A000538 -id:A000538 - cp's OEIS frontend

A101089 Second partial sums of fourth powers (A000583).

1, 18, 116, 470, 1449, 3724, 8400, 17172, 32505, 57838, 97812, 158522, 247793, 375480, 553792, 797640, 1125009, 1557354, 2120020, 2842686, 3759833, 4911236, 6342480, 8105500, 10259145, 12869766, 16011828, 19768546, 24232545, 29506544

Offset: 1

Cecilia Rossiter, Dec 14 2004

a(n) is the n-th antidiagonal sum of the convolution array A213553. - Clark Kimberling, Jun 17 2012
a(n-1)/n^5 is the "retention" of water on a 3 X 3 random surface of n levels - see Knecht et al., 2012, Schrenk et al., 2014. - Robert M. Ziff, Mar 08 2014
The general formula for the second partial sums of m-th powers is: b(n,m) = (n+1)*F(m) - F(m+1), where F(m) is the m-th Faulhaber’s polynomial. - Luciano Ancora, Jan 26 2015

			a(7) = 8400 = 1*(8-1)^4 + 2*(8-2)^4 + 3*(8-3)^4 + 4*(8-4)^4 + 5*(8-5)^4 + 6*(8-6)^4 + 7*(8-7)^4. - _Bruno Berselli_, Jan 31 2014

Vincenzo Librandi, Table of n, a(n) for n = 1..1000
Luciano Ancora, Recurrence relation for the second partial sums of m-th powers.
Luciano Ancora, Second partial sums of the m-th powers.
Craig L. Knecht, Walter Trump, Daniel ben-Avraham and Robert M. Ziff, Retention Capacity of Random Surfaces, Phys. Rev. Lett., Vol. 108 (2012), 045703.
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.
C. P. Neuman and D. I. Schonbach, Evaluation of sums of convolved powers using Bernoulli numbers, SIAM Rev., Vol. 19, No. 1 (1977), pp. 90-99. MR0428678 (55 #1698). See Table 1. - _N. J. A. Sloane_, Mar 23 2014
Claudio de J. Pita Ruiz V., Some Number Arrays Related to Pascal and Lucas Triangles, J. Int. Seq., Vol. 16 (2013), Article 13.5.7.
Cecilia Rossiter, Depictions, Explorations and Formulas of the Euler/Pascal Cube. [Dead link]
Cecilia Rossiter, Depictions, Explorations and Formulas of the Euler/Pascal Cube. [Cached copy, May 15 2013]
K. J. Schrenk, N. A. M. Araújo, R. M. Ziff and H. J. Herrmann Retention Capacity of Correlated Surfaces, arXiv:1403.2082 [cond-mat.stat-mech], 2014.
Index entries for linear recurrences with constant coefficients, signature (7,-21,35,-35,21,-7,1).

Partial sums of A000538.

Cf. A000583, A002415, A101090, A201126, A213553.

List([1..40], n-> (n+1)^2*(2*(n+1)^4-5*(n+1)^2+3)/60); # G. C. Greubel, Jul 31 2019

[(1/60)*n*(n+1)^2*(n+2)*(2*n*(n+2)-1): n in [1..40]]; // Vincenzo Librandi, Mar 24 2014

f:=n->(2*n^6-5*n^4+3*n^2)/60;
[seq(f(n),n=0..50)]; # N. J. A. Sloane, Mar 23 2014

a[n_] := n(n+1)^2(n+2)(2n(n+2) -1)/60; Table[a[n], {n, 40}]
CoefficientList[Series[(1+x)*(1+10*x+x^2)/(1-x)^7, {x,0,40}], x] (* Vincenzo Librandi, Mar 24 2014 *)
Nest[Accumulate[#]&,Range[30]^4,2] (* Harvey P. Dale, Aug 13 2024 *)

a(n)=n*(n+1)^2*(n+2)*(2*n*(n+2)-1)/60 \\ Charles R Greathouse IV, Mar 18 2014

[n*(n+1)^2*(n+2)*(2*n*(n+2)-1)/60 for n in range(1,40)] # Danny Rorabaugh, Apr 20 2015

a(n) = (1/60)*n*(n+1)^2*(n+2)*(2*n*(n+2)-1).
G.f.: x*(1+x)*(1+10*x+x^2)/(1-x)^7. - Colin Barker, Apr 04 2012
a(n) = Sum_{i=1..n} i*(n+1-i)^4, by the definition. - Bruno Berselli, Jan 31 2014
a(n) = 2*a(n-1) - a(n-2) + n^4. - Luciano Ancora, Jan 08 2015
Sum_{n>=1} 1/a(n) = 85/3 + 10*Pi^2/3 - 20*sqrt(2/3)*Pi*cot(sqrt(3/2)*Pi). - Amiram Eldar, Jan 26 2022
a(n) = (1/2)*Sum_{1 <= i, j <= n+1} (i - j)^4 - Peter Bala, Jun 11 2024

Edited by Ralf Stephan, Dec 16 2004

A323542 a(n) = Product_{k=0..n} (k^4 + (n-k)^4).

