A022521 -id:A022521 - cp's OEIS frontend

A254644 Fourth partial sums of fifth powers (A000584).

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

A178487 a(n) = floor(n^(1/5)): integer part of fifth root of n.

Original entry on oeis.org

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

Offset: 0

M. F. Hasler, Oct 09 2010

Each term k appears (k+1)^5 - k^5 times consecutively (A022521). - Bernard Schott, Mar 07 2023

Sequences a(n) = floor(n^(1/k)): A001477 (k=1), A000196 (k=2), A048766 (k=3), A255270 (k=4), this sequence (k= 5), A178489 (k=6), A057427 (k->oo).

Cf. A022521, A253206.

[n eq 0 select 0 else Iroot(n, 5): n in [0..110]]; // Bruno Berselli, Feb 20 2015

seq(floor(n^(1/5)), n=0..100); # Ridouane Oudra, Feb 26 2023

Floor[Range[0,120]^(1/5)] (* Harvey P. Dale, Aug 15 2012 *)

PARI
```
A178487(n)=floor(sqrtn(n+.5,5))
```

a(n) = sqrtnint(n, 5); \\ Michel Marcus, Dec 22 2016

from sympy import integer_nthroot
def A178487(n): return integer_nthroot(n,5)[0] # Chai Wah Wu, Jun 06 2025

G.f.: Sum_{k>=1} x^(k^5)/(1 - x). - Ilya Gutkovskiy, Dec 22 2016

a(n) = Sum_{i=1..n} A253206(i)*floor(n/i). - Ridouane Oudra, Feb 26 2023

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

A198064 Triangle read by rows (n >= 0, 0 <= k <= n, m = 4); T(n,k) = Sum{j=0..m} Sum{i=0..m} (-1)^(j+i)C(i,j)n^j*k^(m-j).

Original entry on oeis.org

0, 1, 1, 16, 5, 16, 81, 31, 31, 81, 256, 121, 80, 121, 256, 625, 341, 211, 211, 341, 625, 1296, 781, 496, 405, 496, 781, 1296, 2401, 1555, 1031, 781, 781, 1031, 1555, 2401, 4096, 2801, 1936, 1441, 1280, 1441, 1936, 2801, 4096, 6561, 4681, 3355, 2511, 2101

Offset: 0

Peter Luschny, Oct 26 2011

			[0]                      0
[1]                     1, 1
[2]                  16, 5, 16
[3]                81, 31, 31, 81
[4]            256, 121, 80, 121, 256
[5]         625, 341, 211, 211, 341, 625
[6]     1296, 781, 496, 405, 496, 781, 1296
[7] 2401, 1555, 1031, 781, 781, 1031, 1555, 2401

Cf. A057427, A003056, A073254, A198063, A198065.

A198064 := (n,k) -> k^4-2*k^3*n+4*k^2*n^2-3*k*n^3+n^4:

T(n,k) = k^4-2*k^3*n+4*k^2*n^2-3*k*n^3+n^4.
T(n,0) = T(n,n) = n^m = n^4 = A000583(n).
T(2n,n) = (m+1)n^m = 5n^4.
T(2n+1,n+1) = (n+1)^(m+1)-n^(m+1) = (n+1)^5-n^5 = A022521(n).
Sum{k=0..n} T(n,k) = (16n^5+30n^4+15n^3-n)/30.
T(n+1,k+1)C(n,k)^5/(k+1)^4 = A197654(n,k).

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.

A254871 Seventh partial sums of fifth powers (A000584).

Original entry on oeis.org

1, 39, 495, 3705, 19995, 85917, 311493, 989235, 2823990, 7383610, 17931498, 40889862, 88304970, 181852230, 359140470, 683363994, 1257722271, 2246496825, 3905261425, 6623425575, 10983195405, 17840105595, 28431558675, 44521334325, 68589834300, 104081944356

Offset: 1

Luciano Ancora, Feb 17 2015

			Second differences:      30, 180,  570,  1320,  2550, ...   (A068236)
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, ...   (A254471)
Seventh partial sums: 1, 39, 495, 3705, 19995, 85917, ... (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 (13,-78,286,-715,1287,-1716,1716,-1287,715,-286,78,-13,1).

Cf. A000539, A000584, A022521, A101092, A101099, A254471, A254644, A254682, A254869, A254870, A254872.

