A029635 -id:A029635 - cp's OEIS frontend

A191302 Denominators in triangle that leads to the Bernoulli numbers.

1, 2, 2, 3, 2, 2, 2, 3, 15, 2, 6, 3, 2, 1, 5, 105, 2, 6, 15, 15, 2, 3, 3, 105, 105, 2, 2, 5, 7, 35, 2, 3, 3, 21, 21, 231, 2, 6, 15, 15, 21, 21, 2, 1, 5, 15, 1, 77, 15015, 2, 6, 3, 35, 15, 33, 1155

Offset: 0

Paul Curtz, May 30 2011

For the definition of the ASPEC array coefficients see the formulas; see also A029635 (Lucas triangle), A097207 and A191662 (k-dimensional square pyramidal numbers).
The antidiagonal row sums of the ASPEC array equal A042950(n) and A098011(n+3).
The coefficients of the T(n,m) array are defined in A190339. We define the coefficients of the SBD array with the aid of the T(n,n+1), see the formulas and the examples.
Multiplication of the coefficients in the rows of the ASPEC array with the coefficients in the columns of the SBD array leads to the coefficients of the BSPEC triangle, see the formulas. The BSPEC triangle can be looked upon as a spectrum for the Bernoulli numbers.
The row sums of the BSPEC triangle give the Bernoulli numbers A164555(n)/A027642(n).
For the numerators of the BSPEC triangle coefficients see A192456.

			The first few rows of the array ASPEC array:
  2, 1,  1,  1,   1,   1,    1,
  2, 3,  4,  5,   6,   7,    8,
  2, 5,  9, 14,  20,  27,   35,
  2, 7, 16, 30,  50,  77,  112,
  2, 9, 25, 55, 105, 182,  294,
The first few T(n,n+1) = T(n,n)/2 coefficients:
1/2, -1/6, 1/15, -4/105, 4/105, -16/231, 3056/15015, ...
The first few rows of the SBD array:
  1/2,   0,   0,     0
  1/2,   0,   0,     0
  1/2, -1/6,  0,     0
  1/2, -1/6,  0,     0
  1/2, -1/6, 1/15,   0
  1/2, -1/6, 1/15,   0
  1/2, -1/6, 1/15, -4/105
  1/2, -1/6, 1/15, -4/105
The first few rows of the BSPEC triangle:
  B(0) =   1   = 1/1
  B(1) =  1/2  = 1/2
  B(2) =  1/6  = 1/2 - 1/3
  B(3) =   0   = 1/2 - 1/2
  B(4) = -1/30 = 1/2 - 2/3 +  2/15
  B(5) =   0   = 1/2 - 5/6 +  1/3
  B(6) =  1/42 = 1/2 - 1/1 +  3/5  - 8/105
  B(7) =   0   = 1/2 - 7/6 + 14/15 - 4/15

Cf. A028246 (Worpitzky), A085737/A085738 (Conway-Sloane) and A051714/A051715 (Akiyama-Tanigawa) for other triangles that lead to the Bernoulli numbers. - Johannes W. Meijer, Jul 02 2011

nmax:=13: mmax:=nmax:
A164555:=proc(n): if n=1 then 1 else numer(bernoulli(n)) fi: end:
A027642:=proc(n): if n=1 then 2 else denom(bernoulli(n)) fi: end:
for m from 0 to 2*mmax do T(0,m):=A164555(m)/A027642(m) od:
for n from 1 to nmax do for m from 0 to 2*mmax do T(n,m):=T(n-1,m+1)-T(n-1,m) od: od:
seq(T(n,n+1),n=0..nmax):
for n from 0 to nmax do ASPEC(n,0):=2: for m from 1 to mmax do ASPEC(n,m):= (2*n+m)*binomial(n+m-1,m-1)/m od: od:
for n from 0 to nmax do seq(ASPEC(n,m),m=0..mmax) od:
for n from 0 to nmax do for m from 0 to 2*mmax do SBD(n,m):=0 od: od:
for m from 0 to mmax do for n from 2*m to nmax do SBD(n,m):= T(m,m+1) od: od:
for n from 0 to nmax do seq(SBD(n,m), m= 0..mmax/2) od:
for n from 0 to nmax do BSPEC(n,2) := SBD(n,2)*ASPEC(2,n-4) od:
for m from 0 to mmax do for n from 0 to nmax do BSPEC(n,m) := SBD(n,m)*ASPEC(m,n-2*m) od: od:
for n from 0 to nmax do seq(BSPEC(n,m), m=0..mmax/2) od:
seq(add(BSPEC(n, k), k=0..floor(n/2)) ,n=0..nmax):
Tx:=0:
for n from 0 to nmax do for m from 0 to floor(n/2) do a(Tx):= denom(BSPEC(n,m)): Tx:=Tx+1: od: od:
seq(a(n),n=0..Tx-1); # Johannes W. Meijer, Jul 02 2011

