A027480 -id:A027480 - cp's OEIS frontend

A108495 a(n) = (n^7 - n)/6.

0, 0, 21, 364, 2730, 13020, 46655, 137256, 349524, 797160, 1666665, 3247860, 5971966, 10458084, 17568915, 28476560, 44739240, 68389776, 102036669, 148978620, 213333330, 300181420, 415726311, 567470904, 764411900, 1017252600

Offset: 0

Henry Bottomley, Jun 06 2005

Also integer sequences for (n^2-n)/1 (A002378 offset), (n^3-n)/2 (A027480 offset), (n^43-n)/42 (A108496) and (n^1807-n)/1806.

			a(2) = (2^7 - 2)/6 = 126/6 = 21.

Vincenzo Librandi, Table of n, a(n) for n = 0..1000
D. Zagier, Problems posed at the St Andrews Colloquium, 1996.
Index entries for linear recurrences with constant coefficients, signature (8,-28,56,-70,56,-28,8,-1).

Cf. A014117, A108497, A108499.

[(n^7-n)/6: n in [0..40]]; // Vincenzo Librandi, May 02 2011

Table[(n^7-n)/6,{n,0,30}] (* or *) LinearRecurrence[ {8,-28,56,-70,56,-28,8,-1},{0,0,21,364,2730,13020,46655,137256},30] (* Harvey P. Dale, Apr 16 2014 *)

[(n**7-n)//6 for n in range(41)] # David Radcliffe, Jun 06 2025

a(n) = (n-1)*A059721(n) = -A024004(n)*n/6.

G.f.: 7*x^2*(3 + 28*x + 58*x^2 + 28*x^3 + 3*x^4)/(1-x)^8. [Colin Barker, May 08 2012]

A108496 a(n) = (n^43 - n)/42.

Original entry on oeis.org

0, 0, 209430786243, 7815642080822311372, 1842172677508006361457030, 27068294695622864223661876860, 68747114771196346634599779308105, 51995580380757061883555053636996008

Offset: 0

Henry Bottomley, Jun 06 2005

nonn

Also integer sequences for (n^2-n)/1 (A002378 offset), (n^3-n)/2 (A027480 offset), (n^7-n)/6 (A108495) and (n^1807-n)/1806.

			a(2) = (2^43 - 2)/42 = 8796093022206/42 = 209430786243.

Vincenzo Librandi, Table of n, a(n) for n = 0..1000
D. Zagier, Problems posed at the St Andrews Colloquium, 1996.

Cf. A014117, A108497, A108499.

[(n^43-n)/42: n in [0..20]]; // Vincenzo Librandi, May 02 2011

Table[(n^43-n)/42,{n,0,10}] (* Harvey P. Dale, Jan 31 2022 *)

a(n) = (n-1)*A108048(n).

A174002 a(n) = n*binomial(n+4, 4).

Original entry on oeis.org

0, 5, 30, 105, 280, 630, 1260, 2310, 3960, 6435, 10010, 15015, 21840, 30940, 42840, 58140, 77520, 101745, 131670, 168245, 212520, 265650, 328900, 403650, 491400, 593775, 712530, 849555, 1006880, 1186680, 1391280, 1623160, 1884960, 2179485

Offset: 0

Reinhard Zumkeller, Mar 05 2010, Mar 17 2010

This sequence can be computed from Pascal's triangle. Find the fifth number in a row and multiply it by the second number of the next row. - Alonso del Arte, Jan 21 2018

Index entries for linear recurrences with constant coefficients, signature (6, -15, 20, -15, 6, -1).

Cf. A033488, A027480, A003506.

