A000537 -id:A000537 - cp's OEIS frontend

A113751 Number of diagonal rectangles with corners on an n X n grid of points.

0, 0, 1, 8, 30, 88, 199, 408, 748, 1280, 2053, 3168, 4666, 6712, 9363, 12728, 16952, 22256, 28681, 36536, 45870, 56936, 69967, 85264, 102860, 123232, 146557, 173128, 203138, 237192, 275243, 318104, 365856, 418912, 477649, 542392, 613406, 691848

Offset: 1

T. D. Noe, Nov 09 2005

nonn

The diagonal rectangles are the ones whose sides are not parallel to the grid axes. All the rectangles can be reflected so that the slope of one side is >= 1. There are a total of A046657(n-1) these slopes. These slopes are the basis of the Mathematica program that counts the rectangles.

			a(3) = 1 because for the 3 X 3 grid, there is only one diagonal rectangle - a square having sides sqrt(2) units.
a(4) = 8 because for the 4 X 4 grid, there are 4 squares having sides sqrt(2) units, 2 squares having sides sqrt(5) units and 2 rectangles that are sqrt(2) by 2*sqrt(2) units.

Jinyuan Wang, Table of n, a(n) for n = 1..1000

Cf. A000537 (parallel rectangles on an n X n grid), A085582 (all rectangles on an n X n grid).

Table[n=m-1; slopes=Union[Flatten[Table[a/b, {b, n}, {a, b, n-b}]]]; rects=0; Do[b=Numerator[slopes[[i]]]; a=Denominator[slopes[[i]]]; base={a+b, a+b}; l=0; While[l++; k=l; While[extent=base+{b, a}(k-1)+{a, b}(l-1); extent[[1]]<=n && extent[[2]]<=n, pos={n+1, n+1}-extent; If[a==b && k==l, fact=1, If[pos[[1]]==pos[[2]], fact=2, fact=4]]; rects=rects+fact*Times@@pos; k++ ]; k>l], {i, Length[slopes]}]; rects, {m, 1, 42}]

a(n) = A085582(n) - A000537(n-1). [corrected by David Radcliffe, Feb 06 2020]

a(1) = 0 prepended by Jinyuan Wang, Feb 06 2020

A131643 Cubes that are also sums of three or more consecutive positive cubes.

Original entry on oeis.org

216, 8000, 64000, 216000, 343000, 5832000, 35937000, 157464000, 1540798875, 3951805941, 22069810125, 23295638016, 58230605376, 170400029184, 4767078987000, 19814511816000, 241152896222784, 565199024832000, 731189187729000, 5399901725184000, 13389040129314816, 15517248640897024

Offset: 1

Tanya Khovanova, Sep 08 2007

nonn

Note that by Fermat's theorem no cube is the sum of two positive cubes.
All entries have the form A000537(j) - A000537(i-1) with 1 <= i < j, for example (j,i) = (5,3), (14,11), (22,3), (30,6), (34,15), (69,6), (109,11). - R. J. Mathar, Sep 14 2007 [Presumably this comment refers just to the terms shown, and not to every term in the sequence. - N. J. A. Sloane, Dec 19 2015]
Subsequence of A265845 (numbers that are sums of consecutive positive cubes in more than one way) which is sparse: among the first 1000 terms of A265845, only 17 are cubes. - Jonathan Sondow, Jan 10 2016

			216 = 27 + 64 + 125.
Note that "positive" is needed in the definition, otherwise the sequence would contain 8 = (-1)^3 + 0^3 + 1^3 + 2^3. - _N. J. A. Sloane_, Dec 19 2015

Donovan Johnson, Table of n, a(n) for n = 1..55 (terms < 2*10^23)

a(n) = A097811(n)^3. - Donovan Johnson, Nov 09 2012

Cf. A000537, A265845.

Select[Union[ Flatten[Table[ Plus @@ Table[i^3, {i, k, j}], {k, 1000}, {j, k + 1, 1000}]]], # <= 1000^3 && IntegerQ[ #^(1/3)] &]

More terms from R. J. Mathar, Sep 14 2007
More terms from Donovan Johnson, Mar 09 2008
Name edited by Jon E. Schoenfield, Dec 07 2015

A154323 Central coefficients of number triangle A113582.

