A000537 -id:A000537 - cp's OEIS frontend

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

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

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

A256188 In positive integers: replace k*(k+1)/2 with the first k numbers.

Original entry on oeis.org

1, 2, 1, 2, 4, 5, 1, 2, 3, 7, 8, 9, 1, 2, 3, 4, 11, 12, 13, 14, 1, 2, 3, 4, 5, 16, 17, 18, 19, 20, 1, 2, 3, 4, 5, 6, 22, 23, 24, 25, 26, 27, 1, 2, 3, 4, 5, 6, 7, 29, 30, 31, 32, 33, 34, 35, 1, 2, 3, 4, 5, 6, 7, 8, 37, 38, 39, 40, 41, 42, 43, 44, 1, 2, 3, 4

Offset: 1

Reinhard Zumkeller, Mar 26 2015

nonn

a(A002061(n)) = 1;
a(A253169(n)) = n and a(m) != n for m < A253169(n);
a(A000537(n)) = A000217(n) and a(m) != A000217(n) for m < A000537(n);
see A004202 and A014132 for record values greater than 1 and where they occur: A014132(n) = a(A004202(n)).

			.  A000217 | 1,  3,      6,          10,                 15,       . . .
.  A000027 | _,2,___,4,5,_____,7,8,9,_______,11,12,13,14,_________,16,...
.  A002260 | 1,  1,2,    1,2,3,      1,2,3,4,            1,2,3,4,5,
.  --------+-------------------------------------------------------------
.     a(n) | 1,2,1,2,4,5,1,2,3,7,8,9,1,2,3,4,11,12,13,14,1,2,3,4,5,16,17,...

Reinhard Zumkeller, Table of n, a(n) for n = 1..10000

Cf. A255437, A000217, A014132, A002260, A000537, A004202, A014132, A002061, A255878 (first differences), A255879 (partial sums).

a256188 n = a256188_list !! (n-1)
a256188_list = f 0 [1..] a002260_tabl where
   f k xs (zs:zss) = us ++ zs ++ f (k + 1) vs zss
                     where (us, v:vs) = splitAt k xs

Table[If[OddQ[Sqrt[8n+1]],Range[(Sqrt[8n+1]-1)/2],n],{n,50}]//Flatten (* Harvey P. Dale, Jun 01 2019 *)

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

A052149 Number of nonsquare rectangles on an n X n board.

Original entry on oeis.org

0, 4, 22, 70, 170, 350, 644, 1092, 1740, 2640, 3850, 5434, 7462, 10010, 13160, 17000, 21624, 27132, 33630, 41230, 50050, 60214, 71852, 85100, 100100, 117000, 135954, 157122, 180670, 206770, 235600, 267344, 302192, 340340, 381990, 427350, 476634

Offset: 1

Ronald Arms (ron.arms(AT)stanfordalumni.org), Jan 23 2000

Partial sums of A045991 (n^3-n^2). - Jeremy Gardiner, Jun 30 2013

			a(10) = 10 * 9 * 11 * 32 / 12 = 2640.
a(5) = 170 and the sum from 1 to 5 is 15, giving 1*(15-1)=14, 2*(15-2)=26, 2*(15-3)=36, 4*(15-4)=44 and 5*(15-5)=50; adding 14+26+36+44+50=170. Do the same for each n and get a(n). - _J. M. Bergot_, Oct 31 2014

Vincenzo Librandi, Table of n, a(n) for n = 1..1000
Project Euler, Sum square difference: Problem 6.
Index entries for linear recurrences with constant coefficients, signature (5,-10,10,-5,1).

Cf. A035291, A045991, A033487, A007290, A000537, A000330, A000914.

I:=[0, 4, 22, 70, 170]; [n le 5 select I[n] else 5*Self(n-1)-10*Self(n-2)+10*Self(n-3)-5*Self(n-4)+Self(n-5): n in [1..45]]; // Vincenzo Librandi, Apr 28 2012

a:=n->sum(j^3-j^2, j=0..n): seq(a(n), n=1..37); # