A017197 -id:A017197 - cp's OEIS frontend

A017203 a(n) = (9*n + 3)^7.

2187, 35831808, 1801088541, 21870000000, 137231006679, 587068342272, 1954897493193, 5455160701056, 13348388671875, 29509034655744, 60170087060757, 114868566764928, 207616015289871, 358318080000000, 594467302491009, 953133216331392, 1483273860320763, 2248392813428736

Offset: 0

N. J. A. Sloane

Vincenzo Librandi, Table of n, a(n) for n = 0..10000
Index entries for linear recurrences with constant coefficients, signature (8,-28,56,-70,56,-28,8,-1).

Cf. A013665, A016783, A017197.

[(9*n+3)^7: n in [0..20]]; // Vincenzo Librandi, Jul 23 2011

Table[(9*n + 3)^7, {n, 0, 20}] (* Amiram Eldar, Oct 03 2024 *)

From Amiram Eldar, Oct 03 2024: (Start)
a(n) = A017197(n)^7 = 3^7 * A016783(n).
Sum_{n>=0} 1/a(n) = 28*Pi^7/(215233605*sqrt(3)) + 1093*zeta(7)/4782969. (End)

A017205 a(n) = (9*n + 3)^9.

Original entry on oeis.org

19683, 5159780352, 794280046581, 19683000000000, 208728361158759, 1352605460594688, 6351461955384057, 23762680013799936, 75084686279296875, 208215748530929664, 520411082988487293, 1195092568622310912, 2558036924386500591, 5159780352000000000, 9892530380752880769

Offset: 0

N. J. A. Sloane

Vincenzo Librandi, Table of n, a(n) for n = 0..10000
Index entries for linear recurrences with constant coefficients, signature (10,-45,120,-210,252,-210,120,-45,10,-1).

Cf. A013667, A016785, A017197.

[(9*n+3)^9: n in [0..20]]; // Vincenzo Librandi, Jul 23 2011

Table[(9*n + 3)^9, {n, 0, 20}] (* Amiram Eldar, Oct 03 2024 *)

From Amiram Eldar, Oct 03 2024: (Start)
a(n) = A017197(n)^9 = 3^9 * A016785(n).
Sum_{n>=0} 1/a(n) = 1618*Pi^9/(1098337086315*sqrt(3)) + 9841*zeta(9)/387420489. (End)

A017207 a(n) = (9*n + 3)^11.

Original entry on oeis.org

177147, 743008370688, 350277500542221, 17714700000000000, 317475837322472439, 3116402981210161152, 20635899893042801193, 103510234140112521216, 422351360321044921875, 1469170321634239709184, 4501035456767426597157, 12433743083946522728448, 31517572945366073781711

Offset: 0

N. J. A. Sloane

Vincenzo Librandi, Table of n, a(n) for n = 0..10000
Index entries for linear recurrences with constant coefficients, signature (12,-66,220,-495,792,-924,792,-495,220,-66,12,-1).

Cf. A013669, A016787, A017197.

[(9*n+3)^11: n in [0..15]]; // Vincenzo Librandi, Jul 23 2011

Table[(9*n + 3)^11, {n, 0, 15}] (* Amiram Eldar, Oct 03 2024 *)

From Amiram Eldar, Oct 03 2024: (Start)
a(n) = A017197(n)^11 = 3^11 * A016787(n).
Sum_{n>=0} 1/a(n) = 7388*Pi^11/(444826519957575*sqrt(3)) + 88573*zeta(11)/31381059609.  (End)

A101468 Triangle read by rows: T(n,k)=(n+1-k)(3k+1).