(* a=ASPEC, b=BSPEC *) nmax = 13; a[n_, 0] = 2; a[n_, m_] := (2n+m)*Binomial[n+m-1, m-1]/m; b[n_] := BernoulliB[n]; b[1]=1/2; bb = Table[b[n], {n, 0, nmax}]; diff = Table[ Differences[bb, n], {n, 1, nmax}]; dd = Diagonal[diff]; sbd[n_, m_] := If[n >= 2m, -dd[[m+1]], 0]; b[n_, m_] := sbd[n, m]*a[m, n-2m]; Table[b[n, m], {n, 0, nmax}, {m, 0, Floor[n/2]}] // Flatten // Denominator (* Jean-François Alcover_, Aug 09 2012 *)

ASPEC(n, 0) = 2 and ASPEC(n, m) = (2*n+m)*binomial(n+m-1, m-1)/m, n >= 0, m >= 1.
ASPEC(n, m) = ASPEC(n-1, m) + ASPEC(n, m-1), n >= 1, m >= 1, with ASPEC(n, 0) = 2, n >= 0, and ASPEC(0,m) = 1, m >= 1.
SBD(n, m) = T(m, m+1), n >= 2*m; see A190339 for the definition of the T(n, m).
BSPEC(n, m) = SBD(n, m)*ASPEC(m, n-2*m)
Sum_{k=0..floor(n/2)} BSPEC(n, k) = A164555(n)/A027642(n).

Edited, Maple program and crossrefs added by Johannes W. Meijer, Jul 02 2011

A098599 Riordan array ((1+2*x)/(1+x), (1+x)).

Original entry on oeis.org

1, 1, 1, -1, 2, 1, 1, 0, 3, 1, -1, 0, 2, 4, 1, 1, 0, 0, 5, 5, 1, -1, 0, 0, 2, 9, 6, 1, 1, 0, 0, 0, 7, 14, 7, 1, -1, 0, 0, 0, 2, 16, 20, 8, 1, 1, 0, 0, 0, 0, 9, 30, 27, 9, 1, -1, 0, 0, 0, 0, 2, 25, 50, 35, 10, 1, 1, 0, 0, 0, 0, 0, 11, 55, 77, 44, 11, 1, -1, 0, 0, 0, 0, 0, 2, 36, 105, 112, 54, 12, 1, 1, 0, 0, 0, 0, 0, 0, 13, 91, 182, 156, 65, 13, 1

Offset: 0

Paul Barry, Sep 17 2004

			Triangle begins as:
   1;
   1, 1;
  -1, 2, 1;
   1, 0, 3, 1;
  -1, 0, 2, 4, 1;
   1, 0, 0, 5, 5,  1;
  -1, 0, 0, 2, 9,  6,  1;
   1, 0, 0, 0, 7, 14,  7,  1;
  -1, 0, 0, 0, 2, 16, 20,  8, 1;
   1, 0, 0, 0, 0,  9, 30, 27, 9, 1;

G. C. Greubel, Table of n, a(n) for n = 0..1325

Cf. A000007, A005408, A029635, A040000, A079757, A098601.
Row sums are A098600.
Diagonal sums are A098601.
Apart from signs, same as A100218.
Very similar to triangle A111125.

A098599:= func< n,k | n eq 0 select 1 else Binomial(k, n-k) + Binomial(k-1, n-k-1) >;
[A098599(n,k): k in [0..n], n in [0..12]]; // G. C. Greubel, Mar 27 2024

T[n_, k_]:= If[n==0, 1, Binomial[k,n-k] +Binomial[k-1,n-k-1]];
Table[T[n,k], {n,0,12}, {k,0,n}]//Flatten (* G. C. Greubel, Mar 27 2024 *)

def A098599(n,k): return 1 if n==0 else binomial(k, n-k) + binomial(k-1, n-k-1)
flatten([[A098599(n,k) for k in range(n+1)] for n in range(13)]) # G. C. Greubel, Mar 27 2024

