A000539 -id:A000539 - cp's OEIS frontend

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

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

A023002 Sum of 10th powers.

Original entry on oeis.org

0, 1, 1025, 60074, 1108650, 10874275, 71340451, 353815700, 1427557524, 4914341925, 14914341925, 40851766526, 102769130750, 240627622599, 529882277575, 1106532668200, 2206044295976, 4222038196425, 7792505423049, 13923571680850

Offset: 0

David W. Wilson

T. D. Noe, Table of n, a(n) for n = 0..1000
Bruno Berselli, A description of the recursive method in Formula lines (second formula): website Matem@ticamente (in Italian).
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. 13.
Eric Weisstein's World of Mathematics, Power Sum.
Index entries for linear recurrences with constant coefficients, signature (12,-66,220,-495,792,-924,792,-495,220,-66,12,-1).

Sequences of the form Sum_{j=0..n} j^m : A000217 (m=1), A000330 (m=2), A000537 (m=3), A000538 (m=4), A000539 (m=5), A000540 (m=6), A000541 (m=7), A000542 (m=8), A007487 (m=9), this sequence (m=10), A123095 (m=11), A123094 (m=12), A181134 (m=13).
Row 10 of array A103438.
Cf. A215083, A069095.

[&+[n^10: n in [0..m]]: m in [0..19]];  // Bruno Berselli, Aug 23 2011

A023002:= n-> bernoulli(11, n+1)/11; seq(A023002(n), n=0..30); # G. C. Greubel, Jul 21 2021

Table[Sum[k^10, {k, n}], {n, 0, 30}] (* Vladimir Joseph Stephan Orlovsky, Aug 14 2008 *)
Accumulate[Range[0,20]^10] (* Harvey P. Dale, Aug 23 2011 *)

a(n)=(6*x^11+33*x^10+55*x^9-66*x^7+66*x^5-33*x^3+5*x)/66 \\ Charles R Greathouse IV, Aug 23 2011

a(n)=sum(i=0,10,binomial(11,i)*bernfrac(i)*n^(11-i))/11+n^10 \\ Charles R Greathouse IV, Aug 23 2011

A023002_list, m = [0], [3628800, -16329600, 30240000, -29635200, 16435440, -5103000, 818520, -55980, 1022, -1, 0 , 0]
for _ in range(20):
    for i in range(11):
        m[i+1]+= m[i]
    A023002_list.append(m[-1])
print(A023002_list) # Chai Wah Wu, Nov 05 2014

[bernoulli_polynomial(n,11)/11 for n in range(2, 21)]# Zerinvary Lajos, May 17 2009

a(n) = n*(n+1)*(2*n+1)*(n^2+n-1)(3*n^6 +9*n^5 +2*n^4 -11*n^3 +3*n^2 +10*n -5)/66 (see MathWorld, Power Sum, formula 40). - Bruno Berselli, Apr 26 2010
a(n) = n*A007487(n) - Sum_{i=0..n-1} A007487(i). - Bruno Berselli, Apr 27 2010
From Bruno Berselli, Aug 23 2011: (Start)
a(n) = -a(-n-1).
G.f.: x*(1+x)*(1 +1012*x +46828*x^2 +408364*x^3 +901990*x^4 +408364*x^5 +46828*x^6 +1012*x^7 +x^8)/(1-x)^12. (End)
a(n) = (-1)*Sum_{j=1..10} j*Stirling1(n+1,n+1-j)*Stirling2(n+10-j,n). - Mircea Merca, Jan 25 2014
a(n) = Sum_{i=1..n} J_10(i)*floor(n/i), where J_10 is A069095. - Ridouane Oudra, Jul 17 2025

A059977 a(n) = binomial(n+2, 2)^4.

Original entry on oeis.org

1, 81, 1296, 10000, 50625, 194481, 614656, 1679616, 4100625, 9150625, 18974736, 37015056, 68574961, 121550625, 207360000, 342102016, 547981281, 855036081, 1303210000, 1944810000, 2847396321, 4097152081, 5802782976, 8100000000, 11156640625, 15178486401

Offset: 0

Robert G. Wilson v, Mar 06 2001

nonn

Number of 4-dimensional cage assemblies.

See Chap. 61, "Hyperspace Prisons", of C. Pickover's book "Wonders of Numbers" for full explanation of "cage numbers."

			1 = (1 + 1)/2, 81 = (33 + 129)/2, 1296 = (276 + 2316)/2, 10000 = (1300 + 18700)/2, ... - _Philippe Deléham_, May 25 2015

Clifford A. Pickover, "Wonders of Numbers, Adventures in Mathematics, Mind and Meaning," Oxford University Press, 2001, p. 325.

Harry J. Smith, Table of n, a(n) for n = 0..1000
Clifford A. Pickover, "Wonders of Numbers, Adventures in Mathematics, Mind and Meaning," Zentralblatt review.
Index entries for linear recurrences with constant coefficients, signature (9,-36,84,-126,126,-84,36,-9,1).

