A002492 -id:A002492 - cp's OEIS frontend

A199394 The number of ways to color the vertices of all (11) simple unlabeled graphs on 4 nodes using at most n colors.

11, 90, 357, 996, 2255, 4446, 7945, 13192, 20691, 31010, 44781, 62700, 85527, 114086, 149265, 192016, 243355, 304362, 376181, 460020, 557151, 668910, 796697, 941976, 1106275, 1291186, 1498365, 1729532, 1986471, 2271030, 2585121, 2930720, 3309867, 3724666

Offset: 1

Geoffrey Critzer, Nov 05 2011

Vincenzo Librandi, Table of n, a(n) for n = 1..1000

Cf. A002492 (3 nodes).

Table[Total[Table[CycleIndex[GraphData[{4,k}, "Automorphisms"], s], {k, 1,11}] /. Table[s[i]-> n, {i, 1, 4}]], {n, 1, 25}]

Vec(-x*(x^3+17*x^2+35*x+11)/(x-1)^5  + O(x^100)) \\ Colin Barker, Sep 22 2013

a(n)= 8*n^4/3 + 4*n^3 + 10*n^2/3 + n.

G.f.: -x*(x^3+17*x^2+35*x+11) / (x-1)^5. - Colin Barker, Sep 22 2013

More terms from Colin Barker, Sep 22 2013

Added a(4) = 996 from Vincenzo Librandi, Sep 23 2013

A270309 Irregular triangle read by rows: T(n,k) = ((n-k)+1)^2 if odd-n and odd-k; T(n,k) = k^2 if odd-n and even-k; T(n,k) = (n/2-(k/2-1/2))^2 if even-n and odd-k; T(n,k) = (k/2+1)^2 if even-n and even-k; where n >= 1, k = 1..2*n.

Original entry on oeis.org

1, 1, 1, 1, 1, 1, 9, 4, 1, 1, 4, 9, 4, 1, 1, 4, 4, 1, 1, 4, 25, 4, 9, 16, 1, 1, 16, 9, 4, 25, 9, 1, 4, 4, 1, 9, 9, 1, 4, 4, 1, 9, 49, 4, 25, 16, 9, 36, 1, 1, 36, 9, 16, 25, 4, 49, 16, 1, 9, 4, 4, 9, 1, 16, 16, 1, 9, 4, 4, 9, 1, 16, 81, 4, 49, 16, 25, 36, 9, 64, 1, 1, 64, 9, 36, 25, 16, 49, 4, 81, 25, 1, 16, 4, 9, 9, 4, 16, 1, 25, 25, 1, 16, 4, 9, 9, 4, 16, 1, 25

Offset: 1

Kival Ngaokrajang, Mar 15 2016

Refer to A269845, but change to n+2 X n instead of n+1 X n.

There are triangles appearing along main diagonal. If the area of the smallest triangles are defined as 1, then the areas of all other triangles seem to be square numbers. Conjectures: (i) Even terms of row sum is A002492. (ii) Odd terms of row sum/2 is A100157. See illustration in links.

			Irregular triangle begins:
n\k  1   2   3   4   5   6   7   8   9  10  11  12  13  14  15  16  ...
1    1,  1
2    1,  1,  1,  1
3    9,  4,  1,  1,  4,  9
4    4,  1,  1,  4,  4,  1,  1,  4
5   25,  4,  9, 16,  1,  1, 16,  9,  4, 25
6    9,  1,  4,  4,  1,  9,  9,  1,  4,  4,  1,  9
7   49,  4, 25, 16,  9, 36,  1,  1, 36,  9, 16, 25,  4, 49
8   16,  1,  9,  4,  4,  9,  1, 16, 16,  1,  9,  4,  4,  9,  1, 16
...

Kival Ngaokrajang, Illustration of initial terms, Row sum

Cf. A002492, A100157, A269845.

A330151 Partial sums of 4th powers of the even numbers.

Original entry on oeis.org

0, 16, 272, 1568, 5664, 15664, 36400, 74816, 140352, 245328, 405328, 639584, 971360, 1428336, 2042992, 2852992, 3901568, 5237904, 6917520, 9002656, 11562656, 14674352, 18422448, 22899904, 28208320, 34458320, 41769936, 50272992, 60107488, 71423984, 84383984

Offset: 0

Assoul Abdelkarim, Dec 03 2019

			a(4) = 0^4 + 2^4 + 4^4 + 6^4 + 8^4 = 5664.

Colin Barker, Table of n, a(n) for n = 0..1000
Abdelkarim Assoul, The sum of the natural numbers peers, odd of p-th degree, hal-01924427, 2015.
Index entries for linear recurrences with constant coefficients, signature (6,-15,20,-15,6,-1).

Cf. A000538, A002309.

