A101095 -id:A101095 - cp's OEIS frontend

A047969 Square array of nexus numbers a(n,k) = (n+1)^(k+1) - n^(k+1) (n >= 0, k >= 0) read by upwards antidiagonals.

1, 1, 1, 1, 3, 1, 1, 5, 7, 1, 1, 7, 19, 15, 1, 1, 9, 37, 65, 31, 1, 1, 11, 61, 175, 211, 63, 1, 1, 13, 91, 369, 781, 665, 127, 1, 1, 15, 127, 671, 2101, 3367, 2059, 255, 1, 1, 17, 169, 1105, 4651, 11529, 14197, 6305, 511, 1, 1, 19, 217, 1695, 9031

Offset: 0

N. J. A. Sloane

If each row started with an initial 0 (i.e., a(n,k) = (n+1)^k - n^k) then each row would be the binomial transform of the preceding row. - Henry Bottomley, May 31 2001
a(n-1, k-1) is the number of ordered k-tuples of positive integers such that the largest of these integers is n. - Alford Arnold, Sep 07 2005
From Alford Arnold, Jul 21 2006: (Start)
The sequences in A047969 can also be calculated using the Eulerian Array (A008292) and Pascal's Triangle (A007318) as illustrated below: (cf. A101095).
  1       1       1       1       1       1
  1       1       1       1       1       1
  -----------------------------------------
  1       2       3       4       5       6
          1       2       3       4       5
  1       3       5       7       9      11
  -----------------------------------------
  1       3       6      10      15      21
          4      12      24      40      60
                  1       3       6      10
  1       7      19      37      61      91
  -----------------------------------------
  1       4      10      20      35      56
         11      44     110     220     385
                 11      44     110     220
                          1       4      10
  1      15      65     175     369     671
  -----------------------------------------  (End)
From Peter Bala, Oct 26 2008: (Start)
The above remarks of Alford Arnold may be summarized by saying that (the transpose of) this array is the Hilbert transform of the triangle of Eulerian numbers A008292 (see A145905 for the definition of the Hilbert transform). In this context, A008292 is best viewed as the array of h-vectors of permutohedra of type A. See A108553 for the Hilbert transform of the array of h-vectors of type D permutohedra. Compare this array with A009998.
The polynomials n^k - (n-1)^k, k = 1,2,3,..., which give the nonzero entries in the columns of this array, satisfy a Riemann hypothesis: their zeros lie on the vertical line Re s = 1/2 in the complex plane. See A019538 for the connection between the polynomials n^k - (n-1)^k and the Stirling polynomials of the simplicial complexes dual to the type A permutohedra.
(End)
Empirical: (n+1)^(k+1) - n^(k+1) is the number of first differences of length k+1 arrays of numbers in 0..n, k > 0. - R. H. Hardin, Jun 30 2013
a(n-1, k-1) is the number of bargraphs of width k and height n. Examples: a(1,2) = 7 because we have [1,1,2], [1,2,1], [2,1,1], [1,2,2], [2,1,2], [2,2,1], and [2,2,2]; a(2,1) = 5 because we have [1,3], [2,3], [3,1], [3,2], and [3,3] (bargraphs are given as compositions). This comment is equivalent to A. Arnold's Sep 2005 comment. - Emeric Deutsch, Jan 30 2017

			Array a begins:
  [n\k][0  1   2    3    4   5  6  ...
  [0]   1  1   1    1    1   1  1  ...
  [1]   1  3   7   15   31  63  ...
  [2]   1  5  19   65  211  ...
  [3]   1  7  37  175  ...
  ...
Triangle T begins:
  n\m   0   1    2     3     4      5      6      7      8     9  10 ...
  0:    1
  1:    1   1
  2:    1   3    1
  3:    1   5    7     1
  4:    1   7   19    15     1
  5:    1   9   37    65    31      1
  6:    1  11   61   175   211     63      1
  7:    1  13   91   369   781    665    127      1
  8:    1  15  127   671  2101   3367   2059    255      1
  9:    1  17  169  1105  4651  11529  14197   6305    511     1
  10:   1  19  217  1695  9031  31031  61741  58975  19171  1023   1
  ...  - _Wolfdieter Lang_, May 07 2021

J. H. Conway and R. K. Guy, The Book of Numbers, Copernicus Press, NY, 1996, p. 54.

T. D. Noe, Rows n = 0..100 of triangle, flattened
A. Blecher, C. Brennan, A. Knopfmacher and Helmut Prodinger, The height and width of bargraphs, Discrete Applied Math. 180, (2015), 36-44.
Eric Weisstein's World of Mathematics, Nexus Number

Cf. A047970.
Cf. A009998, A108553 (Hilbert transform of array of h-vectors of type D permutohedra), A145904, A145905.
Row n sequences of array a: A000012, A000225(k+1), A001047(k+1), A005061(k+1), A005060(k+1), A005062(k+1), A016169(k+1), A016177(k+1), A016185(k+1), A016189(k+1), A016195(k+1), A016197(k+1).
Column k sequences of array a: (nexus numbers): A000012, A005408, A003215, A005917(n+1), A022521, A022522, A022523, A022524, A022525, A022526, A022527, A022528.
Cf. A343237 (row reversed triangle).

Flatten[Table[n = d - e; k = e; (n + 1)^(k + 1) - n^(k + 1), {d, 0, 100}, {e, 0, d}]] (* T. D. Noe, Feb 22 2012 *)

T(n,m):=if m=0 then 1 else sum(k!*(-1)^(m+k)*stirling2(m,k)*binomial(n+k-1,n),k,0,m); /* Vladimir Kruchinin, Jan 28 2018 */

From Vladimir Kruchinin: (Start)
O.g.f. of e.g.f of rows of array: ((1-x)*exp(y))/(1-x*exp(y))^2.
T(n,m) = Sum_{k=0..m} k!*(-1)^(m+k)*Stirling2(m,k)*C(n+k-1,n), T(n,0)=1.(End)
From Wolfdieter Lang, May 07 2021: (Start)
T(n,m) = a(n-m,m) = (n-m+1)^(m+1) - (n-m)^(m+1), n >= 0, m = 0, 1,..., n.
O.g.f. column k of the array: polylog(-(k+1), x)*(1-x)/x. See the Peter Bala comment above, and the Eulerian triangle A008292 formula by Vladeta Jovovic, Sep 02 2002.
E.g.f. of e.g.f. of row of the array: exp(y)*(1 + x*(exp(y) - 1))*exp(x*exp(y)).
O.g.f. of triangle's exponential row polynomials R(n, y) = Sum_{m=0} T(n, m)*(y^m)/m!: G(x, y) = exp(x*y)*(1 - x)/(1 - x*exp(x*y))^2. (End)

A101096 Third differences of fifth powers (A000584).

Original entry on oeis.org

1, 29, 150, 390, 750, 1230, 1830, 2550, 3390, 4350, 5430, 6630, 7950, 9390, 10950, 12630, 14430, 16350, 18390, 20550, 22830, 25230, 27750, 30390, 33150, 36030, 39030, 42150, 45390, 48750, 52230, 55830, 59550, 63390, 67350, 71430, 75630, 79950, 84390, 88950

Offset: 1

Cecilia Rossiter, Dec 15 2004

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

For n>=3 a(n) is equal to the number of functions f:{1,2,3,4,5}->{1,2,...,n} such that Im(f) contains 3 fixed elements. - Aleksandar M. Janjic and Milan Janjic, Feb 24 2007

Danny Rorabaugh, Table of n, a(n) for n = 1..10000
Milan Janjic, Enumerative Formulas for Some Functions on Finite Sets.
David 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, 2003.
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 (3,-3,1).

