A101092 -id:A101092 - cp's OEIS frontend

A000539 Sum of 5th powers: 0^5 + 1^5 + 2^5 + ... + n^5.

0, 1, 33, 276, 1300, 4425, 12201, 29008, 61776, 120825, 220825, 381876, 630708, 1002001, 1539825, 2299200, 3347776, 4767633, 6657201, 9133300, 12333300, 16417401, 21571033, 28007376, 35970000, 45735625, 57617001, 71965908, 89176276, 109687425, 133987425, 162616576

Offset: 0

N. J. A. Sloane

This sequence is related to A000538 by a(n) = n*A000538(n) - Sum_{i=0..n-1} A000538(i). - Bruno Berselli, Apr 26 2010

See comment in A008292 for a formula for r-th successive summation of Sum_{k=1..n} k^j. - Gary Detlefs, Jan 02 2014

M. Abramowitz and I. A. Stegun, eds., Handbook of Mathematical Functions, National Bureau of Standards Applied Math. Series 55, 1964 (and various reprintings), p. 813.
L. Comtet, Advanced Combinatorics, Reidel, 1974, p. 155.
R. L. Graham, D. E. Knuth and O. Patashnik, Concrete Mathematics. Addison-Wesley, Reading, MA, 1991, p. 275.
N. J. A. Sloane, A Handbook of Integer Sequences, Academic Press, 1973 (includes this sequence).
N. J. A. Sloane and Simon Plouffe, The Encyclopedia of Integer Sequences, Academic Press, 1995 (includes this sequence).

T. D. Noe, Table of n, a(n) for n = 0..1000
M. Abramowitz and I. A. Stegun, eds., Handbook of Mathematical Functions, National Bureau of Standards, Applied Math. Series 55, Tenth Printing, 1972 [alternative scanned copy].
Bruno Berselli, A description of the transform in Comments lines: 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.
Simon Plouffe, Approximations de séries génératrices et quelques conjectures, Dissertation, Université du Québec à Montréal, 1992; arXiv:0911.4975 [math.NT], 2009.
Simon Plouffe, 1031 Generating Functions, Appendix to Thesis, Montreal, 1992
Eric Weisstein's World of Mathematics, Faulhaber's Formula
Wikipedia, Faulhaber's formula
Index entries for linear recurrences with constant coefficients, signature (7,-21,35,-35,21,-7,1).

Partial sums of A000584. Row 5 of array A103438.

Cf. A000217, A000537, A000538, A006542, A008292, A059378, A101092.

[n^2*(n+1)^2*(2*n^2+2*n-1)/12: n in [0..30]]; // Vincenzo Librandi, Apr 04 2015

A000539:=-(1+26*z+66*z**2+26*z**3+z**4)/(z-1)**7; # Simon Plouffe in his 1992 dissertation
a[0]:=0:a[1]:=1:for n from 2 to 50 do a[n]:=a[n-1]+n^5 od: seq(a[n], n=0..30); # Zerinvary Lajos, Feb 22 2008
a:=n->sum(j^5,j=0..n): seq(a(n), n=0..30); # Zerinvary Lajos, Jun 05 2008

Accumulate[Range[0, 40]^5]
LinearRecurrence[{7, -21, 35, -35, 21, -7, 1}, {0, 1, 33, 276, 1300, 4425, 12201}, 41] (* Jean-François Alcover, Feb 09 2016 *)

A000539(n):=n^2*(n+1)^2*(2*n^2+2*n-1)/12$ makelist(A000539(n),n,0,30); /* Martin Ettl, Nov 12 2012 */

a(n)=n^2*(n+1)^2*(2*n^2+2*n-1)/12 \\ Charles R Greathouse IV, Jul 15 2011

concat(0, Vec(x*(1+26*x+66*x^2+26*x^3+x^4)/(1-x)^7 + O(x^100))) \\ Altug Alkan, Dec 07 2015

A000539_list, m = [0], [120, -240, 150, -30, 1, 0, 0]
for _ in range(10**2):
    for i in range(6):
        m[i+1] += m[i]
    A000539_list.append(m[-1]) # Chai Wah Wu, Nov 05 2014

def A000539(n): return n**2*(n**2*(n*(n+3<<1)+5)-1)//12 # Chai Wah Wu, Oct 03 2024

a(n) = n^2*(n+1)^2*(2*n^2+2*n-1)/12.
a(n) = sqrt(Sum_{j=1..n}Sum_{i=1..n}(i*j)^5). - Alexander Adamchuk, Oct 26 2004
a(n) = Sum_{i = 1..n} J_5(i)*floor(n/i), where J_5 is A059378. - Enrique Pérez Herrero, Feb 26 2012
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) + 120. - Ant King, Sep 23 2013
a(n) = 120*C(n+3,6) + 30*C(n+2,4) + C(n+1,2) (Knuth). - Gary Detlefs, Jan 02 2014
a(n) = -Sum_{j=1..5} j*Stirling1(n+1,n+1-j)*Stirling2(n+5-j,n). - Mircea Merca, Jan 25 2014
Sum_{n>=1} 1/a(n) = 60 - 4*Pi^2 + 8*sqrt(3)*Pi * tan(sqrt(3)*Pi/2). - Vaclav Kotesovec, Feb 13 2015
a(n) = (n + 1)^2*n^2*(n + 1/2 + sqrt(3/4))*(n + 1/2 - sqrt(3/4))/6. See the Graham et al. reference, p. 275. - Wolfdieter Lang, Apr 02 2015
G.f.: x*(1+26*x+66*x^2+26*x^3+x^4)/(1-x)^7. - Robert Israel, Dec 07 2015
a(n) = (4/3)*A000217(n)^3 - (1/3)*A000217(n)^2. - Michael Raney, Feb 19 2016
a(n) = (binomial(n+1,4) + 6*binomial(n+2,4) + binomial(n+3,4))*(binomial(n+2,3) - binomial(n+1,3)). - Tony Foster III, Oct 21 2018
a(n) = 24*A006542(n+2) + A000537(n). - Yasser Arath Chavez Reyes, May 04 2024
E.g.f.: exp(x)*x*(12 + 186*x + 360*x^2 + 195*x^3 + 36*x^4 + 2*x^5)/12. - Stefano Spezia, May 04 2024

A254640 Third partial sums of sixth powers (A001014).

Original entry on oeis.org

1, 67, 927, 6677, 32942, 126378, 404634, 1129854, 2833479, 6515509, 13947505, 28115451, 53846156, 98669156, 173975076, 296541132, 490504893, 789878583, 1241708083, 1909993393, 2880500634, 4266609710, 6216356510, 8920844010, 12624212835, 17635378761

Offset: 1

Luciano Ancora, Feb 04 2015

This is one of a sequence of arrays that are the convolutions of the zero-padded sequences binomial(2n-1+k,k) with the Eulerian polynomials E(n,x) of A008292, represented by E(n,x) (1-x)^(-2n), which generate increasing partial sums of powers of integers:
n= 2) (1 + 4*x + x^2)/(1-x)^4 is the o.g.f. of A000578, the convolution of (1,4,1) with A000292, giving the powers of m^3.
n= 3) (1 + 11*x + 11*x^2 + x^3)/(1-x)^6 is the o.g.f. of A000538, convolution of (1,11,11,1) with A000389, giving the partial sums of m^4.
n= 4) (1 + 26*x + 66*x^2 + 26*x^3 + x^4)/(1-x)^8, the o.g.f. of A101092, convolution of (1,26,66,26,1) with A000580, the second partial sums of m^5
n= 5) (1 + 57*x + 302*x^2 + 302*x^3 + 57*x^4 + x^5)/(1-x)^10, the o.g.f. of A254460, convolution of (1,57,302,302,57,1) with A000582, giving the third partial sums of m^6. - Tom Copeland, Dec 07 2015

Colin Barker, Table of n, a(n) for n = 1..1000
Luciano Ancora, Partial sums of m-th powers with Faulhaber polynomials
Index entries for linear recurrences with constant coefficients, signature (10,-45,120,-210,252,-210,120,-45,10,-1).

Cf. A008292, A000578, A000292, A000538, A000389, A101092, A000580, A254460, A000582.

List([1..30], n-> Binomial(n+3,4)*(2*n+3)*(5*n^4 +30*n^3 +35*n^2 -30*n +2)/210); # G. C. Greubel, Aug 28 2019

[n*(1+n)*(2+n)*(3+n)*(3+2*n)*(2-30*n+35*n^2+30*n^3+ 5*n^4)/5040: n in [1..30]]; // Vincenzo Librandi, Feb 05 2015

seq(binomial(n+3,4)*(2*n+3)*(5*n^4 +30*n^3 +35*n^2 -30*n +2)/210, n=1..30); # G. C. Greubel, Aug 28 2019

