A024166 -id:A024166 - cp's OEIS frontend

A087110 This table shows the coefficients of combinatorial formulas needed for generating the sequential sums of p-th powers of binomial coefficients C(n,6). The p-th row (p>=1) contains a(i,p) for i=1 to 6p-5, where a(i,p) satisfies Sum_{i=1..n} C(i+5,6)^p = 7 C(n+6,7) * Sum_{i=1..6p-5} a(i,p) C(n-1,i-1)/(i+6).

1, 1, 6, 15, 20, 15, 6, 1, 1, 48, 687, 4850, 20385, 55908, 104959, 137886, 127050, 80640, 33642, 8316, 924, 1, 342, 21267, 527876, 7020525, 58015362, 324610399, 1297791264, 3839203452, 8595153000, 14760228672, 19560928464, 19987430694

Offset: 1

André F. Labossière, Aug 11 2003

From Peter Bala, Mar 11 2018: (Start)

The table entries T(n,k) are the coefficients when expressing the polynomial C(x+6,6)^p of degree 6*p in terms of falling factorials: C(x+6,6)^p = Sum_{k = 0..6*p} T(p,k)*C(x,k). It follows that Sum_{i = 0..n-1} C(i+6,6)^p = Sum_{k = 0..6*p} T(p,k)*C(n,k+1). (End)

			Row 3 contains 1,48,687,...,924, so Sum_{i=1..n} C(i+5,6)^3 = 7 * C(n+6,7) * [ a(1,3)/7 + a(2,3)*C(n-1,1)/8 + a(3,3)*C(n-1,2)/9 + ... + a(13,3)*C(n-1,12)/19 ] = 7 * C(n+6,7) * [ 1/7 + 48*C(n-1,1)/8 + 687*C(n-1,2)/9 + ... + 924*C(n-1,12)/19 ]. Cf. A086028 for more details.

G. C. Greubel, Table of n, a(n) for the first 40 rows, flattened
Dukes, C. D. White, Web Matrices: Structural Properties and Generating Combinatorial Identities, arXiv:1603.01589 [math.CO], 2016.

Cf. A000292, A024166, A087127, A024166, A085438, A085439, A085440, A085441, A085442, A087107, A000332, A086020, A086021, A086022, A087108, A000389, A086023, A086024, A087109, A000579, A086025, A086026, A000580, A086027, A086028, A087111, A027555, A086029, A086030.

Cf. A087127, A237252.

seq(seq(add( (-1)^(k-i)*binomial(k, i)*binomial(i+6, 6)^n, i = 0..k), k = 0..6*n), n = 0..5); # Peter Bala, Mar 11 2018

a[i_, p_] := Sum[Binomial[i - 1, 2*k - 2]*Binomial[i - 2*k + 7, i - 2*k + 1]^(p - 1) - Binomial[i - 1, 2*k - 1]*Binomial[i - 2*k + 6, i - 2*k]^(p - 1), {k, 1, (2*i + 1 + (-1)^(i - 1))/4}]; Table[If[p == 1, 1, a[i, p]], {p, 1, 10}, {i, 1, 6*p - 5}]//Flatten (* G. C. Greubel, Nov 23 2017 *)

{a(i, p) = sum(k=1, (2*i + 1 + (-1)^(i - 1))/4, binomial(i - 1, 2*k - 2)*binomial(i - 2*k + 7, i - 2*k + 1)^(p - 1) - binomial(i - 1, 2*k - 1)*binomial(i - 2*k + 6, i - 2*k)^(p - 1))}; for(p=1,8, for(i=1, 6*p-5, print1(if(p==1,1,a(i,p)), ", "))) \\ G. C. Greubel, Nov 23 2017

a(i, p) = Sum_{k=1..[2*i+1+(-1)^(i-1)]/4} [ C(i-1, 2*k-2)*C(i-2*k+7, i-2*k+1)^(p-1) -C(i-1, 2*k-1)*C(i-2*k+6, i-2*k)^(p-1) ]
From Peter Bala, Mar 11 2018: (Start)
The following remarks assume the row and column indices start at 0.
T(n,k) = Sum_{i = 0..k} (-1)^(k-i)*binomial(k,i) * binomial(i+6,6)^n. Equivalently, let v_n denote the sequence (1, 7^n, 28^n, 84^n, ...) regarded as an infinite column vector, where 1, 7, 28, 84, ... is the sequence binomial(n+6,6) - see A000579. Then the n-th row of this table is determined by the matrix product P^(-1)*v_n, where P denotes Pascal's triangle A007318.
Recurrence: T(n+1,k) = Sum_{i = 0..6} C(6,i)*C(k+6-i,6)*T(n,k-i) with boundary conditions T(n,0) = 1 for all n and T(n,k) = 0 for k > 6*n.
n-th row polynomial R(n,x) = (1 + x)^6 o (1 + x)^6 o ... o (1 + x)^6 (n factors), where o denotes the black diamond product of power series defined in Dukes and White.
R(n+1,x) = 1/6!*(1 + x)^6 * (d/dx)^6(x^6*R(n,x)).
R(n,x) = Sum_{i >= 0} binomial(i+6,6)^n*x^i/(1 + x)^(i+1).
(1 - x)^(6*n)*R(n,x/(1 - x)) appears to equal the n-th row polynomial of A237252. (End)

Edited by Dean Hickerson, Aug 16 2003

A122193 Triangle T(n,k) of number of loopless multigraphs with n labeled edges and k labeled vertices and without isolated vertices, n >= 1; 2 <= k <= 2*n.

