A254471 -id:A254471 - cp's OEIS frontend

A254469 Sixth partial sums of cubes (A000578).

1, 14, 96, 450, 1650, 5082, 13728, 33462, 75075, 157300, 311168, 586092, 1058148, 1841100, 3100800, 5073684, 8090181, 12603954, 19228000, 28778750, 42329430, 61274070, 87403680, 122996250, 170922375, 234768456, 318979584, 429024376, 571584200, 754769400

Offset: 1

Luciano Ancora, Feb 15 2015

			First differences:   1,  7, 19,  37,   61,   91, ... (A003215)
-------------------------------------------------------------------------
The cubes:           1,  8, 27,  64,  125,  216, ... (A000578)
-------------------------------------------------------------------------
First partial sums:  1,  9, 36, 100,  225,  441, ... (A000537)
Second partial sums: 1, 10, 46, 146,  371,  812, ... (A024166)
Third partial sums:  1, 11, 57, 203,  574, 1386, ... (A101094)
Fourth partial sums: 1, 12, 69, 272,  846, 2232, ... (A101097)
Fifth partial sums:  1, 13, 82, 354, 1200, 3432, ... (A101102)
Sixth partial sums:  1, 14, 96, 450, 1650, 5082, ... (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 .
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. 15.
Index entries for linear recurrences with constant coefficients, signature (10,-45,120,-210,252,-210,120,-45,10,-1).

Cf. A000537, A000578, A003215, A024166, A101094, A101097, A101102, A254470, A254471, A254472.

[n*(1+n)^2*(2+n)*(3+n)*(4+n)*(5+n)^2*(6+n)/60480: n in [1..30]]; // Vincenzo Librandi, Feb 15 2015

Table[n (1 + n)^2 (2 + n) (3 + n) (4 + n) (5 + n)^2 (6 + n)/60480, {n, 27}] (* or *) CoefficientList[Series[(1 + 4 x + x^2)/(- 1 + x)^10, {x, 0, 26}], x]
Nest[Accumulate,Range[30]^3,6] (* or *) LinearRecurrence[{10,-45,120,-210,252,-210,120,-45,10,-1},{1,14,96,450,1650,5082,13728,33462,75075,157300},30] (* Harvey P. Dale, Sep 03 2016 *)

a(n)=n*(1+n)^2*(2+n)*(3+n)*(4+n)*(5+n)^2*(6+n)/60480 \\ Charles R Greathouse IV, Oct 07 2015

G.f.: (x + 4*x^2 + x^3)/(- 1 + x)^10.
a(n) = n*(1 + n)^2*(2 + n)*(3 + n)*(4 + n)*(5 + n)^2*(6 + n)/60480.
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^3.
From Amiram Eldar, Jan 26 2022: (Start)
Sum_{n>=1} 1/a(n) = 217/200.
Sum_{n>=1} (-1)^(n+1)/a(n) = 223769/200 - 8064*log(2)/5. (End)

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

A254470 Sixth partial sums of fourth powers (A000583).

Original entry on oeis.org

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

Offset: 1

Luciano Ancora, Feb 15 2015

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

Luciano Ancora, Table of n, a(n) for n = 1..1000
Luciano Ancora, Partial sums of m-th powers with Faulhaber polynomials.
Luciano Ancora, Pascal’s triangle and recurrence relations for partial sums of m-th powers.
Index entries for linear recurrences with constant coefficients, signature (11,-55,165,-330,462,-462,330,-165,55,-11,1).

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

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

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

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

G.f.: (-x - 11*x^2 - 11*x^3 - x^4)/(- 1 + x)^11.
a(n) = n*(1 + n)*(2 + n)*(3 + n)^2*(4 + n)*(5 + n)*(6 + n)*(1 + 12*n + 2*n^2)/302400.
a(n) = 6*a(n-1) - 15*a(n-2) + 20*a(n-3) - 15*a(n-4) + 6*a(n-5) - a(n-6) + n^4.
Sum_{n>=1} 1/a(n) = 3320303/2601 + 1400*Pi^2/17 + (8960/17)*sqrt(2/17)*Pi*cot(sqrt(17/2)*Pi). - Amiram Eldar, Jan 26 2022

A254472 Sixth partial sums of sixth powers (A001014).

Original entry on oeis.org

1, 70, 1134, 9870, 59220, 275562, 1063530, 3552978, 10577385, 28652260, 71725108, 167911380, 371057232, 779831820, 1568210220, 3032733564, 5663906745, 10251608346, 18037546450, 30931714450, 51814612980, 84952851750, 136562787270, 215565263550, 334584493425

Offset: 1

Luciano Ancora, Feb 15 2015

			First differences:   1, 63,  665, 3367, 11529, ... (A022522)
--------------------------------------------------------------------------
The sixth powers:    1, 64,  729, 4096, 15625, ... (A001014)
--------------------------------------------------------------------------
First partial sums:  1, 65,  794, 4890, 20515, ... (A000540)
Second partial sums: 1, 66,  860, 5750, 26265, ... (A101093)
Third partial sums:  1, 67,  927, 6677, 32942, ... (A254640)
Fourth partial sums: 1, 68,  995, 7672, 40614, ... (A254645)
Fifth partial sums:  1, 69, 1064, 8736, 49350, ... (A254683)
Sixth partial sums:  1, 70, 1134, 9870, 59220, ... (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. A000540, A001014, A022522, A101093, A254469, A254470, A254471, A254640, A254645, A254683.

[n*(1+n)*(2+n)*(3+n)^2*(4+n)*(5+n)*(6+n)*(-3+5*n+n^2)* (3+7*n+n^2)/665280: n in [1..30]]; // Vincenzo Librandi, Feb 15 2015

Table[n (1 + n) (2 + n) (3 + n)^2 (4 + n) (5 + n) (6 + n) (- 3 + 5 n + n^2) (3 + 7 n + n^2)/665280, {n, 22}] (* or *) CoefficientList[Series[(- 1 - 57 x - 302 x^2 - 302 x^3 - 57 x^4 - x^5)/(- 1 + x)^13, {x, 0, 28}], x]
Nest[Accumulate,Range[30]^6,6] (* Harvey P. Dale, Oct 02 2015 *)

vector(50,n,n*(1 + n)*(2 + n)*(3 + n)^2*(4 + n)*(5 + n)*(6 + n)*(-3 + 5*n + n^2)*(3 + 7*n + n^2)/665280) \\ Derek Orr, Feb 19 2015

G.f.: (-x - 57*x^2 - 302*x^3 - 302*x^4 - 57*x^5 - x^6)/(- 1 + x)^13.
a(n) = n*(1 + n)*(2 + n)*(3 + n)^2*(4 + n)*(5 + n)*(6 + n)*(-3 + 5*n + n^2)*(3 + 7*n + n^2)/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^6.
Sum_{n>=1} 1/a(n) = 25622179/76545 - 3080*Pi^2/81 + 149600*Pi*tan(sqrt(37)*Pi/2)/(243*sqrt(37)). - Amiram Eldar, Jan 27 2022

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.

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A254469 Sixth partial sums of cubes (A000578).

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A101095 Fourth difference of fifth powers (A000584).

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A254470 Sixth partial sums of fourth powers (A000583).

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A254472 Sixth partial sums of sixth powers (A001014).

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A254871 Seventh partial sums of fifth powers (A000584).

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