Cf. A069477.
Third differences of A000584, second differences of A022521, and first differences of A068236.
Cf. A101095 for other sequences related to MagicNKZ.
Cf. A001844.

m:=40; R:=PowerSeriesRing(Integers(), m); Coefficients(R!( x*(x^4+26*x^3+66*x^2+26*x+1)/(1-x)^3)); // G. C. Greubel, Dec 01 2018

MagicNKZ=Sum[(-1)^j*Binomial[n+1-z, j]*(k-j+1)^n, {j, 0, k+1}];Table[MagicNKZ, {n, 5, 5}, {z, 3, 3}, {k, 0, 34}]
CoefficientList[Series[(-z^4-26z^3-66z^2-26z-1)/(z-1)^3, {z, 0, 200}], z] (* Vladimir Joseph Stephan Orlovsky, Jun 19 2011 *)
Join[{1,29},Differences[Range[0,40]^5,3]] (* or *) LinearRecurrence[{3,-3,1},{1,29,150,390,750},40] (* Harvey P. Dale, Feb 02 2017 *)

a(n)=if(n>2,60*n^2-180*n+150,28*n-27) \\ Charles R Greathouse IV, Oct 11 2015

[sum([(-1)^j*binomial(3, j)*(k-j+1)^5 for j in range(min(k+2,4))]) for k in range(40)] # Danny Rorabaugh, Apr 27 2015

a(k+1) = MagicNKZ(5,k,3) where MagicNKZ(n,k,z) = Sum_{j=0..k+1} (-1)^j*binomial(n+1-z,j)*(k-j+1)^n. (Cf. A101095.)
a(n+1) = 30*(1 - 2*n + 2*n^2) for n>2.
a(n+3) = A069477(n). - Vladimir Joseph Stephan Orlovsky, Jun 19 2011
G.f.: x*(x^4+26*x^3+66*x^2+26*x+1)/(1-x)^3. - Colin Barker, Oct 17 2012
Sum_{n>=1} 1/a(n) = (Pi/60)*tanh(Pi/2) + 871/870. - Amiram Eldar, Jan 27 2022

MagicNKZ material edited and SeriesAtLevelR material removed by Danny Rorabaugh, Apr 27 2015

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

A254644 Fourth partial sums of fifth powers (A000584).

Original entry on oeis.org

1, 36, 381, 2336, 10326, 36552, 110022, 292512, 704847, 1567852, 3263403, 6422208, 12046268, 21675408, 37608828, 63194304, 103199469, 164281524, 255573769, 389409504, 582206130, 855534680, 1237402530, 1763779680, 2480401755, 3444885756, 4729197591, 6422513536, 8634521016, 11499207456

Offset: 1

Luciano Ancora, Feb 05 2015

			Fifth differences:   1, 27,  93,  119,   120, (repeat 120) (A101100)
Fourth differences:  1, 28, 121,  240,   360,   480, ...   (A101095)
Third differences:   1, 29, 150,  390,   750,  1230, ...   (A101096)
Second differences:  1, 30, 180,  570,  1320,  2550, ...   (A101098)
First differences:   1, 31, 211,  781,  2101,  4651, ...   (A022521)
-------------------------------------------------------------------------
The fifth powers:    1, 32, 243, 1024,  3125,  7776, ...   (A000584)
-------------------------------------------------------------------------
First partial sums:  1, 33, 276, 1300,  4425, 12201, ...   (A000539)
Second partial sums: 1, 34, 310, 1610,  6035, 18236, ...   (A101092)
Third partial sums:  1, 35, 345, 1955,  7990, 26226, ...   (A101099)
Fourth partial sums: 1, 36, 381, 2336, 10326, 36552, ...   (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(10,-45,120,-210,252,-210,120,-45,10,-1).

Cf. A000539, A000584, A022521, A101092, A101095, A101096, A101098, A101099, A101100.

Cf. A101091 (fourth partial sums of fourth powers).