Triangle: T(n, k) = binomial(k, n-k) + binomial(k-1, n-k-1), with T(0, 0) = 1.
Sum_{k=0..n} T(n, k) = A098600(n) (row sums).
T(n,k) = T(n-1,k-1) - T(n-1,k) + 2*T(n-2,k-1) + T(n-3,k-1), T(0,0)=1, T(1,0)=1, T(1,1)=1, T(n,k)=0 if k<0 or if k>n. - Philippe Deléham, Jan 09 2014
From G. C. Greubel, Mar 27 2024: (Start)
T(2*n, n) = A040000(n).
T(2*n+1, n) = A000007(n).
T(2*n-1, n) = A005408(n-1), n >= 1.
Sum_{k=0..n} (-1)^k*T(n, k) = A079757(n).
Sum_{k=0..floor(n/2)} T(n-k, k) = A098601(n). (End)

A097207 Triangle read by rows: T(n,k) = binomial(n,k) + 2*binomial(n,k-1).

Original entry on oeis.org

1, 1, 3, 1, 4, 5, 1, 5, 9, 7, 1, 6, 14, 16, 9, 1, 7, 20, 30, 25, 11, 1, 8, 27, 50, 55, 36, 13, 1, 9, 35, 77, 105, 91, 49, 15, 1, 10, 44, 112, 182, 196, 140, 64, 17, 1, 11, 54, 156, 294, 378, 336, 204, 81, 19, 1, 12, 65, 210, 450, 672, 714, 540, 285, 100, 21, 1, 13, 77, 275, 660

Offset: 0

N. J. A. Sloane, Sep 21 2004

			Triangle begins:
1
1 3
1 4 5
1 5 9 7
1 6 14 16 9

Reinhard Zumkeller, Rows n=0..150 of triangle, flattened
H. W. Gould, Power sum identities for arbitrary symmetric arrays, SIAM J. Appl. Math., 17 (1969), 307-316.
Index entries for triangles and arrays related to Pascal's triangle

Cf. A029637, A110813 (row-reversed).

a097207 n k = a097207_tabl !! n !! k
a097207_row n = a097207_tabl !! n
a097207_tabl = map init $ tail a029635_tabl
-- Reinhard Zumkeller, Mar 12 2012

T[n_, k_] := Binomial[n, k] + 2Binomial[n, k - 1]; Flatten[ Table[ T[n, k], {n, 0, 10}, {k, 0, n}]] (* Robert G. Wilson v, Sep 21 2004 *)

T(n,k) = A029635(n+1,k), 0 <= k <= n. - Reinhard Zumkeller, Mar 12 2012

More terms from Robert G. Wilson v, Sep 21 2004

A156886 a(n) = Sum_{k=0..n} C(n,k)C(3n+k,k).

Original entry on oeis.org

1, 5, 43, 416, 4239, 44485, 475780, 5156548, 56437231, 622361423, 6904185523, 76964141600, 861408728964, 9673849095708, 108954068684616, 1230185577016156, 13920106205444335, 157814104889538739

Offset: 0

Paul Barry, Feb 17 2009

a(n)=[x^n](1+5x+9x^2+7x^3+2x^4)^n. The coefficients (1,5,9,7,2) are the 5th row of A029635.

Robert Israel, Table of n, a(n) for n = 0..937
P. Barry, A Note on a Family of Generalized Pascal Matrices Defined by Riordan Arrays, Journal of Integer Sequences, 16 (2013), #13.5.4.

Cf. A001850, A114496, A029635, A156887.

A156886 := proc(n)
    add(binomial(n,k)*binomial(3*n+k,k), k = 0..n);
end proc:
seq(A156886(n), n = 0..20); # Peter Bala, Feb 11 2018

a[n_] := Sum[ Binomial[n, k] Binomial[3n + k, k], {k, 0, n}]; Array[a, 21, 0] (* Robert G. Wilson v, Feb 11 2018 *)

From Peter Bala, Feb 11 2018: (Start)
a(n) = Sum_{k = 0..n} (-1)^(n-k)*C(n,k)*C(3*n+k,n)*2^k.
a(n) = Sum_{k = 0..n} C(n,k)*C(3*n,k)*2^(n-k),
12*n*(3*n-1)*(3*n-2)*(238*n^2 - 663*n + 457)*a(n) = 2*(150416*n^5 - 644640*n^4 + 1020351*n^3 - 734334*n^2 + 237007*n - 26880)*a(n-1) - (3*n-3)*(3*n-4)*(3*n-5)*(238*n^2 - 187*n + 32)*a(n-2). (End)
a(n) = P_n(0,2*n,3) where P_n(a,b,x) is the n-th Jacobi polynomial with parameters a and b. - Robert Israel, Feb 11 2018
a(n) ~ sqrt(1/3 + 11/(12*sqrt(7))) * ((316 + 119*sqrt(7))/54)^n / sqrt(Pi*n). - Vaclav Kotesovec, Jan 09 2023

A156887 a(n) = Sum_{k=0..n} C(n,k)C(4n+k,k).