[n*(1+n)*(2+n)*(3+n)*(4+n)*(5+n)*(6+n)*(7+n)*(-21+49*n +56*n^2+14*n^3+n^4)/3991680: n in [1..30]]; // Vincenzo Librandi, Feb 19 2015

Table[n (1 + n) (2 + n) (3 + n) (4 + n) (5 + n) (6 + n) (7 + n) ((-21 + 49 n + 56 n^2 + 14 n^3 + n^4)/3991680), {n, 23}] (* or *)
CoefficientList[Series[(- 1 - 26 x - 66 x^2 - 26 x^3 - x^4)/(- 1 + x)^13, {x, 0, 22}], x]

vector(50, n, n*(1 + n)*(2 + n)*(3 + n)*(4 + n)*(5 + n)*(6 + n)*(7 + n)*(-21 + 49*n + 56*n^2 + 14*n^3 + n^4)/3991680) \\ Derek Orr, Feb 19 2015

G.f.: (- x - 26*x^2 - 66*x^3 - 26*x^4 - x^5)/(- 1 + x)^13.
a(n) = n*(1 + n)*(2 + n)*(3 + n)*(4 + n)*(5 + n)*(6 + n)*(7 + n)*(-21 + 49*n + 56*n^2 + 14*n^3 + n^4)/3991680.
a(n) = 7*a(n-1) - 21*a(n-2) + 35*a(n-3) - 35*a(n-4) + 21*a(n-5) - 7*a(n-6) + a(n-7) + n^5.

A072025 a(n) = n^4 + 2n^3 + 4n^2 + 3n + 1 = ((n+1)^5+n^5) / (2n+1).

Original entry on oeis.org

1, 11, 55, 181, 461, 991, 1891, 3305, 5401, 8371, 12431, 17821, 24805, 33671, 44731, 58321, 74801, 94555, 117991, 145541, 177661, 214831, 257555, 306361, 361801, 424451, 494911, 573805, 661781, 759511, 867691, 987041, 1118305, 1262251, 1419671, 1591381

Offset: 0

Henry Bottomley, Jun 06 2002

Vincenzo Librandi, Table of n, a(n) for n = 0..10000
Index entries for linear recurrences with constant coefficients, signature (5,-10,10,-5,1).

Cf. A022521, A072024.

[((n+1)^5+n^5)/(2*n+1): n in [0..40]]; // Vincenzo Librandi, Nov 23 2011

Table[((n+1)^5+n^5)/(2n+1),{n,0,30}] (* Vincenzo Librandi, Nov 23 2011 *)
LinearRecurrence[{5,-10,10,-5,1},{1,11,55,181,461},50] (* Harvey P. Dale, Dec 14 2019 *)

a(n)=n^4+2*n^3+4*n^2+3*n+1 \\ Charles R Greathouse IV, Nov 23 2011

Vec((1+x)^2*(1+4*x+x^2)/(1-x)^5 + O(x^100)) \\ Colin Barker, Dec 01 2015

From Colin Barker, Dec 01 2015: (Start)
a(n) = 5*a(n-1)-10*a(n-2)+10*a(n-3)-5*a(n-4)+a(n-5) for n>4.
G.f.: (1+x)^2*(1+4*x+x^2) / (1-x)^5.
(End)

A069477 a(n) = 60n^2 + 180n + 150.

Original entry on oeis.org

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, 93630, 98430, 103350

Offset: 1

Eli McGowan (ejmcgowa(AT)mail.lakeheadu.ca), Apr 11 2002

First differences of A068236, successive differences of (n+1)^5 - n^5 (A022521).

Vincenzo Librandi, Table of n, a(n) for n = 1..10000
Index entries for linear recurrences with constant coefficients, signature (3,-3,1).

Cf. A001844, A022521, A068236, A069477, A069894, A101096.

[30*(2*n^2 + 6*n + 5): n in [1..40]]; // Vincenzo Librandi, Nov 23 2011

Table[30 (2 n^2 + 6 n + 5), {n, 1, 100}] (* Vladimir Joseph Stephan Orlovsky, Jun 19 2011 *)
LinearRecurrence[{3,-3,1},{390,750,1230},40] (* Harvey P. Dale, Apr 06 2012 *)

a(n)=60*n^2+180*n+150 \\ Charles R Greathouse IV, Nov 23 2011

a(n) = 3*a(n-1) - 3*a(n-2) + a(n-3); a(1)=390, a(2)=750, a(3)=1230. - Harvey P. Dale, Apr 06 2012
Sum_{n>=1} 1/a(n) = (Pi/60)*tanh(Pi/2) - 1/25. - Amiram Eldar, Jan 27 2022
From Elmo R. Oliveira, Feb 08 2025: (Start)
G.f.: 30*x*(5*x^2 - 14*x + 13)/(1-x)^3.
E.g.f.: 30*(exp(x)*(2*x^2 + 8*x + 5) - 5).
a(n) = 30*A001844(n+1) = 15*A069894(n+1). (End)

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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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A178487 a(n) = floor(n^(1/5)): integer part of fifth root of n.

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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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A198064 Triangle read by rows (n >= 0, 0 <= k <= n, m = 4); 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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A254471 Sixth partial sums of fifth powers (A000584).

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A198064 Triangle read by rows (n >= 0, 0 <= k <= n, m = 4); T(n,k) = Sum{j=0..m} Sum{i=0..m} (-1)^(j+i)C(i,j)n^j*k^(m-j).

A072025 a(n) = n^4 + 2n^3 + 4n^2 + 3n + 1 = ((n+1)^5+n^5) / (2n+1).

A069477 a(n) = 60n^2 + 180n + 150.