List([1..30], n-> Binomial(n+4,5)*(5*(n+2)^4 -35*(n+2)^2 +36)/126); # G. C. Greubel, Aug 28 2019

[Binomial(n+4,5)*(5*(n+2)^4 -35*(n+2)^2 +36)/126: n in [1..30]]; // G. C. Greubel, Aug 28 2019

seq(binomial(n+4,5)*(5*(n+2)^4 -35*(n+2)^2 +36)/126, n=1..30); # G. C. Greubel, Aug 28 2019

Table[n(1+n)(2+n)(3+n)(4+n)(-24 +20n +85n^2 +40n^3 +5n^4)/15120, {n, 30}] (* or *) Accumulate[Accumulate[Accumulate[Accumulate[Range[24]^5]]]] (* or *) CoefficientList[Series[(1 +26x +66x^2 +26x^3 +x^4)/(1-x)^10, {x, 0, 30}], x]
Nest[Accumulate,Range[30]^5,4] (* or *) LinearRecurrence[{10,-45,120, -210,252,-210,120,-45,10,-1}, {1,36,381,2336,10326,36552,110022,292512, 704847,1567852},30] (* Harvey P. Dale, May 08 2016 *)

vector(30, n, m=n+2; binomial(m+2,5)*(5*m^4 -35*m^2 +36)/126) \\ G. C. Greubel, Aug 28 2019

[binomial(n+4,5)*(5*(n+2)^4 -35*(n+2)^2 +36)/126 for n in (1..30)] # G. C. Greubel, Aug 28 2019

G.f.: x*(1 + 26*x + 66*x^2 + 26*x^3 + x^4)/(1 - x)^10.
a(n) = n*(1 + n)*(2 + n)*(3 + n)*(4 + n)*(-24 + 20*n + 85*n^2 + 40*n^3 + 5*n^4)/15120.
a(n) = 4*a(n-1) - 6*a(n-2) + 4*a(n-3) - a(n-4) + n^5.

Edited by Bruno Berselli, Feb 10 2015

A254682 Fifth partial sums of fifth powers (A000584).

Original entry on oeis.org

1, 37, 418, 2754, 13080, 49632, 159654, 452166, 1157013, 2724865, 5988268, 12410476, 24456744, 46132152, 83740980, 146935284, 250134753, 414416277, 669990046, 1059399550, 1641605680, 2497140360, 3734542890, 5498322570

Offset: 1

Luciano Ancora, Feb 12 2015

			Fifth differences:   1, 27,  93,  119,   120, (repeat 120) (A101100)
Fourth differences:  1, 28, 121,  240,   360,   480, ...   (A101095)
Third differences:   1, 29, 150,  390,   750,  1230, ...   (A101096)
Second differences:  1, 30, 180,  570,  1320,  2550, ...   (A101098)
First differences:   1, 31, 211,  781,  2101,  4651, ...   (A022521)
-------------------------------------------------------------------------
The fifth powers:    1, 32, 243, 1024,  3125,  7776, ...   (A000584)
-------------------------------------------------------------------------
First partial sums:  1, 33, 276, 1300,  4425, 12201, ...   (A000539)
Second partial sums: 1, 34, 310, 1610,  6035, 18236, ...   (A101092)
Third partial sums:  1, 35, 345, 1955,  7990, 26226, ...   (A101099)
Fourth partial sums: 1, 36, 381, 2336, 10326, 36552, ...   (A254644)
Fifth partial sums:  1, 37, 418, 2754, 13080, 49632, ...   (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. A000539, A000584, A022521, A101092, A101095, A101096, A101098, A101099, A101100, A254644, A254681, A254683, A254684.

