A005917 -id:A005917 - cp's OEIS frontend

A198063 Triangle read by rows (n >= 0, 0 <= k <= n, m = 3); T(n,k) = Sum{j=0..m} Sum{i=0..m} (-1)^(j+i)C(i,j)n^j*k^(m-j).

0, 1, 1, 8, 4, 8, 27, 15, 15, 27, 64, 40, 32, 40, 64, 125, 85, 65, 65, 85, 125, 216, 156, 120, 108, 120, 156, 216, 343, 259, 203, 175, 175, 203, 259, 343, 512, 400, 320, 272, 256, 272, 320, 400, 512, 729, 585, 477, 405, 369, 369, 405, 477, 585, 729

Offset: 0

Peter Luschny, Oct 26 2011

Read as an infinite symmetric square array, this is the table A(n,k)=(n+k)(n^2+k^2), cf. A321500 for the triangle with k <= n. - M. F. Hasler, Nov 22 2018

			[0]                   0
[1]                  1, 1
[2]                8, 4, 8
[3]             27, 15, 15, 27
[4]           64, 40, 32, 40, 64
[5]        125, 85, 65, 65, 85, 125
[6]   216, 156, 120, 108, 120, 156, 216
[7] 343, 259, 203, 175, 175, 203, 259, 343
From _M. F. Hasler_, Nov 22 2018: (Start)
Can also be seen as the square array A(n,k)=(n+k)*(n^2 + k^2) read by antidiagonals:
n | k: 0   1   2   3 ...
--+----------------------
0 |    0   1   8  27 ...
1 |    1   4  15  40 ...
2 |    8  15  32  65 ...
3 |   27  40  65 108 ...
...      ...     ...
(End)

G. C. Greubel, Rows n=0..100 of triangle, flattened

Cf. A057427, A003056, A073254, A198064, A198065.

[[2*k^2*n-2*k*n^2+n^3: k in [0..n]]: n in [0..12]]; // G. C. Greubel, Nov 23 2018

A198063 := (n,k) -> 2*k^2*n-2*k*n^2+n^3:

t[n_, k_] := 2 k^2*n - 2 k*n^2 + n^3; Table[t[n, k], {n, 0, 10}, {k, 0, n}] // Flatten (* Amiram Eldar, Nov 22 2018 *)

A198063(n,k)=2*k^2*n-2*k*n^2+n^3 \\ See also A321500. - M. F. Hasler, Nov 22 2018

[[ 2*k^2*n-2*k*n^2+n^3 for k in range(n+1)] for n in range(12)] # G. C. Greubel, Nov 23 2018

T(n,k) = 2*k^2*n - 2*k*n^2 + n^3.
T(n,0) = T(n,n) = n^m = n^3 = A000578(n).
T(2*n,n) = (m+1)n^m = 4*n^3 = A033430(n).
T(2*n+1,n+1) = (n+1)^(m+1) - n^(m+1) = (n+1)^4 - n^4 = A005917(n).
Sum{k=0..n} T(n,k) = (2*n^4 + 3*n^3 + n^2)/3 = A098077(n).
T(n+1,k+1)*C(n,k)^4/(k+1)^3 = A197653(n,k).

A212133 Number of (w,x,y,z) with all terms in {1,...,n} and median=mean.

Original entry on oeis.org

0, 1, 8, 33, 88, 185, 336, 553, 848, 1233, 1720, 2321, 3048, 3913, 4928, 6105, 7456, 8993, 10728, 12673, 14840, 17241, 19888, 22793, 25968, 29425, 33176, 37233, 41608, 46313, 51360, 56761, 62528, 68673, 75208, 82145, 89496, 97273, 105488, 114153, 123280

Offset: 0

Clark Kimberling, May 04 2012

For a guide to related sequences, see A211795.

For n>=1, a(n) is the number of cells in the n-th rhombic-dodecahedral polycube. - George Sicherman, Jan 22 2024

			a(2) counts these 4-tuples:  (1,1,1,1), (1,1,2,2), (1,2,1,2), (2,1,1,2), (1,2,2,1), (2,1,2,1), (2,2,1,1), (2,2,2,2).