[ (n^5+10*n^4+35*n^3+50*n^2+24*n)/24: n in [0..40] ]; // Vincenzo Librandi, Dec 28 2010

Table[n Binomial[n + 4, 4], {n, 0, 40}] (* or *) LinearRecurrence[{6, -15, 20, -15, 6, -1}, {0, 5, 30, 105, 280, 630}, 40] (* Harvey P. Dale, Dec 03 2011 *)

a(n) = (n^5 + 10*n^4 + 35*n^3 + 50*n^2 + 24*n) / 24.
For n > 0: a(n) = A003506(n+4, 5).
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), with a(0)=0, a(1)=5, a(2)=30, a(3)=105, a(4)=280, a(5)=630. - Harvey P. Dale, Dec 03 2011
G.f.: 5*x/(1-x)^6. - Colin Barker, Mar 18 2012

Title switched with first Formula section entry, at the suggestion of Alonso del Arte, by Jon E. Schoenfield, Jan 28 2018

A293617 Array of triangles read by ascending antidiagonals, T(m, n, k) = Pochhammer(m, k) * Stirling2(n + m, k + m) with m >= 0, n >= 0 and 0 <= k <= n.

Original entry on oeis.org

1, 1, 0, 1, 1, 0, 1, 3, 1, 0, 1, 6, 2, 1, 0, 1, 10, 3, 7, 3, 0, 1, 15, 4, 25, 12, 2, 0, 1, 21, 5, 65, 30, 6, 1, 0, 1, 28, 6, 140, 60, 12, 15, 7, 0, 1, 36, 7, 266, 105, 20, 90, 50, 12, 0, 1, 45, 8, 462, 168, 30, 350, 195, 60, 6, 0, 1, 55, 9, 750, 252, 42, 1050, 560, 180, 24, 1, 0

Offset: 0

Peter Luschny, Oct 20 2017

			Array starts:
m\j| 0   1  2     3       4       5       6       7       8       9      10
---|-----------------------------------------------------------------------
m=0| 1,  0, 0,    0,      0,      0,      0,      0,      0,      0,      0
m=1| 1,  1, 1,    1,      3,      2,      1,      7,     12,      6,      1
m=2| 1,  3, 2,    7,     12,      6,     15,     50,     60,     24,     31
m=3| 1,  6, 3,   25,     30,     12,     90,    195,    180,     60,    301
m=4| 1, 10, 4,   65,     60,     20,    350,    560,    420,    120,   1701
m=5| 1, 15, 5,  140,    105,     30,   1050,   1330,    840,    210,   6951
m=6| 1, 21, 6,  266,    168,     42,   2646,   2772,   1512,    336,  22827
m=7| 1, 28, 7,  462,    252,     56,   5880,   5250,   2520,    504,  63987
m=8| 1, 36, 8,  750,    360,     72,  11880,   9240,   3960,    720, 159027
m=9| 1, 45, 9, 1155,    495,     90,  22275,  15345,   5940,    990, 359502
   A000217, A001296,A027480,A002378,A001297,A293475,A033486,A007531,A001298
.
m\j| ...      11      12      13      14
---|-----------------------------------------
m=0| ...,      0,      0,      0,      0, ... [A000007]
m=1| ...,     15,     50,     60,     24, ... [A028246]
m=2| ...,    180,    390,    360,    120, ... [A053440]
m=3| ...,   1050,   1680,   1260,    360, ... [A294032]
m=4| ...,   4200,   5320,   3360,    840, ...
m=5| ...,  13230,  13860,   7560,   1680, ...
m=6| ...,  35280,  31500,  15120,   3024, ...
m=7| ...,  83160,  64680,  27720,   5040, ...
m=8| ..., 178200, 122760,  47520,   7920, ...
m=9| ..., 353925, 218790,  77220,  11880, ...
         A293476,A293608,A293615,A052762, ...
.
The parameter m runs over the triangles and j indexes the triangles by reading them by rows. Let T(m, n) denote the row [T(m, n, k) for 0 <= k <= n] and T(m) denote the triangle [T(m, n) for n >= 0]. Then for instance T(2) is the triangle A053440, T(3, 2) is row 2 of A294032 (which is [25, 30, 12]) and T(3, 2, 1) = 30.
.
Remark: To adapt the sequences A028246 and A053440 to our enumeration use the exponential generating functions exp(x)/(1 - y*(exp(x) - 1)) and exp(x)*(2*exp(x) - y*exp(2*x) + 2*y*exp(x) - 1 - y)/(1 - y*(exp(x) - 1))^2 instead of those indicated in their respective entries.

