A001296 -id:A001296 - cp's OEIS frontend

A317017 Expansion of Product_{k>=1} 1/(1 - x^k)^((3k+1)binomial(k+2,3)/4).

1, 1, 8, 33, 126, 441, 1571, 5338, 17900, 58359, 187134, 588966, 1826537, 5580784, 16831549, 50135506, 147650112, 430187724, 1240908651, 3545808444, 10042128414, 28201458999, 78567720054, 217225969695, 596254164090, 1625343030654, 4401332943214, 11843216471115, 31674767502610

Offset: 0

Ilya Gutkovskiy, Jul 19 2018

nonn

Euler transform of A001296.

N. J. A. Sloane, Transforms

Cf. A000391, A001296, A278768, A279216, A305653, A317019, A317020, A317021.

a:=series(mul(1/(1-x^k)^((3*k+1)*binomial(k+2,3)/4),k=1..100),x=0,29): seq(coeff(a,x,n),n=0..28); # Paolo P. Lava, Apr 02 2019

nmax = 28; CoefficientList[Series[Product[1/(1 - x^k)^((3 k + 1) Binomial[k + 2, 3]/4), {k, 1, nmax}], {x, 0, nmax}], x]
nmax = 28; CoefficientList[Series[Exp[Sum[x^k (1 + 2 x^k)/(k (1 - x^k)^5), {k, 1, nmax}]], {x, 0, nmax}], x]
a[n_] := a[n] = If[n == 0, 1, Sum[Sum[d^2 (d + 1) (d + 2) (3 d + 1)/24, {d, Divisors[k]}] a[n - k], {k, 1, n}]/n]; Table[a[n], {n, 0, 28}]

G.f.: Product_{k>=1} 1/(1 - x^k)^A001296(k).
G.f.: exp(Sum_{k>=1} x^k*(1 + 2*x^k)/(k*(1 - x^k)^5)).
a(n) ~ Pi^(1/288)/(2 * 3^(577/864) * 7^(145/1728) * n^(1009/1728)) * exp(1/144 - (1/12-Zeta'(-1))/12 - (11 * Zeta(3))/(80 * Pi^2) + (1383 * Zeta(5))/(640 * Pi^4) + (11025 * Zeta(3) * Zeta(5)^2)/(2 * Pi^12) - (694575 * Zeta(5)^3)/(2 * Pi^14) + (13127467500 * Zeta(5)^5)/Pi^24 + (5 * Zeta'(-3))/12 + ((-21 * 3^(1/3) * 7^(1/6) * Pi)/6400 - (35 * 3^(1/3) * 7^(1/6) * Zeta(3) * Zeta(5))/(2 * Pi^7) + (15435 * 3^(1/3) * 7^(1/6) * Zeta(5)^2)/(16 * Pi^9) - (175573125 * 3^(1/3) * 7^(1/6) * Zeta(5)^4)/(4 * Pi^19)) * n^(1/6) + (((7/3)^(1/3) * Zeta(3))/(4 * Pi^2) - (21 * 3^(2/3) * 7^(1/3) * Zeta(5))/(8 * Pi^4) + (147000 * 3^(2/3) * 7^(1/3) * Zeta(5)^3)/Pi^14) * n^(1/3) + ((sqrt(7) * Pi)/40 - (1575 * sqrt(7) * Zeta(5)^2)/Pi^9) * sqrt(n) + ((15 * 3^(1/3) * 7^(2/3) * Zeta(5))/(2 * Pi^4)) * n^(2/3) + ((2 * 3^(2/3) * Pi)/(5 * 7^(1/6))) * n^(5/6)). - Vaclav Kotesovec, Jul 28 2018

A321450 Numbers that are sums of consecutive pentagonal pyramidal numbers (A002411).

Original entry on oeis.org

0, 1, 6, 7, 18, 24, 25, 40, 58, 64, 65, 75, 115, 126, 133, 139, 140, 196, 201, 241, 259, 265, 266, 288, 322, 397, 405, 437, 455, 461, 462, 484, 550, 610, 685, 693, 725, 726, 743, 749, 750, 889, 936, 955, 1015, 1090, 1130, 1148, 1154, 1155, 1183, 1243, 1276, 1439

Offset: 1

Ilya Gutkovskiy, Dec 21 2018

nonn

Eric Weisstein's World of Mathematics, Pentagonal Pyramidal Number

Cf. A001296, A002411, A319184, A322468, A322479, A322638, A322651, A322652, A322653.