Original entry on oeis.org

1, 2, 10, 37, 101, 226, 442, 785, 1297, 2026, 3026, 4357, 6085, 8282, 11026, 14401, 18497, 23410, 29242, 36101, 44101, 53362, 64010, 76177, 90001, 105626, 123202, 142885, 164837, 189226, 216226, 246017, 278785, 314722, 354026, 396901, 443557, 494210, 549082, 608401, 672401, 741322, 815410, 894917, 980101

Offset: 0

Paul Barry, Jan 07 2009

a(n) equals n!^3 times the determinant of the n X n matrix whose (i,j)-entry is KroneckerDelta[i, j] (((i^3 + 1)/(i^3)) - 1) + 1. - John M. Campbell, May 20 2011
Let b(0)=b(1)=1; b(n)=max(b(n-1)+(n-1)^3, b(n-2)+(n-2)^3); then a(n)=b(n+1). - Yalcin Aktar, Jul 28 2011
a(n-1) is the number of sets of n words of length n over binary alphabet where the first letter occurs n times. a(2) = 10: {aab,abb,bbb}, {aab,bab,bbb}, {aab,bba,bbb}, {aba,abb,bbb}, {aba,bab,bbb}, {aba,bba,bbb}, {abb,baa,bbb}, {abb,bab,bba}, {baa,bab,bbb}, {baa,bba,bbb}. - Alois P. Heinz, Feb 16 2023

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

Cf. A000537, A113582, A360660.

Main diagonal of A360693.

[(n^4 + 2*n^3 + n^2 + 4)/4: n in [0..40]]; // Vincenzo Librandi, Feb 13 2015

s = 1; lst = {s}; Do[s += n^3; AppendTo[lst, s], {n, 1, 42, 1}]; lst (* Zerinvary Lajos, Jul 12 2009 *)
Table[n!^3*Det[Array[KroneckerDelta[#1,#2](((#1^3+1)/(#1^3))-1)+1&,{n,n}]],{n,1,30}] (* John M. Campbell, May 20 2011 *)
LinearRecurrence[{5, -10, 10, -5, 1}, {1, 2, 10, 37, 101}, 25] (* or *) Table[(n^4 + 2*n^3 + n^2 + 4)/4, {n,0,25}] (* G. C. Greubel, Sep 11 2016 *)

a(n) = (n^4 + 2*n^3 + n^2 + 4)/4.
G.f.: (1 - 3*x + 10*x^2 - 3*x^3 + x^4)/(1-x)^5.
a(n) = 1 + C(n+1,2)^2 = 1 + A000537(n).
From G. C. Greubel, Sep 11 2016: (Start)
a(n) = 5*a(n-1) - 10*a(n-2) + 10*a(n-3) - 5*a(n-4) + a(n-5).
E.g.f.: (1/4)*(4 + 4*x + 14*x^2 + 8*x^3 + x^4)*exp(x). (End)
a(n) = a(n-1)+n^3. - Charles U. Lonappan, Jun 09 2021

A212331 a(n) = 5n(n+5)/2.

Original entry on oeis.org

0, 15, 35, 60, 90, 125, 165, 210, 260, 315, 375, 440, 510, 585, 665, 750, 840, 935, 1035, 1140, 1250, 1365, 1485, 1610, 1740, 1875, 2015, 2160, 2310, 2465, 2625, 2790, 2960, 3135, 3315, 3500, 3690, 3885, 4085, 4290, 4500, 4715, 4935, 5160, 5390, 5625, 5865

Offset: 0

Bruno Berselli, May 30 2012

Numbers of the form n*t(n+5,h)-(n+5)*t(n,h), where t(k,h) = k*(k+2*h+1)/2 for any h. Likewise:
A000217(n) = n*t(n+1,h)-(n+1)*t(n,h),
A005563(n) = n*t(n+2,h)-(n+2)*t(n,h),
A140091(n) = n*t(n+3,h)-(n+3)*t(n,h),
A067728(n) = n*t(n+4,h)-(n+4)*t(n,h) (n>0),
A140681(n) = n*t(n+6,h)-(n+6)*t(n,h).
This is the case r=7 in the formula:
u(r,n) = (P(r, P(n+r, r+6)) - P(n+r, P(r, r+6))) / ((r+5)*(r+6)/2)^2, where P(s, m) is the m-th s-gonal number.
Also, a(k) is a square for k = (5/2)*(A078986(n)-1).
Sum of reciprocals of a(n), for n>0: 137/750.
Also, numbers h such that 8*h/5+25 is a square.
The table given below as example gives the dimensions D(h, n) of the irreducible SU(3) multiplets (h,n). See the triangle A098737 with offset 0, and the comments there, also with a link and the Coleman reference. - Wolfdieter Lang, Dec 18 2020