Original entry on oeis.org

1, 6, 64, 768, 9708, 126386, 1676956, 22548168, 306167324, 4188703512, 57649462164, 797294161824, 11071026740964, 154250752864812, 2155368246401224, 30192512693210888, 423859798484668188, 5961793387214958792, 83998039356129372448, 1185277027372535468544

Offset: 0

Paul Barry, Feb 17 2009

a(n)=[x^n] (1+6x+14x^2+16x^3+9x^4+2x^5)^n. The coefficients (1,6,14,16,9,2) are the 6th row of A029635.

Seiichi Manyama, Table of n, a(n) for n = 0..862
P. Barry, A Note on a Family of Generalized Pascal Matrices Defined by Riordan Arrays, Journal of Integer Sequences, 16 (2013), #13.5.4.

Cf. A001850, A114496, A156886, A359646.

A156887 := proc(n)
    add(binomial(n,k)*binomial(4*n+k,k),k=0..n) ;
end proc: # R. J. Mathar, Feb 25 2015

Table[Sum[Binomial[n,k]Binomial[4n+k,k],{k,0,n}],{n,0,30}] (* Harvey P. Dale, Jul 24 2018 *)

{a(n) = sum(k=0, n, binomial(n, k)*binomial(4*n+k, k))} \\ Seiichi Manyama, Feb 02 2019

Conjecture: 576*n*(32901928701*n-65877527665)*(4*n-3)*(2*n-1)*(4*n-1)*a(n) +(-8795436181229177*n^5 +35251410418024655*n^4 -47934714902592853*n^3 +29414167990853161*n^2 -9060238526902314*n +1466702211905280)*a(n-1) +8*(10299715469615*n^5 -136961193094719*n^4 +872530072905392*n^3 -2699499511785411*n^2 +3902106377543903*n -2123717948975100)*a(n-2) -64*(2*n-5)*(4*n-9)*(n-2)*(27741827*n-2925269736)*(4*n-11)*a(n-3)=0. - R. J. Mathar, Feb 25 2015
From Peter Bala, Feb 11 2018: (Start)
a(n) = Sum_{k = 0..n} (-1)^(n-k)*C(n,k)*C(4*n+k,n)*2^k.
a(n) = Sum_{k = 0..n} C(n,k)*C(4*n,k)*2^(n-k).
8*(4*n)*(4*n-1)*(4*n-2)**(4*n-3)*(4633*n^3-19662*n^2+27593*n-12804)*a(n) = (137604733*n^7-859190528*n^6+2179882848*n^5-2890753162*n^4+2144669963*n^3-880916550*n^2+182941416*n-14515200)*a(n-1) - (4*n-4)*(4*n-5)*(4*n-6)*(4*n-7)*(4633*n^3-5763*n^2+2168*n-240)*a(n-2). A proof of Mathar's conjectured third-order recurrence above follows easily using this second-order recurrence.  (End)
a(n) ~ sqrt(5 + 33/sqrt(41)) * ((29701 + 4633*sqrt(41)))^n / (sqrt(Pi*n) * 2^(12*n + 2)). - Vaclav Kotesovec, Jan 09 2023

A157000 Triangle T(n,k) = (n/k)*binomial(n-k-1, k-1) read by rows.