Reinhard Zumkeller, Table of n, a(n) for n = 0..10000
Z. Janelidze, F. van Niekerk, and J. Viljoen, What is the maximal connected partial symmetry index of a connected graph of a given size?, arXiv:2502.00704 [math.CO], 2025. See p. 4.
Index entries for linear recurrences with constant coefficients, signature (4,-6,4,-1).

Cf. A211795.
Cf. A226449. - Bruno Berselli, Jun 09 2013
Cf. A005917.

a212133 n = if n == 0 then 0 else (a005917 n + 1) `div` 2
-- Reinhard Zumkeller, Nov 13 2014

t = Compile[{{n, _Integer}},
Module[{s = 0}, (Do[If[Apply[Plus, Rest[Most[Sort[{w, x, y, z}]]]]/2 == (w + x + y + z)/4, s = s + 1], {w, 1, #}, {x, 1, #}, {y, 1, #}, {z, 1, #}] &[n]; s)]];
Flatten[Map[{t[#]} &, Range[0, 50]]] (* A212133 *)
(* Peter J. C. Moses, May 01 2012 *)

a(n)=2*n^3-3*n^2+2*n; \\ Joerg Arndt, Jun 22 2012

concat(0, Vec(x*(1 + 4*x + 7*x^2) / (1 - x)^4 + O(x^40))) \\ Colin Barker, Dec 02 2017

a(n) = 4*a(n-1) - 6*a(n-2) + 4*a(n-3) - a(n-4).
a(n) = n * (2*n^2 - 3*n + 2). - J. M. Bergot, Jun 22 2012
a(n) = A000384(n) + n*A000384(n-1). - Bruno Berselli, Jun 07 2013
a(n) = (A005917(n) + 1) / 2 for n > 0. - Reinhard Zumkeller, Nov 13 2014
G.f.: x*(1 + 4*x + 7*x^2) / (1 - x)^4. - Colin Barker, Dec 02 2017

Closed form adapted to the offset by Bruno Berselli, Jun 07 2013

A221967 T(n,k)=Number of -k..k arrays of length n with the sum ahead of each element differing from the sum following that element by k or less.

Original entry on oeis.org

3, 5, 9, 7, 25, 15, 9, 49, 65, 33, 11, 81, 175, 225, 63, 13, 121, 369, 833, 705, 129, 15, 169, 671, 2241, 3647, 2305, 255, 17, 225, 1105, 4961, 12609, 16513, 7425, 513, 19, 289, 1695, 9633, 34111, 73089, 73983, 24065, 1023, 21, 361, 2465, 17025, 78273, 241153

Offset: 1

R. H. Hardin Feb 01 2013

Table starts
....3.......5.........7..........9..........11...........13............15
....9......25........49.........81.........121..........169...........225
...15......65.......175........369.........671.........1105..........1695
...33.....225.......833.......2241........4961.........9633.........17025
...63.....705......3647......12609.......34111........78273........159615
..129....2305.....16513......73089......241153.......653185.......1535745
..255....7425.....73983.....419841.....1690623......5407233......14661375
..513...24065....332801....2419713....11888129.....44890625.....140355585
.1023...77825...1495039...13930497....83512319....372332545....1342437375
.2049..251905...6719489...80230401...586864641...3089205249...12843782145
.4095..815105..30195711..462012417..4123582463..25628045313..122870296575
.8193.2637825.135700481.2660655105.28975366145.212618141697.1175482548225

			Some solutions for n=6 k=4
..4...-2....4....1...-4...-1...-2....1...-2...-1....1....3....4....1...-1...-1
.-4....4...-4....0....4....4....3....2....3....2...-2...-4...-2...-3....3....3
..1...-3....3...-2...-1...-2...-3...-2...-2....2....0....3....1....2....0...-1
..0....2...-1....3...-2....0....2....2....3...-3....4....1...-3...-2...-2....1
..3...-4...-2...-3....3....3...-2....1...-1....0...-1...-3....0....3...-3...-2
..1....1....2....1...-1...-2...-1....1...-1....1....0....1....2...-4....4....2