Original entry on oeis.org

1, 1, 6, 6, 1, 24, 114, 180, 90, 1, 78, 978, 4320, 8460, 7560, 2520, 1, 240, 6810, 63540, 271170, 604800, 730800, 453600, 113400, 1, 726, 43746, 774000, 6075900, 25424280, 61923960, 90720000, 78813000, 37422000, 7484400

Offset: 1

Vladeta Jovovic, Aug 24 2006

T(n,k) equals the number of arrangements on a line of n (nondegenerate) finite closed intervals having k distinct endpoints. See the 'IBM Ponder This' link. An example is given below. - Peter Bala, Jan 28 2018

T(n,k) equals the number of alignments of length k of n strings each of length 2. See Slowinski. Cf. A131689 (alignments of strings of length 1) and A299041 (alignments of strings of length 3). - Peter Bala, Feb 04 2018

			Triangle begins:
  1;
  1,  6,   6;
  1, 24, 114,  180,   90;
  1, 78, 978, 4320, 8460, 7560, 2520;
  ...
From _Francisco Santos_, Nov 17 2017: (Start)
For n=3 edges and k=4 vertices there are three loopless multigraphs without isolated vertices: a path, a Y-graph, and the multigraph {12, 34, 34}. The number of labelings in each is 3!4!/a, where a is the number of automorphisms. This gives respectively 3!4!/2 = 72, 3!4!/6 = 24 and 3!4!/8 = 18, adding up to 72 + 24 + 18 = 114. (End)
From _Peter Bala_, Jan 28 2018: (Start)
T(2,3) = 6: Consider 2 (nondegenerate) finite closed intervals [a, b] and [c, d]. There are 6 arrangements of these two intervals with 3 distinct endpoints:
  ...a--b--d....  a < b = c < d
  ...a...c--b...  a < c < b = d
  ...a--d...b...  a = c < d < b
  ...a--b...d...  a = c < b < d
  ...c...a--d...  c < a < b = d
  ...c--a--b....  c < a = d < b
T(2,4) = 6: There are 6 arrangements of the two intervals with 4 distinct endpoints:
  ...a--b...c--d.....  no intersection a < b < c < d
  ...a...c...b...d...  a < c < b < d
  ...a...c--d...b....  [c,d] is a proper subset of [a,b]
  ...c...a...d...b...  c < a < d < b
  ...c...a--b...d... [a,b] is a proper subset of [c,d]
  ...c--d...a--b.....  no intersection c < d < a < b.
Sums of powers of triangular numbers:
Row 2: Sum_{i = 2..n-1} C(i,2)^2 = C(n,3) + 6*C(n,4) + 6*C(n,5);
Row 3: Sum_{i = 2..n-1} C(i,2)^3 = C(n,3) + 24*C(n,4) + 114*C(n,5) + 180*C(n,6) + 90*C(n,7). See A024166 and A085438.
exp( Sum_{n >= 1} R(n,2)*x^n/n ) = (1 + x + 19*x^2 + 1147*x^3 + 145606*x^4 + 31784062*x^5 + ... )^4
exp( Sum_{n >= 1} R(n,3)*x^n/n ) = (1 + x + 37*x^2 + 4453*x^3 + 1126375*x^4 + 489185863*x^5 + ... )^9
exp( Sum_{n >= 1} R(n,4)*x^n/n ) = (1 + x + 61*x^2 + 12221*x^3 + 5144411*x^4 + 3715840571*x^5 + ... )^16 (End)
From _Peter Bala_, Feb 04 2018: (Start)
T(3,3) = 24 alignments of length 3 of 3 strings each of length 2. Examples include
  (i) A B -    (ii) A - B
      - C D         - C D
      - E F         E F -
There are 18 alignments of type (i) with two gap characters in one of the columns (3 ways of putting 2 gap characters in a column x 2 ways to place the other letter in the row which doesn't yet have a gap character x 3 columns: there are 6 alignments of type (ii) with a single gap character in each column (3 ways to put a single gap character in the first column x 2 ways to then place a single gap character in the second column). (End)

Michael De Vlieger, Table of n, a(n) for n = 1..10000 (rows 1 <= n <= 100).
Peter Bala, Deformations of the Hadamard product of power series
Peter Bala, Notes on A122193
S. Eger, On the Number of Many-to-Many Alignments of N Sequences, arXiv:1511.00622 [math.CO], 2015.
J. Engbers and C. Stocker, Two Combinatorial Proofs of Identities Involving Sums of Powers of Binomial Coefficients, Integers 16 (2016), #A58.
M. Dukes and C. D. White, Web Matrices: Structural Properties and Generating Combinatorial Identities, arXiv:1603.01589 [math.CO], 2016.
IBM Ponder This, Jan 01 2001
J. B. Slowinski, The Number of Multiple Alignments, Molecular Phylogenetics and Evolution 10:2 (1998), 264-266. doi:10.1006/mpev.1998.0522

Row sums give A055203.
Cf. A078739, A005799, A019538, A131689, A154283, A299041, A087127.
For Sum_{i = 2..n} C(i,2)^k see A024166 (k = 2), A085438 - A085442 ( k = 3 thru 7).

