A180324 -id:A180324 - cp's OEIS frontend

A213500 Rectangular array T(n,k): (row n) = bc, where b(h) = h, c(h) = h + n - 1, n >= 1, h >= 1, and = convolution.

1, 4, 2, 10, 7, 3, 20, 16, 10, 4, 35, 30, 22, 13, 5, 56, 50, 40, 28, 16, 6, 84, 77, 65, 50, 34, 19, 7, 120, 112, 98, 80, 60, 40, 22, 8, 165, 156, 140, 119, 95, 70, 46, 25, 9, 220, 210, 192, 168, 140, 110, 80, 52, 28, 10, 286, 275, 255, 228, 196, 161, 125, 90

Offset: 1

Clark Kimberling, Jun 14 2012

Principal diagonal: A002412.
Antidiagonal sums: A002415.
Row 1:  (1,2,3,...)**(1,2,3,...) = A000292.
Row 2:  (1,2,3,...)**(2,3,4,...) = A005581.
Row 3:  (1,2,3,...)**(3,4,5,...) = A006503.
Row 4:  (1,2,3,...)**(4,5,6,...) = A060488.
Row 5:  (1,2,3,...)**(5,6,7,...) = A096941.
Row 6:  (1,2,3,...)**(6,7,8,...) = A096957.
...
In general, the convolution of two infinite sequences is defined from the convolution of two n-tuples:  let X(n) = (x(1),...,x(n)) and Y(n)=(y(1),...,y(n)); then X(n)**Y(n) = x(1)*y(n)+x(2)*y(n-1)+...+x(n)*y(1); this sum is the n-th term in the convolution of infinite sequences:(x(1),...,x(n),...)**(y(1),...,y(n),...), for all n>=1.
...
In the following guide to related arrays and sequences, row n of each array T(n,k) is the convolution b**c of the sequences b(h) and c(h+n-1). The principal diagonal is given by T(n,n) and the n-th antidiagonal sum by S(n). In some cases, T(n,n) or S(n) differs in offset from the listed sequence.
b(h)........ c(h)........ T(n,k) .. T(n,n) .. S(n)
h .......... h .......... A213500 . A002412 . A002415
h .......... h^2 ........ A212891 . A213436 . A024166
h^2 ........ h .......... A213503 . A117066 . A033455
h^2 ........ h^2 ........ A213505 . A213546 . A213547
h .......... h*(h+1)/2 .. A213548 . A213549 . A051836
h*(h+1)/2 .. h .......... A213550 . A002418 . A005585
h*(h+1)/2 .. h*(h+1)/2 .. A213551 . A213552 . A051923
h .......... h^3 ........ A213553 . A213554 . A101089
h^3 ........ h .......... A213555 . A213556 . A213547
h^3 ........ h^3 ........ A213558 . A213559 . A213560
h^2 ........ h*(h+1)/2 .. A213561 . A213562 . A213563
h*(h+1)/2 .. h^2 ........ A213564 . A213565 . A101094
2^(h-1) .... h .......... A213568 . A213569 . A047520
2^(h-1) .... h^2 ........ A213573 . A213574 . A213575
h .......... Fibo(h) .... A213576 . A213577 . A213578
Fibo(h) .... h .......... A213579 . A213580 . A053808
Fibo(h) .... Fibo(h) .... A067418 . A027991 . A067988
Fibo(h+1) .. h .......... A213584 . A213585 . A213586
Fibo(n+1) .. Fibo(h+1) .. A213587 . A213588 . A213589
h^2 ........ Fibo(h) .... A213590 . A213504 . A213557
Fibo(h) .... h^2 ........ A213566 . A213567 . A213570
h .......... -1+2^h ..... A213571 . A213572 . A213581
-1+2^h ..... h .......... A213582 . A213583 . A156928
-1+2^h ..... -1+2^h ..... A213747 . A213748 . A213749
h .......... 2*h-1 ...... A213750 . A007585 . A002417
2*h-1 ...... h .......... A213751 . A051662 . A006325
2*h-1 ...... 2*h-1 ...... A213752 . A100157 . A071238
2*h-1 ...... -1+2^h ..... A213753 . A213754 . A213755
-1+2^h ..... 2*h-1 ...... A213756 . A213757 . A213758
2^(n-1) .... 2*h-1 ...... A213762 . A213763 . A213764
2*h-1 ...... Fibo(h) .... A213765 . A213766 . A213767
Fibo(h) .... 2*h-1 ...... A213768 . A213769 . A213770
Fibo(h+1) .. 2*h-1 ...... A213774 . A213775 . A213776
Fibo(h) .... Fibo(h+1) .. A213777 . A001870 . A152881
h .......... 1+[h/2] .... A213778 . A213779 . A213780
1+[h/2] .... h .......... A213781 . A213782 . A005712
1+[h/2] .... [(h+1)/2] .. A213783 . A213759 . A213760
h .......... 3*h-2 ...... A213761 . A172073 . A002419
3*h-2 ...... h .......... A213771 . A213772 . A132117
3*h-2 ...... 3*h-2 ...... A213773 . A214092 . A213818
h .......... 3*h-1 ...... A213819 . A213820 . A153978
3*h-1 ...... h .......... A213821 . A033431 . A176060
3*h-1 ...... 3*h-1 ...... A213822 . A213823 . A213824
3*h-1 ...... 3*h-2 ...... A213825 . A213826 . A213827
3*h-2 ...... 3*h-1 ...... A213828 . A213829 . A213830
2*h-1 ...... 3*h-2 ...... A213831 . A213832 . A212560
3*h-2 ...... 2*h-1 ...... A213833 . A130748 . A213834
h .......... 4*h-3 ...... A213835 . A172078 . A051797
4*h-3 ...... h .......... A213836 . A213837 . A071238
4*h-3 ...... 2*h-1 ...... A213838 . A213839 . A213840
2*h-1 ...... 4*h-3 ...... A213841 . A213842 . A213843
2*h-1 ...... 4*h-1 ...... A213844 . A213845 . A213846
4*h-1 ...... 2*h-1 ...... A213847 . A213848 . A180324
[(h+1)/2] .. [(h+1)/2] .. A213849 . A049778 . A213850
h .......... C(2*h-2,h-1) A213853
...
Suppose that u = (u(n)) and v = (v(n)) are sequences having generating functions U(x) and V(x), respectively. Then the convolution u**v has generating function U(x)*V(x). Accordingly, if u and v are homogeneous linear recurrence sequences, then every row of the convolution array T satisfies the same homogeneous linear recurrence equation, which can be easily obtained from the denominator of U(x)*V(x). Also, every column of T has the same homogeneous linear recurrence as v.