Eric Weisstein's World of Mathematics, Nørlund Polynomial.

A000217(n) = T(n, 1, 0), A001296(n) = T(n, 2, 0), A027480(n) = T(n, 2, 1),
A002378(n) = T(n, 2, 2), A001297(n) = T(n, 3, 0), A293475(n) = T(n, 3, 1),
A033486(n) = T(n, 3, 2), A007531(n) = T(n, 3, 3), A001298(n) = T(n, 4, 0),
A293476(n) = T(n, 4, 1), A293608(n) = T(n, 4, 2), A293615(n) = T(n, 4, 3),
A052762(n) = T(n, 4, 4), A052787(n) = T(n, 5, 5), A000225(n) = T(1, n, 1),
A028243(n) = T(1, n, 2), A028244(n) = T(1, n, 3), A028245(n) = T(1, n, 4),
A032180(n) = T(1, n, 5), A228909(n) = T(1, n, 6), A228910(n) = T(1, n, 7),
A000225(n) = T(2, n, 0), A007820(n) = T(n, n, 0).
A028246(n,k) = T(1, n, k), A053440(n,k) = T(2, n, k), A294032(n,k) = T(3, n, k),
A293926(n,k) = T(n, n, k), A124320(n,k) = T(n, k, k), A156991(n,k) = T(k, n, n).
Cf. A293616.

A293617 := proc(m, n, k) option remember:
if m = 0 then 0^n elif k < 0 or k > n then 0 elif n = 0 then 1 else
(k+m)*A293617(m,n-1,k) + k*A293617(m,n-1,k-1) + A293617(m-1,n,k) fi end:
for m in [$0..4] do for n in [$0..6] do print(seq(A293617(m, n, k), k=0..n)) od od;
# Sample uses:
A027480 := n -> A293617(n, 2, 1): A293608 := n -> A293617(n, 4, 2):
# Flatten:
a := proc(n) local w; w := proc(k) local t, s; t := 1; s := 1;
while t <= k do s := s + 1; t := t + s od; [s - 1, s - t + k] end:
seq(A293617(n - k, w(k)[1], w(k)[2]), k=0..n) end: seq(a(n), n = 0..11);

T[m_, n_, k_] := Pochhammer[m, k] StirlingS2[n + m, k + m];
For[m = 0, m < 7, m++, Print[Table[T[m, n, k], {n,0,6}, {k,0,n}]]]
A293617Row[m_, n_] := Table[T[m, n, k], {k,0,n}];
(* Sample use: *)
A293926Row[n_] := A293617Row[n, n];

T(m,n,k) = (k + m)*T(m, n-1, k) + k*T(m, n-1, k-1) + T(m-1, n, k) with boundary conditions T(0, n, k) = 0^n; T(m, n, k) = 0 if k<0 or k>n; and T(m, 0, k) = 0^k.

T(m,n,k) = Pochhammer(m, k)*binomial(n + m, k + m)*NorlundPolynomial(n - k, -k - m).

A316224 a(n) = n(2n + 1)(4n + 1).

Original entry on oeis.org

0, 15, 90, 273, 612, 1155, 1950, 3045, 4488, 6327, 8610, 11385, 14700, 18603, 23142, 28365, 34320, 41055, 48618, 57057, 66420, 76755, 88110, 100533, 114072, 128775, 144690, 161865, 180348, 200187, 221430, 244125, 268320, 294063, 321402, 350385, 381060, 413475, 447678, 483717

Offset: 0

Bruno Berselli, Jun 27 2018

Sums of the consecutive integers from A000384(n) to A000384(n+1)-1. This is the case s=6 of the formula n*(n*(s-2) + 1)*(n*(s-2) + 2)/2 related to s-gonal numbers.

