A002411 -id:A002411 - cp's OEIS frontend

A164948 Fibonacci matrix read by antidiagonals. (Inverse of A136158.)

1, 1, -1, 3, -4, 1, 9, -15, 7, -1, 27, -54, 36, -10, 1, 81, -189, 162, -66, 13, -1, 243, -648, 675, -360, 105, -16, 1, 729, -2187, 2673, -1755, 675, -153, 19, -1, 2187, -7290, 10206, -7938, 3780, -1134, 210, -22, 1, 6561, -24057, 37908, -34020, 19278, -7182, 1764, -276, 25, -1, 19683, -78732, 137781, -139968, 91854, -40824, 12474, -2592, 351, -28, 1

Offset: 0

Mark Dols, Sep 01 2009

Triangle, read by rows, given by [1,2,0,0,0,0,0,0,0,...] DELTA [-1,0,0,0,0,0,0,0,0,...] where DELTA is the operator defined in A084938. - Philippe Deléham, Sep 02 2009

			As triangle:
    1;
    1,   -1;
    3,   -4,    1;
    9,  -15,    7,   -1;
   27,  -54,   36,  -10,    1;
   81, -189,  162,  -66,   13,   -1;
  243, -648,  675, -360,  105,  -16,    1;

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

Cf. A000007, A000326, A001296, A001519, A002411, A003688, A006234.
Cf. A016777, A038763, A080420, A080421, A081294, A084938, A098399.
Cf. A133494, A136158, A164942.

A164948:= func< n,k | n eq 0 select 1 else (-1)^k*3^(n-k-1)*(n+2*k)*Binomial(n,k)/n >;
[A164948(n, k): k in [0..n], n in [0..12]]; // G. C. Greubel, Dec 26 2023

A164948[n_,k_]:= If[n==0,1,(-1)^k*3^(n-k-1)*(n+2*k)*Binomial[n,k]/n];
Table[A164948[n,k], {n,0,12}, {k,0,n}]//Flatten (* G. C. Greubel, Dec 26 2023 *)

def A164948(n,k): return 1 if (n==0) else (-1)^k*3^(n-k-1)*((n+2*k)/n)*binomial(n, k)
flatten([[A164948(n, k) for k in range(n+1)] for n in range(13)]) # G. C. Greubel, Dec 26 2023

Sum_{k=0..n} T(n, k) = A000007(n).
Sum_{k=0..floor(n/2)} T(n-k, k) = A001519(n).
From Philippe Deléham, Oct 09 2011: (Start)
T(n,k) = 3*T(n-1,k) - T(n-1,k-1) with T(0,0)=1, T(1,0)=1, T(1,1)=-1.
Row n: Expansion of (1-x)*(3-x)^(n-1), n>0. (End)
G.f.: (1-2*x)/(1-3*x+x*y). - R. J. Mathar, Aug 12 2015
From G. C. Greubel, Dec 26 2023: (Start)
T(n, k) = (-1)^k * A136158(n, k).
T(n, k) = (-1)^k*3^(n-k-1)*((n+2*k)/n)*binomial(n, k), for n > 0, with T(0, 0) = 1.
T(n, 0) = A133494(n).
T(n, 1) = -A006234(n+2), n >= 1.
T(n, 2) = A080420(n-2), n >= 2.
T(n, 3) = -A080421(n-3), n >= 3.
T(2*n, n) = 4*(-1)^n*A098399(n-1) - (1/3)*[n=0].
T(n, n-4) = 27*(-1)^n*A001296(n-3), n >= 4.
T(n, n-3) = 9*(-1)^(n-1)*A002411(n-2), n >= 3.
T(n, n-2) = 3*(-1)^n*A000326(n-1) = (-1)^n*A062741(n-1), n >= 2.
T(n, n-1) = (-1)^(n-1)*A016777(n-1), n >= 1.
T(n, n) = (-1)^n.
Sum_{k=0..n} (-1)^k*T(n, k) = A081294(n).
Sum_{k=0..n} (-1)^k*T(n-k, k) = A003688(n). (End)

More terms from Philippe Deléham, Oct 09 2011

A185732 Accumulation array of the polygonal number array (A086270), by antidiagonals.