Table[n (1 + n) (2 + n) (3 + n) (4 + n) (5 + n) (- 2 + 5 n + n^2) (9 + 10 n + 2 n^2)/60480, {n,24}] (* or *)
CoefficientList[Series[(- 1 - 26 x - 66 x^2 - 26 x^3 - x^4)/(- 1 + x)^11, {x,0,23}], x]
Nest[Accumulate,Range[30]^5,5] (* or *) LinearRecurrence[{11,-55,165,-330,462,-462,330,-165,55,-11,1},{1,37,418,2754,13080,49632,159654,452166,1157013,2724865,5988268},30] (* Harvey P. Dale, Jan 30 2019 *)

a(n)=n*(1+n)*(2+n)*(3+n)*(4+n)*(5+n)*(-2+5*n+n^2)*(9+10*n+2*n^2)/60480 \\ Charles R Greathouse IV, Oct 07 2015

G.f.: (- x - 26*x^2 - 66*x^3 - 26*x^4 - x^5)/(- 1 + x)^11.
a(n) = n*(1 + n)*(2 + n)*(3 + n)*(4 + n)*(5 + n)*(- 2 + 5*n + n^2)*(9 + 10*n + 2*n^2)/60480.
a(n) = 5*a(n-1) - 10*a(n-2) + 10*a(n-3) - 5*a(n-4) + a(n-5) + n^5.
Sum_{n>=1} 1/a(n) = 475867/180 - (2560/13)*sqrt(7)*Pi*tan(sqrt(7)*Pi/2) + (210/13)*sqrt(3/11)*Pi*tan(sqrt(33)*Pi/2). - Amiram Eldar, Jan 27 2022

A101100 The first summation of row 5 of Euler's triangle - a row that will recursively accumulate to the power of 5.

Original entry on oeis.org

1, 27, 93, 119, 120, 120, 120, 120, 120, 120, 120, 120, 120, 120, 120, 120, 120, 120, 120, 120, 120, 120, 120, 120, 120, 120, 120, 120, 120, 120, 120, 120, 120, 120, 120

Offset: 1

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

Miklos Bona, editor, Handbook of Enumerative Combinatorics, CRC Press, 2015, page 533.

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.
Eric Weisstein's World of Mathematics Worpitzky's Identity of 1883.
Eric Weisstein's World of Mathematics Eulerian Number.
Eric Weisstein's World of Mathematics Nexus number.
Eric Weisstein's World of Mathematics Finite Differences.
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, A101095, A101096, A101098, A022521, A000584, A000539, A101092, A101099. Above: A101104, this sequence.

Within the "cube" of related sequences with construction based upon SeriesAtLevelR formula, with n downward, x rightward and r backward: Before: this sequence, A101095, A101096, A101098, A022521, A000584, A000539, A101092, A101099.

R:=PowerSeriesRing(Integers(), 40); Coefficients(R!( x*(1+26*x+66*x^2+26*x^3+x^4)/(1-x) )); // G. C. Greubel, May 07 2019

MagicNKZ=Sum[(-1)^j*Binomial[n+1-z, j]*(k-j+1)^n, {j, 0, k+1}]; Table[MagicNKZ, {n, 5, 5}, {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, 5, 5}, {r, -5, -5}, {x, 5, 35}]

{a(n) = if(n==1, 1, if(n==2, 27, if(n==3, 93, if(n==4, 119, 120))) )}; \\ G. C. Greubel, May 07 2019

a=(x*(1+26*x+66*x^2+26*x^3+x^4)/(1-x)).series(x, 40).coefficients(x, sparse=False); a[1:] # G. C. Greubel, May 07 2019

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

cp's OEIS Frontend

A047969 Square array of nexus numbers a(n,k) = (n+1)^(k+1) - n^(k+1) (n >= 0, k >= 0) read by upwards antidiagonals.

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A101096 Third differences of fifth powers (A000584).

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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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A254644 Fourth partial sums of fifth powers (A000584).

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A254682 Fifth partial sums of fifth powers (A000584).

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A101100 The first summation of row 5 of Euler's triangle - a row that will recursively accumulate to the power of 5.

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