A365668 G.f. A(x) satisfies: A(x) = x * (1 + A(x))^5 / (1 - 2 * A(x)).

Original entry on oeis.org

0, 1, 7, 73, 905, 12354, 179305, 2715192, 42414021, 678476755, 11058588574, 182999237590, 3066447596459, 51926183715280, 887204891847960, 15276037569668880, 264797324173666845, 4617195655522976361, 80930337327794271445, 1425171253004955494215, 25202145191953299213490

Offset: 0

Ilya Gutkovskiy, Sep 26 2023

nonn

Reversion of g.f. for 4-dimensional figurate numbers A001296 (with signs).

Eric Weisstein's World of Mathematics, Series Reversion

Cf. A001296, A002294, A064063, A365755, A366014, A366035, A366036, A366037.

nmax = 20; A[] = 0; Do[A[x] = x (1 + A[x])^5/(1 - 2 A[x]) + O[x]^(nmax + 1) // Normal, nmax + 1]; CoefficientList[A[x], x]
CoefficientList[InverseSeries[Series[x (1 - 2 x)/(1 + x)^5, {x, 0, 20}], x], x]	
Join[{0}, Table[1/n Sum[Binomial[n + k - 1, k] Binomial[5 n, n - k - 1] 2^k, {k, 0, n - 1}], {n, 1, 20}]]

a(n) = (1/n) * Sum_{k=0..n-1} binomial(n+k-1,k) * binomial(5*n,n-k-1) * 2^k for n > 0.

a(n) ~ sqrt(32 - 19*sqrt(5/2)) * 3^(4*n - 3/2) * 5^(3*n) / (sqrt(Pi) * n^(3/2) * 2^(2*n + 3/2) * (25 + 34*sqrt(10))^n). - Vaclav Kotesovec, Sep 27 2023

A373449 Number A(n,k) of (binary) heaps of length n whose element set is a subset of [k]; square array A(n,k), n>=0, k>=0, read by antidiagonals.

Original entry on oeis.org

1, 1, 0, 1, 1, 0, 1, 2, 1, 0, 1, 3, 3, 1, 0, 1, 4, 6, 5, 1, 0, 1, 5, 10, 14, 7, 1, 0, 1, 6, 15, 30, 25, 11, 1, 0, 1, 7, 21, 55, 65, 53, 16, 1, 0, 1, 8, 28, 91, 140, 173, 100, 26, 1, 0, 1, 9, 36, 140, 266, 448, 400, 222, 36, 1, 0, 1, 10, 45, 204, 462, 994, 1225, 1122, 386, 56, 1, 0

Offset: 0

Alois P. Heinz, Jun 05 2024

These heaps may contain repeated elements.

			A(3,1) = 1: 111.
A(3,2) = 5: 111, 211, 212, 221, 222.
A(3,3) = 14: 111, 211, 212, 221, 222, 311, 312, 313, 321, 322, 323, 331, 332, 333.
(The examples use max-heaps.)
Square array A(n,k) begins:
  1, 1,  1,   1,    1,     1,     1,     1,      1, ...
  0, 1,  2,   3,    4,     5,     6,     7,      8, ...
  0, 1,  3,   6,   10,    15,    21,    28,     36, ...
  0, 1,  5,  14,   30,    55,    91,   140,    204, ...
  0, 1,  7,  25,   65,   140,   266,   462,    750, ...
  0, 1, 11,  53,  173,   448,   994,  1974,   3606, ...
  0, 1, 16, 100,  400,  1225,  3136,  7056,  14400, ...
  0, 1, 26, 222, 1122,  4147, 12428, 32028,  73644, ...
  0, 1, 36, 386, 2336, 10036, 34242, 98922, 251922, ...

Alois P. Heinz, Antidiagonals n = 0..200, flattened
Eric Weisstein's World of Mathematics, Heap
Wikipedia, Binary heap

Columns k=0-2 give: A000007, A000012, A091980(n+1).
Rows n=0-6 give: A000012, A001477, A000217, A000330, A001296, A207361, A001249(k-1).
Main diagonal gives A373450.
Cf. A373451.