			Northwest corner (the array is read by southwest falling antidiagonals):
  1,  4, 10, 20,  35,  56,  84, ...
  2,  7, 16, 30,  50,  77, 112, ...
  3, 10, 22, 40,  65,  98, 140, ...
  4, 13, 28, 50,  80, 119, 168, ...
  5, 16, 34, 60,  95, 140, 196, ...
  6, 19, 40, 70, 110, 161, 224, ...
T(6,1) = (1)**(6) = 6;
T(6,2) = (1,2)**(6,7) = 1*7+2*6 = 19;
T(6,3) = (1,2,3)**(6,7,8) = 1*8+2*7+3*6 = 40.

Clark Kimberling, Antidiagonals n = 1..60, flattened

Cf. A000027.

b[n_] := n; c[n_] := n
t[n_, k_] := Sum[b[k - i] c[n + i], {i, 0, k - 1}]
TableForm[Table[t[n, k], {n, 1, 10}, {k, 1, 10}]]
Flatten[Table[t[n - k + 1, k], {n, 12}, {k, n, 1, -1}]]
r[n_] := Table[t[n, k], {k, 1, 60}]  (* A213500 *)

t(n,k) = sum(i=0, k - 1, (k - i) * (n + i));
tabl(nn) = {for(n=1, nn, for(k=1, n, print1(t(k,n - k + 1),", ");); print(););};
tabl(12) \\ Indranil Ghosh, Mar 26 2017

def t(n, k): return sum((k - i) * (n + i) for i in range(k))
for n in range(1, 13):
    print([t(k, n - k + 1) for k in range(1, n + 1)]) # Indranil Ghosh, Mar 26 2017

