A051923 -id:A051923 - 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.

A093560 (3,1) Pascal triangle.

Original entry on oeis.org

1, 3, 1, 3, 4, 1, 3, 7, 5, 1, 3, 10, 12, 6, 1, 3, 13, 22, 18, 7, 1, 3, 16, 35, 40, 25, 8, 1, 3, 19, 51, 75, 65, 33, 9, 1, 3, 22, 70, 126, 140, 98, 42, 10, 1, 3, 25, 92, 196, 266, 238, 140, 52, 11, 1, 3, 28, 117, 288, 462, 504, 378, 192, 63, 12, 1, 3, 31, 145, 405, 750, 966, 882, 570, 255, 75, 13, 1

Offset: 0

Wolfdieter Lang, Apr 22 2004

The array F(3;n,m) gives in the columns m >= 1 the figurate numbers based on A016777, including the pentagonal numbers A000326 (see the W. Lang link).
This is the third member, d=3, in the family of triangles of figurate numbers, called (d,1) Pascal triangles: A007318 (Pascal (d=1), A029653 (d=2).
This is an example of a Riordan triangle (see A053121 for a comment and the 1991 Shapiro et al. reference on the Riordan group) with o.g.f. of column no. m of the type g(x)*(x*f(x))^m with f(0)=1. Therefore the o.g.f. for the row polynomials p(n,x):=Sum_{m=0..n} a(n,m)*x^m is G(z,x)=g(z)/(1-x*z*f(z)). Here: g(x)=(1+2*x)/(1-x), f(x)=1/(1-x), hence G(z,x)=(1+2*z)/(1-(1+x)*z).
The SW-NE diagonals give the Lucas numbers A000032: L(n) = Sum_{k=0..ceiling((n-1)/2)} a(n-1-k,k), n >= 1, with L(0)=2. Observation by Paul Barry, Apr 29 2004. Proof via recursion relations and comparison of inputs.
Triangle T(n,k), read by rows, given by [3,-2,0,0,0,0,0,0,...] DELTA [1,0,0,0,0,0,0,0,...] where DELTA is the operator defined in A084938. - Philippe Deléham, Sep 17 2009
For a closed-form formula for generalized Pascal's triangle see A228576. - Boris Putievskiy, Sep 09 2013
From Wolfdieter Lang, Jan 09 2015: (Start)
The signed lower triangular matrix (-1)^(n-1)*a(n,m) is the inverse of the Riordan matrix A106516; that is Riordan ((1-2*x)/(1+x),x/(1+x)).
See the Peter Bala comment from Dec 23 2014 in A106516 for general Riordan triangles of the type (g(x), x/(1-x)): exp(x)*r(n,x) = d(n,x) with the e.g.f. r(n,x) of row n and the e.g.f. of diagonal n.
Similarly, for general Riordan triangles of the type (g(x), x/(1+x)): exp(x)*r(n,-x) = d(n,x). (End)
The n-th row polynomial is (3 + x)*(1 + x)^(n-1) for n >= 1. More generally, the n-th row polynomial of the Riordan array ( (1-a*x)/(1-b*x), x/(1-b*x) ) is (b - a + x)*(b + x)^(n-1) for n >= 1. - Peter Bala, Mar 02 2018
Binomial(n-2,k)+2*Binomial(n-3,k) is also the number of permutations avoiding both 123 and 132 with k double descents, i.e., positions with w[i]>w[i+1]>w[i+2]. - Lara Pudwell, Dec 19 2018

			Triangle begins
  1,
  3,  1,
  3,  4,  1,
  3,  7,  5,   1,
  3, 10, 12,   6,   1,
  3, 13, 22,  18,   7,   1,
  3, 16, 35,  40,  25,   8,   1,
  3, 19, 51,  75,  65,  33,   9,  1,
  3, 22, 70, 126, 140,  98,  42, 10,  1,
  3, 25, 92, 196, 266, 238, 140, 52, 11, 1,

Kurt Hawlitschek, Johann Faulhaber 1580-1635, Veroeffentlichung der Stadtbibliothek Ulm, Band 18, Ulm, Germany, 1995, Ch. 2.1.4. Figurierte Zahlen.
Ivo Schneider, Johannes Faulhaber 1580-1635, Birkhäuser, Basel, Boston, Berlin, 1993, ch.5, pp. 109-122.

Reinhard Zumkeller, Rows n = 0..125 of triangle, flattened
Peter Bala, A note on the diagonals of a proper Riordan Array
M. Bukata, R. Kulwicki, N. Lewandowski, L. Pudwell, J. Roth, and T. Wheeland, Distributions of Statistics over Pattern-Avoiding Permutations, arXiv preprint arXiv:1812.07112 [math.CO], 2018.
W. Lang, First 10 rows and array of figurate numbers .