			From the first and second comment derives the following table:
----------------------------------------------------------------
h \ n | 0   1    2    3    4    5    6    7    8    9    10
------|---------------------------------------------------------
0     | 0,  1,   3,   6,  10,  15,  21,  28,  36,  45,   55, ...  (A000217)
1     | 0,  3,   8,  15,  24,  35,  48,  63,  80,  99,  120, ...  (A005563)
2     | 0,  6,  15,  27,  42,  60,  81, 105, 132, 162,  195, ...  (A140091)
3     | 0, 10,  24,  42,  64,  90, 120, 154, 192, 234,  280, ...  (A067728)
4     | 0, 15,  35,  60,  90, 125, 165, 210, 260, 315,  375, ...  (A212331)
5     | 0, 21,  48,  81, 120, 165, 216, 273, 336, 405,  480, ...  (A140681)
6     | 0, 28,  63, 105, 154, 210, 273, 343, 420, 504,  595, ...
7     | 0, 36,  80, 132, 192, 260, 336, 420, 512, 612,  720, ...
8     | 0, 45,  99, 162, 234, 315, 405, 504, 612, 729,  855, ...
9     | 0, 55, 120, 195, 280, 375, 480, 595, 720, 855, 1000, ...
with the formula n*(h+1)*(h+n+1)/2. See also A098737.

Bruno Berselli, Table of n, a(n) for n = 0..1000
Index entries for linear recurrences with constant coefficients, signature (3,-3,1).

Cf. A000217, A000537, A005563, A055998, A067728, A098737, A140091, A140681.

Magma
```
[5*n*(n+5)/2: n in [0..46]];
```
Mathematica
```
Table[(5/2) n (n + 5), {n, 0, 46}]
```

a(n)=5*n*(n+5)/2 \\ Charles R Greathouse IV, Oct 07 2015

G.f.: 5*x*(3-2*x)/(1-x)^3.
a(n) = a(-n-5) = 5*A055998(n).
E.g.f.: (5/2)*x*(x + 6)*exp(x). - G. C. Greubel, Jul 21 2017
Sum_{n>=1} (-1)^(n+1)/a(n) = 4*log(2)/25 - 47/750. - Amiram Eldar, Feb 26 2022

Extended by Bruno Berselli, Aug 05 2015

A098928 Number of cubes that can be formed from the points of a cubical grid of n X n X n points.

Original entry on oeis.org

0, 1, 9, 36, 100, 229, 473, 910, 1648, 2795, 4469, 6818, 10032, 14315, 19907, 27190, 36502, 48233, 62803, 80736, 102550, 128847, 160271, 197516, 241314, 292737, 352591, 421764, 501204, 592257, 696281, 814450, 948112, 1098607, 1267367

Offset: 1

Ignacio Larrosa Cañestro, Oct 19 2004, Sep 29 2009

nonn

Skew cubes are allowed.

			For n = 3 there are 8 cubes of volume 1 and 1 cube of volume 8; thus a(3)=9. - _José María Grau Ribas_, Mar 15 2014
a(6)=229 because we can place 15^2 cubes in a 6 X 6 X 6 cubical grid with their edges parallel to the faces of the grid, plus 4 cubes of edge 3 with a vertex in each face of the lattice and the other two vertices on a diagonal.

Baitian Li, Table of n, a(n) for n = 1..10000 (terms 1..101 from E. J. Ionascu and R. A. Obando)
E. J. Ionascu and R. A. Obando, Counting all cubes in {0,1,...,n}^3, arXiv:1003.4569 [math.NT], 2010.
Eugen J. Ionascu and Andrei Markov, Platonic solids in Z^3, Journal of Number Theory, Volume 131, Issue 1, January 2011, Pages 138-145.
Eugen J. Ionascu and R. A. Obando, Cubes in {0,1,...,N}^3, INTEGERS, 12A (2012), #A9. - From _N. J. A. Sloane_, Feb 05 2013
I. Larrosa, SMSU Problem Corner.
Baitian Li, C++ program for A098928

Cf. A103158.

Cf. A000537 (without skew cubes), A002415 (number of squares with corners on an n X n grid), A108279, A102698.