Partial sums of A016744.

a[n_] := (8/15)*n*(6*n^4 + 15*n^3 + 10*n^2 - 1); Array[a, 31, 0] (* Amiram Eldar, Dec 08 2019 *)

a(n) = sum(i=0, n, 16*i^4); \\ Jinyuan Wang, Dec 07 2019

concat(0, Vec(16*x*(1 + x)*(1 + 10*x + x^2) / (1 - x)^6 + O(x^30))) \\ Colin Barker, Dec 08 2019

def A330151(n): return 8*n*(n**2*(n*(6*n + 15) + 10) - 1)//15 # Chai Wah Wu, Dec 07 2021

a(n) = Sum_{k=1..n} (2*k)^4 = (8/15)*n*(6*n^4 + 15*n^3 + 10*n^2 - 1).
a(n) = 16*A000538(n).
From Colin Barker, Dec 08 2019: (Start)
G.f.: 16*x*(1 + x)*(1 + 10*x + x^2) / (1 - x)^6.
a(n) = 6*a(n-1) - 15*a(n-2) + 20*a(n-3) - 15*a(n-4) + 6*a(n-5) - a(n-6) for n>5.
(End)
E.g.f.: (8/15)*exp(x)*x*(30 + 225*x + 250*x^2 + 75*x^3 + 6*x^4). - Stefano Spezia, Dec 08 2019
a(n+1) = 12*A002299(n) + A002492(n+1). - Yasser Arath Chavez Reyes, Mar 07 2024

More terms from Jinyuan Wang, Dec 07 2019

A338046 G.f.: Sum_{k>=0} x^(2^k) / (1 - x^(2^k))^4.

Original entry on oeis.org

1, 5, 10, 25, 35, 66, 84, 145, 165, 255, 286, 430, 455, 644, 680, 961, 969, 1305, 1330, 1795, 1771, 2310, 2300, 3030, 2925, 3731, 3654, 4704, 4495, 5640, 5456, 6945, 6545, 8109, 7770, 9741, 9139, 11210, 10660, 13275, 12341, 15015, 14190, 17490, 16215, 19596, 18424, 22630

Offset: 1

Ilya Gutkovskiy, Oct 08 2020

nonn

Cf. A000292, A000335, A000447, A001511, A002492, A129527, A209229, A328407, A338045.

nmax = 48; CoefficientList[Series[Sum[x^(2^k) /(1 - x^(2^k))^4, {k, 0, Floor[Log[2, nmax]] + 1}], {x, 0, nmax}], x] // Rest
a[n_] := If[EvenQ[n], a[n/2] + n (n + 1) (n + 2)/6, n (n + 1) (n + 2)/6]; Table[a[n], {n, 1, 48}]
Table[(1/6) DivisorSum[n, Boole[IntegerQ[Log[2, n/#]]] # (# + 1) (# + 2) &], {n, 1, 48}]

G.f. A(x) satisfies: A(x) = A(x^2) + x / (1 - x)^4.
a(2*n) = a(n) + A002492(n), a(2*n+1) = A000447(n+1).
a(n) = (1/6) * Sum_{d|n} A209229(n/d) * d * (d + 1) * (d + 2).
Product_{n>=1} (1 + x^n)^a(n) = g.f. for A000335.

A349682 a(n) = A000292(6*n + 1) where A000292 are the tetrahedral numbers.

Original entry on oeis.org

1, 84, 455, 1330, 2925, 5456, 9139, 14190, 20825, 29260, 39711, 52394, 67525, 85320, 105995, 129766, 156849, 187460, 221815, 260130, 302621, 349504, 400995, 457310, 518665, 585276, 657359, 735130, 818805, 908600, 1004731, 1107414, 1216865, 1333300, 1456935, 1587986

Offset: 0

Ralf Steiner, Nov 25 2021

Euclid of Alexandria, Elements, VII Def. 17, p. 194, 300 BCE.
Index entries for linear recurrences with constant coefficients, signature (4,-6,4,-1).

Cf. A000292, A000447, A002492, A016921.

nterms=50;Table[36n^3+36n^2+11n+1,{n,0,nterms-1}] (* Paolo Xausa, Nov 25 2021 *)

a(n) = subst(m*(m+1)*(m+2)/6, 'm, 6*n+1); \\ Michel Marcus, Dec 16 2021

def A349682(n): return n*(n*(36*n + 36) + 11) + 1 # Chai Wah Wu, Dec 27 2021

a(n) = 1 + 11*n + 36*n^2 + 36*n^3 = (1 + 2*n)*(1 + 3*n)*(1 + 6*n).
G.f.: (1 + 80*x + 125*x^2 + 10*x^3)/(1 - x)^4. - Stefano Spezia, Nov 29 2021
From Elmo R. Oliveira, Aug 22 2025: (Start)
E.g.f.: exp(x)*(1 + 83*x + 144*x^2 + 36*x^3).
a(n) = 4*a(n-1) - 6*a(n-2) + 4*a(n-3) - a(n-4). (End)
From Amiram Eldar, Aug 31 2025: (Start)
Sum_{n>=0} 1/a(n) = Pi/(4*sqrt(3)) + 2*log(2) - 3*log(3)/4.
Sum_{n>=0} (-1)^n/a(n) = (3/4 - 1/sqrt(3))*Pi + sqrt(3)*log(2 + sqrt(3))/2 - log(2). (End)

A361226 Square array T(n,k) = k((1+2n)*k - 1)/2; n>=0, k>=0, read by antidiagonals upwards.