# Note that the function implements the full triangle because it can be
# much better reused and referenced in this form.
A122193 := (n,k) -> A078739(n,k)*k!/2^n:
# Displays the truncated triangle from the definition:
seq(print(seq(A122193(n,k),k=2..2*n)),n=1..6); # Peter Luschny, Mar 25 2011

t[n_, k_] := Sum[(-1)^(n - r) Binomial[n, r] StirlingS2[n + r, k], {r, 0, n}]; Table[t[n, k] k!/2^n, {n, 6}, {k, 2, 2 n}] // Flatten (* Michael De Vlieger, Nov 18 2017, after Jean-François Alcover at A078739 *)

Double e.g.f.: exp(-x)*Sum_{n>=0} exp(binomial(n,2)*y)*x^n/n!.
T(n,k) = S_{2,2}(n,k)*k!/2^n; S_{2,2} the generalized Stirling numbers A078739. - Peter Luschny, Mar 25 2011
From Peter Bala, Jan 28 2018: (Start)
T(n,k) = Sum_{i = 0..k} (-1)^(k-i)*binomial(k,i)*(i*(i-1)/2)^n.
T(n,k) = k*(k-1)/2*( T(n-1,k) + 2*T(n-1,k-1) + T(n-1,k-2) ) for 2 < k <= 2*n with boundary conditions T(n,2) = 1 for n >= 1 and T(n,k) = 0 if (k < 2) or (k > 2*n).
n-th row polynomial R(n,x) = Sum_{i >= 2} (i*(i-1)/2)^n * x^i/(1+x)^(i+1) for n >= 1.
1/(1-x)*R(n,x/(1-x)) = Sum_{i >= 2} (i*(i-1)/2)^n*x^i for n >= 1.
R(n,x) = 1/2^n*Sum_{k = 0..n} (-1)^(n-k)*binomial(n,k)*F(n+k,x), where F(n,x) = Sum_{k = 0..n} k!*Stirling2(n,k)*x^k is the n-th Fubini polynomial, the n-th row polynomial of A131689.
R(n,x) = x/(1+x)*1/2^n*Sum_{k = 0..n} binomial(n,k)*F(n+k,x) for n >= 1.
The polynomials Sum_{k = 2..2*n} T(n,k)*x^(k-2)*(1-x)^(2*n-k) are the row polynomials of A154283.
A154283 * A007318 equals the row reverse of this array.
Sum_{k = 2..2*n} T(n,k)*binomial(x,k) = ( binomial(x,2) )^n. Equivalently, Sum_{k = 2..2*n} (-1)^k*T(n,k)*binomial(x+k,k) = ( binomial(x+2,2) )^n. Cf. the Worpitzky-type identity Sum_{k = 1..n} A019538(n,k)*binomial(x,k) = x^n.
Sum_{i = 2..n-1} (i*(i-1)/2)^m = Sum_{k = 2..2*m} T(m,k) * binomial(n,k+1) for m >= 1. See Examples below.
R(n,x) = x^2 o x^2 o ... o x^2 (n factors), where o is the black diamond product of power series defined in Dukes and White. Note the polynomial x o x o ... o x (n factors) is the n-th row polynomial of A019538.
x^2*R(n,-1-x) = (1+x)^2*R(n,x) for n >= 1.
R(n+1,x) = 1/2*x^2*(d/dx)^2 ((1+x)^2*R(n,x)).
The zeros of R(n,x) belong to the interval [-1, 0].
Alternating row sums equal 1, that is R(n,-1) = 1.
R(n,-2) = 4*R(n,1) = 4*A055203(n).
4^n*Sum_{k = 2..2*n} T(n,k)*(-1/2)^k appears to equal (-1)^(n+1)*A005799(n) for n >= 1.
For k a nonzero integer, the power series A(k,x) := exp( Sum_{n >= 1} 1/k^2*R(n,k)*x^n/n ) appear to have integer coefficients. See the Example section.
Sum_{k = 2..2*n} T(n,k)*binomial(x,k-2) = binomial(x,2)^n - 2*binomial(x+1,2)^n + binomial(x+2,2)^n. These polynomials have their zeros on the vertical line Re x = -1/2 in the complex plane (the corresponding property also holds for the row polynomials of A019538 with a factor of x removed). (End)
From Peter Bala, Mar 08 2018: (Start)
n-th row polynomial R(n,x) = coefficient of (z_1)^2 * ... * (z_n)^2 in the expansion of the rational function 1/(1 + x - x*(1 + z_1)*...*(1 + z_n)).
The n-th row of the table is given by the matrix product P^(-1)*v_n, where P denotes Pascal's triangle A007318 and v_n is the sequence (0, 0, 1, 3^n, 6^n, 10^n, ...) regarded as an infinite column vector, where 1, 3, 6, 10, ... is the sequence of triangular numbers A000217. Cf. A087127. (End)

Definition corrected by Francisco Santos, Nov 17 2017

A101104 a(1)=1, a(2)=12, a(3)=23, and a(n)=24 for n>=4.

Original entry on oeis.org

1, 12, 23, 24, 24, 24, 24, 24, 24, 24, 24, 24, 24, 24, 24, 24, 24, 24, 24, 24, 24, 24, 24, 24, 24, 24, 24, 24, 24, 24, 24, 24, 24, 24, 24, 24, 24, 24, 24, 24, 24, 24, 24, 24, 24, 24, 24, 24, 24, 24, 24, 24, 24, 24, 24, 24, 24, 24, 24

Offset: 1

Cecilia Rossiter, Dec 15 2004

Original name: The first summation of row 4 of Euler's triangle - a row that will recursively accumulate to the power of 4.

D. J. Pengelley, The bridge between the continuous and the discrete via original sources in Study the Masters: The Abel-Fauvel Conference [pdf], Kristiansand, 2002, (ed. Otto Bekken et al), National Center for Mathematics Education, University of Gothenburg, Sweden, in press.
C. Rossiter, Depictions, Explorations and Formulas of the Euler/Pascal Cube [Dead link]
C. Rossiter, Depictions, Explorations and Formulas of the Euler/Pascal Cube [Cached copy, May 15 2013]
Eric Weisstein, Link to section of MathWorld: Worpitzky's Identity of 1883
Eric Weisstein, Link to section of MathWorld: Eulerian Number
Eric Weisstein, Link to section of MathWorld: Nexus number
Eric Weisstein, Link to section of MathWorld: Finite Differences
Index entries for linear recurrences with constant coefficients, signature (1).