The inverse binomial transform is 0, 15, 60, 48, 0, ... (0 continued).

			Row sums of the triangle:
|  0 |  ................................................................. 0
|  1 |  2  3  4  5  .................................................... 15
|  6 |  7  8  9 10 11 12 13 14  ........................................ 90
| 15 | 16 17 18 19 20 21 22 23 24 25 26 27  ........................... 273
| 28 | 29 30 31 32 33 34 35 36 37 38 39 40 41 42 43 44  ............... 612
| 45 | 46 47 48 49 50 51 52 53 54 55 56 57 58 59 60 61 62 63 64 65  .. 1155
...
where:
. first column is A000384,
. second column is A130883 (without 1),
. third column is A033816,
. diagonal is A014106,
. 0, 2, 8, 18, 32, 50, ... are in A001105.

Index entries for linear recurrences with constant coefficients, signature (4,-6,4,-1).

First bisection of A059270 and subsequence of A034828, A047866, A109900, A290168.

Sums of the consecutive integers from P(s,n) to P(s,n+1)-1, where P(s,k) is the k-th s-gonal number: A027480 (s=3), A055112 (s=4), A228888 (s=5).

GAP
```
List([0..40], n -> n*(2*n+1)*(4*n+1));
```

[n*(2*n+1)*(4*n+1) for n in 0:40] |> println

Magma
```
[n*(2*n+1)*(4*n+1): n in [0..40]];
```

seq(n*(2*n+1)*(4*n+1),n=0..40); # Muniru A Asiru, Jun 27 2018

Table[n (2 n + 1) (4 n + 1), {n, 0, 40}]

Maxima
```
makelist(n*(2*n+1)*(4*n+1), n, 0, 40);
```
PARI
```
vector(40, n, n--; n*(2*n+1)*(4*n+1))
```
Python
```
[n*(2*n+1)*(4*n+1) for n in range(40)]
```
Sage
```
[n*(2*n+1)*(4*n+1) for n in (0..40)]
```

O.g.f.: 3*x*(5 + 10*x + x^2)/(1 - x)^4.
E.g.f.: x*(15 + 30*x + 8*x^2)*exp(x).
a(n) = 4*a(n-1) - 6*a(n-2) + 4*a(n-3) - a(n-4).
a(n) =  3*A258582(n).
a(n) = -3*A100157(-n).
Sum_{n>0} 1/a(n) = 2*(3 - log(4)) - Pi.
Sum_{n>=1} (-1)^(n+1)/a(n) = log(2) + 2*sqrt(2)*log(1+sqrt(2)) + (sqrt(2)-1/2)*Pi - 6. - Amiram Eldar, Sep 17 2022

A008670 Molien series for Weyl group F_4.

Original entry on oeis.org

1, 1, 1, 2, 3, 3, 5, 6, 7, 9, 11, 12, 16, 18, 20, 24, 28, 30, 36, 40, 44, 50, 56, 60, 69, 75, 81, 90, 99, 105, 117, 126, 135, 147, 159, 168, 184, 196, 208, 224, 240, 252, 272, 288, 304, 324, 344, 360, 385, 405, 425, 450, 475, 495, 525, 550, 575, 605, 635, 660, 696, 726, 756

Offset: 0

N. J. A. Sloane

Number of partitions of n into parts 1, 3, 4 and 6. - Ilya Gutkovskiy, May 24 2017

Coxeter and Moser, Generators and Relations for Discrete Groups, Table 10.
L. Smith, Polynomial Invariants of Finite Groups, Peters, 1995, p. 199 (No. 28).

Alois P. Heinz, Table of n, a(n) for n = 0..1000
INRIA Algorithms Project, Encyclopedia of Combinatorial Structures 236
G. Nebe, E. M. Rains and N. J. A. Sloane, Self-Dual Codes and Invariant Theory, Springer, Berlin, 2006.
Index entries for Molien series
Index entries for linear recurrences with constant coefficients, signature (1,0,1,0,-1,1,-2,1,-1,0,1,0,1,-1).