Original entry on oeis.org

0, 0, 0, 0, 1, 1, 0, 2, 5, 3, 0, 3, 9, 12, 6, 0, 4, 13, 21, 22, 10, 0, 5, 17, 30, 38, 35, 15, 0, 6, 21, 39, 54, 60, 51, 21, 0, 7, 25, 48, 70, 85, 87, 70, 28, 0, 8, 29, 57, 86, 110, 123, 119, 92, 36, 0, 9, 33, 66, 102, 135, 159, 168, 156, 117, 45

Offset: 0

Paul Curtz, Mar 05 2023

The main diagonal is A002414.
The first upper diagonal is A160378(n+1).
The antidiagonals sums are A034827(n+2).
b(n) = (A034827(n+3) = 0, 2, 10, 30, 70, ...) - (A002414(n) = 0, 1, 9, 30, 70, ...) = 0, 1, 1, 0, 0, 5, 21, 56, ... .
b(n+2) = A299120(n). b(n+4) = A033275(n). b(n+4) - b(n) = A002492(n).

			The rows are
  0, 0,  1,  3,  6,  10,  15,  21, ...   = A161680
  0, 1,  5, 12, 22,  35,  51,  70, ...   = A000326
  0, 2,  9, 21, 38,  60,  87, 119, ...   = A005476
  0, 3, 13, 30, 54,  85, 123, 168, ...   = A022264
  0, 4, 17, 39, 70, 110, 159, 217, ...   = A022266
  ... .
Columns: A000004, A001477, A016813, A017197=3*A016777, 2*A017101, 5*A016873, 3*A017581, 7*A017017, ... (coefficients from A026741).
Difference between two consecutive rows: A000290. Hence A143844.
This square array read by antidiagonals leads to the triangle
  0
  0   0
  0   1   1
  0   2   5   3
  0   3   9  12   6
  0   4  13  21  22  10
  0   5  17  30  38  35  15
  ... .

Cf. A000004, A000290, A000326, A001477, A002414, A005476, A016777, A016813, A016873, A017017, A017101, A017197, A017581, A022264, A022266, A026741, A034827, A160378, A161680, A360962.

Cf. A002492, A033275, A143844, A299120.

T[n_, k_] := k*((2*n + 1)*k - 1)/2; Table[T[n - k, k], {n, 0, 10}, {k, 0, n}] // Flatten (* Amiram Eldar, Mar 05 2023 *)

a(n) = { my(row = (sqrtint(8*n+1)-1)\2, column = n - binomial(row + 1, 2)); binomial(column, 2) + column^2 * (row - column) } \\ David A. Corneth, Mar 05 2023

# Seen as a triangle:
from functools import cache
@cache
def Trow(n: int) -> list[int]:
    if n == 0: return [0]
    r = Trow(n - 1)
    return [r[k] + k * k if k < n else r[n - 1] + n - 1 for k in range(n + 1)]
for n in range(7): print(Trow(n)) # Peter Luschny, Mar 05 2023

Take successively sequences n*(n-1)/2, n*(3*n-1)/2, n*(5*n-1)/2, ... listed in the EXAMPLE section.
G.f.: y*(x + y)/((1 - y)^3*(1 - x)^2). - Stefano Spezia, Mar 06 2023
E.g.f.: exp(x+y)*y*(2*x + y + 2*x*y)/2. - Stefano Spezia, Feb 21 2024

cp's OEIS Frontend

A199394 The number of ways to color the vertices of all (11) simple unlabeled graphs on 4 nodes using at most n colors.

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A270309 Irregular triangle read by rows: T(n,k) = ((n-k)+1)^2 if odd-n and odd-k; T(n,k) = k^2 if odd-n and even-k; T(n,k) = (n/2-(k/2-1/2))^2 if even-n and odd-k; T(n,k) = (k/2+1)^2 if even-n and even-k; where n >= 1, k = 1..2*n.

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A330151 Partial sums of 4th powers of the even numbers.

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A338046 G.f.: Sum_{k>=0} x^(2^k) / (1 - x^(2^k))^4.

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A349682 a(n) = A000292(6*n + 1) where A000292 are the tetrahedral numbers.

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A361226 Square array T(n,k) = k*((1+2*n)*k - 1)/2; n>=0, k>=0, read by antidiagonals upwards.

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A361226 Square array T(n,k) = k((1+2n)*k - 1)/2; n>=0, k>=0, read by antidiagonals upwards.