Needs["Quaternions`"];
(* Initialize variables *)
R = 20;
NN = 1010;
(* Quaternion operations *)
test[q_Quaternion] :=
  Module[{unit, res, a, b, c, u, v, w, p},
   If[Round[Norm[q]] > R, Return[]];
   If[q == Quaternion[0, 0, 0, 0], Return[]];
   unit = Quaternion[0, 1, 0, 0];
   res = q ** unit ** Conjugate[q];
   a = Abs[res[[2]]] + Abs[res[[3]]] + Abs[res[[4]]];
   unit = Quaternion[0, 0, 1, 0];
   res = q ** unit ** Conjugate[q];
   b = Abs[res[[2]]] + Abs[res[[3]]] + Abs[res[[4]]];
   unit = Quaternion[0, 0, 0, 1];
   res = q ** unit ** Conjugate[q];
   c = Abs[res[[2]]] + Abs[res[[3]]] + Abs[res[[4]]];
   For[i = 1, i <= (R - 1)/Max[a, b, c], i++,
    If[SquareFreeQ[i], {u = a*i;
      v = b*i;
      w = c*i;
      p = Max[u, v, w] + 1;
      coe[[p + 1, 4]] += (1);
      coe[[p + 1, 3]] -= (u + v + w);
      coe[[p + 1, 2]] += (u*v + v*w + w*u);
      coe[[p + 1, 1]] -= (u*v*w)}]]];
(* Set up coefficient matrix *)
coe = ConstantArray[0, {NN, 4}];
(* Loop through quaternions *)
rt = Ceiling[Sqrt[R]] + 1;
For[s = -rt, s <= rt, s++,
  For[x = -rt, x <= rt, x++,
   For[y = -rt, y <= rt, y++,
    For[z = -rt, z <= rt, z++, test[Quaternion[s, x, y, z]];
     test[Quaternion[s + 0.5, x + 0.5, y + 0.5, z + 0.5]];]]]];
newCoe = coe;
newCoe[[2 ;; ;; 2]] = coe[[2 ;; ;; 2]]/2;
(* Calculate and output results *)
For[i = 2, i <= R + 1, i++, ans = 0;
  For[j = 4, j >= 1, j--, newCoe[[i, j]] += newCoe[[i - 1, j]];
   ans = ans*(i - 1) + newCoe[[i, j]];
   ];
  Print[i - 1, " ", ans/24];];
(* Haomin Yang, Aug 29 2023 *)

Edited by Ray Chandler, Apr 05 2010

Further edited by N. J. A. Sloane, Mar 31 2016

A132117 Binomial transform of [1, 7, 17, 17, 6, 0, 0, 0, ...].

Original entry on oeis.org

1, 8, 32, 90, 205, 406, 728, 1212, 1905, 2860, 4136, 5798, 7917, 10570, 13840, 17816, 22593, 28272, 34960, 42770, 51821, 62238, 74152, 87700, 103025, 120276, 139608, 161182, 185165, 211730, 241056, 273328, 308737, 347480, 389760, 435786, 485773, 539942, 598520

Offset: 1

Gary W. Adamson, Aug 10 2007

Equals row sums of triangle A178067. - Gary W. Adamson, May 18 2010
Antidiagonal sums of the convolution array A213771. - Clark Kimberling, Jul 04 2012
Partial sums of A081436. - J. M. Bergot, Jun 20 2013

			a(3) = 32 = (1, 2, 1) dot (1, 7, 17) = (1 + 14 + 17).
a(5) = 15^2 - (10+6+3+1) = A000537(5) - A000292(4) = 225 - 20 = 205. - _Bruno Berselli_, May 01 2010

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

Cf. A178067. - Gary W. Adamson, May 18 2010

a:= n-> (Matrix([[0,0,2,13,46]]). Matrix(5, (i,j)-> if (i=j-1) then 1 elif j=1 then [5,-10,10,-5,1][i] else 0 fi)^n)[1,1]: seq(a(n), n=1..29); # Alois P. Heinz, Aug 07 2008
a:= n-> (4+(6+(8+6*n)*n)*n)*n/24: seq(a(n),n=1..40); # Alois P. Heinz, Aug 07 2008

Table[(4 n + 6 n^2 + 8 n^3 + 6 n^4) / 24, {n, 50}] (* Vincenzo Librandi, Jun 21 2013 *)

a(n) = (4*n+6*n^2+8*n^3+6*n^4)/24 \\ Charles R Greathouse IV, Sep 03 2011

Let M = the infinite lower triangular matrix of the natural numbers: [1; 2,3; 4,5,6; ...]; and V = [1, 2, 3, ...]. Then M*V = A132117.
O.g.f.: -x(1+x)(2x+1)/(-1+x)^5. - R. J. Mathar, Apr 02 2008
a(n) = (4*n + 6*n^2 + 8*n^3 + 6*n^4)/24. - Alois P. Heinz, Aug 07 2008
a(n) = A000217(n)^2 - Sum_{i=1..n-1} A000217(i) = n*(n+1)*(3*n^2+n+2)/12. - Bruno Berselli, May 01 2010

More terms from R. J. Mathar, Apr 02 2008

A154229 Triangle T(n, k) = T(n-1, k) + T(n-1, k-1) + ((n+1)(n+2)/2)^2T(n-2, k-1), read by rows.

Original entry on oeis.org

1, 1, 1, 1, 38, 1, 1, 139, 139, 1, 1, 365, 8828, 365, 1, 1, 807, 70492, 70492, 807, 1, 1, 1592, 357459, 7062136, 357459, 1592, 1, 1, 2889, 1404923, 98777227, 98777227, 1404923, 2889, 1, 1, 4915, 4631612, 824036625, 14498379854, 824036625, 4631612, 4915, 1

Offset: 0

Roger L. Bagula, Jan 05 2009

Row sums are: {1, 2, 40, 280, 9560, 142600, 7780240, 200370080, 16155726160, ...}.