R. H. Hardin, Table of n, a(n) for n = 1..334

Column 1 is A062510(n+1)
Column 2 is A189318
Row 2 is A016754
Row 3 is A005917(n+1)
Row 4 is A142993

Empirical for column k:
k=1: a(n) = a(n-1) +2*a(n-2)
k=2: a(n) = 3*a(n-1) +2*a(n-2) -4*a(n-3)
k=3: a(n) = 3*a(n-1) +8*a(n-2) -4*a(n-3) -8*a(n-4)
k=4: a(n) = 5*a(n-1) +8*a(n-2) -20*a(n-3) -8*a(n-4) +16*a(n-5)
k=5: a(n) = 5*a(n-1) +18*a(n-2) -20*a(n-3) -48*a(n-4) +16*a(n-5) +32*a(n-6)
k=6: a(n) = 7*a(n-1) +18*a(n-2) -56*a(n-3) -48*a(n-4) +112*a(n-5) +32*a(n-6) -64*a(n-7)
k=7: a(n) = 7*a(n-1) +32*a(n-2) -56*a(n-3) -160*a(n-4) +112*a(n-5) +256*a(n-6) -64*a(n-7) -128*a(n-8)
Empirical for row n:
n=1: a(n) = 2*n + 1
n=2: a(n) = 4*n^2 + 4*n + 1
n=3: a(n) = 4*n^3 + 6*n^2 + 4*n + 1
n=4: a(n) = (16/3)*n^4 + (32/3)*n^3 + (32/3)*n^2 + (16/3)*n + 1
n=5: a(n) = (20/3)*n^5 + (50/3)*n^4 + 20*n^3 + (40/3)*n^2 + (16/3)*n + 1
n=6: a(n) = (128/15)*n^6 + (128/5)*n^5 + (112/3)*n^4 + 32*n^3 + (272/15)*n^2 + (32/5)*n + 1
n=7: a(n) = (488/45)*n^7 + (1708/45)*n^6 + (2912/45)*n^5 + (602/9)*n^4 + (2072/45)*n^3 + (952/45)*n^2 + (32/5)*n + 1

A344332 Side s of squares of type 2 that can be tiled with squares of two different sizes so that the number of large or small squares is the same.

Original entry on oeis.org

15, 30, 45, 60, 65, 75, 90, 105, 120, 130, 135, 150, 165, 175, 180, 195, 210, 225, 240, 255, 260, 270, 285, 300, 315, 325, 330, 345, 350, 360, 369, 375, 390, 405, 420, 435, 450, 455, 465, 480, 495, 510, 520, 525, 540, 555, 570, 585, 600, 615, 630, 645, 650, 660, 671, 675, 690, 700, 705, 715, 720, 735, 738, 750, 765, 780, 795, 810, 825, 840, 845, 855, 870, 875, 885, 900