A:= proc(n, k) option remember; `if`(n=0, 1,
     (g-> (f-> add(A(f, j)*A(n-1-f, j), j=1..k)
             )(min(g-1, n-g/2)))(2^ilog2(n)))
    end:
seq(seq(A(n, d-n), n=0..d), d=0..12);

A[n_, k_] := A[n, k] = If[n == 0, 1,
   Function[g, Function[f, Sum[A[f, j]*A[n-1-f, j], {j, 1, k}]][
   Min[g-1, n-g/2]]][2^(Length[IntegerDigits[n, 2]]-1)]];
Table[Table[A[n, d-n], {n, 0, d}], {d, 0, 12}] // Flatten (* Jean-François Alcover, Jun 08 2024, after Alois P. Heinz *)

A(n,k) = Sum_{j=0..k} binomial(k,j) * A373451(n,k-j).

A104634 Triangle T(n,k) = (k-1-n)(k-2-n)(k+2*n)/6, 1<=k<=n.

Original entry on oeis.org

1, 5, 2, 14, 8, 3, 30, 20, 11, 4, 55, 40, 26, 14, 5, 91, 70, 50, 32, 17, 6, 140, 112, 85, 60, 38, 20, 7, 204, 168, 133, 100, 70, 44, 23, 8, 285, 240, 196, 154, 115, 80, 50, 26, 9, 385, 330, 276, 224, 175, 130, 90, 56, 29, 10, 506, 440, 375, 312, 252, 196, 145, 100, 62, 32, 11, 650, 572, 495, 420, 348, 280, 217, 160, 110, 68, 35, 12, 819, 728, 638, 550, 465, 384

Offset: 1

Gary W. Adamson, Mar 18 2005

			The first few rows are:
1;
5, 2;
14, 8, 3;
30, 20, 11, 4;
55, 40, 26, 14, 5;
91, 70, 50, 32, 17, 6;
...

G. C. Greubel, Rows n=1..100 of triangle, flattened

Cf. A000330 (column 1), A007290 (column 2), A051925 (column 3), A001296 (row sums), A104633, A000332.

[[(k-1-n)*(k-2-n)*(k+2*n)/6: k in [1..n]]: n in [1..20]]; // G. C. Greubel, Aug 12 2018

A104634 := proc(n,k) (k-1-n)*(k-2-n)*(k+2*n)/6 ; end proc:
seq(seq(A104634(n,k),k=1..n),n=1..15) ; # R. J. Mathar, Aug 31 2011

Table[(k-1-n)*(k-2-n)*(k+2*n)/6, {n, 1, 20}, {k, 1, n}] // Flatten (* G. C. Greubel, Aug 12 2018 *)

for(n=1,20, for(k=1,n, print1((k-1-n)*(k-2-n)*(k+2*n)/6, ", "))) \\ G. C. Greubel, Aug 12 2018

The triangle is created by the matrix product A002260 * A004736, both infinite lower triangular matrices.

Definition in closed form provided by R. J. Mathar, Aug 31 2011

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

A228317 The hyper-Wiener index of the triangular graph T(n) (n >= 1).

Original entry on oeis.org

0, 0, 3, 21, 75, 195, 420, 798, 1386, 2250, 3465, 5115, 7293, 10101, 13650, 18060, 23460, 29988, 37791, 47025, 57855, 70455, 85008, 101706, 120750, 142350, 166725, 194103, 224721, 258825, 296670, 338520, 384648, 435336, 490875, 551565, 617715, 689643

Offset: 1

Emeric Deutsch, Aug 26 2013

The triangular graph T(n) is the graph whose vertices represent the 2-subsets of {1,2,...,n} and two vertices are adjacent provided the corresponding 2-subsets have a nonempty intersection.

The triangular graph T(n) is a strongly regular graph with parameters n*(n-1)/2, 2*(n-2), n-2, and 4 (see the Brualdi and Ryser reference, Theorem 5.2.4).

R. A. Brualdi and H. J. Ryser, Combinatorial Matrix Theory, Cambridge Univ. Press, 1992.

G. G. Cash, Relationship between the Hosoya polynomial and the hyper-Wiener index, Applied Mathematics Letters, 15(7) (2002), 893-895.
Eric Weisstein's World of Mathematics, Triangular Graph.
Index entries for linear recurrences with constant coefficients, signature (5,-10,10,-5,1).