T(n,k) = 4*T(n,k-1) - 6*T(n,k-2) + 4*T(n,k-3) - T(n,k-4).
T(n,k) = 2*T(n-1,k) - T(n-2,k).
G.f. for row n:  x*(n - (n - 1)*x)/(1 - x)^4.

A213847 Rectangular array: (row n) = b**c, where b(h) = 4h-1, c(h) = 2n-3+2*h, n>=1, h>=1, and ** = convolution.

Original entry on oeis.org

3, 16, 9, 47, 36, 15, 104, 89, 56, 21, 195, 176, 131, 76, 27, 328, 305, 248, 173, 96, 33, 511, 484, 415, 320, 215, 116, 39, 752, 721, 640, 525, 392, 257, 136, 45, 1059, 1024, 931, 796, 635, 464, 299, 156, 51, 1440, 1401

Offset: 1

Clark Kimberling, Jul 05 2012

Principal diagonal: A213848.
Antidiagonal sums: A180324.
Row 1, (3,7,11,15,...)**(1,3,5,7,...): A172482.
Row 2, (3,7,11,15,...)**(3,5,7,9,...): (4*k^3 + 15*k^2 + 8*k)/3.
Row 3, (3,7,11,15,...)**(5,7,9,13,...): (4*k^3 + 27*k^2 + 14*k)/3.
For a guide to related arrays, see A212500.

			Northwest corner (the array is read by falling antidiagonals):
3....16...47....104...195...328
9....36...89....176...305...484
15...56...131...248...415...640
21...76...173...320...525...796

Clark Kimberling, Antidiagonals n = 1..60, flattened

Cf. A212500.

b[n_]:=4n-1;c[n_]:=2n-1;
t[n_,k_]:=Sum[b[k-i]c[n+i],{i,0,k-1}]
TableForm[Table[t[n,k],{n,1,10},{k,1,10}]]
Flatten[Table[t[n-k+1,k],{n,12},{k,n,1,-1}]]
r[n_]:=Table[t[n,k],{k,1,60}] (* A213847 *)
Table[t[n,n],{n,1,40}] (* A213848 *)
s[n_]:=Sum[t[i,n+1-i],{i,1,n}]
Table[s[n],{n,1,50}] (* A180324 *)

T(n,k) = 4*T(n,k-1)-6*T(n,k-2)+4*T(n,k-3)-T(n,k-4).

G.f. for row n: f(x)/g(x), where f(x) = x*(6*n-3 + 4*(n-2)x - (2*n-3)*x^2) and g(x) = (1-x)^4.

A322677 a(n) = 16n(n+1)(2n+1)^2.

Original entry on oeis.org

0, 288, 2400, 9408, 25920, 58080, 113568, 201600, 332928, 519840, 776160, 1117248, 1560000, 2122848, 2825760, 3690240, 4739328, 5997600, 7491168, 9247680, 11296320, 13667808, 16394400, 19509888, 23049600, 27050400, 31550688, 36590400, 42211008, 48455520

Offset: 0

Seiichi Manyama, Dec 23 2018

			(sqrt(2) - sqrt(1))^4 = (sqrt(9) - sqrt(8))^2 = sqrt(289) - sqrt(288). So a(1) = 288.

Colin Barker, Table of n, a(n) for n = 0..1000
Index entries for linear recurrences with constant coefficients, signature (5,-10,10,-5,1).

sqrt(a(n)+1) + sqrt(a(n)) = (sqrt(n+1) + sqrt(n))^k: A033996(n) (k=2), A322675 (k=3), this sequence (k=4).

Cf. A180324, A185096, A339483.