For other sequences based upon MagicNKZ(n,k,z):
..... |  n = 1  |  n = 2  |  n = 3  |  n = 4  |  n = 5  |  n = 6  |  n = 7
---------------------------------------------------------------------------
z = 0 | A000007 | A019590 | .......MagicNKZ(n,k,0) = A008292(n,k+1) .......
z = 1 | A000012 | A040000 | A101101 | thisSeq | A101100 | ....... | .......
z = 2 | A000027 | A005408 | A008458 | A101103 | A101095 | ....... | .......
z = 3 | A000217 | A000290 | A003215 | A005914 | A101096 | ....... | .......
z = 4 | A000292 | A000330 | A000578 | A005917 | A101098 | ....... | .......
z = 5 | A000332 | A002415 | A000537 | A000583 | A022521 | ....... | A255181
z = 6 | A000389 | A005585 | A024166 | A000538 | A000584 | A022522 | A255177
z = 7 | A000579 | A040977 | A101094 | A101089 | A000539 | A001014 | A022523
z = 8 | A000580 | A050486 | A101097 | A101090 | A101092 | A000540 | A001015
z = 9 | A000581 | A053347 | A101102 | A101091 | A101099 | A101093 | A000541
Cf. A101095 for an expanded table and more about MagicNKZ.

MagicNKZ = Sum[(-1)^j*Binomial[n+1-z, j]*(k-j+1)^n, {j, 0, k+1}];Table[MagicNKZ, {n, 4, 4}, {z, 1, 1}, {k, 0, 34}]
Join[{1, 12, 23},LinearRecurrence[{1},{24},56]] (* Ray Chandler, Sep 23 2015 *)

a(k) = MagicNKZ(4,k,1) where MagicNKZ(n,k,z) = Sum_{j=0..k+1} (-1)^j*binomial(n+1-z,j)*(k-j+1)^n (cf. A101095). That is, a(k) = Sum_{j=0..k+1} (-1)^j*binomial(4, j)*(k-j+1)^4.
a(1)=1, a(2)=12, a(3)=23, and a(n)=24 for n>=4. - Joerg Arndt, Nov 30 2014
G.f.: x*(1+11*x+11*x^2+x^3)/(1-x). - Colin Barker, Apr 16 2012

New name from Joerg Arndt, Nov 30 2014

Original Formula edited and Crossrefs table added by Danny Rorabaugh, Apr 22 2015

A222716 Numbers which are both the sum of n+1 consecutive triangular numbers and the sum of the n-1 immediately following triangular numbers.

Original entry on oeis.org

0, 10, 100, 460, 1460, 3710, 8120, 15960, 28920, 49170, 79420, 122980, 183820, 266630, 376880, 520880, 705840, 939930, 1232340, 1593340, 2034340, 2567950, 3208040, 3969800, 4869800, 5926050, 7158060, 8586900, 10235260, 12127510, 14289760, 16749920, 19537760, 22684970, 26225220, 30194220, 34629780, 39571870, 45062680, 51146680

Offset: 1

Jonathan Sondow, Mar 02 2013

The n+1 consecutive triangular numbers start with the A028387(n-2)-th triangular number A000217(n^2-n-1), while the n-1 consecutive triangular numbers start with the A000290(n)-th triangular number A000217(n^2).
Similar sums of consecutive integers are A059270.
Similar sums of consecutive squares are A059255.
Berselli points out that a(n) = 10*A024166(n-1) = A000292(n-1)*(3*n^2 - 2). Since a(n) is a sum of triangular numbers, 10=1+2+3+4 is the 4th triangular number, A024166 is a sum of cubes, and A000292 is a tetrahedral number, is there a geometric proof of Berselli's formula? (Compare Nelsen and Unal's "Proof Without Words: Runs of Triangular Numbers.") [Jonathan Sondow, Mar 04 2013]

			T(1) + T(2) + T(3) = 1 + 3 + 6 = 10 = T(4) and 4 = 2^2, so a(2) = 10.
T(5) + T(6) + T(7) + T(8) = 15 + 21 + 28 + 36 = 100 = 45 + 55 = T(9) + T(10) and 9 = 3^2, so a(3) = 100.

Seiichi Manyama, Table of n, a(n) for n = 1..10000
Roger B. Nelsen and Hasan Unal, Proof Without Words: Runs of Triangular Numbers, Math. Mag., 85 (2012), 373.
Index entries for linear recurrences with constant coefficients, signature (6,-15,20,-15,6,-1).

Cf. A000217, A000292, A024166, A059255, A059270.

Table[ n/6 (2 - 5 n^2 + 3 n^4), {n, 1, 40}]
LinearRecurrence[{6,-15,20,-15,6,-1},{0,10,100,460,1460,3710},40] (* Harvey P. Dale, Apr 19 2016 *)

a(n) = T(n^2-n-1)+T(n^2-n)+...+T(n^2-1) = T(n^2)+T(n^2+1)+...+T(n^2+n-2), where T = A000217.
a(n) = (3*n^5 - 5*n^3 + 2*n)/6 = (n-1)*n*(n+1)*(3*n^2 - 2)/6.
G.f.: 10*x^2*(1+4*x+x^2)/(1-x)^6. [Bruno Berselli, Mar 04 2013]
a(n) = -a(-n) = 10*A024166(n-1) = A000292(n-1)*A100536(n). [Bruno Berselli, Mar 04 2013]
a(n) = TP(n^2-1)-TP(n^2-n-2) = TP(n^2+n-2)-TP(n^2-1) = TP(n-1)*(3*n^2-2), where TP = A000292. [Jonathan Sondow, Mar 04 2013]

A254469 Sixth partial sums of cubes (A000578).