Cf. Column sequences for m=1..9: A016777, A000326 (pentagonal), A002411, A001296, A051836, A051923, A050494, A053367, A053310;

A007318 (Pascal's triangle), A029653 ((2,1) Pascal triangle), A093561 ((4,1) Pascal triangle), A228196, A228576.

Concatenation([1],Flat(List([1..11],n->List([0..n],k->Binomial(n,k)+2*Binomial(n-1,k))))); # Muniru A Asiru, Dec 20 2018

a093560 n k = a093560_tabl !! n !! k
a093560_row n = a093560_tabl !! n
a093560_tabl = [1] : iterate
               (\row -> zipWith (+) ([0] ++ row) (row ++ [0])) [3, 1]
-- Reinhard Zumkeller, Aug 31 2014

from math import comb, isqrt
def A093560(n): return comb(r:=(m:=isqrt(k:=n+1<<1))-(k<=m*(m+1)),a:=n-comb(r+1,2))*(r+(r-a<<1))//r if n else 1 # Chai Wah Wu, Nov 12 2024

a(n, m)=F(3;n-m, m) for 0<= m <= n, otherwise 0, with F(3;0, 0)=1, F(3;n, 0)=3 if n>=1 and F(3;n, m):=(3*n+m)*binomial(n+m-1, m-1)/m if m>=1.
G.f. column m (without leading zeros): (1+2*x)/(1-x)^(m+1), m>=0.
Recursion: a(n, m)=0 if m>n, a(0, 0)= 1; a(n, 0)=3 if n>=1; a(n, m)= a(n-1, m) + a(n-1, m-1).
T(n, k) = C(n, k) + 2*C(n-1, k). - Philippe Deléham, Aug 28 2005
Equals M * A007318, where M = an infinite triangular matrix with all 1's in the main diagonal and all 2's in the subdiagonal. - Gary W. Adamson, Dec 01 2007
Sum_{k=0..n} T(n,k) = A151821(n+1). - Philippe Deléham, Sep 17 2009
exp(x) * e.g.f. for row n = e.g.f. for diagonal n. For example, for n = 3 we have exp(x)*(3 + 7*x + 5*x^2/2! + x^3/3!) = 3 + 10*x + 22*x^2/2! + 40*x^3/3! + 65*x^4/4! + .... The same property holds more generally for Riordan arrays of the form ( f(x), x/(1 - x) ). - Peter Bala, Dec 22 2014
G.f.: (-1-2*x)/(-1+x+x*y). - R. J. Mathar, Aug 11 2015

Incorrect connection with A046055 deleted by N. J. A. Sloane, Jul 08 2009

A125232 Triangle T(n,k) read by rows: the (n-k)-th term of the k-fold iterated partial sum of the pentagonal numbers.

Original entry on oeis.org

1, 5, 1, 12, 6, 1, 22, 18, 7, 1, 35, 40, 25, 8, 1, 51, 75, 65, 33, 9, 1, 70, 126, 140, 98, 42, 10, 1, 92, 196, 266, 238, 140, 52, 11, 1, 117, 288, 462, 504, 378, 192, 63, 12, 1, 145, 405, 750, 966, 882, 570, 255, 75, 13, 1, 176, 550, 1155, 1716, 1848, 1452, 825, 330, 88, 14, 1

Offset: 1

Gary W. Adamson, Nov 24 2006

			First few rows of the triangle are:
   1;
   5,   1;
  12,   6,   1;
  22,  18,   7,   1;
  35,  40,  25,   8,   1;
  51,  75,  65,  33,   9,   1;
  70, 126, 140,  98,  42,  10,   1;
  ...
Example: (5,3) = 65 = 25 + 40 = (4,3) + (4,2).

Albert H. Beiler, "Recreations in the Theory of Numbers", Dover, 1966, p 189.

Robert Israel, Table of n, a(n) for n = 1..10011(rows 0 to 140, flattened)

Columns: A000326 (pentagonal numbers), A002411, A001296, A051836, A051923.

Cf. A095264 (row sums).