Original entry on oeis.org

0, 1, 512, 1896129, 14101250048, 242755875390625, 7888809923487203328, 452522453429009743939201, 42521926771106843499966758912, 6212193882217859346149080691430849, 1350441156698962215630405632000000000000, 421551664651621436548685508587919503984205889

Offset: 0

Vaclav Kotesovec, Jan 17 2019

nonn

Seiichi Manyama, Table of n, a(n) for n = 0..125

Cf. A272247, A323497, A323540, A323541, A323543, A323544, A323545, A323546.

Cf. 2*A000538 and A259108 (with sum instead of product).

[(&*[(k^4 + (n-k)^4): k in [0..n]]): n in [0..15]]; // Vincenzo Librandi, Jan 18 2019

Table[Product[k^4+(n-k)^4, {k, 0, n}], {n, 0, 15}]

m=4; vector(15, n, n--; prod(k=0,n, k^m + (n-k)^m)) \\ G. C. Greubel, Jan 18 2019

m=4; [product(k^m +(n-k)^m for k in (0..n)) for n in (0..15)] # G. C. Greubel, Jan 18 2019

a(n) ~ exp((Pi*(sqrt(2) - 1/2) - 4)*n) * n^(4*n + 4).

A119617 Integers of the form c(n)/b(n) where c(n+1)=c(n)+(n+1)^4 with c(0)=1 and b(n+1)=b(n)+(n+1)^2 with b(0)=1.

Original entry on oeis.org

1, 7, 25, 43, 79, 109, 163, 205, 277, 331, 421, 487, 595, 673, 799, 889, 1033, 1135, 1297, 1411, 1591, 1717, 1915, 2053, 2269, 2419, 2653, 2815, 3067, 3241, 3511, 3697, 3985, 4183, 4489, 4699, 5023, 5245, 5587, 5821, 6181, 6427, 6805, 7063, 7459, 7729

Offset: 1

Paolo P. Lava and Giorgio Balzarotti, Jun 06 2006

The sequence is the union of A134153 and A134154 (without the first term of A134154): A134153(0)=1, A134154(1)=7, A134153(1)=25, A134154(2)=43, A134153(2)=79 and so on.

			c(0)/b(0) = 1/1 =1.
c(3)/b(3) = (1+2^4+3^4)/(1+2^2+3^2)= (1+16+81)/(1+4+9) = 98/14 = 7.

Vincenzo Librandi, Table of n, a(n) for n = 1..10000
Index entries for linear recurrences with constant coefficients, signature (1,2,-2,-1,1).

Cf. A000538, A000330.

[(30*n*(n-1)-3*(2*n-1)*(-1)^n+5)/8: n in [1..46]]; // Bruno Berselli, Jun 27 2011

P:=proc(n) local f,i,j,nu,de; nu:=0; de:=0; for i from 1 by 1 to n do nu:=nu+i^4; de:=de+i^2; f:=nu/de; if trunc(f)=f then print(f); fi; od; end: P(1000);

LinearRecurrence[{1,2,-2,-1,1},{1,7,25,43,79},50] (* Harvey P. Dale, Jan 21 2017 *)

makelist((30*n*(n-1)-3*(2*n-1)*(-1)^n+5)/8,n,1,46); /* Bruno Berselli, Jun 27 2011 */

for(n=1,46, print1((30*n*(n-1)-3*(2*n-1)*(-1)^n+5)/8", ")); \\ Bruno Berselli, Jun 27 2011

From Bruno Berselli, Jun 27 2011: (Start)
G.f.: x*(1+6*x+16*x^2+6*x^3+x^4)/((1+x)^2*(1-x)^3).
a(n) = (30*n*(n-1)-3*(2*n-1)*(-1)^n+5)/8. (End)

A134153 a(n) = 15n^2 + 9n + 1.