Table[n(1+n)(2+n)(3+n)(3+2n)(2 -30n +35n^2 +30n^3 +5n^4)/5040, {n, 30}] (* or *) CoefficientList[Series[(x+1)(x^4 +56x^3 +246x^2 +56x +1)/(x - 1)^10, {x, 0, 30}], x] (* Vincenzo Librandi, Feb 05 2015 *)

vector(30, n, n*(1+n)*(2+n)*(3+n)*(3+2*n)*(2-30*n+35*n^2+30*n^3+5*n^4)/5040) \\ Colin Barker, Feb 04 2015

def A254640(n): return n*(n*(n*(n*(n*(n*(n*(n*(10*n + 135) + 720) + 1890) + 2394) + 945) - 640) - 450) + 36)//5040 # Chai Wah Wu, Dec 07 2021

[binomial(n+3,4)*(2*n+3)*(5*n^4 +30*n^3 +35*n^2 -30*n +2)/210 for n in (1..30)] # G. C. Greubel, Aug 28 2019

a(n) = n*(1+n)*(2+n)*(3+n)*(3+2*n)*(2 -30*n +35*n^2 +30*n^3 +5*n^4)/5040.

G.f.: x*(1+x)*(1 +56*x +246*x^2 +56*x^3 +x^4)/(1-x)^10. - Colin Barker, Feb 04 2015

A253636 Second partial sums of eighth powers (A001016).

Original entry on oeis.org

1, 258, 7076, 79430, 542409, 2685004, 10592400, 35277012, 103008345, 270739678, 652829892, 1464901802, 3092704433, 6196296120, 11862778432, 21824228040, 38761435089, 66718602714, 111659333380, 182200064046, 290563654073, 453803117636, 695353566480, 1046979329500

Offset: 1

Luciano Ancora, Jan 07 2015

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) are the m-th Faulhaber’s formulas.

G. C. Greubel, 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
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.
Index entries for linear recurrences with constant coefficients, signature (11,-55,165,-330,462,-462,330,-165,55,-11,1).

Cf. A001016, A101089, A101092, A101093.

List([1..30], n-> n*(n+1)^2*(n+2)*(2*n^6 +12*n^5 +17*n^4 -12*n^3 -19*n^2 +18*n -3)/180); # G. C. Greubel, Aug 28 2019

[n*(n+1)^2*(n+2)*(2*n^6+12*n^5+17*n^4-12*n^3-19*n^2+18*n-3)/180: n in [1..25]]; // Bruno Berselli, Jan 08 2015

seq(n*(n+1)^2*(n+2)*(2*n^6 +12*n^5 +17*n^4 -12*n^3 -19*n^2 +18*n -3))/180, n=1..30); # G. C. Greubel, Aug 28 2019

Table[n(n+1)^2(n+2)(2n^6 +12n^5 +17n^4 -12n^3 -19n^2 +18n -3)/180, {n,30}] (* Bruno Berselli, Jan 08 2015 *)
Nest[Accumulate,Range[30]^8,2] (* or *) LinearRecurrence[{11,-55,165,-330,462,-462,330,-165,55,-11,1},{1,258,7076,79430,542409,2685004,10592400, 35277012, 103008345,270739678,652829892},30] (* Harvey P. Dale, Jul 02 2017 *)

a(n)=(2*n^10+20*n^9+75*n^8+120*n^7+42*n^6-84*n^5-50*n^4+40*n^3+21*n^2-6*n)/180 \\ Charles R Greathouse IV, Sep 08 2015

[(2*n^10+20*n^9+75*n^8+120*n^7+42*n^6-84*n^5-50*n^4+40*n^3+21*n^2-6*n)/180 for n in [1..24]] # Tom Edgar, Jan 07 2015

a(n) = (2*n^10 + 20*n^9 + 75*n^8 + 120*n^7 + 42*n^6 - 84*n^5 - 50*n^4 + 40*n^3 + 21*n^2 - 6*n)/180.
a(n) = 2*a(n-1) - a(n-2) + n^8. - Robert Israel, Jan 07 2015
G.f.: x*(1 + x)*(1 + 246*x + 4047*x^2 + 11572*x^3 + 4047*x^4 + 246*x^5 + x^6) / (1 - x)^11. - Bruno Berselli, Jan 08 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

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

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.

cp's OEIS Frontend

A000539 Sum of 5th powers: 0^5 + 1^5 + 2^5 + ... + n^5.

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A254640 Third partial sums of sixth powers (A001014).

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A253636 Second partial sums of eighth powers (A001016).

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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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