Offset: 1

Bernard Schott, May 20 2021

nonn

This sequence is relative to the generalization of the 4th problem proposed for the pupils in grade 6 during the 19th Mathematical Festival at Moscow in 2008 (see A344330).
There are two types of solutions, the second one is proposed here, while type 1 is described in A344331.
If m is a term and k > 1, k * m is another term.
Every term (primitive or not primitive) is the side of an elementary square of type 2 (see A346263).
Some notations: s = side of the tiled square, a = side of small squares, b = side of large squares, and z = number of small squares = number of large squares.
-> Primitive squares
Side s of primitive squares of type 2 must satisfy the Diophantine equation s^2 = z * (a^2+b^2) with the conditions a^2+b^2 = c^2 and gcd(a, b, c) = 1.
In this case, q = a/(c-b) must be odd, and side s = q*c = a*c/(c-b) = (a+b)*c/a with a number of squares z = q^2 = (a/(c-b))^2 = ((b+c)/a)^2.
Indeed, these conditions give exactly the following solutions for n >= 2: s = n^4-(n-1)^4 (A005917), a = 2*n-1 (A005408), b = 2*n*(n-1) (A046092), c = 2*n*(n-1)+1 (A001844), z = (2*n-1)^2 (A016754); this results come from the identity:
   (n^4 - (n-1)^4)^2 = (2*n-1)^2 * ((2*n-1)^2 + (2*n*(n-1))^2).
For n >= 2, every primitive square is composed by a square S1 of z = (2*n-1)^2 large squares with side b = 2*n*(n-1), then an edge on two sides of this square S1 of z = (2*n-1)^2 small squares with side a = 2*n-1.
See example with design of square of side s = 15 with a = 3, b = 4, c = 5, q = 3, z = 9, obtained with n= 2.
-> Non-primitive squares
If s is the side of a primitive square of type 2, then every k * s, k > 1 is a non-primitive term that gives two distinct tilings of type 2.
The square ks X ks can be tiled with z = q^2 = (2n-1)^2 = (a/(c-b))^2 = ((b+c)/a)^2 squares of side ka and of side kb, but also,
The square ks X ks can be tiled with z = k^2*q^2 = ((2n-1)*k)^2 = (k*a/(c-b))^2 = (k*(b+c)/a)^2 squares of side a and of side b (see example).

			Primitive square with s = 15:
  a = 3, b = 4, c = 5, s = 15, z = 9; s = 15 is the side of primitive square, with  z = 9 squares of size 3 x 3 and 9 squares of size 4 x 4
Non-primitive square k*s = 2*15 = 30:
  a = 3, b = 4, c = 5, s = 30, z = 36, this square is obtained with 4 copies of the primitive square as below.
  a = 6, b = 8, c = 10, s = 30, z = 9, this square and its tiling are exactly as the primitive square with scale 2.
               b = 4 (or = 8)     a = 3 (or = 6)
            ________ ________ ________ ______ ______________________________
           |        |        |        |      |                              |
           |        |        |        |      |                              |
           |        |        |        |______|                              |
           |_______ |________|________|      |                              |
           |        |        |        |      |                              |
           |        |        |        |______|                              |
           |        |        |        |      |                              |
           |________|________|________|      |                              |
           |        |        |        |______|                              |
           |        |        |        |      |                              |
           |        |        |        |      |                              |
           |_____ __|___ ____|_ ______|______|                              |
           |     |      |      |      |      |                              |
           |     |      |      |      |      |                              |
           |_____|______|______|______|______|______________________________|
           |                                 |                              |
           |                                 |                              |
           |                                 |                              |
           |                                 |                              |
           |                                 |                              |
           |                                 |                              |
           |                                 |                              |
           |                                 |                              |
           |                                 |                              |
           |                                 |                              |
           |                                 |                              |
           |                                 |                              |
           |                                 |                              |
           |                                 |                              |
           |_________________________________|______________________________|
                      s = 15               s = 30

Ivan Yashchenko, Invitation to a Mathematical Festival, pp. 10 and 102, MSRI, Mathematical Circles Library, 2013.

Index to sequences related to Olympiads.

Cf. A001844, A005408, A005917, A046092, A046754, A344330, A344331.

pts(lim) = my(v=List(), m2, s2, h2, h); for(middle=4, lim-1, m2=middle^2; for(small=1, middle, s2=small^2; if(issquare(h2=m2+s2, &h), if(h>lim, break); listput(v, [small, middle, h])))); vecsort(Vec(v)); \\ A009000
isdp4(s) = my(k=1, x); while(((x=k^4 - (k-1)^4) <= s), if (x == s, return (1)); k++); return(0);
isokp2(s) = {if (!isdp4(s), return(0)); if (s % 2, my(vp = pts(s)); for (i=1, #vp, my(vpi = vp[i], a = vpi[1], b = vpi[2], c = vpi[3]); if (a*c/(c-b) == s, return(1)); ); ); }
isok2(s) = {if (isokp2(s), return (1)); fordiv(s, d, if ((d>1) || (dMichel Marcus, Jun 04 2021

A255270 Integer part of fourth root of n.