Original entry on oeis.org

2, 3, 4, 2, 5, 5, 6, 9, 2, 7, 14, 7, 8, 20, 16, 2, 9, 27, 30, 9, 10, 35, 50, 25, 2, 11, 44, 77, 55, 11, 12, 54, 112, 105, 36, 2, 13, 65, 156, 182, 91, 13, 14, 77, 210, 294, 196, 49, 2, 15, 90, 275, 450, 378, 140, 15, 16, 104, 352, 660, 672, 336, 64, 2, 17, 119, 442, 935, 1122, 714, 204, 17

Offset: 2

Roger L. Bagula, Feb 20 2009

Row sums are A001610(n-1).
Triangle A034807 (coefficients of Lucas polynomials) with the first column omitted. - Philippe Deléham, Mar 17 2013
T(n,k) is the number of ways to select k knights from a round table of n knights, no two adjacent. - Bert Seghers, Mar 02 2014

			The table starts in row n=2, column k=1 as:
   2;
   3;
   4,  2;
   5,  5;
   6,  9,   2;
   7, 14,   7;
   8, 20,  16,   2;
   9, 27,  30,   9;
  10, 35,  50,  25,  2;
  11, 44,  77,  55, 11;
  12, 54, 112, 105, 36, 2;

J. Riordan, An Introduction to Combinatorial Analysis, Wiley, 1958, pp. 199

G. C. Greubel, Rows n = 2..100 of triangle, flattened

Cf. A132460, A113279, A082985, A029635.

[[n*Binomial(n-k-1,k-1)/k: k in [1..Floor(n/2)]]: n in [2..20]]; // G. C. Greubel, Apr 25 2019

Table[(n/k)*Binomial[n-k-1, k-1], {n,2,20}, {k,1,Floor[n/2]}]//Flatten (* modified by G. C. Greubel, Apr 25 2019 *)

a(n,k)=n*binomial(n-k-1,k-1)/k; \\ Charles R Greathouse IV, Jul 10 2011

[[n*binomial(n-k-1,k-1)/k for k in (1..floor(n/2))] for n in (2..20)] # G. C. Greubel, Apr 25 2019

T(n,k) = binomial(n-k,k) + binomial(n-k-1,k-1). - Bert Seghers, Mar 02 2014

Offset 2, keyword:tabf, more terms by the Assoc. Eds. of the OEIS, Nov 01 2010

A164851 Generalized Lucas-Pascal triangle; (11*10^n, 1).

Original entry on oeis.org

1, 11, 1, 110, 12, 1, 1100, 122, 13, 1, 11000, 1222, 135, 14, 1, 110000, 12222, 1357, 149, 15, 1, 1100000, 122222, 13579, 1506, 164, 16, 1, 11000000, 1222222, 135801, 15085, 1670, 180, 17, 1

Offset: 0

Mark Dols, Aug 28 2009

			Triangle begins:
         1;
        11,      1;
       110,     12,      1;
      1100,    122,     13,     1;
     11000,   1222,    135,    14,    1;
    110000,  12222,   1357,   149,   15,   1;
   1100000, 122222,  13579,  1506,  164,  16,  1;
  11000000,1222222, 135801, 15085, 1670, 180, 17, 1;
  ...

Robert Israel, Table of n, a(n) for n = 0..10010(rows 0 to 140, flattened)

Cf. A029635, A164844, A228196.

G[0]:= 1;
G[1]:= 11+x;
G[2]:= 110+12*x+x^2;
for nn from 3 to 20 do
  G[nn]:= expand((x+11)*G[nn-1]-10*(x+1)*G[nn-2]);
od:
seq(seq(coeff(G[n],x,j),j=0..n),n=0..20); # Robert Israel, Jul 17 2017

T[0, 0] := 1; T[n_, n_] := 1; T[n_, 0] := 11*10^(n - 1); T[n_, k_] := T[n - 1, k - 1] + T[n - 1, k];  Table[T[n, k], {n, 0, 10}, {k, 0, n}] //Flatten (* G. C. Greubel, Dec 22 2017 *)

T(0,0)=1, T(n+1,0)=11*10^n, T(n,n)=1, T(n,k)=T(n-1,k-1)+T(n-1,k) for 0Philippe Deléham, Dec 27 2013

G.f. as triangle: (1-x^2)/((1-10*x)*(1-x-x*y)). - Robert Israel, Jul 17 2017

Initial 1 added by Philippe Deléham, Dec 27 2013

A207628 Triangle of coefficients of polynomials v(n,x) jointly generated with A207627; see the Formula section.