Original entry on oeis.org

1, 4, 2, 10, 9, 3, 20, 24, 15, 4, 35, 50, 42, 22, 5, 56, 90, 90, 64, 30, 6, 84, 147, 165, 140, 90, 39, 7, 120, 224, 273, 260, 200, 120, 49, 8, 165, 324, 420, 434, 375, 270, 154, 60, 9, 220, 450, 612, 672, 630, 510, 350, 192, 72, 10, 286, 605, 855, 984, 980, 861, 665, 440, 234, 85, 11, 364, 792, 1155, 1380, 1440, 1344, 1127, 840, 540, 280, 99, 12, 455, 1014, 1518, 1870, 2025, 1980, 1764, 1428, 1035, 650, 330, 114, 13, 560, 1274, 1950, 2464, 2750, 2790, 2604, 2240

Offset: 1

Clark Kimberling, Feb 01 2011

This is the (first) accumulation array of A086270; the second is A185733. See A144112 for the definition of accumulation array.

			Northwest corner:
1....4....10...20...35
2....9....24...50...90
3....15...42...90...165
4....22...64...140..260
5....30...90...200..375

G. C. Greubel, Table of n, a(n) for the first 50 rows, flattened

Rows 1 to 5: A000292, A006002, A059270, A177814, 5*A002411.
Columns 1 to 4: A000027, A055999, A067728, 10*A000096.
Cf. A144112, A086270, A185732.

f[n_,k_]:=k+n*k(k-1)/2;
TableForm[Table[f[n,k],{n,1,10},{k,1,15}]]  (* Array A086270 *)
Table[f[n-k+1,k],{n,14},{k,n,1,-1}]//Flatten  (* A086270 *)
s[n_,k_]:=Sum[f[i,j],{i,1,n},{j,1,k}]; (* acc. arr. of {f(n,k)} *)
Factor[s[n,k]]  (* formula for A185732 *)
TableForm[Table[s[n,k],{n,1,10},{k,1,15}]] (* acc. arr. A185732 *)
Table[s[n-k+1,k],{n,14},{k,n,1,-1}]//Flatten (* A185732 *)

T(n,k) = k*(k+1)*n*(n+1)*(k*n-n+k+5)/12.

A211791 a(n) = Sum_{y=1..n} Sum_{x=1..n} floor((x^k + y^k)^(1/k)) with k = 2.

Original entry on oeis.org

1, 7, 23, 54, 103, 175, 276, 409, 579, 791, 1050, 1360, 1724, 2149, 2640, 3198, 3832, 4543, 5337, 6217, 7192, 8265, 9437, 10716, 12103, 13609, 15231, 16978, 18857, 20869, 23018, 25307, 27745, 30337, 33084, 35992, 39066, 42309, 45728

Offset: 1

Clark Kimberling, Apr 26 2012

nonn

Row 2 of A211798.

			For a(3) we get the floor() values (1+2+3) + (2+2+3) + (3+3+4) = 23.

Cf. A211792, A211798.

f[x_, y_, k_] := f[x, y, k] = Floor[(x^k + y^k)^(1/k)]
t[k_, n_] := Sum[Sum[f[x, y, k], {x, 1, n}], {y, 1, n}]
Table[t[1, n], {n, 1, 45}]  (* 2*A002411 *)
Table[t[2, n], {n, 1, 45}]  (* A211791 *)
Table[t[3, n], {n, 1, 45}]  (* A211792 *)
TableForm[Table[t[k, n], {k, 1, 12},
                 {n, 1, 16}]] (* A211798 *)
Flatten[Table[t[k, n - k + 1], {n, 1, 12}, {k, 1, n}]]

a(n) = Sum_{y=1..n} Sum_{x=1..n} floor(sqrt(x^2 + y^2)).