Cf. A002411, A006002, A010875, A011934, A027480, A055232, A182260.

MolienSeries(CoxeterGroup("F4")); // Sergei Haller (sergei(AT)sergei-haller.de), Dec 21 2006

R:=PowerSeriesRing(Integers(), 70); Coefficients(R!( 1/((1-x)*(1-x^3)*(1-x^4)*(1-x^6)) )); // G. C. Greubel, Sep 08 2019

a:= proc(n) local m, r; m := iquo (n, 12, 'r'); r:= r+1; ([4, 5, 6, 8, 10, 11, 14, 16, 18, 21, 24, 26][r]+ (6+r+4*m)*m)*m+ [1$3, 2, 3$2, 5, 6, 7, 9, 11, 12][r] end: seq(a(n), n=0..100); # Alois P. Heinz, Oct 06 2008

Take[CoefficientList[Series[1/((1-x^2)(1-x^6)(1-x^8)(1-x^12)),{x,0,130}], x], {1,-1,2}] (* or *) LinearRecurrence[ {1,0,1,0,-1,1,-2,1,-1,0,1,0,1,-1},{1,1,1,2,3,3,5,6,7,9,11,12,16,18},70] (* Harvey P. Dale, Feb 07 2012 *)

my(x='x+O('x^70)); Vec(1/((1-x)*(1-x^3)*(1-x^4)*(1-x^6))) \\ G. C. Greubel, Sep 08 2019

def A008670_list(prec):
    P. = PowerSeriesRing(ZZ, prec)
    return P(1/((1-x)*(1-x^3)*(1-x^4)*(1-x^6))).list()
A008670_list(70) # G. C. Greubel, Sep 08 2019

G.f.: 1/((1-x)*(1-x^3)*(1-x^4)*(1-x^6)). [Corrected by Ralf Stephan, Apr 29 2014]
a(n) = a(n-1) + a(n-3) - a(n-5) + a(n-6) - 2*a(n-7) + a(n-8) - a(n-9) + a(n-11) + a(n-13) - a(n-14), with a(0)=1, a(1)=1, a(2)=1, a(3)=2, a(4)=3, a(5)=3, a(6)=5, a(7)=6, a(8)=7, a(9)=9, a(10)=11, a(11)=12, a(12)=16, a(13)=18. - Harvey P. Dale, Feb 07 2012
a(n) ~ (1/432)*n^3. - Ralf Stephan, Apr 29 2014
a(n) = (120*floor(n/6)^3 + 60*(m+7)*floor(n/6)^2 + 2*(m^5-15*m^4+75*m^3-135*m^2+134*m+240)*floor(n/6) + 3*(m^5-15*m^4+75*m^3-135*m^2+84*m+70) + (m^5-15*m^4+75*m^3-135*m^2+44*m+30)*(-1)^floor(n/6))/240 where m = (n mod 6). - Luce ETIENNE, Aug 14 2018
a(n) = 1 + floor((2*n^3 + 42*n^2 + n*(279 + 9*(-1)^n - 48*[(n mod 3)=2]))/864) where [] is the Iverson bracket. - Hoang Xuan Thanh, Jun 22 2025

A056923 Write the integers in groups: 0; 1,2; 3,4,5; 6,7,8,9; ... and form the product of the members of each group.

Original entry on oeis.org

0, 2, 60, 3024, 240240, 27907200, 4475671200, 948964262400, 257256702743040, 86839771951296000, 35728290125079552000, 17602963463032472448000, 10233395250958706770944000, 6932022668773077815267328000

Offset: 0

Robert G. Wilson v, Sep 09 2000

nonn

Each group begins with a triangular number and proceeds until one short of the next triangular number.

