A220101 -id:A220101 - cp's OEIS frontend

A051666 Rows of triangle formed using Pascal's rule except begin and end n-th row with n^2.

0, 1, 1, 4, 2, 4, 9, 6, 6, 9, 16, 15, 12, 15, 16, 25, 31, 27, 27, 31, 25, 36, 56, 58, 54, 58, 56, 36, 49, 92, 114, 112, 112, 114, 92, 49, 64, 141, 206, 226, 224, 226, 206, 141, 64, 81, 205, 347, 432, 450, 450, 432, 347, 205, 81, 100, 286, 552, 779, 882, 900, 882, 779

Offset: 0

Asher Auel

Row sums give 6*2^n - 4*n - 6 (A051667).
Central terms: T(2*n,n) = 2 * A220101(n). - Reinhard Zumkeller, Aug 05 2013
For a closed-form formula for arbitrary left and right borders of Pascal like triangle see A228196. - Boris Putievskiy, Aug 19 2013
For a closed-form formula for generalized Pascal's triangle see A228576. - Boris Putievskiy, Sep 09 2013

			Triangle begins:
   0;
   1,  1;
   4,  2,  4;
   9,  6,  6,  9;
  16, 15, 12, 15, 16;
  ...

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

Cf. A007318, A000290, A005408, A228196, A228576.

a051666 n k = a051666_tabl !! n !! k
a051666_row n = a051666_tabl !! n
a051666_tabl = map fst $ iterate
   (\(vs, w:ws) -> (zipWith (+) ([w] ++ vs) (vs ++ [w]), ws))
   ([0], [1, 3 ..])
-- Reinhard Zumkeller, Aug 05 2013

T[n_, 0] := n^2; T[n_, n_] := n^2;
T[n_, k_] := T[n, k] = T[n-1, k-1] + T[n-1, k];
Table[T[n, k], {n, 0, 10}, {k, 0, n}] // Flatten (* Jean-François Alcover, Jul 13 2018 *)

More terms from James Sellers

A265612 a(n) = CatalanNumber(n+1)n(1+3n)/(6+2n).

Original entry on oeis.org

0, 1, 7, 35, 156, 660, 2717, 11011, 44200, 176358, 700910, 2778446, 10994920, 43459650, 171655785, 677688675, 2674776720, 10555815270, 41656918050, 164401379610, 648887951400, 2561511781920, 10113397410402, 39937416869070, 157743149913776, 623178050662300

Offset: 0

Peter Luschny, Dec 15 2015

nonn

This is row n=7 in the array A(n,k) = (rf(k+n-2,k-1)-(k-1)*(k-2)*rf(k+n-2, k-3))/ (k-1)! if n>=3 and A(n,0)=0, A(n,1)=1, A(n,2)=n; rf(n,k) denotes the rising factorial. See the cross-references for other values of n and the table in A264357.

G. C. Greubel, Table of n, a(n) for n = 0..1000

Cf. A000108, A007946, A097613, A051960, A029651, A051924, A129869, A220101, A264357, A265613.

A265612 := n -> 2*4^n*GAMMA(3/2+n)*n*(1+3*n)/(sqrt(Pi)*GAMMA(4+n)):
seq(simplify(A265612(n)), n=0..25);

Table[SeriesCoefficient[(5 x + (I (x - 1) (7 x - 2))/Sqrt[4 x - 1] - 2 - x^2)/(2 x^3), {x, 0, n}], {n, 0, 25}] (* or *)
Table[2*4^n Gamma[3/2 + n] n (1 + 3 n)/(Sqrt[Pi] Gamma[4 + n]), {n, 0, 25}] (* or *)
Table[CatalanNumber[n + 1] n ((1 + 3 n)/(6 + 2 n)), {n, 0, 25}] (* Michael De Vlieger, Dec 15 2015 *)

for(n=0,25, print1(round(2*4^n*gamma(3/2+n)*n*(1+3*n)/(sqrt(Pi)*gamma(4+n))), ", ")) \\ G. C. Greubel, Feb 06 2017

a = lambda n: catalan_number(n+1)*n*(1+3*n)/(6+2*n)
[a(n) for n in range(26)]