Definition corrected by Georg Fischer, Sep 10 2022

A211792 a(n) = Sum_{y=1..n} Sum_{x=1..n} floor((x^k + y^k)^(1/k)) with k = 3.

Original entry on oeis.org

1, 7, 22, 51, 97, 164, 258, 382, 541, 741, 982, 1271, 1611, 2008, 2466, 2986, 3577, 4241, 4982, 5807, 6715, 7714, 8808, 10000, 11297, 12701, 14217, 15848, 17600, 19477, 21482, 23620, 25895, 28313, 30879, 33592, 36460, 39487, 42678, 46036

Offset: 1

Clark Kimberling, Apr 26 2012

nonn

Row 3 of A211798.

			For a(3) we get the floor() values (1+2+3) + (2+2+3) + (3+3+3) = 22.

David A. Corneth, Table of n, a(n) for n = 1..10000

Cf. A211791, A211798.

f[x_, y_, k_] := f[x, y, k] = Floor[(x^k + y^k)^(1/k)]
t[k_, n_] := Sum[Sum[f[x, y, k], {x, 1, n}], {y, 1, n}]
Table[t[1, n], {n, 1, 45}]  (* 2*A002411 *)
Table[t[2, n], {n, 1, 45}]  (* A211791 *)
Table[t[3, n], {n, 1, 45}]  (* A211792 *)
TableForm[Table[t[k, n], {k, 1, 12},
                 {n, 1, 16}]] (* A211798 *)
Flatten[Table[t[k, n - k + 1], {n, 1, 12}, {k, 1, n}]]

first(n) = { res = vector(n); res[1] = 1; for(i = 2, n, i3 = i^3; s = sum(j = 1, i-1, sqrtnint(i3 + j^3, 3)); res[i] = res[i-1] + sqrtnint(2*i3, 3) + 2*s; ); res } \\ David A. Corneth, Sep 12 2022

a(n) = Sum_{y=1..n} Sum_{x=1..n} floor((x^3 + y^3)^(1/3)).

a(n) = a(n-1) + floor((2*n^3)^(1/3)) + 2*Sum_{i = 1..n-1} floor((n^3 + i^3)^(1/3)) for n >= 2 and a(1) = 1. - David A. Corneth, Sep 12 2022

Definition changed by Georg Fischer, Sep 10 2022

A211798 R(k,n) = Sum_{y=1..n} Sum_{x=1..n} floor((x^k + y^k)^(1/k)), square array read by descending antidiagonals.

Original entry on oeis.org

2, 12, 1, 36, 7, 1, 80, 23, 7, 1, 150, 54, 22, 7, 1, 252, 103, 51, 22, 7, 1, 392, 175, 97, 50, 22, 7, 1, 576, 276, 164, 95, 50, 22, 7, 1, 810, 409, 258, 162, 95, 50, 22, 7, 1, 1100, 579, 382, 254, 161, 95, 50, 22, 7, 1, 1452, 791, 541, 375, 253, 161, 95, 50, 22

Offset: 1

Clark Kimberling, Apr 26 2012

			Northwest corner:
  2  12  36  80 150 252 392
  1   7  23  54 103 175 276
  1   7  22  51  97 164 258
  1   7  22  50  95 162 254
  1   7  22  50  95 161 254
  1   7  22  50  95 161 253

Cf. A002411 ((1/2) * row 1), A002412 (limiting row), A211791 (row 2), A211792 (row 3).

f[x_, y_, k_] := f[x, y, k] = Floor[(x^k + y^k)^(1/k)]
t[k_, n_] := Sum[Sum[f[x, y, k], {x, 1, n}], {y, 1, n}]
Table[t[1, n], {n, 1, 45}]  (* 2*A002411 *)
Table[t[2, n], {n, 1, 45}]  (* A211791 *)
Table[t[3, n], {n, 1, 45}]  (* A211792 *)
TableForm[Table[t[k, n], {k, 1, 12},
                 {n, 1, 16}]] (* A211798 *)
Flatten[Table[t[k, n - k + 1], {n, 1, 12}, {k, 1, n}]]

R(k,n) = Sum_{y=1..n} Sum_{x=1..n} floor((x^k + y^k)^(1/k)).