Also, the number under the radical using Brahmagupta's formula for an n-sided cyclic quadrilateral with sides 1..n. - Ben Paul Thurston, Dec 05 2006

Vincenzo Librandi, Table of n, a(n) for n = 0..100
Nick Hobson, Solution to puzzle 10: Farmer's enclosure

Cf. A000217, A027480.

a:= n-> mul(n*(n+1)/2+j, j=0..n):
seq(a(n), n=0..15);  # Alois P. Heinz, Feb 02 2019

Table[(n (n + 3)/2)!/((n - 1)(n + 2)/2)!, {n, 0, 15}]
Times@@Range[First[#],Last[#]-1]&/@Partition[Accumulate[Range[0,15]],2,1] (* Harvey P. Dale, Apr 25 2014 *)

a(n) = (n (n + 3)/2)!/((n - 1)(n + 2)/2)!.

a(n) = Product_{j=1..n+1} ((n+2)*(n+1)/2-j). - Ben Paul Thurston, Dec 05 2006

A074246 Triangle of coefficients, read by rows, where the n-th row forms the polynomial P(n,x) = {Sum_{k=1..n} 1/(k+x)}*{Product_{k=1..n} (k+x)}.

Original entry on oeis.org

1, 3, 2, 11, 12, 3, 50, 70, 30, 4, 274, 450, 255, 60, 5, 1764, 3248, 2205, 700, 105, 6, 13068, 26264, 20307, 7840, 1610, 168, 7, 109584, 236248, 201852, 89796, 22680, 3276, 252, 8, 1026576, 2345400, 2171040, 1077300, 316365, 56700, 6090, 360, 9

Offset: 1

Paul D. Hanna, Sep 19 2002

The n-th row polynomial, P(n,x), has ordered zeros {z_k < z_(k+1), 0

The higher-order exponential integrals E(x,m,n) are defined in A163931 and the asymptotic expansion of E(x,m=2,n) can be found in A028421. We determined with the latter that E(x,m=2,n+1) = (exp(-x)/x^2)*(1 - (3+2*n)/x + (11+12*n+3*n^2)/x^2 - (50+70*n+30*n^2+ 4*n^3)/x^3 + .... ). The polynomial coefficients in the numerators lead to the coefficients of the triangle given above. The numerators of the o.g.f.s of the right hand columns of this triangle lead for z = 1 to A001147. - Johannes W. Meijer, Oct 16 2009

			Polynomials begin:
P(1,x) = 1,
P(2,x) = 3 + 2x,
P(3,x) = 11 + 12x + 3x^2,
P(4,x) = 50 + 70x + 30x^2 + 4x^3,
P(5,x) = 274 + 450x + 255x^2 + 60x^3 + 5x^4,
P(6,x) = 1764 + 3248x + 2205x^2 + 700x^3 + 105x^4 + 6x^5,
P(7,x) = 13068 + 26264x + 20307x^2 + 7840x^3 + 1610x^4 + 168x^5 + 7x^6,
P(8,x) = 109584 + 236248x + 201852x^2 + 89796x^3 + 22680x^4 + 3276x^5 + 252x^6 + 8x^7,
P(9,x) = 1026576 + 2345400x + 2171040x^2 + 1077300x^3 + 316365x^4 + 56700x^5 + 6090x^6 + 360x^7 + 9x^8,
P(10,x) = 10628640 + 25507152x + 25228500x^2 + 13667720x^3 + 4510275x^4 + 946638x^5 + 127050x^6 + 10560x^7 + 495x^8 + 10x^9, ...

G. C. Greubel, Table of n, a(n) for the first 50 rows, flattened

See references and formulas at A000254, A001705. Cf. A028421.