Original entry on oeis.org

1, 2, 4, 3, 8, 7, 4, 12, 14, 10, 5, 16, 21, 20, 13, 6, 20, 28, 30, 26, 16, 7, 24, 35, 40, 39, 32, 19, 8, 28, 42, 50, 52, 48, 38, 22, 9, 32, 49, 60, 65, 64, 57, 44, 25, 10, 36, 56, 70, 78, 80, 76, 66, 50, 28, 11, 40, 63, 80, 91, 96, 95, 88, 75, 56, 31, 12, 44, 70, 90, 104, 112, 114

Offset: 0

Lambert Klasen (lambert.klasen(AT)gmx.de) and Gary W. Adamson, Jan 21 2005

The triangle is generated from the product A*B
of the infinite lower triangular matrices A =
1 0 0 0...
1 1 0 0...
1 1 1 0...
1 1 1 1...
... and B =
1 0 0 0...
1 4 0 0...
1 4 7 0...
1 4 7 10...
...
Row sums give pentagonal pyramidal numbers A002411 T(n+0,0)= 1*n=A000027(n) T(n+0,1)= 4*n=A008586(n) T(n+1,2)= 7*n=A008589(n) T(n+2,3)=10*n=A008592(n) ...
so for example T(n+1,n-0)=6*n+2=A016933(n) T(n+1,n-1)=9*n+3=A017197(n) T(n+2,n-1)=12*n+4=A017569(n)
T(n,0)*T(n,1) = A033996(n) (8 times triangular numbers)
T(n,n)*T(n,0) = A000567(n+1) (Octagonal numbers) etc.

			Triangle begins:
1,
2,  4,
3,  8,  7,
4,  12, 14, 10,
5,  16, 21, 20, 13,
6,  20, 28, 30, 26, 16,
7,  24, 35, 40, 39, 32, 19,
8,  28, 42, 50, 52, 48, 38, 22,
9,  32, 49, 60, 65, 64, 57, 44, 25,
10, 36, 56, 70, 78, 80, 76, 66, 50, 28,
11, 40, 63, 80, 91, 96, 95, 88, 75, 56, 31, etc.
[_Bruno Berselli_, Feb 10 2014]

Cf. A095871 (product B*A), A002411.

t[n_, k_] := If[n < k, 0, (3*k + 1)*(n - k + 1)]; Flatten[ Table[ t[n, k], {n, 0, 11}, {k, 0, n}]] (* Robert G. Wilson v, Jan 21 2005 *)

T(n,k)=if(k>n,0,(n-k+1)*(3*k+1)) for(i=0,10, for(j=0,i,print1(T(i,j),", "));print())

A168410 a(n) = 3 + 9*floor(n/2).

Original entry on oeis.org

3, 12, 12, 21, 21, 30, 30, 39, 39, 48, 48, 57, 57, 66, 66, 75, 75, 84, 84, 93, 93, 102, 102, 111, 111, 120, 120, 129, 129, 138, 138, 147, 147, 156, 156, 165, 165, 174, 174, 183, 183, 192, 192, 201, 201, 210, 210, 219, 219, 228, 228, 237, 237, 246, 246, 255, 255

Offset: 1

Vincenzo Librandi, Nov 25 2009

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

Cf. A017197, A168233.

[3+9*Floor(n/2): n in [1..70]]; // Vincenzo Librandi, Sep 19 2013

Table[3 + 9 Floor[n/2], {n, 70}] (* or *) CoefficientList[Series[3 (1 + 3 x - x^2)/((1 + x) (x - 1)^2), {x, 0, 70}], x] (* Vincenzo Librandi, Sep 19 2013 *)

a(n) = 9*n - a(n-1) - 3, with n>1, a(1)=3.
G.f.: 3*x*(1 + 3*x - x^2) / ( (1+x)*(x-1)^2 ). - R. J. Mathar, Jul 10 2011
a(n) = 3*A168233(n). - R. J. Mathar, Jul 10 2011
a(n) = a(n-1) +a(n-2) -a(n-3). - Vincenzo Librandi, Sep 19 2013
E.g.f.: (3/4)*(3 - 4*exp(x) + (6*x + 1)*exp(2*x))*exp(-x). - G. C. Greubel, Jul 21 2016

New definition by Vincenzo Librandi, Sep 19 2013

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

A017203 a(n) = (9*n + 3)^7.

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A017205 a(n) = (9*n + 3)^9.

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Magma

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A017207 a(n) = (9*n + 3)^11.

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A101468 Triangle read by rows: T(n,k)=(n+1-k)*(3*k+1).

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A168410 a(n) = 3 + 9*floor(n/2).

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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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A101468 Triangle read by rows: T(n,k)=(n+1-k)(3k+1).

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