Original entry on oeis.org

1, 14, 96, 450, 1650, 5082, 13728, 33462, 75075, 157300, 311168, 586092, 1058148, 1841100, 3100800, 5073684, 8090181, 12603954, 19228000, 28778750, 42329430, 61274070, 87403680, 122996250, 170922375, 234768456, 318979584, 429024376, 571584200, 754769400

Offset: 1

Luciano Ancora, Feb 15 2015

			First differences:   1,  7, 19,  37,   61,   91, ... (A003215)
-------------------------------------------------------------------------
The cubes:           1,  8, 27,  64,  125,  216, ... (A000578)
-------------------------------------------------------------------------
First partial sums:  1,  9, 36, 100,  225,  441, ... (A000537)
Second partial sums: 1, 10, 46, 146,  371,  812, ... (A024166)
Third partial sums:  1, 11, 57, 203,  574, 1386, ... (A101094)
Fourth partial sums: 1, 12, 69, 272,  846, 2232, ... (A101097)
Fifth partial sums:  1, 13, 82, 354, 1200, 3432, ... (A101102)
Sixth partial sums:  1, 14, 96, 450, 1650, 5082, ... (this sequence)

Luciano Ancora, Table of n, a(n) for n = 1..1000
Luciano Ancora, Partial sums of m-th powers with Faulhaber polynomials.
Luciano Ancora, Pascal’s triangle and recurrence relations for partial sums of m-th powers .
Feihu Liu, Guoce Xin, and Chen Zhang, Ehrhart Polynomials of Order Polytopes: Interpreting Combinatorial Sequences on the OEIS, arXiv:2412.18744 [math.CO], 2024. See p. 15.
Index entries for linear recurrences with constant coefficients, signature (10,-45,120,-210,252,-210,120,-45,10,-1).

Cf. A000537, A000578, A003215, A024166, A101094, A101097, A101102, A254470, A254471, A254472.

[n*(1+n)^2*(2+n)*(3+n)*(4+n)*(5+n)^2*(6+n)/60480: n in [1..30]]; // Vincenzo Librandi, Feb 15 2015

Table[n (1 + n)^2 (2 + n) (3 + n) (4 + n) (5 + n)^2 (6 + n)/60480, {n, 27}] (* or *) CoefficientList[Series[(1 + 4 x + x^2)/(- 1 + x)^10, {x, 0, 26}], x]
Nest[Accumulate,Range[30]^3,6] (* or *) LinearRecurrence[{10,-45,120,-210,252,-210,120,-45,10,-1},{1,14,96,450,1650,5082,13728,33462,75075,157300},30] (* Harvey P. Dale, Sep 03 2016 *)

a(n)=n*(1+n)^2*(2+n)*(3+n)*(4+n)*(5+n)^2*(6+n)/60480 \\ Charles R Greathouse IV, Oct 07 2015

G.f.: (x + 4*x^2 + x^3)/(- 1 + x)^10.
a(n) = n*(1 + n)^2*(2 + n)*(3 + n)*(4 + n)*(5 + n)^2*(6 + n)/60480.
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) + n^3.
From Amiram Eldar, Jan 26 2022: (Start)
Sum_{n>=1} 1/a(n) = 217/200.
Sum_{n>=1} (-1)^(n+1)/a(n) = 223769/200 - 8064*log(2)/5. (End)

A101095 Fourth difference of fifth powers (A000584).

Original entry on oeis.org

1, 28, 121, 240, 360, 480, 600, 720, 840, 960, 1080, 1200, 1320, 1440, 1560, 1680, 1800, 1920, 2040, 2160, 2280, 2400, 2520, 2640, 2760, 2880, 3000, 3120, 3240, 3360, 3480, 3600, 3720, 3840, 3960, 4080, 4200, 4320, 4440, 4560, 4680, 4800, 4920, 5040, 5160, 5280

Offset: 1

Cecilia Rossiter, Dec 15 2004

Original Name: Shells (nexus numbers) of shells of shells of shells of the power of 5.

The (Worpitzky/Euler/Pascal Cube) "MagicNKZ" algorithm is: MagicNKZ(n,k,z) = Sum_{j=0..k+1} (-1)^j*binomial(n + 1 - z, j)*(k - j + 1)^n, with k>=0, n>=1, z>=0. MagicNKZ is used to generate the n-th accumulation sequence of the z-th row of the Euler Triangle (A008292). For example, MagicNKZ(3,k,0) is the 3rd row of the Euler Triangle (followed by zeros) and MagicNKZ(10,k,1) is the partial sums of the 10th row of the Euler Triangle. This sequence is MagicNKZ(5,k-1,2).

Danny Rorabaugh, Table of n, a(n) for n = 1..10000
D. J. Pengelley, The bridge between the continuous and the discrete via original sources in Study the Masters: The Abel-Fauvel Conference [pdf], Kristiansand, 2002, (ed. Otto Bekken et al), National Center for Mathematics Education, University of Gothenburg, Sweden, in press.
Cecilia Rossiter, Depictions, Explorations and Formulas of the Euler/Pascal Cube [Archive Machine link]
Cecilia Rossiter, Depictions, Explorations and Formulas of the Euler/Pascal Cube [Cached copy, May 15 2013]
Eric Weisstein, Link to section of MathWorld: Worpitzky's Identity of 1883
Eric Weisstein, Link to section of MathWorld: Eulerian Number
Eric Weisstein, Link to section of MathWorld: Nexus number
Eric Weisstein, Link to section of MathWorld: Finite Differences
Index entries for linear recurrences with constant coefficients, signature (2,-1).