G.f.: (5*x+(I*(x-1)*(7*x-2))/sqrt(4*x-1)-2-x^2)/(2*x^3).
a(n) = 2*4^n*Gamma(3/2+n)*n*(1+3*n)/(sqrt(Pi)*Gamma(4+n)).
a(n) = (rf(5+n, n-1)-(n-1)*(n-2)*rf(5+n, n-3))/(n-1)! for n>=3, rf(n,k) the rising factorial.
a(n) = a(n-1)*(2*n*(1+3*n)*(1+2*n)/((n-1)*(3*n-2)*(3+n))) for n>=2.
a(n) ~  4^n*(6-(127/4)/n+(7995/64)/n^2-(223405/512)/n^3+(23501457/16384)/n^4-...) /sqrt(n*Pi).
a(n) = [x^n] x*(1 + x)/(1 - x)^(n+4). - Ilya Gutkovskiy, Oct 09 2017

A264357 Array A(r, n) of number of independent components of a symmetric traceless tensor of rank r and dimension n, written as triangle T(n, r) = A(r, n-r+2), n >= 1, r = 2..n+1.

Original entry on oeis.org

0, 2, 0, 5, 2, 0, 9, 7, 2, 0, 14, 16, 9, 2, 0, 20, 30, 25, 11, 2, 0, 27, 50, 55, 36, 13, 2, 0, 35, 77, 105, 91, 49, 15, 2, 0, 44, 112, 182, 196, 140, 64, 17, 2, 0, 54, 156, 294, 378, 336, 204, 81, 19, 2, 0

Offset: 1

Wolfdieter Lang, Dec 10 2015

A (totally) symmetric traceless tensor of rank r >= 2 and dimension n >= 1 is irreducible.
The array of the number of independent components of a rank r symmetric traceless tensor A(r, n), for r >= 2 and n >=1, is given by risefac(n,r)/r! - risefac(n,r-2)/(r-2)!, where the first term gives the number of independent components of a symmetric tensors of rank r (see a Dec 10 2015 comment under A135278) and the second term is the number of constraints from the tracelessness requirement. The tensor has to be traceless in each pair of indices.
The first rows of the array A, or the first columns (without the first r-2 zeros) of the triangle T are for r = 2..6: A000096, A005581, A005582, A005583, A005584.
Equals A115241 with the first column of positive integers removed. - Georg Fischer, Jul 26 2023

			The array A(r, n) starts:
   r\n 1 2  3   4   5    6    7     8     9    10 ...
   2:  0 2  5   9  14   20   27    35    44    54
   3:  0 2  7  16  30   50   77   112   156   210
   4:  0 2  9  25  55  105  182   294   450   660
   5:  0 2 11  36  91  196  378   672  1122  1782
   6:  0 2 13  49 140  336  714  1386  2508  4290
   7:  0 2 15  64 204  540 1254  2640  5148  9438
   8:  0 2 17  81 285  825 2079  4719  9867 19305
   9:  0 2 19 100 385 1210 3289  8008 17875 37180
  10:  0 2 21 121 506 1716 5005 13013 30888 68068
  ...
The triangle T(n, r) starts:
   n\r  2   3   4   5   6   7  8  9 10 11 ...
   1:   0
   2:   2   0
   3:   5   2   0
   4:   9   7   2   0
   5:  14  16   9   2   0
   6:  20  30  25  11   2   0
   7:  27  50  55  36  13   2  0
   8:  35  77 105  91  49  15  2  0
   9:  44 112 182 196 140  64 17  2  0
  10:  54 156 294 378 336 204 81 19  2  0
  ...
A(r, 1) = 0 , r >= 2, because a symmetric rank r tensor t of dimension one has one component t(1,1,...,1) (r 1's) and if the traces vanish then t vanishes.
A(3, 2) = 2 because a symmetric rank 3 tensor t with three indices taking values from 1 or 2 (n=2) has the four independent components t(1,1,1), t(1,1,2), t(1,2,2), t(2,2,2), and (invoking symmetry) the vanishing traces are Sum_{j=1..2} t(j,j,1) = 0 and Sum_{j=1..2} t(j,j,2) = 0. These are two constraints, which can be used to eliminate, say, t(1,1,1) and t(2,2,2), leaving 2 = A(3, 2) independent components, say, t(1,1,2) and t(1,2,2).
From _Peter Luschny_, Dec 14 2015: (Start)
The diagonals diag(n, k) start:
   k\n  0       1       2       3       4      5       6
   0:   0,      2,      9,     36,    140,   540,   2079, ... A007946
   1:   2,      7,     25,     91,    336,  1254,   4719, ... A097613
   2:   5,     16,     55,    196,    714,  2640,   9867, ... A051960
   3:   9,     30,    105,    378,   1386,  5148,  19305, ... A029651
   4:  14,     50,    182,    672,   2508,  9438,  35750, ... A051924
   5:  20,     77,    294,   1122,   4290, 16445,  63206, ... A129869
   6:  27,    112,    450,   1782,   7007, 27456, 107406, ... A220101
   7:  35,    156,    660,   2717,  11011, 44200, 176358, ... A265612
   8:  44,    210,    935,    4004, 16744, 68952, 281010, ... A265613
  A000096,A005581,A005582,A005583,A005584.
(End)

Cf. A000096, A005581, A005582, A005583, A005584, A007946, A024482, A029651, A051924, A051960, A097613, A115241, A129869, A135278, A220101, A265612, A265613.