Cf. A000217, A000539, A000541, A059827, A059860.

Cf. A168364 (first differences).

with (combinat):seq(mul(stirling2(n+1,n),k=1..4),n=1..24); # Zerinvary Lajos, Dec 16 2007

m = 4; Table[ ( (n^m)(n + 1)^m )/(2^m), {n, 1, 30} ]

a(n) = { ((n + 1)*(n + 2)/2)^4 } \\ Harry J. Smith, Jun 30 2009

[stirling_number2(n+1,n)^4for n in range(1,25)] # Zerinvary Lajos, Mar 14 2009

L(n) = ((n^m)(n + 1)^m)/(2^m) where m is the dimension, which in this case is 4.
O.g.f.: -(1+72*x+603*x^2+1168*x^3+603*x^4+72*x^5+x^6)/(-1+x)^9. - R. J. Mathar, Mar 31 2008
a(n) = A000217(n+1)^4. - R. J. Mathar, Dec 13 2011
a(n) = (A000539(n+1) + A000541(n+1))/2. - Philippe Deléham, May 25 2015
From Amiram Eldar, May 15 2022: (Start)
Sum_{n>=0} 1/a(n) = 160*Pi^2/3 + 16*Pi^4/45 - 560.
Sum_{n>=0} (-1)^n/a(n) = 560 - 640*log(2) - 96*zeta(3). (End)

Better definition from Zerinvary Lajos, May 23 2006

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

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

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

A154231 Triangle T(n, k) = T(n-1, k) + T(n-1, k-1) + ((n+1)^2(n+2)^2(2n^2 +6n +3)/12)*T(n-2, k-1), read by rows.

Original entry on oeis.org

1, 1, 1, 1, 278, 1, 1, 1579, 1579, 1, 1, 6005, 1233308, 6005, 1, 1, 18207, 20504692, 20504692, 18207, 1, 1, 47216, 194715939, 35816807848, 194715939, 47216, 1, 1, 108993, 1319518787, 1302709376779, 1302709376779, 1319518787, 108993, 1, 1, 229819, 7024500980, 24830582225241, 4330171226988158, 24830582225241, 7024500980, 229819, 1

Offset: 0

Roger L. Bagula, Jan 05 2009

Row sums are: {1, 2, 280, 3160, 1245320, 41045800, 36206334160, ...}.

The row sums of this class of sequences (see cross-references) is given by the following. Let S(n) be the row sum then S(n) = 2*S(n-1) + f(n)*S(n-2) for a given f(n). For this sequence f(n) = (n+1)^2*(n+2)^2*(2*n^2 +6*n +3)/12 = A000539(n+1). - G. C. Greubel, Mar 02 2021

			Triangle begins as:
  1;
  1,      1;
  1,    278,          1;
  1,   1579,       1579,             1;
  1,   6005,    1233308,          6005,             1;
  1,  18207,   20504692,      20504692,         18207,          1;
  1,  47216,  194715939,   35816807848,     194715939,      47216,      1;
  1, 108993, 1319518787, 1302709376779, 1302709376779, 1319518787, 108993, 1;

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

Cf. A154227, A154228, A154229, A154230, A154233.

Cf. A000539 (powers of 5).

f:= func< n | Binomial(n+2,2)^2*(2*n^2+6*n+3)/3 >;
function T(n,k)
  if k eq 0 or k eq n then return 1;
  else return T(n-1, k) + T(n-1, k-1) + f(n)*T(n-2, k-1);
  end if; return T;
end function;
[T(n,k): k in [0..n], n in [0..12]]; // G. C. Greubel, Mar 02 2021

T:= proc(n, k) option remember;
      if k=0 or k=n then 1
    else T(n-1, k) + T(n-1, k-1) + ((n+1)^2*(n+2)^2*(2*n^2+6*n+3)/12)*T(n-2, k-1)
      fi; end:
seq(seq(T(n, k), k=0..n), n=0..12); # G. C. Greubel, Mar 02 2021

T[n_, k_]:= T[n,k]= If[k==0 || k==n, 1, T[n-1, k] + T[n-1, k-1] + ((n+1)^2*(n+2)^2*(2*n^2+6*n+3)/12)*T[n-2, k-1] ];
Table[T[n, k], {n,0,12}, {k,0,n}]//Flatten (* modified by G. C. Greubel, Mar 02 2021 *)

def f(n): return binomial(n+2,2)^2*(2*n^2+6*n+3)/3
def T(n,k):
    if (k==0 or k==n): return 1
    else: return T(n-1, k) + T(n-1, k-1) + f(n)*T(n-2, k-1)
flatten([[T(n,k) for k in (0..n)] for n in (0..12)]) # G. C. Greubel, Mar 02 2021

T(n, k) = T(n-1, k) + T(n-1, k-1) + ((n+1)^2*(n+2)^2*(2*n^2 +6*n +3)/12)*T(n-2, k-1) with T(n, 0) = T(n, n) = 1.

Edited by G. C. Greubel, Mar 02 2021

A254471 Sixth partial sums of fifth powers (A000584).