The row sums of this class of sequences (see cross-references) is given by the following. Let S(n) be the row sum then S(n) = 2*S(n-1) + f(n)*S(n-2) for a given f(n). For this sequence f(n) = binomial(n+2, 2)^2 = A000537(n+1). - G. C. Greubel, Mar 02 2021

			Triangle begins as:
  1;
  1,    1;
  1,   38,       1;
  1,  139,     139,         1;
  1,  365,    8828,       365,           1;
  1,  807,   70492,     70492,         807,         1;
  1, 1592,  357459,   7062136,      357459,      1592,       1;
  1, 2889, 1404923,  98777227,    98777227,   1404923,    2889,    1;
  1, 4915, 4631612, 824036625, 14498379854, 824036625, 4631612, 4915, 1;

G. C. Greubel, Rows n = 0..50 of the triangle, flattened

Cf. A154227, A154228, A154230, A154231, A154233.

Cf. A000537.

f:= func< n | Binomial(n+2,2)^2 >;
function T(n,k)
  if k eq 0 or k eq n then return 1;
  else return T(n-1, k) + T(n-1, k-1) + f(n)*T(n-2, k-1);
  end if; return T;
end function;
[T(n,k): k in [0..n], n in [0..12]]; // G. C. Greubel, Mar 02 2021

T:= proc(n, k) option remember;
      if k=0 or k=n then 1
    else T(n-1, k) + T(n-1, k-1) + binomial(n+2,2)^2*T(n-2, k-1)
      fi; end:
seq(seq(T(n, k), k=0..n), n=0..12); # G. C. Greubel, Mar 02 2021

T[n_, k_]:= T[n, k]= If[k==0 || k==n, 1, T[n-1, k] + T[n-1, k-1] + Binomial[n+2, 2]^2*T[n-2, k-1]];
Table[T[n, k], {n,0,12}, {k,0,n}]//Flatten (* modified by G. C. Greubel, Mar 02 2021 *)

def f(n): return binomial(n+2,2)^2
def T(n,k):
    if (k==0 or k==n): return 1
    else: return T(n-1, k) + T(n-1, k-1) + f(n)*T(n-2, k-1)
flatten([[T(n,k) for k in (0..n)] for n in (0..12)]) # G. C. Greubel, Mar 02 2021

T(n, k) = T(n-1, k) + T(n-1, k-1) + ((n+1)*(n+2)/2)^2*T(n-2, k-1) with T(n, 0) = T(n, n) = 1.

Edited by G. C. Greubel, Mar 02 2021

A181144 G.f.: A(x,y) = exp( Sum_{n>=1} [Sum_{k=0..n} C(n,k)^4y^k] x^n/n ) = Sum_{n>=0,k=0..n} T(n,k)x^ny^k, as a triangle of coefficients T(n,k) read by rows.

Original entry on oeis.org

1, 1, 1, 1, 9, 1, 1, 36, 36, 1, 1, 100, 419, 100, 1, 1, 225, 2699, 2699, 225, 1, 1, 441, 12138, 35052, 12138, 441, 1, 1, 784, 42865, 286206, 286206, 42865, 784, 1, 1, 1296, 127191, 1696820, 3932898, 1696820, 127191, 1296, 1, 1, 2025, 330903, 7958563

Offset: 0

Paul D. Hanna, Oct 13 2010

Compare g.f. to that of the following triangle variants:
* Pascal's: exp( Sum_{n>=1} [Sum_{k=0..n} C(n,k)*y^k] * x^n/n );
* Narayana: exp( Sum_{n>=1} [Sum_{k=0..n} C(n,k)^2*y^k] * x^n/n );
* A181143: exp( Sum_{n>=1} [Sum_{k=0..n} C(n,k)^3*y^k] * x^n/n );
* A218115: exp( Sum_{n>=1} [Sum_{k=0..n} C(n,k)^5*y^k] * x^n/n );
* A218116: exp( Sum_{n>=1} [Sum_{k=0..n} C(n,k)^6*y^k] * x^n/n ).