A027480 is the second right hand column. - Johannes W. Meijer, Oct 16 2009

with(combinat): A074246 := proc(n,m): (-1)^(n+m)*binomial(m,1)*stirling1(n+1,m+1) end: seq(seq(A074246(n,m),m=1..n),n=1..9); # Johannes W. Meijer, Oct 16 2009, Revised Sep 09 2012

p[n_, x_] := Sum[1/(k+x), {k, 1, n}] Product[k+x, {k, 1, n}] ; Flatten[Table[ CoefficientList[ p[n, x] // Simplify[#, ComplexityFunction -> Length] &, x], {n, 1, 9}]] (* Jean-François Alcover, May 04 2011 *)

P(n) = Vecrev(sum(k=1, n, prod(k=1, n, (k+x))/(k+x)));
for (n=1, 10, print(P(n))) \\ Michel Marcus, Jan 22 2017

First column is A000254 (Stirling numbers of first kind s(n, 2): a(n+1)=(n+1)*a(n)+n!), while sum of rows is A001705 (generalized Stirling numbers). Also related to Harmonic numbers: P(n, 0)=n!*H(n), H(n)=harmonic number.
T(n,k) = (-1)^(n+k)*k*Stirling1(n+1,k+1). - Johannes W. Meijer, Oct 16 2009
E.g.f.: 1/(1 - z)^(x+1)*log(1/(1 - z)). Cf. A028421. - Peter Bala, Jan 06 2015

A225413 Triangle read by rows: T(n,k) = (A101164(n,k) - A014473(n,k))/2.

Original entry on oeis.org

0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 1, 0, 0, 0, 0, 3, 3, 0, 0, 0, 0, 6, 12, 6, 0, 0, 0, 0, 10, 30, 30, 10, 0, 0, 0, 0, 15, 60, 91, 60, 15, 0, 0, 0, 0, 21, 105, 215, 215, 105, 21, 0, 0, 0, 0, 28, 168, 435, 590, 435, 168, 28, 0, 0, 0, 0, 36, 252, 791, 1365, 1365, 791, 252, 36, 0, 0

Offset: 0

Jeremy Gardiner, Jul 28 2013

Has opposite parity to A140356, A155454.

			Triangle begins as:
  0;
  0,  0;
  0,  0,  0;
  0,  0,  0,   0;
  0,  0,  1,   0,    0;
  0,  0,  3,   3,    0,    0;
  0,  0,  6,  12,    6,    0,    0;
  0,  0, 10,  30,   30,   10,    0,    0;
  0,  0, 15,  60,   91,   60,   15,    0,    0;
  0,  0, 21, 105,  215,  215,  105,   21,    0,    0;
  0,  0, 28, 168,  435,  590,  435,  168,   28,    0,   0;
  0,  0, 36, 252,  791, 1365, 1365,  791,  252,   36,   0,  0;
  0,  0, 45, 360, 1330, 2800, 3571, 2800, 1330,  360,  45,  0,  0;
  0,  0, 55, 495, 2106, 5250, 8197, 8197, 5250, 2106, 495, 55,  0,  0;

Reinhard Zumkeller, Rows n = 0..100 of table, flattened

Cf. A000129, A007318, A008288, A014473, A101164, A121262, A140356, A155454.
3rd column = A000217 (triangular numbers).
4th column = A027480 (n(n+1)(n+2)/2).

a225413 n k = a225413_tabl !! n !! k
a225413_row n = a225413_tabl !! n
a225413_tabl = map (map (`div` 2)) $
               zipWith (zipWith (-)) a101164_tabl a014473_tabl
-- Reinhard Zumkeller, Jul 30 2013

A008288:= func< n,k | (&+[Binomial(n-j, j)*Binomial(n-2*j, k-j): j in [0..k]]) >;
A225413:= func< n,k | (A008288(n,k) - 2*Binomial(n,k) + 1)/2 >;
[A225413(n,k): k in [0..n], n in [0..12]]; // G. C. Greubel, Apr 08 2024

T[n_, k_]:= ((-1)^(n-k)*Hypergeometric2F1[-n+k,k+1,1,2] - 2*Binomial[n, k] +1)/2;
Table[T[n,k], {n,0,12}, {k,0,n}]//Flatten (* G. C. Greubel, Apr 08 2024 *)

def A008288(n,k): return sum(binomial(n-j,j)*binomial(n-2*j,k-j) for j in range(k+1))
def A225413(n,k): return (A008288(n,k) -2*binomial(n,k) +1)//2
flatten([[A225413(n,k) for k in range(n+1)] for n in range(13)]) # G. C. Greubel, Apr 08 2024