Fourth differences of A000584, third differences of A022521, second differences of A101098, and first differences of A101096.
For other sequences based upon MagicNKZ(n,k,z):
...... |  n = 1  |  n = 2  |  n = 3  |  n = 4  |  n = 5  |  n = 6  |  n = 7  |  n = 8
--------------------------------------------------------------------------------------
z =  0 | A000007 | A019590 | .......  MagicNKZ(n,k,0) = T(n,k+1) from A008292  .......
z =  1 | A000012 | A040000 | A101101 | A101104 | A101100 | ....... | ....... | .......
z =  2 | A000027 | A005408 | A008458 | A101103 | thisSeq | ....... | ....... | .......
z =  3 | A000217 | A000290 | A003215 | A005914 | A101096 | ....... | ....... | .......
z =  4 | A000292 | A000330 | A000578 | A005917 | A101098 | ....... | ....... | .......
z =  5 | A000332 | A002415 | A000537 | A000583 | A022521 | ....... | A255181 | .......
z =  6 | A000389 | A005585 | A024166 | A000538 | A000584 | A022522 | A255177 | A255182
z =  7 | A000579 | A040977 | A101094 | A101089 | A000539 | A001014 | A022523 | A255178
z =  8 | A000580 | A050486 | A101097 | A101090 | A101092 | A000540 | A001015 | A022524
z =  9 | A000581 | A053347 | A101102 | A101091 | A101099 | A101093 | A000541 | A001016
z = 10 | A000582 | A054333 | A254469 | A254681 | A254644 | A254640 | A250212 | A000542
z = 11 | A001287 | A054334 | A254869 | A254470 | A254682 | A254645 | A254641 | A253636
z = 12 | A001288 | A057788 | ....... | A254870 | A254471 | A254683 | A254646 | A254642
z = 13 | A010965 | ....... | ....... | ....... | A254871 | A254472 | A254684 | A254647
z = 14 | A010966 | ....... | ....... | ....... | ....... | A254872 | ....... | .......
--------------------------------------------------------------------------------------
Cf. A047969.

I:=[1,28,121,240,360]; [n le 5 select I[n] else 2*Self(n-1)-Self(n-2): n in [1..50]]; // Vincenzo Librandi, May 07 2015

MagicNKZ=Sum[(-1)^j*Binomial[n+1-z, j]*(k-j+1)^n, {j, 0, k+1}];Table[MagicNKZ, {n, 5, 5}, {z, 2, 2}, {k, 0, 34}]
CoefficientList[Series[(1 + 26 x + 66 x^2 + 26 x^3 + x^4)/(1 - x)^2, {x, 0, 50}], x] (* Vincenzo Librandi, May 07 2015 *)
Join[{1,28,121,240},Differences[Range[50]^5,4]] (* or *) LinearRecurrence[{2,-1},{1,28,121,240,360},50] (* Harvey P. Dale, Jun 11 2016 *)

a(n)=if(n>3, 120*n-240, 33*n^2-72*n+40) \\ Charles R Greathouse IV, Oct 11 2015

[1,28,121]+[120*(k-2) for k in range(4,36)] # Danny Rorabaugh, Apr 23 2015

a(k+1) = Sum_{j=0..k+1} (-1)^j*binomial(n + 1 - z, j)*(k - j + 1)^n; n = 5, z = 2.
For k>3, a(k) = Sum_{j=0..4} (-1)^j*binomial(4, j)*(k - j)^5 = 120*(k - 2).
a(n) = 2*a(n-1) - a(n-2), n>5. G.f.: x*(1+26*x+66*x^2+26*x^3+x^4) / (1-x)^2. - Colin Barker, Mar 01 2012

MagicNKZ material edited, Crossrefs table added, SeriesAtLevelR material removed by Danny Rorabaugh, Apr 23 2015

Name changed and keyword 'uned' removed by Danny Rorabaugh, May 06 2015

A334781 Array read by antidiagonals: T(n,k) = Sum_{i=1..n} binomial(1+i,2)^k.

Original entry on oeis.org

0, 0, 1, 0, 1, 2, 0, 1, 4, 3, 0, 1, 10, 10, 4, 0, 1, 28, 46, 20, 5, 0, 1, 82, 244, 146, 35, 6, 0, 1, 244, 1378, 1244, 371, 56, 7, 0, 1, 730, 8020, 11378, 4619, 812, 84, 8, 0, 1, 2188, 47386, 108020, 62003, 13880, 1596, 120, 9, 0, 1, 6562, 282124, 1047386, 867395, 256484, 35832, 2892, 165, 10

Offset: 0

Andrew Howroyd, May 15 2020

			Array begins:
===============================================================
n\k | 0  1    2     3      4        5         6           7
----|----------------------------------------------------------
  0 | 0  0    0     0      0        0         0           0 ...
  1 | 1  1    1     1      1        1         1           1 ...
  2 | 2  4   10    28     82      244       730        2188 ...
  3 | 3 10   46   244   1378     8020     47386      282124 ...
  4 | 4 20  146  1244  11378   108020   1047386    10282124 ...
  5 | 5 35  371  4619  62003   867395  12438011   181141499 ...
  6 | 6 56  812 13880 256484  4951496  98204132  1982230040 ...
  7 | 7 84 1596 35832 871140 22161864 580094436 15475158552 ...
  ...