Original entry on oeis.org

0, 1, 1, 1, 1, 1, 1, 1, 1, 1, 1, 1, 1, 1, 1, 1, 2, 2, 2, 2, 2, 2, 2, 2, 2, 2, 2, 2, 2, 2, 2, 2, 2, 2, 2, 2, 2, 2, 2, 2, 2, 2, 2, 2, 2, 2, 2, 2, 2, 2, 2, 2, 2, 2, 2, 2, 2, 2, 2, 2, 2, 2, 2, 2, 2, 2, 2, 2, 2, 2, 2, 2, 2, 2, 2, 2, 2, 2, 2, 2, 2, 3, 3, 3, 3, 3, 3

Offset: 0

Bruno Berselli, Feb 20 2015

n appears (n+1)^4 - n^4 times (A005917).

Bruno Berselli, Table of n, a(n) for n = 0..1000

Cf. A005917.
Cf. sequences of the type floor(n^(1/k)): A000196 (k=2), A048766 (k=3), this sequence (k=4), A178487 (k=5), A178489 (k=6).
Cf. A219009.

[IsZero(n) select 0 else Iroot(n, 4): n in [0..100]];

[Floor(n^(1/4)): n in [0..100]]; // Vincenzo Librandi, Feb 20 2015

A255270 := proc(n)
    floor( n^(1/4)) ;
end proc:
seq(A255270(n),n=0..100) ; # R. J. Mathar, May 08 2020

Mathematica
```
Floor[Range[0, 100]^(1/4)]
```
PARI
```
vector(100, n, n--; floor(n^(1/4)))
```

a(n) = sqrtnint(n, 4); \\ Michel Marcus, Dec 22 2016

from sympy import integer_nthroot
def A255270(n): return integer_nthroot(n,4)[0] # Chai Wah Wu, Jun 06 2025

Sage
```
[floor(n^(1/4)) for n in (0..100)]
```

a(n) = floor(n^(1/4)) = floor(sqrt(A000196(n))).
G.f.: Sum_{k>=1} x^(k^4)/(1 - x). - Ilya Gutkovskiy, Dec 22 2016
a(n) = Sum_{i=1..n} A219009(i)*floor(n/i). - Ridouane Oudra, Feb 26 2023

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

A045975 Take the first odd integer and multiple of 1, the next 2 even integers and multiples of 2, the next 3 odd integers and multiples of 3, the next 4 even integers and multiples of 4, ...

Original entry on oeis.org

1, 2, 4, 9, 15, 21, 24, 28, 32, 36, 45, 55, 65, 75, 85, 90, 96, 102, 108, 114, 120, 133, 147, 161, 175, 189, 203, 217, 224, 232, 240, 248, 256, 264, 272, 280, 297, 315, 333, 351, 369, 387, 405, 423, 441, 450, 460, 470, 480, 490, 500, 510, 520, 530, 540, 561, 583, 605, 627, 649, 671, 693

Offset: 1

Fang-kuo Huang (gsyps(AT)ms17.hinet.net)

A generalized Connell sequence.

			Triangle begins:
    1;
    2,   4;
    9,  15,  21;
   24,  28,  32,  36;
   45,  55,  65,  75,  85;
   90,  96, 102, 108, 114, 120;
  133, 147, 161, 175, 189, 203, 217;
  ...

Reinhard Zumkeller, Rows n=1..150 of triangle, flattened

Cf. A001614, A033291.