Original entry on oeis.org

1, 1, 4, 1, 6, 8, 1, 8, 20, 16, 1, 10, 36, 56, 32, 1, 12, 56, 128, 144, 64, 1, 14, 80, 240, 400, 352, 128, 1, 16, 108, 400, 880, 1152, 832, 256, 1, 18, 140, 616, 1680, 2912, 3136, 1920, 512, 1, 20, 176, 896, 2912, 6272, 8960, 8192, 4352, 1024, 1, 22, 216

Offset: 1

Clark Kimberling, Feb 21 2012

Column n is divisible by 2^(n-1); row n ends with 2^(n-1) for n > 2.
Also triangle T(n,k), k=0..n, read by rows, given by (1, 0, 0, 0, 0, 0, 0, 0, ...) DELTA (4, -2, 0, 0, 0, 0, 0, 0, 0, ...) where DELTA is the operator defined in A084938. - Philippe Deléham, Feb 25 2012
Also seems to be square array of unsigned coefficients of 3U-2T (with T and U the two sequences of Chebyshev polynomials). - Thomas Baruchel, Jun 03 2018

			First five rows:
  1;
  1,  4;
  1,  6,  8;
  1,  8, 20, 16;
  1, 10, 36, 56, 32;

Cf. A207627.

u[1, x_] := 1; v[1, x_] := 1; z = 16;
u[n_, x_] := u[n - 1, x] + v[n - 1, x]
v[n_, x_] := 2 x*u[n - 1, x] + 2 x*v[n - 1, x] + 1
Table[Factor[u[n, x]], {n, 1, z}]
Table[Factor[v[n, x]], {n, 1, z}]
cu = Table[CoefficientList[u[n, x], x], {n, 1, z}];
TableForm[cu]
Flatten[%]    (* A207627 *)
Table[Expand[v[n, x]], {n, 1, z}]
cv = Table[CoefficientList[v[n, x], x], {n, 1, z}];
TableForm[cv]
Flatten[%]    (* A207628 *)

u(n,x) = u(n-1,x) + v(n-1,x),
v(n,x) = 2x*u(n-1,x) + 2x*v(n-1,x) + 1,
where u(1,x)=1, v(1,x)=1.
The polynomials v(n,x) seem to be v(n)=sum(k=0,n-1, (-1)^(k+n+1) * x^k * polcoeff( 3*polchebyshev(2*n-k-2,2)-2*polchebyshev(2*n-k-2,1), k)) by using the PARI syntax. - Thomas Baruchel, Jun 05 2018
As triangle T(n,k), k=0..n:
G.f.: (1+2*y*x)/(1-(1+2*y)*x). - Philippe Deléham, Feb 25 2012
T(n,k) = 2*T(n-1,k-1) + T(n-1,k) with T(0,0) = T(1,0) = 1, T(1,1) = 4. - Philippe Deléham, Feb 25 2012
As triangle T(n,k), k=0..n, it is given by T(n,k) = A029635(n,k)*2^k with T(0,0) = 1. - Philippe Deléham, Feb 25 2012

A293600 G.f. A(x,y) = Sum_{-oo..+oo} (x - y^n)^(n+1), as a flattened rectangular array of coefficients T(n,k) of x^n * y^(k*(n+k-1)) in A(x,y) for n>=1.

Original entry on oeis.org

1, 1, -2, 1, -3, 2, 1, -4, 5, -2, 1, -5, 9, -7, 2, 1, -6, 14, -16, 9, -2, 1, -7, 20, -30, 25, -11, 2, 1, -8, 27, -50, 55, -36, 13, -2, 1, -9, 35, -77, 105, -91, 49, -15, 2, 1, -10, 44, -112, 182, -196, 140, -64, 17, -2, 1, -11, 54, -156, 294, -378, 336, -204, 81, -19, 2, 1, -12, 65, -210, 450, -672, 714, -540, 285, -100, 21, -2, 1, -13, 77, -275, 660, -1122, 1386, -1254, 825, -385, 121, -23, 2, 1, -14, 90, -352, 935, -1782, 2508, -2640, 2079, -1210, 506, -144, 25, -2, 1, -15, 104, -442, 1287, -2717, 4290, -5148, 4719, -3289, 1716, -650, 169, -27, 2

Offset: 1

Paul D. Hanna, Oct 16 2017

Compare g.f. to the identity: Sum_{-oo..+oo} (x - y^n)^(n-1) = 0.

The Lucas triangle, A029635, consists of essentially the same coefficients, but differs in signs and initial term.