Definition changed by Georg Fischer, Sep 10 2022

A292474 Number of solutions to +-1 +- 5 +- 12 +- ... +- n(3n-1)/2 = 0.

Original entry on oeis.org

1, 0, 0, 0, 0, 0, 2, 2, 4, 0, 2, 4, 4, 0, 30, 46, 78, 0, 210, 366, 644, 0, 2032, 3696, 6694, 0, 21936, 39886, 73098, 0, 246172, 454074, 841714, 0, 2899542, 5401222, 10073398, 0, 35282910, 66213604, 124427582, 0, 441326270, 832775792, 1573861942, 0, 5642205488

Offset: 0

Seiichi Manyama, Sep 17 2017

nonn

			For n=6 the 2 solutions are +1+5-12+22+35-51 = 0 and -1-5+12-22-35+51 = 0.

Seiichi Manyama, Table of n, a(n) for n = 0..200

Cf. A000326, A002411, A158092, A158380, A292475.

{a(n) = polcoeff(prod(k=1, n, x^(k*(3*k-1)/2)+1/x^(k*(3*k-1)/2)), 0)}

Constant term in the expansion of Product_{k=1..n} (x^(k*(3*k-1)/2)+1/x^(k*(3*k-1)/2)).

a(4*k+1) = 0 for k >= 0.

A008670 Molien series for Weyl group F_4.

Original entry on oeis.org

1, 1, 1, 2, 3, 3, 5, 6, 7, 9, 11, 12, 16, 18, 20, 24, 28, 30, 36, 40, 44, 50, 56, 60, 69, 75, 81, 90, 99, 105, 117, 126, 135, 147, 159, 168, 184, 196, 208, 224, 240, 252, 272, 288, 304, 324, 344, 360, 385, 405, 425, 450, 475, 495, 525, 550, 575, 605, 635, 660, 696, 726, 756

Offset: 0

N. J. A. Sloane

Number of partitions of n into parts 1, 3, 4 and 6. - Ilya Gutkovskiy, May 24 2017

Coxeter and Moser, Generators and Relations for Discrete Groups, Table 10.
L. Smith, Polynomial Invariants of Finite Groups, Peters, 1995, p. 199 (No. 28).

Alois P. Heinz, Table of n, a(n) for n = 0..1000
INRIA Algorithms Project, Encyclopedia of Combinatorial Structures 236
G. Nebe, E. M. Rains and N. J. A. Sloane, Self-Dual Codes and Invariant Theory, Springer, Berlin, 2006.
Index entries for Molien series
Index entries for linear recurrences with constant coefficients, signature (1,0,1,0,-1,1,-2,1,-1,0,1,0,1,-1).

Cf. A002411, A006002, A010875, A011934, A027480, A055232, A182260.

MolienSeries(CoxeterGroup("F4")); // Sergei Haller (sergei(AT)sergei-haller.de), Dec 21 2006

R:=PowerSeriesRing(Integers(), 70); Coefficients(R!( 1/((1-x)*(1-x^3)*(1-x^4)*(1-x^6)) )); // G. C. Greubel, Sep 08 2019

a:= proc(n) local m, r; m := iquo (n, 12, 'r'); r:= r+1; ([4, 5, 6, 8, 10, 11, 14, 16, 18, 21, 24, 26][r]+ (6+r+4*m)*m)*m+ [1$3, 2, 3$2, 5, 6, 7, 9, 11, 12][r] end: seq(a(n), n=0..100); # Alois P. Heinz, Oct 06 2008