Seen as a triangle read by rows: cf. A204558 (row sums), A005917 (central terms), A204556 (left edge), A204557 (right edge).

a045975 n k = a045975_tabl !! (n-1) !! (k-1)
a045975_row n = a045975_tabl !! (n-1)
a045975_tabl = f 1 [1..] where
   f k xs = ys : f (k+1) (dropWhile (<= last ys) xs) where
     ys | even k    = take k ms
        | otherwise = take k $ filter odd ms
     ms = filter ((== 0) . (`mod` k)) xs
-- Reinhard Zumkeller, Jan 18 2012

first[n_?EvenQ] := (n - 1)*n^2/2; first[n_?OddQ] := n*(n^2 - 2n + 3)/2; row[n_] := (ro = {first[n]}; next = first[n] + n; While[ Length[ro] < n, If[Mod[next , 2] == Mod[n, 2], AppendTo[ro, next]]; next = next + n]; ro); Flatten[ Table[row[n], {n, 1, 11}]](* Jean-François Alcover, Jun 08 2012 *)

More terms from James Sellers

Keyword tabl added by Reinhard Zumkeller, Jan 18 2012

A254470 Sixth partial sums of fourth powers (A000583).

Original entry on oeis.org

1, 22, 198, 1134, 4884, 17226, 52338, 141570, 348777, 795652, 1701700, 3444948, 6651216, 12321804, 22011804, 38073948, 63985977, 104782986, 167620090, 262495090, 403165620, 608300550, 902911230, 1320114510, 1903286385, 2708672616, 3808530792, 5294887048

Offset: 1

Luciano Ancora, Feb 15 2015

			First differences:   1, 15,  65, 175,  369,   671, ... (A005917)
-------------------------------------------------------------------------
The fourth powers:   1, 16,  81, 256,  625,  1296, ... (A000583)
-------------------------------------------------------------------------
First partial sums:  1, 17,  98, 354,  979,  2275, ... (A000538)
Second partial sums: 1, 18, 116, 470, 1449,  3724, ... (A101089)
Third partial sums:  1, 19, 135, 605, 2054,  5778, ... (A101090)
Fourth partial sums: 1, 20, 155, 760, 2814,  8592, ... (A101091)
Fifth partial sums:  1, 21, 176, 936, 3750, 12342, ... (A254681)
Sixth partial sums:  1, 22, 198,1134, 4884, 17226, ... (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.
Index entries for linear recurrences with constant coefficients, signature (11,-55,165,-330,462,-462,330,-165,55,-11,1).

Cf. A000538, A000583, A005917, A101089, A101090, A101091, A254469, A254471, A254472, A254681.

[n*(1+n)*(2+n)*(3+n)^2*(4+n)*(5+n)*(6+n)*(1+12*n+ 2*n^2)/302400: n in [1..30]]; // Vincenzo Librandi, Feb 15 2015

Table[n (1 + n) (2 + n) (3 + n)^2 (4 + n) (5 + n) (6 + n) (1 + 12 n + 2 n^2)/302400,{n, 25}] (* or *) CoefficientList[Series[(- 1 - 11 x - 11 x^2 - x^3)/(- 1 + x)^11, {x, 0, 24}], x]
Nest[Accumulate,Range[30]^4,6] (* or *) LinearRecurrence[{11,-55,165,-330,462,-462,330,-165,55,-11,1},{1,22,198,1134,4884,17226,52338,141570,348777,795652,1701700},30] (* Harvey P. Dale, Apr 23 2016 *)

vector(50,n,n*(1 + n)*(2 + n)*(3 + n)^2*(4 + n)*(5 + n)*(6 + n)*(1 + 12*n + 2*n^2)/302400) \\ Derek Orr, Feb 19 2015

G.f.: (-x - 11*x^2 - 11*x^3 - x^4)/(- 1 + x)^11.
a(n) = n*(1 + n)*(2 + n)*(3 + n)^2*(4 + n)*(5 + n)*(6 + n)*(1 + 12*n + 2*n^2)/302400.
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^4.
Sum_{n>=1} 1/a(n) = 3320303/2601 + 1400*Pi^2/17 + (8960/17)*sqrt(2/17)*Pi*cot(sqrt(17/2)*Pi). - Amiram Eldar, Jan 26 2022

A348210 Varma's Kosta numbers of semi-standard tableaux: array A(n>=2, k>=0) read by rising antidiagonals.