Andrew Howroyd, Table of n, a(n) for n = 0..1325

Columns k=0..7 are A001477, A000292, A024166, A085438, A085439, A085440, A085441, A085442.
Rows n=0..3 are A000004, A000012, A034472, A074508.
Main diagonal is A249564(n > 0).
Cf. A154283 (coefficients).

T(n,k) = {sum(i=1, n, binomial(1+i,2)^k)}

T(n,k) = Sum_{i=0..2*(k-1)} A154283(k,i) * binomial(n+2+i, 2*k+i) for k > 0.

A213553 Rectangular array: (row n) = bc, where b(h) = h, c(h) = (n-1+h)^3, n>=1, h>=1, and = convolution.

Original entry on oeis.org

1, 10, 8, 46, 43, 27, 146, 142, 118, 64, 371, 366, 334, 253, 125, 812, 806, 766, 658, 466, 216, 1596, 1589, 1541, 1406, 1150, 775, 343, 2892, 2884, 2828, 2666, 2346, 1846, 1198, 512, 4917, 4908, 4844, 4655, 4271, 3646, 2782, 1753, 729, 7942

Offset: 1

Clark Kimberling, Jun 17 2012

Principal diagonal: A213554
Antidiagonal sums: A101089
row 1, (1,2,3,...)**(1,8,27,...): A024166
row 2, (1,2,3,...)**(8,27,64,...): (3*k^5 + 30*k^4 + 115*k^3 + 210*k^2 + 122*k)/60
row 3, (1,2,3,...)**(27,64,125,...): (3*k^5 + 45*k^4 + 265*k^3 + 765*k^2 + 542*k)/120
For a guide to related arrays, see A213500.

			Northwest corner (the array is read by falling antidiagonals):
1.....10....46.....146....371
8.....43....142....366....806
27....118...334....766....1541
64....253...658....1406...2666
125...466...1150...2346...4271

G. C. Greubel, Antidiagonals n = 1..100, flattened

Cf. A213500.

Flat(List([1..12], n-> List([1..n], k-> Binomial(n-k+2, 2)*(12*k^3 +9*k^2*n -9*k^2 +6*k*n^2 +3*k*n -k +n*(3*n^2 +6*n +1))/30 ))); # G. C. Greubel, Jul 31 2019

[Binomial(n-k+2, 2)*(12*k^3 +9*k^2*n -9*k^2 +6*k*n^2 +3*k*n -k + n*(3*n^2 +6*n +1))/30: k in [1..n], n in [1..12]]; // G. C. Greubel, Jul 31 2019

(* First program *)
b[n_]:= n; c[n_]:= n^3;
T[n_, k_]:= Sum[b[k-i] c[n+i], {i, 0, k-1}]
TableForm[Table[T[n, k], {n, 1, 10}, {k, 1, 10}]]
Flatten[Table[T[n-k+1, k], {n, 12}, {k, n, 1, -1}]]
r[n_]:= Table[T[n, k], {k, 1, 60}]  (* A213553 *)
d = Table[T[n, n], {n, 1, 40}] (* A213554 *)
s[n_]:= Sum[T[i, n+1-i], {i, 1, n}]
s1 = Table[s[n], {n, 1, 50}] (* A101089 *)
(* Second program *)
Table[Binomial[n-k+2, 2]*(12*k^3 +9*k^2*n -9*k^2 +6*k*n^2 +3*k*n -k +n*(3*n^2 +6*n +1))/30, {n, 12}, {k, n}]//Flatten (* G. C. Greubel, Jul 31 2019 *)

t(n,k) = binomial(n-k+2, 2)*(12*k^3 +9*k^2*n -9*k^2 +6*k*n^2 +3*k*n -k +n*(3*n^2 +6*n +1))/30;
for(n=1,12, for(k=1,n, print1(t(n,k), ", "))) \\ G. C. Greubel, Jul 31 2019

[[binomial(n-k+2, 2)*(12*k^3 +9*k^2*n -9*k^2 +6*k*n^2 +3*k*n -k + n*(3*n^2 +6*n +1))/30 for k in (1..n)] for n in (1..12)] # G. C. Greubel, Jul 31 2019

T(n,k) = 6*T(n,k-1) - 15*T(n,k-2) + 20*T(n,k-3) - 15*T(n,k-4) + 6*T(n,k-5) -T(n,k-6).
G.f. for row n:  f(x)/g(x), where f(x) = n^3 + (-3*n^3 + 3*n^2 + 3*n + 1)*x + (3*n^3 - 6*n^2 + 4)*x^2 - ((n-1)^3)*x^3 and g(x) = (1 - x)^6.
T(n,k) = k*((3*k^4 - 5*k^2 + 2) + 15*k*(k^2 - 1)*n + 30*(k^2 - 1)*n^2 + 30*(k + 1)*n^3)/60. - G. C. Greubel, Jul 31 2019

A101101 a(1)=1, a(2)=5, and a(n)=6 for n >= 3.

Original entry on oeis.org

1, 5, 6, 6, 6, 6, 6, 6, 6, 6, 6, 6, 6, 6, 6, 6, 6, 6, 6, 6, 6, 6, 6, 6, 6, 6, 6, 6, 6, 6, 6, 6, 6, 6, 6, 6, 6, 6, 6, 6, 6, 6, 6, 6, 6, 6, 6, 6, 6, 6, 6, 6, 6, 6, 6, 6, 6, 6, 6, 6, 6, 6, 6, 6, 6, 6, 6, 6, 6, 6, 6, 6, 6, 6, 6, 6, 6, 6, 6, 6

Offset: 1

Cecilia Rossiter (cecilia(AT)noticingnumbers.net), Dec 15 2004

Previous name was: The first summation of row 3 of Euler's triangle - a row that will recursively accumulate to the power of 3.