			G.f.: A(x,y) = 1 + (1+y)*x + (1+9*y+y^2)*x^2 + (1+36*y+36*y^2+y^3)*x^3 + (1+100*y+419*y^2+100*y^3+y^4)*x^4 +...
The logarithm of the g.f. equals the series:
log(A(x,y)) = (1 + y)*x
+ (1 + 2^4*y + y^2)*x^2/2
+ (1 + 3^4*y + 3^4*y^2 + y^3)*x^3/3
+ (1 + 4^4*y + 6^4*y^2 + 4^4*y^3 + y^4)*x^4/4
+ (1 + 5^4*y + 10^4*y^2 + 10^4*y^3 + 5^4*y^4 + y^5)*x^5/5 +...
Triangle begins:
1;
1, 1;
1, 9, 1;
1, 36, 36, 1;
1, 100, 419, 100, 1;
1, 225, 2699, 2699, 225, 1;
1, 441, 12138, 35052, 12138, 441, 1;
1, 784, 42865, 286206, 286206, 42865, 784, 1;
1, 1296, 127191, 1696820, 3932898, 1696820, 127191, 1296, 1;
1, 2025, 330903, 7958563, 36955542, 36955542, 7958563, 330903, 2025, 1;
1, 3025, 776688, 31205941, 261852055, 525079969, 261852055, 31205941, 776688, 3025, 1; ...
Note that column 1 forms the sum of cubes (A000537), and forms the squares of the triangular numbers.
Inverse binomial transform of columns begins:
[1];
[1, 8, 19, 18, 6];
[1, 35, 348, 1549, 3713, 5154, 4161, 1818, 333];
[1, 99, 2500, 27254, 161793, 589819, 1409579, 2282850, 2529900, 1893972, 917349, 259854, 32726]; ...

Cf. A000537 (column 1), A166992 (row sums), A166898 (antidiagonal sums), A218140.

Cf. variants: A001263 (Narayana), A181143, A218115, A218116.

{T(n,k)=polcoeff(polcoeff(exp(sum(m=1,n,sum(j=0,m,binomial(m,j)^4*y^j)*x^m/m)+O(x^(n+1))),n,x),k,y)}
for(n=0, 10, for(k=0, n, print1(T(n, k), ", ")); print(""))

A286724 Triangle read by rows. A generalization of unsigned Lah numbers, called L[2,1].

Original entry on oeis.org

1, 2, 1, 8, 8, 1, 48, 72, 18, 1, 384, 768, 288, 32, 1, 3840, 9600, 4800, 800, 50, 1, 46080, 138240, 86400, 19200, 1800, 72, 1, 645120, 2257920, 1693440, 470400, 58800, 3528, 98, 1, 10321920, 41287680, 36126720, 12042240, 1881600, 150528, 6272, 128, 1, 185794560, 836075520, 836075520, 325140480, 60963840, 6096384, 338688, 10368, 162, 1, 3715891200, 18579456000, 20901888000, 9289728000, 2032128000, 243855360, 16934400, 691200, 16200, 200, 1