A[r_, n_] := Pochhammer[n, r]/r! - Pochhammer[n, r-2]/(r-2)!;
T[n_, r_] := A[r, n-r+2];
Table[T[n, r], {n, 1, 10}, {r, 2, n+1}] (* Jean-François Alcover, Jun 28 2019 *)

A = lambda r, n: rising_factorial(n,r)/factorial(r) - rising_factorial(n,r-2)/factorial(r-2)
for r in (2..10): [A(r,n) for n in (1..10)] # Peter Luschny, Dec 13 2015

T(n, r) = A(r, n-r+2) with the array A(r, n) = risefac(n,r)/r! - risefac(n,r-2)/(r-2)! where the rising factorial risefac(n,k) = Product_{j=0..k-1} (n+j) and risefac(n,0) = 1.
From Peter Luschny, Dec 14 2015: (Start)
A(n+2, n+1) = A007946(n-1) = CatalanNumber(n)*3*n*(n+1)/(n+2) for n>=0.
A(n+2, n+2) = A024482(n+2) = A097613(n+2) = CatalanNumber(n+1)*(3*n+4)/2 for n>=0.
A(n+2, n+3) = A051960(n+1) = CatalanNumber(n+1)*(3*n+5) for n>=0.
A(n+2, n+4) = A029651(n+2) = CatalanNumber(n+1)*(6*n+9) for n>=0.
A(n+2, n+5) = A051924(n+3) = CatalanNumber(n+2)*(3*n+7) for n>=0.
A(n+2, n+6) = A129869(n+4) = CatalanNumber(n+2)*(3*n+8)*(2*n+5)/(n+4) for n>=0.
A(n+2, n+7) = A220101(n+4) = CatalanNumber(n+3)*(3*(n+3)^2)/(n+5) for n>=0.
A(n+2, n+8) = CatalanNumber(n+4)*(n+3)*(3*n+10)/(2*n+12) for n>=0.
Let for n>=0 and k>=0 diag(n,k) = A(k+2,n+k+1) and G(n,k) = 2^(k+2*n)*Gamma((3-(-1)^k+2*k+4*n)/4)/(sqrt(Pi)*Gamma(k+n+0^k)) then
diag(n,0) = G(n,0)*(n*3)/(n+2),
diag(n,1) = G(n,1)*(3*n+4)/((n+1)*(n+2)),
diag(n,2) = G(n,2)*(3*n+5)/(n+2),
diag(n,3) = G(n,3)*3,
diag(n,4) = G(n,4)*(3*n+7),
diag(n,5) = G(n,5)*(3*n+8),
diag(n,6) = G(n,6)*3*(3+n)^2,
diag(n,7) = G(n,7)*(3+n)*(10+3*n). (End)

A265613 a(n) = CatalanNumber(n+1)n(3n^2+5n+2)/((4+n)*(3+n)).

Original entry on oeis.org

0, 1, 8, 44, 210, 935, 4004, 16744, 68952, 281010, 1136960, 4576264, 18349630, 73370115, 292746300, 1166182800, 4639918800, 18443677230, 73261092240, 290845019400, 1154169552900, 4578702310182, 18159992594568, 72014135814704, 285542883894800, 1132125641947300

Offset: 0

Peter Luschny, Dec 15 2015

nonn

This is row n=8 in the array A(n,k) = (rf(k+n-2,k-1)-(k-1)*(k-2)*rf(k+n-2, k-3))/ (k-1)! if n>=3 and A(n,0)=0, A(n,1)=1, A(n,2)=n; rf(n,k) denotes the rising factorial. See the cross-references for other values of n and the table in A264357.

Robert Israel, Table of n, a(n) for n = 0..1630

Cf. A000108, A007946, A097613, A051960, A029651, A051924, A129869, A220101, A264357, A265612.