Original entry on oeis.org

1, 38, 456, 3210, 16290, 65922, 225576, 677742, 1834755, 4559620, 10547888, 22958364, 47415108, 93547260, 177288240, 324223524, 574358277, 988774554, 1658764600, 2718164150, 4359769830, 6856910190, 10591453080, 16089775650, 24068499975, 35492110056

Offset: 1

Luciano Ancora, Feb 15 2015

			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, ... (A254682)
Sixth partial sums:  1, 38, 456, 3210, 16290, 65922, ... (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 (12,-66,220,-495,792,-924,792,-495,220,-66,12,-1).

Cf. A000539, A000584, A022521, A101092, A101099, A254469, A254470, A254472, A254644, A254682.

[n*(1+n)*(2+n)*(3+n)*(4+n)*(5+n)*(6+n)*(-29+54*n+ 81*n^2+24*n^3+2*n^4)/665280: n in [1..30]]; // Vincenzo Librandi, Feb 15 2015

Table[n (1 + n) (2 + n) (3 + n) (4 + n) (5 + n) (6 + n) (- 29 + 54 n + 81 n^2 + 24 n^3 + 2 n^4)/665280, {n, 23}] (* or *) CoefficientList[Series[(1 + 26 x + 66 x^2 + 26 x^3 + x^4)/(- 1 + x)^12, {x, 0, 28}], x]

vector(50,n,n*(1 + n)*(2 + n)*(3 + n)*(4 + n)*(5 + n)*(6 + n)*(-29 + 54*n + 81*n^2 + 24*n^3 + 2*n^4)/665280) \\ Derek Orr, Feb 19 2015

G.f.: (x + 26*x^2 + 66*x^3 + 26*x^4 + x^5)/(- 1 + x)^12.
a(n) = n*(1 + n)*(2 + n)*(3 + n)*(4 + n)*(5 + n)*(6 + n)*(-29 + 54*n + 81*n^2 + 24*n^3 + 2*n^4)/665280.
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^5.

A062393 a(n) = n^5 - (n-1)^5 + (n-2)^5 - ... +(-1)^n*0^5.

Original entry on oeis.org

0, 1, 31, 212, 812, 2313, 5463, 11344, 21424, 37625, 62375, 98676, 150156, 221137, 316687, 442688, 605888, 813969, 1075599, 1400500, 1799500, 2284601, 2869031, 3567312, 4395312, 5370313, 6511063, 7837844, 9372524, 11138625, 13161375

Offset: 0

Henry Bottomley, Jun 21 2001

The general formula for alternating sums of powers is in terms of the Swiss-Knife polynomials P(n,x) A153641 2^(-n-1)(P(n,1)-(-1)^k P(n,2k+1)). Thus a(k) = |2^(-6)(P(5,1)-(-1)^k P(5,2k+1))|. - Peter Luschny, Jul 12 2009

Harry J. Smith, Table of n, a(n) for n = 0..1000
Index entries for linear recurrences with constant coefficients, signature (5,-9,5,5,-9,5,-1).

Cf. A000539, A000584. A062392 for 4th powers, A152725 for 6th powers.

a := n -> (1-(-1)^n+n^2*(n^2*(2*n+5)-5))/4; # Peter Luschny, Jul 12 2009

k=0;lst={k};Do[k=n^5-k;AppendTo[lst, k], {n, 1, 5!}];lst (* Vladimir Joseph Stephan Orlovsky, Dec 11 2008 *)
Table[Total[(Times@@@Partition[Riffle[Range[n,0,-1],{1,-1},{2,-1,2}],2])^5],{n,0,30}] (* or *) LinearRecurrence[ {5,-9,5,5,-9,5,-1},{0,1,31,212,812,2313,5463},40] (* Harvey P. Dale, Feb 01 2013 *)

{ a=0; for (n=0, 1000, write("b062393.txt", n, " ", a=n^5 - a) ) } \\ Harry J. Smith, Aug 07 2009

a(n) = (2*n^5 + 5*n^4 - 5*n^2 + 1 - (-1)^n)/4 = n^5 - a(n-1).
G.f.: x*(x^4 + 26*x^3 + 66*x^2 + 26*x + 1)/((x-1)^6*(x+1)). - Colin Barker, Sep 19 2012
a(0)=0, a(1)=1, a(2)=31, a(3)=212, a(4)=812, a(5)=2313, a(6)=5463, a(n) = 5*a(n-1) - 9*a(n-2) + 5*a(n-3) + 5*a(n-4) - 9*a(n-5) + 5*a(n-6) - a(n-7). - Harvey P. Dale, Feb 01 2013

cp's OEIS Frontend

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

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A023002 Sum of 10th powers.

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A059977 a(n) = binomial(n+2, 2)^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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A101095 Fourth difference of fifth powers (A000584).

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A154231 Triangle T(n, k) = T(n-1, k) + T(n-1, k-1) + ((n+1)^2(n+2)^2(2n^2 +6n +3)/12)*T(n-2, k-1), read by rows.