Original entry on oeis.org

0, 1, 0, 1, 1, 0, 1, 3, 1, 0, 1, 6, 5, 1, 0, 1, 15, 16, 7, 1, 0, 1, 36, 65, 31, 9, 1, 0, 1, 91, 260, 175, 51, 11, 1, 0, 1, 232, 1085, 981, 369, 76, 13, 1, 0, 1, 603, 4600, 5719, 2661, 671, 106, 15, 1, 0, 1, 1585, 19845, 33922, 19929, 5916, 1105, 141, 17, 1, 0, 1, 4213, 86725, 204687, 151936, 54131, 11516, 1695, 181, 19, 1, 0

Offset: 2

R. J. Mathar, Oct 07 2021

(More characteristic NAME desired.)

Each row is a polynomial in k, which implies that the inverse binomial transformation of each row is a finite sequence and that the row can be represented by a rational generating function (A348211).

			The array starts in row n=2 with columns k>=0 as:
  0   0    0    0     0     0      0      0 ...
  1   1    1    1     1     1      1      1 ...
  1   3    5    7     9    11     13     15 ...
  1   6   16   31    51    76    106    141 ...
  1  15   65  175   369   671   1105   1695 ...
  1  36  260  981  2661  5916  11516  20385 ...
  1  91 1085 5719 19929 54131 124501 254255 ...
Antidiagonal rows begin as:
  0;
  1,   0;
  1,   1,    0;
  1,   3,    1,    0;
  1,   6,    5,    1,    0;
  1,  15,   16,    7,    1,    0;
  1,  36,   65,   31,    9,    1,   0;
  1,  91,  260,  175,   51,   11,   1,   0;
  1, 232, 1085,  981,  369,   76,  13,   1,  0;
  1, 603, 4600, 5719, 2661,  671, 106,  15,  1,  0;

G. C. Greubel, Antidiagonals n = 2..52, flattened
D.-N. Verma, Towards Classifying Finite Point-Set Configurations, 1997, Unpublished. [Scanned copy of annotated version of preprint given to me by the author in 1997. - _N. J. A. Sloane_, Oct 03 2021]

Cf. A005043 (column k=1), A007043 (k=2), A264608 (k=3), A272393 (k=4), A005408 (row n=4), A005891 (n=5), A005917 (n=6), A348211 (condensed g.f.)

A:= func< n,k | (&+[(-1)^(j+1)*Binomial(n,j)*Binomial((n-2*j)*k+n-j-2,n-3)/2 : j in [0..Floor((n-1)/2)]]) >;
A348210:= func< n,k | A(n-k,k) >;
[A348210(n,k): k in [0..n-2], n in [2..13]]; // G. C. Greubel, Feb 28 2024

A348210 := proc(n,k)
    local a,j ;
    a := 0 ;
    for j from 0 to floor((n-1)/2) do
            a := a+ (-1)^j *binomial(n,j) *binomial( (n-2*j)*k+n-j-2,n-3) ;
    end do:
    -a/2 ;
end proc:
seq( seq( A348210(d-k,k),k=0..d-2),d=2..12) ;

A[n_, k_] := (-1/2)*Sum[(-1)^j*Binomial[n, j]*Binomial[(n - 2*j)*k + n - j - 2, n - 3], {j, 0, Floor[(n - 1)/2]}];
Table[A[n - k, k], {n, 2, 13}, {k, 0, n - 2}] // Flatten (* Jean-François Alcover, Mar 06 2023 *)

def A(n,k): return sum( (-1)^(j+1)*binomial(n,j)*binomial((n-2*j)*k+n-j-2,n-3) for j in range(1+(n-1)//2) )/2
def A348210(n,k): return A(n-k, k)
flatten([[A348210(n,k) for k in range(n-1)] for n in range(2,13)]) # G. C. Greubel, Feb 28 2024

A(n,k) = (-1/2)*Sum_{j=0..floor((n-1)/2)} (-1)^j *binomial(n,j) *binomial((n-2*j)*k+n-j-2,n-3).
A(7,k) = 1 + 7*k*(k+1)*(11*k^2+11*k+8)/12.
A(8,k) = (2*k+1)*(4*k^2+6*k+3)*(4*k^2+2*k+1)/3.
A(9,k) = 1 + k*(k+1)*(289*k^4+578*k^3+581*k^2+292*k+108)/16.