			G.f. A(x,y) = Sum_{n>=1} x^n * Sum_{k>=0} T(n,k) * y^(k*(n+k-1))
such that A(x,y) = Sum_{-oo..+oo} (x - y^n)^(n+1).
Explicitly, the g.f. of this array begins:
A(x,y) = x*(1 - 2*y + 2*y^4 - 2*y^9 + 2*y^16 - 2*y^25 + 2*y^36 +...)
+ x^2*(1 - 3*y^2 + 5*y^6 - 7*y^12 + 9*y^20 - 11*y^30 + 13*y^42 +...)
+ x^3*(1 - 4*y^3 + 9*y^8 - 16*y^15 + 25*y^24 - 36*y^35 + 49*y^48 +...)
+ x^4*(1 - 5*y^4 + 14*y^10 - 30*y^18 + 55*y^28 - 91*y^40 + 140*y^54 +...)
+ x^5*(1 - 6*y^5 + 20*y^12 - 50*y^21 + 105*y^32 - 196*y^45 + 336*y^60 +...)
+ x^6*(1 - 7*y^6 + 27*y^14 - 77*y^24 + 182*y^36 - 378*y^50 + 714*y^66 +...)
+ x^7*(1 - 8*y^7 + 35*y^16 - 112*y^27 + 294*y^40 - 672*y^55 + 1386*y^72 +...)
+ x^8*(1 - 9*y^8 + 44*y^18 - 156*y^30 + 450*y^44 - 1122*y^60 + 2508*y^78 +...)
+...
Summing along columns gives the alternate g.f.:
A(x,y) = x/(1-x) + Sum_{n>=1} (-1)^n * x * y^(n^2) * (2 - x*y^n)/(1 - x*y^n)^(n+1).
Note that the coefficient of x in A(x,y) is Jacobi's theta_4 function of y.
Also, the coefficient of x^2 in A(x,y) equals Product_{n>=1} (1 - y^(2*n))^3.
RECTANGULAR ARRAY.
This array of coefficients T(n,k) of x^n * y^(k*(n+k-1)) in A(x,y) begins:
n=1: [1, -2, 2, -2, 2, -2, 2, -2, 2, -2, 2, ...];
n=2: [1, -3, 5, -7, 9, -11, 13, -15, 17, -19, 21, ...];
n=3: [1, -4, 9, -16, 25, -36, 49, -64, 81, -100, 121, ...];
n=4: [1, -5, 14, -30, 55, -91, 140, -204, 285, -385, 506, ...];
n=5: [1, -6, 20, -50, 105, -196, 336, -540, 825, -1210, 1716, ...];
n=6: [1, -7, 27, -77, 182, -378, 714, -1254, 2079, -3289, 5005, ...];
n=7: [1, -8, 35, -112, 294, -672, 1386, -2640, 4719, -8008, 13013, ...];
n=8: [1, -9, 44, -156, 450, -1122, 2508, -5148, 9867, -17875, 30888, ...];
n=9: [1, -10, 54, -210, 660, -1782, 4290, -9438, 19305, -37180, 68068, ...]; ...
where row n has g.f.: (1 - z) / (1 + z)^n.
The array has the alternate g.f.: (1 - z) / (1 - x + z).
RELATED SERIES.
We may also write A(x,y) = P(x,y) + Q(x,y) where
P(x,y) = -1 + Sum_{n>=0} (-1)^n * y^(n*(n-1)) / (1 - x*y^n)^(n+1),
Q(x,y) = Sum_{n>=0} (-1)^n * y^(n*(n+1)) / (1 - x*y^(n+1))^n.
These series begin as follows:
P(x,y) = (-1 + y^2 - y^6 + y^12 - y^20 + y^30 - y^42 + y^56 - y^72 +...)
+ x*(1 - 2*y + 3*y^4 - 4*y^9 + 5*y^16 - 6*y^25 + 7*y^36 - 8*y^49 +...)
+ x^2*(1 - 3*y^2 + 6*y^6 - 10*y^12 + 15*y^20 - 21*y^30 + 28*y^42 +...)
+ x^3*(1 - 4*y^3 + 10*y^8 - 20*y^15 + 35*y^24 - 56*y^35 + 84*y^48 +...)
+ x^4*(1 - 5*y^4 + 15*y^10 - 35*y^18 + 70*y^28 - 126*y^40 + 210*y^54 +...)
+ x^5*(1 - 6*y^5 + 21*y^12 - 56*y^21 + 126*y^32 - 252*y^45 + 462*y^60 +...)
+ x^6*(1 - 7*y^6 + 28*y^14 - 84*y^24 + 210*y^36 - 462*y^50 + 924*y^66 +...)
+ x^7*(1 - 8*y^7 + 36*y^16 - 120*y^27 + 330*y^40 - 792*y^55 + 1716*y^72 +...)
+...
Q(x,y) = (1 - y^2 + y^6 - y^12 + y^20 - y^30 + y^42 - y^56 + y^72 +...)
+ x*(-y^4 + 2*y^9 - 3*y^16 + 4*y^25 - 5*y^36 + 6*y^49 - 7*y^64 +...)
+ x^2*(-y^6 + 3*y^12 - 6*y^20 + 10*y^30 - 15*y^42 + 21*y^56 +...)
+ x^3*(-y^8 + 4*y^15 - 10*y^24 + 20*y^35 - 35*y^48 + 56*y^63 +...)
+ x^4*(-y^10 + 5*y^18 - 15*y^28 + 35*y^40 - 70*y^54 + 126*y^70 +...)
+ x^5*(-y^12 + 6*y^21 - 21*y^32 + 56*y^45 - 126*y^60 + 252*y^77 +...)
+ x^6*(-y^14 + 7*y^24 - 28*y^36 + 84*y^50 - 210*y^66 + 462*y^84 +...)
+ x^7*(-y^16 + 8*y^27 - 36*y^40 + 120*y^55 - 330*y^72 + 792*y^91 +...)
+...