Decimal expansion of 47/30. - Elmo R. Oliveira, Aug 09 2024

D. J. Pengelley, The bridge between the continuous and the discrete via original sources in Study the Masters: The Abel-Fauvel Conference [pdf], Kristiansand, 2002, (ed. Otto Bekken et al.), National Center for Mathematics Education, University of Gothenburg, Sweden, in press.
C. Rossiter, Depictions, Explorations and Formulas of the Euler/Pascal Cube. [broken link: domain now owned by a domain grabber]
Eric Weisstein, Link to section of MathWorld: Worpitzky's Identity of 1883.
Eric Weisstein, Link to section of MathWorld: Eulerian Number.
Eric Weisstein, Link to section of MathWorld: Nexus number.
Eric Weisstein, Link to section of MathWorld: Finite Differences.
Dominika Závacká, Cristina Dalfó, and Miquel Angel Fiol, Integer sequences from k-iterated line digraphs, CEUR: Proc. 24th Conf. Info. Tech. - Appl. and Theory (ITAT 2024) Vol 3792, 156-161. See p. 161, Table 2.
Index entries for linear recurrences with constant coefficients, signature (1).

Within the "cube" of related sequences with construction based upon MaginNKZ formula, with n downward, k rightward and z backward:
Before: this_sequence, A008458, A003215, A000578, A000537, A024166 or A024166, A101094, A101097, A101102.
Above: this_sequence, below: A101104, A101100.
Within the "cube" of related sequences with construction based upon SeriesAtLevelR formula, with n downward, x rightward and r backward:
Above: this_sequence, below: A101103, A101096.

MagicNKZ=Sum[(-1)^j*Binomial[n+1-z, j]*(k-j+1)^n, {j, 0, k+1}];Table[MagicNKZ, {n, 3, 3}, {z, 1, 1}, {k, 0, 34}] (* OR *)
SeriesAtLevelR = Sum[Eulerian[n, i - 1]*Binomial[n + x - i + r, n + r], {i, 1, n}]; Table[SeriesAtLevelR, {n, 3, 3}, {r, -3, -3}, {x, 4, 35}]
Join[{1, 5},LinearRecurrence[{1},{6},78]] (* Ray Chandler, Sep 23 2015 *)

G.f.: x*(1+4*x+x^2)/(1-x). - L. Edson Jeffery, Jan 29 2012

I wish the sequence was as interesting as the list of references! - N. J. A. Sloane

New name from Joerg Arndt, Nov 30 2014

A185957 Second accumulation array of the array min{n,k}, by antidiagonals.

Original entry on oeis.org

1, 3, 3, 6, 10, 6, 10, 21, 21, 10, 15, 36, 46, 36, 15, 21, 55, 81, 81, 55, 21, 28, 78, 126, 146, 126, 78, 28, 36, 105, 181, 231, 231, 181, 105, 36, 45, 136, 246, 336, 371, 336, 246, 136, 45, 55, 171, 321, 461, 546, 546, 461, 321, 171, 55, 66, 210, 406, 606, 756, 812, 756

Offset: 1

Clark Kimberling, Feb 07 2011

A member of the accumulation chain
... < A003982 < A003783 < A115262 < A185957 <...,
where A003783(n,k)=min{n,k}.  See A144112 for the definition of accumulation array.
A185957 also gives the symmetric matrix based on the triangular numbers s=(1,3,6,10,15,....; viz, let T be the infinite square matrix whose n-th row is formed by putting n-1 zeros before the terms of s.  Let T' be the transpose of T.  Then A185957 represents the matrix product M=T'*T.  M is the self-fusion matrix of s, as defined at A193722.  See A202678 for characteristic polynomials of principal submatrices of M.

			Northwest corner:
1....3....6....10...15
3....10...21...36...55
6....21...46...81...126
10...36...81...146..231

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

Cf. A144112, A003783, A115262.

U = NestList[Most[Prepend[#, 0]] &, #, Length[#] - 1] &[Table[k (k + 1)/2, {k, 1, 15}]];
L = Transpose[U]; M = L.U; TableForm[M]
m[i_, j_] := M[[i]][[j]];
Flatten[Table[m[i, n + 1 - i], {n, 1, 12}, {i, 1, n}]]
f[n_] := Sum[m[i, n], {i, 1, n}] + Sum[m[n, j], {j, 1, n - 1}]
Table[f[n], {n, 1, 12}]
Table[Sqrt[f[n]], {n, 1, 12}] (* A000292 *)
Table[m[1, j], {j, 1, 12}] (* A000217 *)
Table[m[2, j], {j, 1, 12}] (* A014105 *)
Table[m[j, j], {j, 1, 12}] (* A024166 *)
Table[m[j, j + 1], {j, 1, 12}] (* A112851 *)
Table[Sum[m[i, n + 1 - i], {i, 1, n}], {n, 1, 12}] (* A001769 *)

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A122193 Triangle T(n,k) of number of loopless multigraphs with n labeled edges and k labeled vertices and without isolated vertices, n >= 1; 2 <= k <= 2*n.

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A101104 a(1)=1, a(2)=12, a(3)=23, and a(n)=24 for n>=4.

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A222716 Numbers which are both the sum of n+1 consecutive triangular numbers and the sum of the n-1 immediately following triangular numbers.

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A254469 Sixth partial sums of cubes (A000578).

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A101095 Fourth difference of fifth powers (A000584).

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A334781 Array read by antidiagonals: T(n,k) = Sum_{i=1..n} binomial(1+i,2)^k.

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A213553 Rectangular array: (row n) = bc, where b(h) = h, c(h) = (n-1+h)^3, n>=1, h>=1, and = convolution.