A265613 := n -> (4*4^n*n*(n+1)*(3*n+2)*GAMMA(n+3/2))/(sqrt(Pi)*GAMMA(n+5)):
seq(simplify(A265613(n)), n=0..25);

Table[SeriesCoefficient[I (14 x^2 + I Sqrt[4 x - 1] (4 x^2 - 7 x + 2) - 11 x + 2 (1 - x^3))/(2 x^4 Sqrt[4 x - 1]), {x, 0, n}], {n, 0, 25}]
(* or *)
Table[(4^(n + 1) n (n + 1) (3 n + 2) Gamma[n + 3/2])/(Sqrt[Pi] Gamma[n + 5]), {n, 0, 25}] (* or *)
Table[CatalanNumber(n+1) n (3 n^2 + 5 n + 2)/((4 + n) (3 + n)), {n, 0, 25}] (* Michael De Vlieger, Dec 15 2015 *)

a = lambda n: catalan_number(n+1)*n*(3*n^2+5*n+2)/((4+n)*(3+n))
[a(n) for n in range(26)]

G.f.: I*(14*x^2+I*sqrt(4*x-1)*(4*x^2-7*x+2)-11*x+2*(1-x^3))/(2*x^4*sqrt(4*x-1)).
a(n) = (4^(n+1)*n*(n+1)*(3*n+2)*Gamma(n+3/2))/(sqrt(Pi)*Gamma(n+5)).
a(n) = (rf(n+6, n-1)-(n-1)*(n-2)*rf(n+6, n-3))/(n-1)! for n>=3, rf(n,k) the rising factorial.
a(n) = a(n-1)*((2*(n+1))*(3*n+2)*(1+2*n)/((n-1)*(3*n-1)*(4+n))) for n>=2.
a(n) ~ 4^n*(12-(191/2)/n+(17595/32)/n^2-(705005/256)/n^3+(104705937/8192)/ n^4-...)/sqrt(n*Pi).
a(n) = [x^n] x*(1 + x)/(1 - x)^(n+5). - Ilya Gutkovskiy, Oct 09 2017

A259775 Stepped path in P(k,n) array of k-th partial sums of squares (A000290).

Original entry on oeis.org

1, 5, 6, 20, 27, 77, 112, 294, 450, 1122, 1782, 4290, 7007, 16445, 27456, 63206, 107406, 243542, 419900, 940576, 1641486, 3640210, 6418656, 14115100, 25110020, 54826020, 98285670, 213286590, 384942375

Offset: 1

Luciano Ancora, Jul 05 2015

The term "stepped path" in the name field is the same used in A001405.

Interleaving of terms of the sequences A220101 and A129869. - Michel Marcus, Jul 05 2015

			The array of k-th partial sums of squares begins:
[1], [5],  14,   30,    55,     91,  ...  A000330
1,   [6], [20],  50,   105,    196,  ...  A002415
1,    7,  [27], [77],  182,    378,  ...  A005585
1,    8,   35, [112], [294],   672,  ...  A040977
1,    9,   44,  156,  [450], [1122], ...  A050486
1,   10,   54,  210,   660,  [1782], ...  A053347
This is essentially A110813 without its first two columns.

Cf. A000330, A002415, A005585, A040977, A050486, A053347.

Table[DifferenceRoot[Function[{a, n}, {(-9168 - 14432*n - 8412*n^2 - 2152*n^3 - 204*n^4)*a[n] +(-1332 - 1902*n - 792*n^2 - 102*n^3)*a[1 + n] + (2100 + 3884*n + 2493*n^2 + 640*n^3 + 51*n^4)*a[2 + n] == 0, a[1] == 1 , a[2] == 5}]][n], {n, 29}]

Conjecture: -(n+5)*(13*n-11)*a(n) +(8*n^2+39*n-35)*a(n-1) +2*(26*n^2+48*n+25)*a(n-2) -4*(8*n+5)*(n-1)*a(n-3)=0. - R. J. Mathar, Jul 16 2015

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A051666 Rows of triangle formed using Pascal's rule except begin and end n-th row with n^2.

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A265612 a(n) = CatalanNumber(n+1)*n*(1+3*n)/(6+2*n).

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A264357 Array A(r, n) of number of independent components of a symmetric traceless tensor of rank r and dimension n, written as triangle T(n, r) = A(r, n-r+2), n >= 1, r = 2..n+1.

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A265613 a(n) = CatalanNumber(n+1)*n*(3*n^2+5*n+2)/((4+n)*(3+n)).

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A259775 Stepped path in P(k,n) array of k-th partial sums of squares (A000290).

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A265612 a(n) = CatalanNumber(n+1)n(1+3n)/(6+2n).

A265613 a(n) = CatalanNumber(n+1)n(3n^2+5n+2)/((4+n)*(3+n)).