Original entry on oeis.org

1, 25, 79, 163, 277, 421, 595, 799, 1033, 1297, 1591, 1915, 2269, 2653, 3067, 3511, 3985, 4489, 5023, 5587, 6181, 6805, 7459, 8143, 8857, 9601, 10375, 11179, 12013, 12877, 13771, 14695, 15649, 16633, 17647, 18691, 19765, 20869, 22003, 23167

Offset: 0

Artur Jasinski, Oct 10 2007

A119617 is union of A134153 and A134154. A000538(n) is divisible by A000330(n) if and only n is congruent to {1, 3} mod 5 (see A047219). A134154(n) is case when n is congruent to 1 mod 5 see cases 2)

Index entries for linear recurrences with constant coefficients, signature (3,-3,1).

Cf. A119617, A134154.

Table[1 + 9 n + 15 n^2, {n, 0, 50}]
Table[Sum[k^4, {k, 1, 5m + 1}]/Sum[k^2, {k, 1, 5m + 1}], {m, 0, 10}]

a(n)=15*n^2+9*n+1 \\ Charles R Greathouse IV, Jun 17 2017

a(n) = 15*n^2 + 9*n + 1.
a(n) = (3*(5*n + 1)^2 + 3*(5*n + 1) - 1)/5.
a(n) = (Sum_{k=1..5*n+1} k^4) / (Sum_{k=1..5*n+1} k^2).
G.f.: -(1+22*x+7*x^2)/(-1+x)^3. - R. J. Mathar, Nov 14 2007

Offset corrected and some punctuation added by R. J. Mathar, Jul 09 2009

A134154 a(n) = 15n^2 - 9n + 1.

Original entry on oeis.org

1, 7, 43, 109, 205, 331, 487, 673, 889, 1135, 1411, 1717, 2053, 2419, 2815, 3241, 3697, 4183, 4699, 5245, 5821, 6427, 7063, 7729, 8425, 9151, 9907, 10693, 11509, 12355, 13231, 14137, 15073, 16039, 17035, 18061, 19117, 20203, 21319, 22465, 23641

Offset: 0

Artur Jasinski, Oct 10 2007

A119617 is union of A134153 and A134154 A000538(n) is divisible by A000330(n) if and only n is congruent to {1, 3} mod 5 (see A047219) A134154(n) is case when n is congruent to 3 mod 5 see cases 2)

Index entries for linear recurrences with constant coefficients, signature (3,-3,1).

Cf. A000538, A119617, A134153.

Table[1 - 9 n + 15 n^2, {n, 0, 50}]
Table[Sum[k^4, {k, 1, 5m + 3}]/Sum[k^2, {k, 1, 5m + 3}], {m, 0, 30}]

a(n)=15*n^2-9*n+1 \\ Charles R Greathouse IV, Jun 17 2017

a(n) = 15*n^2 - 9*n + 1.
a(n+1) = (3*(5*n + 3)^2 + 3*(5*n + 3) - 1)/5.
a(n+1) = (Sum_{k=1..5*n+3} k^4) / (Sum_{k=1..5*n+3} k^2).
G.f.: -(1+4*x+25*x^2)/(-1+x)^3. - R. J. Mathar, Nov 14 2007

A215083 Triangle T(n,k) = sum of the k first n-th powers.

Original entry on oeis.org

0, 0, 1, 0, 1, 5, 0, 1, 9, 36, 0, 1, 17, 98, 354, 0, 1, 33, 276, 1300, 4425, 0, 1, 65, 794, 4890, 20515, 67171, 0, 1, 129, 2316, 18700, 96825, 376761, 1200304, 0, 1, 257, 6818, 72354, 462979, 2142595, 7907396, 24684612, 0, 1, 513, 20196, 282340, 2235465, 12313161, 52666768, 186884496, 574304985, 0, 1, 1025, 60074, 1108650, 10874275, 71340451, 353815700, 1427557524, 4914341925, 14914341925

Offset: 0

Olivier Gérard, Aug 02 2012

First term T(0,0) = 0 can be computed as 1 if one starts the sum at j=0 and take the convention 0^0 = 1.

			Triangle starts (using the convention 0^0 = 1, see the first comment):
[0] 1
[1] 0, 1
[2] 0, 1,  5
[3] 0, 1,  9,  36
[4] 0, 1, 17,  98,  354
[5] 0, 1, 33, 276, 1300,  4425
[6] 0, 1, 65, 794, 4890, 20515, 67171

Vincenzo Librandi, Table of n, a(n) for n = 0..1000

Row sums are A215083.
A215078 is the product of this array with the binomial array.
T(3,k) is the beginning of A000537.
T(4,k) is the beginning of A000538.
T(5,k) is the beginning of A000539.
Cf. A103438.