A322677[n_] := 16*n*(n + 1)*(2*n + 1)^2; Array[A322677, 50, 0] (* or *)
LinearRecurrence[{5, -10, 10, -5, 1}, {0, 288, 2400, 9408, 25920}, 50] (* Paolo Xausa, Aug 26 2025 *)

PARI
```
{a(n) = 16*n*(n+1)*(2*n+1)^2}
```

concat(0, Vec(96*x*(3 + x)*(1 + 3*x) / (1 - x)^5 + O(x^40))) \\ Colin Barker, Dec 23 2018

sqrt(a(n)+1) + sqrt(a(n)) = (sqrt(n+1) + sqrt(n))^4.
sqrt(a(n)+1) - sqrt(a(n)) = (sqrt(n+1) - sqrt(n))^4.
a(n) = A033996(A033996(n)).
Sum_{n>=1} 1/a(n) = (5 - Pi^2/2)/16 = 0.004074862465957543161422156253870277... - Vaclav Kotesovec, Dec 23 2018
From Colin Barker, Dec 23 2018: (Start)
G.f.: 96*x*(3 + x)*(1 + 3*x)/(1 - x)^5.
a(n) = 5*a(n-1) - 10*a(n-2) + 10*a(n-3) - 5*a(n-4) + a(n-5) for n>4. (End)
From Elmo R. Oliveira, Aug 20 2025: (Start)
E.g.f.: 16*x*(2 + x)*(9 + 24*x + 4*x^2)*exp(x).
a(n) = 96*A180324(n) = 32*A339483(n) = 8*A185096(n). (End)

A339483 Number of regular polygons that can be drawn with vertices on a centered hexagonal grid with side length n.

Original entry on oeis.org

0, 9, 75, 294, 810, 1815, 3549, 6300, 10404, 16245, 24255, 34914, 48750, 66339, 88305, 115320, 148104, 187425, 234099, 288990, 353010, 427119, 512325, 609684, 720300, 845325, 985959, 1143450, 1319094, 1514235, 1730265, 1968624, 2230800, 2518329, 2832795

Offset: 0

Peter Kagey, Dec 06 2020

The only regular polygons that can be drawn with vertices on the centered hexagonal grid are equilateral triangles and regular hexagons.

			There are a(2) = 75 regular polygons that can be drawn on the centered hexagonal grid with side length 2: A000537(2) = 9 regular hexagons and A008893(n) = 66 equilateral triangles.
The nine hexagons are:
    * . *       . * .       * * .
   . . . .     * . . *     * . * .
  * . . . *   . . . . .   . * * . .
   . . . .     * . . *     . . . .
    * . *       . * .       . . .
      1           1           7
which are marked with the number of ways to draw the hexagons up to translation.
The 66 equilateral triangles are:
    * . .       * . .       * . .       * . *       * . .       . . .
   * * . .     . . * .     . . . .     . . . .     . . . .     * . . *
  . . . . .   . * . . .   . . . * .   . . * . .   . . . . *   . . . . .
   . . . .     . . . .     * . . .     . . . .     . . . .     . . . .
    . . .       . . .       . . .       . . .       * . .       . * .
     24          14          12          12           2           2
which are marked with the number of ways to draw the triangles up to translation and dihedral action of the hexagon.

Peter Kagey, Table of n, a(n) for n = 0..10000
Burkard Polster, What does this prove? Some of the most gorgeous visual "shrink" proofs ever invented, Mathologer video (2020).
Index entries for linear recurrences with constant coefficients, signature (5,-10,10,-5,1).