Cf. A001296, A006011.

a := proc (n) options operator, arrow: (1/8)*n*(n-1)*(n-2)*(3*n-5) end proc: seq(a(n), n = 1 .. 38);

LinearRecurrence[{5,-10,10,-5,1},{0,0,3,21,75},40] (* Harvey P. Dale, Feb 23 2023 *)

a(n) = n*(n - 1)*(n - 2)*(3*n - 5)/8.
G.f.: 3*x^3*(1 + 2*x)/(1 - x)^5.
The Hosoya-Wiener polynomial of T(n) is (1/8)*n*(n - 1)*(4 + 4*(n-2)*t + (n - 2)*(n - 3)*t^2).
a(n) = 3*A001296(n-2) for n >= 2. - R. J. Mathar, Mar 05 2017

A086689 a(n) = Sum_{i=1..n} i^2*t(i), where t = A000217.

Original entry on oeis.org

1, 13, 67, 227, 602, 1358, 2730, 5034, 8679, 14179, 22165, 33397, 48776, 69356, 96356, 131172, 175389, 230793, 299383, 383383, 485254, 607706, 753710, 926510, 1129635, 1366911, 1642473, 1960777, 2326612, 2745112

Offset: 1

Jon Perry, Jul 28 2003

This sequence is related to A001296 by a(n) = n*A001296(n) - Sum_{i=0..n-1} A001296(i) with n>0. - Bruno Berselli, Jan 21 2013

			a(4) = 227 = 1^2*A000217(1)+2^2*A000217(2)+3^2*A000217(3)+4^2*A000217(4).

Vincenzo Librandi, Table of n, a(n) for n = 1..1000
Index entries for linear recurrences with constant coefficients, signature (6,-15,20,-15,6,-1).

Cf. A001296.

[n*(n+1)*(n+2)*(12*n^2+9*n-1)/120 : n in [1..40]]; // Wesley Ivan Hurt, Nov 19 2014

A086689:=n->n*(n+1)*(n+2)*(12*n^2+9*n-1)/120: seq(A086689(n), n=1..40); # Wesley Ivan Hurt, Nov 19 2014

Table[n (n + 1) (n + 2) (12 n^2 + 9 n - 1)/120, {n, 40}] (* Wesley Ivan Hurt, Nov 19 2014 *)
CoefficientList[Series[(1 + 7 x + 4 x^2) / (x - 1)^6, {x, 0, 50}], x] (* Vincenzo Librandi, Nov 20 2014 *)

t(n)=n*(n+1)/2 for(i=1,30,print1(","sum(j=1,i,j^2*t(i))))

a(n) = n*(n+1)*(n+2)*(12*n^2+9*n-1)/120.
G.f.: x*(1+7*x+4*x^2) / (x-1)^6. - R. J. Mathar, Sep 15 2012
a(n) = 6*a(n-1)-15*a(n-2)+20*a(n-3)-15*a(n-4)+6*a(n-5)-a(n-6). - Wesley Ivan Hurt, Nov 19 2014
a(n) = Sum_{i=1..n} ( i*Sum_{k=1..i} i*k ). - Wesley Ivan Hurt, Nov 19 2014

A095729 A002260 squared, as an infinite lower triangular matrix, read by rows.

Original entry on oeis.org

1, 3, 4, 6, 10, 9, 10, 18, 21, 16, 15, 28, 36, 36, 25, 21, 40, 54, 60, 55, 36, 28, 54, 75, 88, 90, 78, 49, 36, 70, 99, 120, 130, 126, 105, 64, 45, 88, 126, 156, 175, 180, 168, 136, 81, 55, 108, 156, 196, 225, 240, 238, 216, 171, 100, 66, 130, 189, 240, 280, 306, 315, 304

Offset: 1

Gary W. Adamson, Jun 05 2004, Feb 17 2007

Sum of terms in n-th row = A001296(n-1).
By columns, k; even columns sequences as f(x), x = 1, 2, 3...; = (k/2)x^2 + (k^2 - k/2)x. For example, terms in row 2, (A028552): 4, 10, 18, 28, 40...= x^2 + 3x; row 4 = 2x^2 + 14x, row 6 = 3x^2 + 33x, row 8 = 4x^2 + 60x...etc.
The number in the i-th row and j-th column (j<=i) of the squared matrix is j*(binomial[i + 1, 2] - binomial[j, 2]). - Keith Schneider (schneidk(AT)email.unc.edu), Jul 23 2007

			First few rows of the triangle are
  1;
  3, 4;
  6, 10, 9;
  10, 18, 21, 16;
  15, 28, 36, 36, 25;
  21, 40, 54, 60, 55, 36,
  ...
[1 0 0 / 1 2 0 / 1 2 3]^2 = [1 0 0 / 3 4 0 / 6 10 9].
Next higher order matrix generates rows of the one lower order, plus the next row.
For example, the 4 X 4 matrix [1 0 0 0 / 1 2 0 0 / 1 2 3 0 / 1 2 3 4]^2 = [1 0 0 0 / 3 4 0 0 / 6 10 9 0 / 10 18 21 16].