A125232 := proc(n,k) option remember ; if k = 0 then A000326(n) ; elif k = n-1 then 1 ; else procname(n-1,k)+procname(n-1,k-1) ; fi : end: # R. J. Mathar, Jun 09 2008

nmax = 11; col[1] = Table[n(3n-1)/2, {n, 1, nmax}]; col[k_] := col[k] = Prepend[Accumulate[col[k-1]], 0]; Table[col[k][[n]], {n, 1, nmax}, {k, 1, n}] // Flatten (* Jean-François Alcover, Mar 25 2019 *)

T(n,0)=A000326(n). T(n,k)=T(n-1,k) + T(n-1,k-1), k>0. - R. J. Mathar, Jun 09 2008

G.f. as triangle: (1+2*x)/((1-x)^2*(1-x-x*y)). - Robert Israel, Nov 07 2016

Edited and extended by R. J. Mathar, Jun 09 2008

A213551 Rectangular array: (row n) = b**c, where b(h) = h*(h+1)/2, c(h) = b(n-1+h), n>=1, h>=1, and ** = convolution.

Original entry on oeis.org

1, 6, 3, 21, 15, 6, 56, 46, 28, 10, 126, 111, 81, 45, 15, 252, 231, 186, 126, 66, 21, 462, 434, 371, 281, 181, 91, 28, 792, 756, 672, 546, 396, 246, 120, 36, 1287, 1242, 1134, 966, 756, 531, 321, 153, 45, 2002, 1947, 1812, 1596, 1316, 1001, 686, 406

Offset: 1

Clark Kimberling, Jun 17 2012

Principal diagonal:  A213552
Antidiagonal sums:  A051923
Row 1,  (1,3,6,...)**(1,3,6,...): A000389
Row 2,  (1,3,6,...)**(3,6,10,...): (k^5 + 15*k^4 + 85*k^3 + 165*k^2 + 94*k)/120
Row 3,  (1,3,6,...)**(6,10,15,...): (k^5 + 20*k^4 + 155*k^3 + 340*k^2 + 204*k)/120
For a guide to related arrays, see A213500.

			Northwest corner (the array is read by falling antidiagonals):
1....6....21....56....126....252
3....15...46....111...231....434
6....28...81....186...371....672
10...45...126...281...546....966
15...66...181...396...756....1316
21...91...246...531...1001...1722

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

Cf. A213500.

b[n_] := n (n + 1)/2; c[n_] := n (n + 1)/2
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}]  (* A213551 *)
d = Table[t[n, n], {n, 1, 40}] (* A213552 *)
s[n_] := Sum[t[i, n + 1 - i], {i, 1, n}]
s1 = Table[s[n], {n, 1, 50}] (* A051923 *)

T(n,k) = 6*T(n,k-1) - 15*T(n,k-2) + 20*T(n,k-3) - 15*T(n,k-4) + 6*T(n,k-5) - T(n,k-6).

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

A027801 a(n) = 5(n+1)binomial(n+4,5)/2.

Original entry on oeis.org

5, 45, 210, 700, 1890, 4410, 9240, 17820, 32175, 55055, 90090, 141960, 216580, 321300, 465120, 658920, 915705, 1250865, 1682450, 2231460, 2922150, 3782350, 4843800, 6142500, 7719075, 9619155, 11893770, 14599760, 17800200, 21564840, 25970560, 31101840, 37051245

Offset: 1

Thi Ngoc Dinh (via R. K. Guy)

Number of 10-subsequences of [ 1, n ] with just 4 contiguous pairs.

Harvey P. Dale, Table of n, a(n) for n = 1..1000
Index entries for linear recurrences with constant coefficients, signature (7,-21,35,-35,21,-7,1).

Cf. A051923.

Table[5(n+1) Binomial[n+4,5]/2,{n,30}] (* or *) LinearRecurrence[{7,-21,35,-35,21,-7,1},{5,45,210,700,1890,4410,9240},30] (* Harvey P. Dale, Dec 13 2014 *)

G.f.: 5*(1+2x)*x/(1-x)^7.
a(n) = C(n+1, 2)*C(n+4, 4) - Zerinvary Lajos, May 10 2005, corrected by R. J. Mathar, Feb 13 2016
a(n) = 5*A051923(n-1).
From Amiram Eldar, Jan 28 2022: (Start)
Sum_{n>=1} 1/a(n) = 241/18 - 4*Pi^2/3.
Sum_{n>=1} (-1)^(n+1)/a(n) = 2*Pi^2/3 - 64*log(2)/3 + 151/18. (End)

A135857 Partial sums triangle based on A016777. Riordan convolution triangle ((1 + 2*x)/(1-x)^2, x/(1-x)).