Cf. A000537 (regular hexagons), A008893 (equilateral triangles).
Cf. A338323 (cubic grid).
Cf. A003215, A180324, A185096, A322677, A348670.

a[n_] := n*(n+1)*(2*n+1)^2/2; Array[a, 35, 0] (* Amiram Eldar, Jun 20 2025 *)

a(n) = A000537(n) + A008893(n).
a(n) = (1/2)*(n+1)*n*(2*n+1)^2.
a(n) = 3*A180324(n).
Sum_{n>=1} 1/a(n) = 10 - Pi^2 (A348670). - Amiram Eldar, Jun 20 2025
From Elmo R. Oliveira, Aug 20 2025: (Start)
G.f.: -3*x*(x + 3)*(3*x + 1)/(x - 1)^5.
E.g.f.: exp(x)*x*(2 + x)*(9 + 24*x + 4*x^2)/2.
a(n) = 5*a(n-1) - 10*a(n-2) + 10*a(n-3) - 5*a(n-4) + a(n-5).
a(n) = A185096(n)/4 = A322677(n)/32. (End)

A302758 a(n) = n^2(n(4n + 3) + 3n*(-1)^n - 4)/96.

Original entry on oeis.org

0, 1, 3, 14, 25, 66, 98, 200, 270, 475, 605, 966, 1183, 1764, 2100, 2976, 3468, 4725, 5415, 7150, 8085, 10406, 11638, 14664, 16250, 20111, 22113, 26950, 29435, 35400, 38440, 45696, 49368, 58089, 62475, 72846, 78033, 90250, 96330, 110600, 117670, 134211, 142373, 161414

Offset: 1

Wesley Ivan Hurt, Apr 12 2018

Consider the partitions of n into two parts (p,q). Then 2*a(n) represents the total volume of the family of rectangular prisms with dimensions p, q, and (p + q).

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

Cf. A302647.
First bisection is A180324.
After 0, subsequence of A107972.

List([1..50], n -> n^2*(n*(4*n+3)+3*n*(-1)^n-4)/96); # Bruno Berselli, Apr 13 2018

[div(n^2*(n*(4*n+3)+3*n*(-1)^n-4), 96) for n in 1:50] |> println # Bruno Berselli, Apr 13 2018

[n^2*(n*(4*n+3)+3*n*(-1)^n-4)/96: n in [1..50]]; // Bruno Berselli, Apr 13 2018

Table[n*Floor[n/2]*(1 + Floor[n/2])*(3 n - 1 - 2*Floor[n/2])/12, {n, 50}]
Table[n^2 (n (4 n + 3) + 3 n (-1)^n - 4)/96, {n, 1, 50}] (* Bruno Berselli, Apr 13 2018 *)

vector(50, n, nn; n^2*(n*(4*n+3)+3*n*(-1)^n-4)/96) \\ Bruno Berselli, Apr 13 2018

[n**2*(n*(4*n+3)+3*n*(-1)**n-4)/96 for n in range(1, 50)] # Bruno Berselli, Apr 13 2018

[n^2*(n*(4*n+3)+3*n*(-1)^n-4)/96 for n in (1..50)] # Bruno Berselli, Apr 13 2018

a(n) = (n/2) * Sum_{i=1..floor(n/2)} i*(n - i).
a(n) = n*floor(n/2)*(1 + floor(n/2))*(3*n - 1 - 2*floor(n/2))/12.
From Bruno Berselli, Apr 13 2018: (Start)
O.g.f.: x^2*(1 + 2*x + 7*x^2 + 3*x^3 + 3*x^4)/((1 - x)^5*(1 + x)^4).
E.g.f.: x*(-3*(1 - exp(2*x))*exp(-x) + 3*(3 + 11*exp(2*x))*exp(-x)*x - 3*(1 - 9*exp(2*x))*exp(-x)*x^2 + 4*exp(x)*x^3)/96.
a(n) = n^2*(n*(4*n + 3) + 3*n*(-1)^n - 4)/96. Therefore:
a(n) = n^2*(2*n - 1)*(n + 2)/48 for n even; otherwise:
a(n) = n^2*(n - 1)*(n + 1)/24.
n^2*(4*n^2 + n - 1)*a(n+2) - 4*n^2*(n + 2)*a(n+1) - (n + 2)*(n + 3)*(4*n^2 + 9*n + 4)*a(n) = 0. (End)
Sum_{n>=2} 1/a(n) = 156/5 - 48*Pi/5 - 4*Pi^2 + 288*log(2)/5. - Amiram Eldar, Jun 20 2025

A232535 Triangle T(n,k), 0 <= k <= n, read by rows defined by: T(n,k) = (binomial(2n,2k) + binomial(2n+1,2k))/2.