Take[CoefficientList[Series[1/((1-x^2)(1-x^6)(1-x^8)(1-x^12)),{x,0,130}], x], {1,-1,2}] (* or *) LinearRecurrence[ {1,0,1,0,-1,1,-2,1,-1,0,1,0,1,-1},{1,1,1,2,3,3,5,6,7,9,11,12,16,18},70] (* Harvey P. Dale, Feb 07 2012 *)

my(x='x+O('x^70)); Vec(1/((1-x)*(1-x^3)*(1-x^4)*(1-x^6))) \\ G. C. Greubel, Sep 08 2019

def A008670_list(prec):
    P. = PowerSeriesRing(ZZ, prec)
    return P(1/((1-x)*(1-x^3)*(1-x^4)*(1-x^6))).list()
A008670_list(70) # G. C. Greubel, Sep 08 2019

G.f.: 1/((1-x)*(1-x^3)*(1-x^4)*(1-x^6)). [Corrected by Ralf Stephan, Apr 29 2014]
a(n) = a(n-1) + a(n-3) - a(n-5) + a(n-6) - 2*a(n-7) + a(n-8) - a(n-9) + a(n-11) + a(n-13) - a(n-14), with a(0)=1, a(1)=1, a(2)=1, a(3)=2, a(4)=3, a(5)=3, a(6)=5, a(7)=6, a(8)=7, a(9)=9, a(10)=11, a(11)=12, a(12)=16, a(13)=18. - Harvey P. Dale, Feb 07 2012
a(n) ~ (1/432)*n^3. - Ralf Stephan, Apr 29 2014
a(n) = (120*floor(n/6)^3 + 60*(m+7)*floor(n/6)^2 + 2*(m^5-15*m^4+75*m^3-135*m^2+134*m+240)*floor(n/6) + 3*(m^5-15*m^4+75*m^3-135*m^2+84*m+70) + (m^5-15*m^4+75*m^3-135*m^2+44*m+30)*(-1)^floor(n/6))/240 where m = (n mod 6). - Luce ETIENNE, Aug 14 2018
a(n) = 1 + floor((2*n^3 + 42*n^2 + n*(279 + 9*(-1)^n - 48*[(n mod 3)=2]))/864) where [] is the Iverson bracket. - Hoang Xuan Thanh, Jun 22 2025

A085789 Partial sums of n 3-spaced triangular numbers beginning with t(2), e.g., a(2) = t(2) + t(5) = 3 + 15 = 18.

Original entry on oeis.org

3, 18, 54, 120, 225, 378, 588, 864, 1215, 1650, 2178, 2808, 3549, 4410, 5400, 6528, 7803, 9234, 10830, 12600, 14553, 16698, 19044, 21600, 24375, 27378, 30618, 34104, 37845, 41850, 46128, 50688, 55539, 60690, 66150, 71928, 78033, 84474, 91260, 98400, 105903

Offset: 1

Jon Perry, Jul 23 2003

Sums of rows of triangle A100345 (n>0).

Vincenzo Librandi, Table of n, a(n) for n = 1..1000
Index entries for linear recurrences with constant coefficients, signature (4,-6,4,-1).

Cf. A000217, A002411, A004188, A100345.