Offset: 0

Wolfdieter Lang, Jun 16 2017

These generalized unsigned Lah numbers are the instance L[2,1] of the Sheffer triangles called L[d,a], with integers d >= 1 and integers 0 <= a < d with gcd(d,a) = 1. The standard unsigned Lah numbers are L[1,0] = A271703.
The Sheffer structure of L[d,a] is ((1 - d*t)^(-2*a/d), t/(1 - d*t)). This follows from the defining property
  risefac[d,a](x, n) = Sum_{m=0..n} L[d,a](n, m)*fallfac[d,a](x, m), where risefac[d,a](x, n):= Product_{0..n-1} (x  + (a+d*j)) for n >= 1 and risefac[d,a](x, 0) := 1, and fallfac[d,a](x, n):= Product_{0..n-1} (x - (a+d*j)) = for n >= 1 and fallfac[d,a](x, 0) := 1. Such rising and falling factorials arise in the generalization of Stirling numbers of both kinds S2[d,a] and S1[d,a]. See the Peter Bala link under A143395 for these falling factorials called there [t;a,b,c]_n with t=x, a=d, b=0, c=a.
In matrix notation: L[d,a] = S1phat[d,a].S2hat[d,a] with the unsigned scaled Stirling1 and the scaled Stirling2 generalizations with Sheffer structures S1phat[d,a] = ((1 - d*t)^(-a/d), -(1/d)*(log(1 - d*t))) and S2hat[d,a] = (exp(a*t), (1/d)*(exp(d*t) - 1). See, e.g., S1phat[2,1] = A028338 and S2hat[2,1] = A039755.
The a- and z-sequences for these Sheffer matrices have e.g.f.s 1 + d*t and ((1 + d*t)/t)*(1 - (1 + d*t)^(-2*a/d)), respectively. See a W. Lang link under A006232 for these types of sequences.
E.g.f. of row polynomials R[d,a](n, x) := Sum_{m=0..n} L[d,a](n, m)*x^m
  (1 - d*x)^(-2*a/d)*exp(t*x/(1 - d*x)) (this is the e.g.f. for the triangle).
E.g.f. of column m: (1 - d*t)^(-2*a/d)*(t/(1 - d*t))^m/m, m >= 0.
Meixner type identity for (monic) row polynomials: (D_x/(1 + d*D_x)) * R[d,a](n, x) = n*R[d,a](n-1, x), n >= 1, with R[d,a](0, x) = 1. The series in the differentiations D_x = d/dx terminates.
General Sheffer recurrence for row polynomials (see the Roman reference, p. 50, Corollary 3.7.2, rewritten for the present Sheffer notation):
  R[d,a](n, x) = [(2*a+x)*1 + 2*d*(a + x)*D_x + d^2*x*(D_x)^2]*R[d,a](n-1, x), n >= 1, with R[d,a](0, x) = 1.
The inverse matrix L^(-1)[d,a] is Sheffer (g[d,a](-t), -f[d,a](-t)) with L[d,a] Sheffer (g[d,a](t), f[d,a](t)) from above. This means (see the column e.g.f. of Sheffer matrices) that L^(-1)[d,a](n, m) = (-1)^(n-m)*L[d,a](n, m). Therefore, the recurrence relations can easily be rewritten for L^(-1)[d,a] by replacing a -> -a and d -> -d.
fallfac[d,a](x, n) = Sum_{m=0..n} L^(-1)[d,a](n, m)*risefac[d,a](x, m), n >= 0.
From Wolfdieter Lang, Aug 12 2017: (Start)
The Sheffer row polynomials R[d,a](n, x) belong to the Boas-Buck class and satisfy therefore the Boas-Buck identity (see the reference, and we use the notation of Rainville, Theorem 50, p. 141, adapted to an exponential generating function) (E_x - n*1)*R[d,a](n, x) = - n*(2*a*1 + d*E_x) * Sum_{k=0..n-1} d^k*R(d,a;n-1-k,x)/(n-1-k)!, with E_x = x*d/dx (Euler operator).
This implies a recurrence for the sequence of column m: L[d,a](n, m) = (n!*(2*a + d*m)/(n-m))*Sum_{p=0..n-1-m} d^p*L[d,a](n-1-p, m)/(n-1-p)!, for n > m>=0, and input L[d,a](m, m) = 1. For the present [d,a] = [2,1] instance see the formula and example sections. (End)
From Wolfdieter Lang, Sep 14 2017: (Start)
The diagonal sequences are 2^D*D!*(binomial(m+D, m))^2, m >= 0, for D >= 0 (main diagonal D = 0). From the o.g.f.s obtained via Lagrange's theorem. See the second W. Lang link below for the general Sheffer case.
The o.g.f. of the diagonal D sequence is 2^D*D!*Sum_{m=0..D} A008459(D, m)*x^m /(1- x)^(2*D + 1),  D >= 0. (End)
It appears that this is also the matrix square of unsigned triangle of coefficients of Laguerre polynomials n!*L_n(x), abs(A021009(n, k)). - Ali Pourzand, Mar 10 2025 [This observation is correct. - Peter Luschny, Mar 10 2025]

			The triangle T(n, m) begins:
  n\m        0         1         2         3        4       5      6     7   8 9
  0:         1
  1:         2         1
  2:         8         8         1
  3:        48        72        18         1
  4:       384       768       288        32        1
  5:      3840      9600      4800       800       50       1
  6:     46080    138240     86400     19200     1800      72      1
  7:    645120   2257920   1693440    470400    58800    3528     98     1
  8:  10321920  41287680  36126720  12042240  1881600  150528   6272   128   1
  9: 185794560 836075520 836075520 325140480 60963840 6096384 338688 10368 162 1
  ...
From _Wolfdieter Lang_, Aug 12 2017: (Start)
Recurrence for column elements with m >= 1, and input column m = 0: T(3, 2) = (3/2)*T(2, 1) + 2*3*T(2, 2) = (3/2)*8 + 6 = 18.
Four term recurrence: T(3, 2) = T(2, 1) + 2*5*T(2, 2) - 4*2^2*T(1, 2) = 8 + 10 + 0 = 18.
Meixner type identity, n=2: 2*R(1, x) = (D_x - 2*(D_x)^2)*R(2, x), 2*(2 + x) = (8 + 2*x) - 2*2.
Sheffer recurrence: R(2, x) = (2 + x)*(2 + x) + 4*(1 + x)*1 + 0 = 8 + 8*x + x^2.
Boas-Buck recurrence for column m = 2 and n = 4: T(4, 2) = (2*4!*3/2)*(1*T(3, 2)/3! + 2*T(2, 2)/2!) = 4!*3*(18/3! + 1) = 288. (End)
Diagonal sequence D = 1: o.g.f. 2*1!*(1 + 1*x)/(1- x)^3 generating
{2*(binomial(m+1, m))^2}_{m >= 0} = {2, 8, 18, 32, ...}. - _Wolfdieter Lang_, Sep 14 2017

Ralph P. Boas, jr. and R. Creighton Buck, Polynomial Expansions of analytic functions, Springer, 1958, pp. 17 - 21, (last sign in eq. (6.11) should be -).
Earl D. Rainville, Special Functions, The Macmillan Company, New York, 1960, ch. 8, sect. 76, 140 - 146.
Steven Roman, The Umbral Calculus, Academic press, Orlando, London, 1984, p. 50.