Cf. A001296, A028552, A002260.

FindRow[n_] := Module[{i = 0}, While[Binomial[i, 2] < n, i++ ]; i - 1]; FindCol[n_] := n - Binomial[FindRow[n], 2]; A095729[n_] := FindCol[n](Binomial[FindRow[n]+1, 2] - Binomial[FindCol[n], 2]); Table[A095729[i], {i, 1, 91}] (* Keith Schneider (schneidk(AT)email.unc.edu), Jul 23 2007 *)

More terms from Keith Schneider (schneidk(AT)email.unc.edu), Jul 23 2007

Edited by N. J. A. Sloane, Jul 03 2008 at the suggestion of R. J. Mathar

A104727 Triangle T(n,k) = (k-1-n)(k-2-n)(k^2+k+2kn+3n^2+5n)/24 read by rows, 1<=k<=n.

Original entry on oeis.org

1, 7, 3, 25, 15, 6, 65, 45, 26, 10, 140, 105, 71, 40, 15, 266, 210, 155, 103, 57, 21, 462, 378, 295, 215, 141, 77, 28, 750, 630, 511, 395, 285, 185, 100, 36, 1155, 990, 826, 665, 510, 365, 235, 126, 45, 1705, 1485, 1266, 1050, 840, 640, 455, 291, 155, 55, 2431, 2145, 1860, 1578, 1302, 1036, 785, 555, 353, 187, 66, 3367, 3003

Offset: 1

Gary W. Adamson, Mar 20 2005

The triangle is created by multiplying the lower triangular matrix A(n,k) = A000217(k) (1<=k<=n) by the lower triangular matrix B(n,k) = n-k+1 (1<=k<=n): T(n,k) = sum_{j=k..n} A(n,j)*B(j,k).

The commuted product B * A generates triangle A098358.

			First few rows of the triangle are:
1;
7, 3;
25, 15, 6;
665, 45, 26, 10;
140, 105, 71, 40, 15;
266, 210, 155, 103, 57, 21;
...

Cf. A098358, A104727, A024166 (row sums).

T(n,1) = A001296(n). - R. J. Mathar, Oct 29 2011

cp's OEIS Frontend

A317017 Expansion of Product_{k>=1} 1/(1 - x^k)^((3*k+1)*binomial(k+2,3)/4).

Views

Author

Keywords

Comments

Links

Crossrefs

Programs

Maple

Mathematica

Formula

A321450 Numbers that are sums of consecutive pentagonal pyramidal numbers (A002411).

Views

Author

Keywords

Links

Crossrefs

A365668 G.f. A(x) satisfies: A(x) = x * (1 + A(x))^5 / (1 - 2 * A(x)).

Views

Author

Keywords

Comments

Links

Crossrefs

Programs

Mathematica

Formula

A373449 Number A(n,k) of (binary) heaps of length n whose element set is a subset of [k]; square array A(n,k), n>=0, k>=0, read by antidiagonals.

Views

Author

Keywords

Comments

Examples

Links

Crossrefs

Programs

Maple

Mathematica

Formula

A104634 Triangle T(n,k) = (k-1-n)*(k-2-n)*(k+2*n)/6, 1<=k<=n.

Views

Author

Keywords

Examples

Links

Crossrefs

Programs

Magma

Maple

Mathematica

PARI

Formula

Extensions

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

Views

Author

Keywords

Comments

Examples

Crossrefs

Formula

Extensions

A228317 The hyper-Wiener index of the triangular graph T(n) (n >= 1).

Views

Author

Keywords

Comments

References

Links

Crossrefs

Programs

Maple

Mathematica

Formula

A086689 a(n) = Sum_{i=1..n} i^2*t(i), where t = A000217.

Views

Author

Keywords

Comments

A317017 Expansion of Product_{k>=1} 1/(1 - x^k)^((3k+1)binomial(k+2,3)/4).

A104634 Triangle T(n,k) = (k-1-n)(k-2-n)(k+2*n)/6, 1<=k<=n.

A104727 Triangle T(n,k) = (k-1-n)(k-2-n)(k^2+k+2kn+3n^2+5n)/24 read by rows, 1<=k<=n.