T(n, k) = (A101164(n,k) - A014473(n,k))/2.
T(n, k) = (A008288(n,k) - 2*A007318(n,k) + 1)/2.
From G. C. Greubel, Apr 08 2024: (Start)
T(n, n-k) = T(n, k).
Sum_{k=0..n} T(n, k) = (A000129(n+1) + n + 1 - 2^(n+1))/2.
Sum_{k=0..n} (-1)^k*T(n, k) = A121262(n) - [n=0]. (End)

A281258 Digital root of n(n+1)(n+2)/2.

Original entry on oeis.org

0, 3, 3, 3, 6, 6, 6, 9, 9, 9, 3, 3, 3, 6, 6, 6, 9, 9, 9, 3, 3, 3, 6, 6, 6, 9, 9, 9, 3, 3, 3, 6, 6, 6, 9, 9, 9, 3, 3, 3, 6, 6, 6, 9, 9, 9, 3, 3, 3, 6, 6, 6, 9, 9, 9, 3, 3, 3, 6, 6, 6, 9, 9, 9, 3, 3, 3, 6, 6, 6, 9, 9, 9, 3, 3, 3, 6, 6, 6, 9, 9, 9, 3, 3, 3, 6, 6, 6, 9, 9, 9, 3, 3, 3, 6, 6, 6, 9, 9, 9, 3, 3, 3, 6, 6, 6, 9, 9, 9, 3, 3, 3, 6, 6, 6, 9, 9, 9

Offset: 0

Peter M. Chema, Jan 18 2017

Also zero together with period 9: repeat [3, 3, 3, 6, 6, 6, 9, 9, 9].

Also the decimal expansion of 334001/1001001.

Index entries for linear recurrences with constant coefficients, signature (1,0,-1,1,0,-1,1).

Cf. A010888, A027480.

PadRight[{0},120,{9,3,3,3,6,6,6,9,9}] (* Harvey P. Dale, Mar 22 2018 *)

a(n) = A010888(A027480(n)).
From Stefano Spezia, Aug 30 2022: (Start)
G.f.: 3*x*(1 + 2*x^3 + 3*x^6)/((1 - x)*(1 + x^3 + x^6)).
a(n) = a(n-1) - a(n-3) + a(n-4) - a(n-6) + a(n-7) for n > 7. (End)

Previous Showing 31-40 of 59 results. Next

cp's OEIS Frontend

A108495 a(n) = (n^7 - n)/6.

Views

Author

Keywords

Comments

Examples

Links

Crossrefs

Programs

Magma

Mathematica

Python

Formula

A108496 a(n) = (n^43 - n)/42.

Views

Author

Keywords

Comments

Examples

Links

Crossrefs

Programs

Magma

Mathematica

Formula

A174002 a(n) = n*binomial(n+4, 4).

Views

Author

Keywords

Comments

Links

Crossrefs

Programs

Magma

Mathematica

Formula

Extensions

A293617 Array of triangles read by ascending antidiagonals, T(m, n, k) = Pochhammer(m, k) * Stirling2(n + m, k + m) with m >= 0, n >= 0 and 0 <= k <= n.

Views

Author

Keywords

Examples

Links

Crossrefs

Programs

Maple

Mathematica

Formula

A316224 a(n) = n*(2*n + 1)*(4*n + 1).

Views

Author

Keywords

Comments

Examples

Links

Crossrefs

Programs

GAP

Julia

Magma

Maple

Mathematica

Maxima

PARI

Python

Sage

Formula

A008670 Molien series for Weyl group F_4.

Views

Author

Keywords

Comments

References

Links

Crossrefs

Programs

Magma

Magma

Maple

A316224 a(n) = n(2n + 1)(4n + 1).

A281258 Digital root of n(n+1)(n+2)/2.