 Number of partitions of n with rank 3 or higher (the rank of a partition is the largest part minus the number of parts).

Original entry on oeis.org

0, 0, 0, 1, 1, 2, 3, 5, 7, 11, 14, 21, 28, 39, 51, 70, 90, 120, 154, 201, 256, 330, 415, 529, 662, 833, 1035, 1293, 1595, 1976, 2425, 2982, 3640, 4449, 5401, 6565, 7935, 9592, 11543, 13891, 16645, 19943, 23808, 28408, 33792, 40172, 47619, 56413, 66661, 78708, 92724, 109149, 128213, 150486, 176293

Offset: 1

Seiichi Manyama, May 22 2023

nonn

			a(6) = 2 counts these partitions: 6, 5+1.

Vaclav Kotesovec, Table of n, a(n) for n = 1..10000

With rank r or higher: A064174 (r=0), A064173 (r=1), A123975 (r=2), this sequence (r=3), A363231 (r=4).

Cf. A000041, A101200, A115067.

a(n) = sum(k=1, sqrtint(n), (-1)^(k-1)*numbpart(n-k*(3*k+5)/2));

G.f.: (1/Product_{k>=1} (1-x^k)) * Sum_{k>=1} (-1)^(k-1) * x^(k*(3*k+5)/2).
a(n) = p(n-4) - p(n-11) + p(n-21) - ... + (-1)^(k-1) * p(n-k*(3*k+5)/2) + ..., where p() is A000041().
a(n) ~ exp(Pi*sqrt(2*n/3)) / (8*n*sqrt(3)) * (1 - (1/(2*Pi) + 31*Pi/144) / sqrt(n/6)). - Vaclav Kotesovec, May 26 2023

A370238 a(n) = n(3n + 23)/2.

Original entry on oeis.org

0, 13, 29, 48, 70, 95, 123, 154, 188, 225, 265, 308, 354, 403, 455, 510, 568, 629, 693, 760, 830, 903, 979, 1058, 1140, 1225, 1313, 1404, 1498, 1595, 1695, 1798, 1904, 2013, 2125, 2240, 2358, 2479, 2603, 2730, 2860, 2993, 3129, 3268, 3410, 3555, 3703, 3854, 4008

Offset: 0

Torlach Rush, Feb 12 2024

a(a(1)) = A000566(a(1)). This is also true for each of the sequences provided in the formulae below; e.g., A151542(A151542(1)) = A000566(A151542(1)).

Michael De Vlieger, Table of n, a(n) for n = 0..10000
Sela Fried, Counting r X s rectangles in nondecreasing and Smirnov words, arXiv:2406.18923 [math.CO], 2024. See p. 9.
Index entries for linear recurrences with constant coefficients, signature (3,-3,1).

Cf. A000326, A000566, A008594, A005449, A045943, A059845, A115067, A140090, A140091, A140672, A140673, A140674, A140675, A151542, A277976.

Table[n(3n+23)/2,{n,0,48}] (* James C. McMahon, Feb 20 2024 *)

Python
```
def a(n): return n*(3*n+23)//2
```

a(n) = n*(3*n + 23)/2 = A277976(n)/2.
G.f.: x*(13-10*x)/(1-x)^3.
a(n) = A151542(n) + n.
a(n) = A140675(n) + 2*n.
a(n) = A140674(n) + 3*n.
a(n) = A140673(n) + 4*n.
a(n) = A140672(n) + 5*n.
a(n) = A059845(n) + 6*n.
a(n) = A140091(n) + 7*n.
a(n) = A140090(n) + 8*n.
a(n) = A115067(n) + 9*n.
a(n) = A045943(n) + 10*n.
a(n) = A005449(n) + 11*n.
a(n) = A000326(n) + A008594(n).
Sum_{n>=1} 1/a(n) = 823467/2769844 + sqrt(3)*Pi/69 -3*log(3)/23 = 0.2328608... - R. J. Mathar, Apr 23 2024
E.g.f.: exp(x)*x*(26 + 3*x)/2. - Stefano Spezia, Apr 26 2024

cp's OEIS Frontend

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A077414 a(n) = n(n - 1)(n + 2)/2.

A027625 Numerator of n(n+5)/((n+2)(n+3)).

A370238 a(n) = n(3n + 23)/2.