Original entry on oeis.org

1, 4, 1, 7, 5, 1, 10, 12, 6, 1, 13, 22, 18, 7, 1, 16, 35, 40, 25, 8, 1, 19, 51, 75, 65, 33, 9, 1, 22, 70, 126, 140, 98, 42, 10, 1, 25, 92, 196, 266, 238, 140, 52, 11, 1, 28, 117, 288, 462, 504, 378, 192, 63, 12, 1

Offset: 0

Gary W. Adamson, Dec 01 2007

A007318 * a bidiagonal matrix with all 1's in the main diagonal and all 3's in the subdiagonal.
Row sums give A036563(n+2), n >= 0.
From Wolfdieter Lang, Mar 23 2015: (Start)
This is the triangle of iterated partial sums of A016777. Such iterated partial sums of arithmetic progression sequences have been considered by Narayana Pandit (see the Mar 20 2015 comment on A000580 where the MacTutor History of Mathematics archive link and the Gottwald et al. reference, p. 338, are given).
This is therefore the Riordan triangle ((1+2*x)/(1-x)^2, x/(1-x)) with o.g.f. of the columns ((1+2*x)/(1-x)^2)*(x/(1-x))^k, k >= 0.
The column sequences are A016777, A000326, A002411, A001296, A051836, A051923, A050494, A053367, A053310, for k = 0..8.
The alternating row sums are A122553(n) = {1, repeat(3)}.
The Riordan A-sequence is A(y) = 1 + y (implying the Pascal triangle recurrence for k >= 1).
The Riordan Z-sequence is A256096, leading to a recurrence for T(n,0) given in the formula section. See the link "Sheffer a- and z-sequences" under A006232 also for Riordan A- and Z-sequences with references. (End)
When the first column (k = 0) is removed from this triangle, the result is A125232. - Georg Fischer, Jul 26 2023

			The triangle T(n, k) begins:
n\k  0   1   2    3    4    5    6   7   8  9 10 11
0:   1
1:   4   1
2:   7   5   1
3:  10  12   6    1
4:  13  22  18    7    1
5:  16  35  40   25    8    1
6:  19  51  75   65   33    9    1
7:  22  70 126  140   98   42   10   1
8:  25  92 196  266  238  140   52  11   1
9:  28 117 288  462  504  378  192  63  12  1
10: 31 145 405  750  966  882  570 255  75 13  1
11: 34 176 550 1155 1716 1848 1452 825 330 88 14  1
... reformatted and extended by _Wolfdieter Lang_, Mar 23 2015
From _Wolfdieter Lang_, Mar 23 2015: (Start)
T(3, 1) = T(2, 0) + T(2, 1) = 7 + 5 = 12 (Pascal, from the A-sequence given above).
T(4, 0) = 4*T(3, 0) - 9*T(3, 1) + 27*T(3, 2) - 81* T(3, 3) = 4*10 - 9*12 + 27*6 - 81*1 = 13, from the Z-sequence given above and in A256096.
T(4, 0) = 2*T(3, 0) - T(2, 0) = 2*10 - 7 = 13.
(End)

Columns k=0-8 give: A016777, A000326, A002411, A001296, A051836, A051923, A050494, A053367, A053310.
Cf. A036563, A125232.
Cf. A110813. - Philippe Deléham, Feb 08 2009

Binomial transform of an infinite lower triangular matrix with all 1's in the main diagonal and all 3's in the subdiagonal; i.e., by columns - every column = (1, 3, 0, 0, 0, ...).
T(n,k) = (3n-2k+1)*binomial(n+1,k+1)/(n+1). - Philippe Deléham, Feb 08 2009
From Wolfdieter Lang, Mar 23 2015: (Start)
O.g.f. for row polynomials: (1 + 2*z)/((1- z*(1 + x))*(1 - z)) (see the Riordan property from the comment).
O.g.f. for column k (without leading zeros): (1 + 2*x)/(1-x)^(2+k), k >= 0, (Riordan property).
T(n, k) = T(n-1, k-1) + T(n-1, k) for k >= 1. From the Riordan A-sequence given above in a comment.
T(n, 0) = Sum_{j=0..n} Z(j)*T(n-1, j), for n >= 1, from the Riordan Z-sequence A256096 mentioned above in a comment. Of course, T(n, 0) = 2*T(n-1, 0) - T(n-2, 0) for n >= 2 (see A016777).
(End)

Edited. Offset is 0 from the old name and the Philippe Deléham formula. New name, old name as first comment. - Wolfdieter Lang, Mar 23 2015

A301972 a(n) = n(n^2 - 2n + 4)binomial(2n,n)/((n + 1)*(n + 2)).