A215083 := (n, k) -> add(i^n, i=0..k):
for n from 0 to 8 do seq(A215083(n, k), k=0..n) od; # Peter Luschny, Oct 02 2017

Flatten[Table[Table[Sum[j^n, {j, 1, k}], {k, 0, n}], {n, 0, 10}], 1]
Table[ HarmonicNumber[k, -n], {n, 0, 10}, {k, 0, n}] // Flatten (* Jean-François Alcover, Mar 05 2013 *)

T(n, k) = Sum_{j=1..k} j^n

Sum_{j=0..n}((-1)^(n-j)/(j+1)*binomial(n+1,j+1)*T(n,j)) are the Bernoulli numbers B(n) = B(n, 1) by a formula of L. Kronecker. - Peter Luschny, Oct 02 2017

A254681 Fifth partial sums of fourth powers (A000583).

Original entry on oeis.org

1, 21, 176, 936, 3750, 12342, 35112, 89232, 207207, 446875, 906048, 1743248, 3206268, 5670588, 9690000, 16062144, 25912029, 40797009, 62837104, 94875000, 140670530, 205134930, 294610680, 417203280, 583171875, 805386231

Offset: 1

Luciano Ancora, Feb 12 2015

			Fourth differences:  1, 12,  23,  24, (repeat 24)  ...   (A101104)
Third differences:   1, 13,  36,  60,   84,   108, ...   (A101103)
Second differences:  1, 14,  50, 110,  194,   302, ...   (A005914)
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, ...   (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. A000538, A000583, A005914, A005917, A101089, A101090, A101091, A101103, A101104, A254682, A254683, A254684.

[Binomial(n+5,6)*n*(n+5)*(2*n+5)/42: n in [1..30]]; // G. C. Greubel, Dec 01 2018

seq(coeff(series((x+11*x^2+11*x^3+x^4)/(1-x)^10,x,n+1), x, n), n = 1 .. 30); # Muniru A Asiru, Dec 02 2018

Table[n^2(1+n)(2+n)(3+n)(4+n)(5+n)^2(5+2n)/30240, {n,26}] (* or *)
CoefficientList[Series[(1 + 11 x + 11 x^2 + x^3)/(1-x)^10, {x,0,25}], x]
CoefficientList[Series[(1/30240)E^x (30240 + 604800 x + 2041200 x^2 + 2368800 x^3 + 1233540 x^4 + 326592 x^5 + 46410 x^6 + 3540 x^7 + 135 x^8 + 2 x^9), {x, 0, 50}], x]*Table[n!, {n, 0, 50}] (* Stefano Spezia, Dec 02 2018 *)
Nest[Accumulate[#]&,Range[30]^4,5] (* Harvey P. Dale, Jan 03 2022 *)

my(x='x+O('x^30)); Vec((x+11*x^2+11*x^3+x^4)/(1-x)^10) \\ G. C. Greubel, Dec 01 2018

[binomial(n+5,6)*n*(n+5)*(2*n+5)/42 for n in (1..30)] # G. C. Greubel, Dec 01 2018

G.f.:(x + 11*x^2 + 11*x^3 + x^4)/(1 - x)^10.
a(n) = n^2*(1 + n)*(2 + n)*(3 + n)*(4 + n)*(5 + n)^2*(5 + 2*n)/30240.
a(n) = 5*a(n-1) - 10*a(n-2) + 10*a(n-3) - 5*a(n-4) + a(n-5) + n^4.
E.g.f.: (1/30240)*exp(x)*(30240 + 604800*x + 2041200*x^2 + 2368800*x^3 + 1233540*x^4 + 326592*x^5 + 46410*x^6 + 3540*x^7 + 135*x^8 + 2*x^9). - Stefano Spezia, Dec 02 2018
From Amiram Eldar, Jan 26 2022: (Start)
Sum_{n>=1} 1/a(n) = 172032*log(2)/125 - 2382233/2500.
Sum_{n>=1} (-1)^(n+1)/a(n) = 42*Pi^2/25 - 43008*Pi/125 + 2663213/2500. (End)

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

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

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

cp's OEIS Frontend

A101089 Second partial sums of fourth powers (A000583).

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A323542 a(n) = Product_{k=0..n} (k^4 + (n-k)^4).

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A119617 Integers of the form c(n)/b(n) where c(n+1)=c(n)+(n+1)^4 with c(0)=1 and b(n+1)=b(n)+(n+1)^2 with b(0)=1.

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A134153 a(n) = 15*n^2 + 9*n + 1.

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A134154 a(n) = 15*n^2 - 9*n + 1.

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A215083 Triangle T(n,k) = sum of the k first n-th powers.

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A134153 a(n) = 15n^2 + 9n + 1.

A134154 a(n) = 15n^2 - 9n + 1.