Peter Bala, The white diamond product of power series
Wolfdieter Lang, On Sums of Powers of Arithmetic Progressions, and Generalized Stirling, Eulerian and Bernoulli Numbers, arXiv:math/1707.04451 [math.NT], July 2017.
Wolfdieter Lang, On Generating functions of Diagonal Sequences of Sheffer and Riordan Number Triangles, arXiv:1708.01421 [math.NT], August 2017.
Emanuele Munarini, Combinatorial identities involving the central coefficients of a Sheffer matrix, Applicable Analysis and Discrete Mathematics (2019) Vol. 13, 495-517.

Cf. A006232, A025167, A028338, A039755, A143395, A271703, A021009.
Column sequences (no leading zeros): A000165, A014479, A286725.
Diagonal sequences: A000012, 2*A000290(m+1), 8*A000537(n+1), 48*A001249, 384*A288876. - Wolfdieter Lang, Sep 14 2017
Row sums are A025167. - Michael Somos, Sep 27 2017

T := (n, k) -> ifelse(n < k, 0, ifelse(k = 0, n!*2^n, (n/k)*T(n-1, k-1) + 2*n*T(n-1, k))): seq(seq(T(n, k), k = 0..n), n = 0..10);  # Peter Luschny, Mar 10 2025

T[ n_, k_] := Coefficient[ Integrate[ Exp[-x^2 - y x] HermiteH[n, x]^2, {x, -Infinity, Infinity}] / (Sqrt[Pi] Exp[y^2 / 4]), y, 2 k]; (* Michael Somos, Sep 27 2017 *)

# Using the function A021009_triangle, displays as a matrix. Following the observation of Ali Pourzand.
print(A021009_triangle(9)^2)  # Peter Luschny, Mar 10 2025

T(n, m) = L[2,1](n, m) = Sum_{k=m..n} A028338(n, k)*A039755(k, m).
Three term recurrence for column elements with m >= 1: T(n, m) = (n/m)*T(n-1, m-1) + 2*n*T(n-1, m) with T(n, m) = 0 for n < m and the column m = 0 is T(n, 0) = (2*n)!! = n*2^n = A000165(n). (From the a- and z-sequences {1, 2, repeat(0)} and {2, repeat(0)}, respectively.)
Four term recurrence: T(n, m) = T(n-1, m-1) + 2*(2*n-1)*T(n-1, m) - 4*(n-1)^2*T(n-2, m), n >= m >= 0, with T(0, 0) = 1, T(-1, m) = 0, T(n, -1) = 0 and T(n, m) = 0 if n < m.
E.g.f. of row polynomials R(n, x) = R[2,1](n, x) (i.e., e.g.f. of the triangle): (1/(1-2*t))*exp(x*t/(1-2*t)).
E.g.f. of column m sequences: (t^m/(1-2*t)^(m+1))/m!, m >= 0.
Meixner type identity: Sum_{k=0..n-1} (-1)^k*2^k*(D_x)^(k+1)*R(n, x) = n*R(n-1, x), n >= 1, with R(0, x) = 1 and D_x = d/dx.
Sheffer recurrence: R(n, x) = [(2 + x)*1 + 4*(1 + x)*D_x + 4*x*(D_x)^2]*R(n-1, x), n >= 1, and R(0, x) = 1.
Boas-Buck recurrence for column m (see a comment above): T(n, m) = (2*n!*(1 + m)/(n-1))*Sum_{p=0..n-1-m} 2^p*T(n-1-p, m)/(n-1-p)!, for n > m >= 0, and input T(m, m) = 1. - Wolfdieter Lang, Aug 12 2017
Explicit form (from the diagonal sequences with the o.g.f.s given as a comment above): T(n, m) = 2^(n-m)*(n-m)!*(binomial(n, n-m))^2 for n >= m >= 0. - Wolfdieter Lang, Sep 23 2017
Let R(n,x) denote the n-th row polynomial. Then x^n*R(n,x) = x^n o x^n, where o denotes the deformed Hadamard product of power series defined in Bala, Section 3.1. - Peter Bala, Jan 18 2018

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