[3/2*n^2*(n+1): n in [1..40]]; // Vincenzo Librandi, Aug 14 2017

CoefficientList[Series[3 (1 + 2 x) / (1 - x)^4, {x, 0, 40}], x](* Vincenzo Librandi, Aug 14 2017 *)
LinearRecurrence[{4,-6,4,-1},{3,18,54,120},50] (* Harvey P. Dale, May 14 2023 *)

a(n) = 3/2 * n^2*(n+1).
a(n) = 3*n*binomial(n+1,2) = 3*n*A000217(n) = 3*A002411(n). - Arkadiusz Wesolowski, Feb 10 2012
G.f.: 3*(x + 2*x^2)/(1 - x)^4. - Arkadiusz Wesolowski, Feb 11 2012
From Amiram Eldar, Jun 29 2025: (Start)
Sum_{n>=1} 1/a(n) = Pi^2/9 - 2/3.
Sum_{n>=1} (-1)^(n+1)/a(n) = Pi^2/18 - 4*log(2)/3 + 2/3. (End)

More terms from Reinhard Zumkeller, Nov 18 2004

A132118 Triangle read by rows: T(n,k) = n(n-1)/2 + 2k - 1.

Original entry on oeis.org

1, 2, 4, 4, 6, 8, 7, 9, 11, 13, 11, 13, 15, 17, 19, 16, 18, 20, 22, 24, 26, 22, 24, 26, 28, 30, 32, 34, 29, 31, 33, 35, 37, 39, 41, 43, 37, 39, 41, 43, 45, 47, 49, 51, 53, 46, 48, 50, 52, 54, 56, 58, 60, 62, 64, 56, 58, 60, 62, 64, 66, 68, 70, 72, 74, 76

Offset: 1

Gary W. Adamson, Aug 10 2007

			First few rows of the triangle are:
   1;
   2,  4;
   4,  6,  8;
   7,  9, 11, 13;
  11, 13, 15, 17, 19;
  16, 18, 20, 22, 24, 26;
  22, 24, 26, 28, 30, 32, 34;
  29, 31, 33, 35, 37, 39, 41, 43;
  ...

Andrew Howroyd, Table of n, a(n) for n = 1..1275 (rows 1..50)

Column 1 is A000124(n-1).
Main diagonal is A034856.
Row sums are A002411.

Table[(n(n-1))/2+2k-1,{n,20},{k,n}]//Flatten (* Harvey P. Dale, Mar 26 2022 *)

a(16), a(17) corrected by Georg Fischer, Jul 01 2020

Name changed and terms a(56) and beyond from Andrew Howroyd, Apr 17 2021

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

cp's OEIS Frontend

A164948 Fibonacci matrix read by antidiagonals. (Inverse of A136158.)

Views

Author

Keywords

Comments

Examples

Links

Crossrefs

Programs

Magma

Mathematica

SageMath

Formula

Extensions

A185732 Accumulation array of the polygonal number array (A086270), by antidiagonals.

Views

Author

Keywords

Comments

Examples

Links

Crossrefs

Programs

Mathematica

Formula

A211791 a(n) = Sum_{y=1..n} Sum_{x=1..n} floor((x^k + y^k)^(1/k)) with k = 2.

Views

Author

Keywords

Comments

Examples

Crossrefs

Programs

Mathematica

Formula

Extensions

A211792 a(n) = Sum_{y=1..n} Sum_{x=1..n} floor((x^k + y^k)^(1/k)) with k = 3.

Views

Author

Keywords

Comments

Examples

Links

Crossrefs

Programs

Mathematica

PARI

Formula

Extensions

A211798 R(k,n) = Sum_{y=1..n} Sum_{x=1..n} floor((x^k + y^k)^(1/k)), square array read by descending antidiagonals.

Views

Author

Keywords

Examples

Crossrefs

Programs

Mathematica

Formula

Extensions

A292474 Number of solutions to +-1 +- 5 +- 12 +- ... +- n*(3*n-1)/2 = 0.

Views

Author

Keywords

Examples

Links

Crossrefs

Programs

PARI

Formula

A008670 Molien series for Weyl group F_4.

Views

Author

Keywords

Comments

References

Links

Crossrefs

Programs

Magma

A292474 Number of solutions to +-1 +- 5 +- 12 +- ... +- n(3n-1)/2 = 0.

A132118 Triangle read by rows: T(n,k) = n(n-1)/2 + 2k - 1.