A152729 a(n) = (n-2)^4 - a(n-1) - a(n-2), with a(1) = a(2) = 0.

Original entry on oeis.org

0, 0, 1, 15, 65, 176, 384, 736, 1281, 2079, 3201, 4720, 6720, 9296, 12545, 16575, 21505, 27456, 34560, 42960, 52801, 64239, 77441, 92576, 109824, 129376, 151425, 176175, 203841, 234640, 268800, 306560, 348161, 393855, 443905, 498576, 558144

Offset: 1

Vladimir Joseph Stephan Orlovsky, Dec 11 2008

a(n+2) - a(n-1) = n^4 - (n-1)^4 = A005917(n) for all n in Z. - Michael Somos, Sep 02 2018

			0 + 0 + 1 = 1^4; 0 + 1 + 15 = 2^4; 1 + 15 + 65 = 3^4; ...
G.f. = x^3 + 15*x^4 + 65*x^5 + 176*x^6 + 384*x^7 + 736*x^8 + 1281*x^9 + ... - _Michael Somos_, Sep 02 2018

G. C. Greubel, Table of n, a(n) for n = 1..5000
Index entries for linear recurrences with constant coefficients, signature (4,-6,5,-5,6,-4,1).

Cf. A005917, A152728, A152725, A152726, A000212.

m:=50; R:=PowerSeriesRing(Integers(), m); [0,0] cat Coefficients(R!(x^3*(x+1)*(x^2+10*x+1)/((1-x)^5*(x^2+x+1)))); // G. C. Greubel, Sep 01 2018

k0=k1=0;lst={k0,k1};Do[kt=k1;k1=n^4-k1-k0;k0=kt;AppendTo[lst,k1],{n,1,4!}];lst
LinearRecurrence[{4,-6,5,-5,6,-4,1}, {0,0,1,15,65,176,384}, 50] (* G. C. Greubel, Sep 01 2018 *)
a[ n_] := With[ {m = Max[n, 2 - n]}, SeriesCoefficient[ x^3 (1 + x) (1 + 10 x + x^2) / ((1 - x)^5 (1 + x + x^2)), {x , 0, m}]]; (* Michael Somos, Sep 02 2018 *)

concat([0,0], Vec(-x^3*(x+1)*(x^2+10*x+1)/((x-1)^5*(x^2+x+1)) + O(x^100))) \\ Colin Barker, Oct 28 2014

{a(n) = my(m = max(n, 2 - n)); polcoeff( x^3 * (1 + x) * (1 + 10*x + x^2) / ((1 - x)^5 * (1 + x + x^2)) + x * O(x^m), m)}; /* Michael Somos, Sep 02 2018 */

G.f.: -x^3*(x+1)*(x^2+10*x+1) / ((x-1)^5*(x^2+x+1)). - Colin Barker, Oct 28 2014

a(n) = a(2 - n) for all n in Z. - Michael Somos, Sep 02 2018

Definition adapted to offset by Georg Fischer, Jun 18 2021

cp's OEIS Frontend

A198063 Triangle read by rows (n >= 0, 0 <= k <= n, m = 3); T(n,k) = Sum{j=0..m} Sum{i=0..m} (-1)^(j+i)*C(i,j)*n^j*k^(m-j).

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A212133 Number of (w,x,y,z) with all terms in {1,...,n} and median=mean.

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A221967 T(n,k)=Number of -k..k arrays of length n with the sum ahead of each element differing from the sum following that element by k or less.

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A344332 Side s of squares of type 2 that can be tiled with squares of two different sizes so that the number of large or small squares is the same.

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A255270 Integer part of fourth root of n.

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A101095 Fourth difference of fifth powers (A000584).

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A198063 Triangle read by rows (n >= 0, 0 <= k <= n, m = 3); T(n,k) = Sum{j=0..m} Sum{i=0..m} (-1)^(j+i)C(i,j)n^j*k^(m-j).