Cf. A292929, A293385, A029635.

{ T(n,k) = my(z=x+x*O(x^k)); polcoeff( (1-z)/(1+z)^n, k) }
/* Print as a rectangular array: */
for(n=1,10,for(k=0,10,print1(T(n,k),", "));print(""))
/* Print as a triangle: */
for(n=0,14,for(k=0,n,print1(T(n-k+1,k),", "));print(""))
/* Print as a flattened array: */
for(n=0,14,for(k=0,n,print1(T(n-k+1,k),", "));)

G.f. A(x,y) = Sum_{-oo..+oo} (x - y^n)^(n+1).
G.f. A(x,y) = x * Sum_{-oo..+oo} (x - y^n)^n.
G.f. A(x,y) = x/(1-x) + Sum_{n>=1} (-1)^n*x*y^(n^2)*(2 - x*y^n)/(1 - x*y^n)^(n+1).
G.f. A(x,y) = P(x,y) + Q(x,y) where
P(x,y) = Sum_{n>=0} (x - y^n)^(n+1),
P(x,y) = -1 + Sum_{n>=0} (-1)^n * y^(n*(n-1)) / (1 - x*y^n)^(n+1),
Q(x,y) = Sum_{n>=0} (-1)^n * y^(n*(n+1)) / (1 - x*y^(n+1))^n.

A081496 Start with Pascal's triangle; a(n) is the sum of the numbers on the periphery of the n-th central rhombus containing exactly 4 numbers.

Original entry on oeis.org

5, 14, 46, 160, 574, 2100, 7788, 29172, 110110, 418132, 1595620, 6113744, 23505356, 90633800, 350351640, 1357278300, 5268292830, 20483876820, 79765662900, 311038321440, 1214362277700, 4746455801880, 18570960418920, 72728638093800

Offset: 1

Amarnath Murthy, Mar 25 2003

nonn

			The first three rhombuses are
...1...........2.........6
.1...1.......3...3.....10..10
...2......,....6.....,...20
and the corresponding sums are a(1) =5, a(2) =14 and a(3) =46.

Qi Wang, Tau-tilting finite simply connected algebras, arXiv:1910.01937 [math.RT], 2019. See Example 5.1 and table page 17.

Cf. A081494, A081495, A081497, A029635.

seq((9*n-4)*binomial(2*(n-1),(n-1))/n,n=1..26); # C. Ronaldo, Dec 20 2004

{ A029635(n, k) = if( k<0 || k>n, 0, (n==0) + binomial(n, k) + binomial(n-1, k-1))}; \\ program from Michael Somos in A029635
{a(n) = sum(k=0,n,A029635(n, k)^2)} \\ Paul D. Hanna, Oct 17 2017
for(n=1,30,print1(a(n),", "))

a(n) = (9*n-4)*Catalan(n-1) = (9*n-4)*binomial(2*(n-1), (n-1))/n. - C. Ronaldo (aga_new_ac(AT)hotmail.com), Dec 20 2004

a(n) = Sum_{k=0..n} A029635(n,k)^2 for n>=1, where A029635 is the Lucas triangle. - Paul D. Hanna, Oct 17 2017

More terms from C. Ronaldo (aga_new_ac(AT)hotmail.com), Dec 20 2004

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