Original entry on oeis.org

1, 1, 2, 1, 8, 3, 1, 18, 25, 4, 1, 32, 98, 56, 5, 1, 50, 270, 336, 105, 6, 1, 72, 605, 1320, 891, 176, 7, 1, 98, 1183, 4004, 4719, 2002, 273, 8, 1, 128, 2100, 10192, 18590, 13728, 4004, 400, 9, 1, 162, 3468, 22848, 59670, 68068, 34476, 7344, 561, 10, 1, 200, 5415

Offset: 0

Philippe Deléham, Nov 25 2013

Sum_{k=0..n}T(n,k)*x^k = A164111(n), A000012(n), A002001(n), A001653(n+1), A001019(n), A166965(n) for x =-1, 0, 1, 2, 4, 9 respectively.

			Triangle begins:
1
1, 2
1, 8, 3
1, 18, 25, 4
1, 32, 98, 56, 5
1, 50, 270, 336, 105, 6
1, 72, 605, 1320, 891, 176, 7
1, 98, 1183, 4004, 4719, 2002, 273, 8
1, 128, 2100, 10192, 18590, 13728, 4004, 400, 9

Cf. A086645, A091042
Cf. Columns : A000012, A001105, A180324 ; Diagonals: A000027, A131423
Cf. T(2*n,n): A228329, Row sums : A002001

T := (n,k) -> binomial(2*n, 2*k)*(2*n+1-k)/(2*n+1-2*k);
seq(seq(T(n,k), k=0..n), n=0..9); # Peter Luschny, Nov 26 2013

Flatten[Table[(Binomial[2n,2k]+Binomial[2n+1,2k])/2,{n,0,10},{k,0,n}]] (* Harvey P. Dale, Jul 05 2015 *)

G.f.: (1-x)/(1-2*x*(1+y)+x^2*(1-y)^2).
T(n,k) = 2*T(n-1,k)+2*T(n-1,k-1)+2*T(n-2,k-1)-T(n-2,k)-T(n-2,k-2), T(0,0)=T(1,0)=1, T(1,1)=2, T(n,k)=0 if k<0 or if k>n.
T(n,k) = (A086645(n,k) + A091042(n,k))/2.
T(n,k) = binomial(2*n,2*k)*(2*n+1-k)/(2*n+1-2*k). - Peter Luschny, Nov 26 2013

cp's OEIS Frontend

A213500 Rectangular array T(n,k): (row n) = b**c, where b(h) = h, c(h) = h + n - 1, n >= 1, h >= 1, and ** = convolution.

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A213847 Rectangular array: (row n) = b**c, where b(h) = 4*h-1, c(h) = 2*n-3+2*h, n>=1, h>=1, and ** = convolution.

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A322677 a(n) = 16*n*(n+1)*(2*n+1)^2.

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A339483 Number of regular polygons that can be drawn with vertices on a centered hexagonal grid with side length n.

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A302758 a(n) = n^2*(n*(4*n + 3) + 3*n*(-1)^n - 4)/96.

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A232535 Triangle T(n,k), 0 <= k <= n, read by rows defined by: T(n,k) = (binomial(2*n,2*k) + binomial(2*n+1,2*k))/2.

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A213500 Rectangular array T(n,k): (row n) = bc, where b(h) = h, c(h) = h + n - 1, n >= 1, h >= 1, and = convolution.

A213847 Rectangular array: (row n) = b**c, where b(h) = 4h-1, c(h) = 2n-3+2*h, n>=1, h>=1, and ** = convolution.

A322677 a(n) = 16n(n+1)(2n+1)^2.

A302758 a(n) = n^2(n(4n + 3) + 3n*(-1)^n - 4)/96.

A232535 Triangle T(n,k), 0 <= k <= n, read by rows defined by: T(n,k) = (binomial(2n,2k) + binomial(2n+1,2k))/2.