Original entry on oeis.org

0, 1, 4, 21, 112, 570, 2772, 13013, 59488, 266526, 1175720, 5123426, 22108704, 94645460, 402503220, 1702300725, 7165821120, 30043474230, 125523450360, 522857438070, 2172127120800, 9002522512620, 37233403401480, 153704429299746, 633442159732032, 2606543487445100, 10710790748646352, 43957192722175908

Offset: 0

Ilya Gutkovskiy, Mar 29 2018

nonn

For n > 2, a(n) is the n-th term of the main diagonal of iterated partial sums array of n-gonal numbers (in other words, a(n) is the n-th (n+2)-dimensional n-gonal number, see also example).

			For n = 5 we have:
----------------------------
0   1    2    3     4    [5]
----------------------------
0,  1,   5,  12,   22,   35,  ... A000326 (pentagonal numbers)
0,  1,   6,  18,   40,   75,  ... A002411 (pentagonal pyramidal numbers)
0,  1,   7,  25,   65,  140,  ... A001296 (4-dimensional pyramidal numbers)
0,  1,   8,  33,   98,  238,  ... A051836 (partial sums of A001296)
0,  1,   9,  42,  140,  378,  ... A051923 (partial sums of A051836)
0,  1,  10,  52,  192, [570], ... A050494 (partial sums of A051923)
----------------------------
therefore a(5) = 570.

Index to sequences related to polygonal numbers

Cf. A000984, A002457, A006484, A057145, A060354, A080851, A080852, A100119, A180266, A275490, A292551.

Table[n (n^2 - 2 n + 4) Binomial[2 n, n]/((n + 1) (n + 2)), {n, 0, 27}]
nmax = 27; CoefficientList[Series[(-4 + 31 x - 66 x^2 + 28 x^3 + (4 - 7 x) (1 - 4 x)^(3/2))/(2 x^2 (1 - 4 x)^(3/2)), {x, 0, nmax}], x]
nmax = 27; CoefficientList[Series[Exp[2 x] (4 - x + 2 x^2) BesselI[1, 2 x]/x - 2 Exp[2 x] (2 - x) BesselI[0, 2 x], {x, 0, nmax}], x] Range[0, nmax]!
Table[SeriesCoefficient[x (1 - 3 x + n x)/(1 - x)^(n + 3), {x, 0, n}], {n, 0, 27}]

O.g.f.: (-4 + 31*x - 66*x^2 + 28*x^3 + (4 - 7*x)*(1 - 4*x)^(3/2))/(2*x^2*(1 - 4*x)^(3/2)).
E.g.f.: exp(2*x)*(4 - x + 2*x^2)*BesselI(1,2*x)/x - 2*exp(2*x)*(2 - x)*BesselI(0,2*x).
a(n) = [x^n] x*(1 - 3*x + n*x)/(1 - x)^(n+3).
a(n) ~ 4^n*sqrt(n)/sqrt(Pi).
D-finite with recurrence: -(n+2)*(961*n-3215)*a(n) +4*(2081*n^2-4414*n-4668)*a(n-1) -28*(320*n-389)*(2*n-3)*a(n-2)=0. - R. J. Mathar, Jan 27 2020

cp's OEIS Frontend

A050494 Partial sums of A051923.

Views

Author

Keywords

Comments

References

Links

Crossrefs

Programs

Mathematica

Formula

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.

Views

Author

Keywords

Comments

Examples

Links

Crossrefs

Programs

Mathematica

PARI

Python

Formula

A093560 (3,1) Pascal triangle.

Views

Author

Keywords

Comments

Examples

References

Links

Crossrefs

Programs

GAP

Haskell

Python

Formula

Extensions

A125232 Triangle T(n,k) read by rows: the (n-k)-th term of the k-fold iterated partial sum of the pentagonal numbers.

Views

Author

Keywords

Examples

References

Links

Crossrefs

Programs

Maple

Mathematica

Formula

Extensions

A213551 Rectangular array: (row n) = b**c, where b(h) = h*(h+1)/2, c(h) = b(n-1+h), n>=1, h>=1, and ** = convolution.

Views

Author

Keywords

Comments

Examples

Links

Crossrefs

Programs

Mathematica

Formula

A027801 a(n) = 5*(n+1)*binomial(n+4,5)/2.

Views

Author

Keywords

Comments

Links

Crossrefs

Programs

Mathematica

Formula

A135857 Partial sums triangle based on A016777. Riordan convolution triangle ((1 + 2*x)/(1-x)^2, x/(1-x)).

Views

Author

Keywords

Comments

Examples

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

A027801 a(n) = 5(n+1)binomial(n+4,5)/2.

A301972 a(n) = n(n^2 - 2n + 4)binomial(2n,n)/((n + 1)*(n + 2)).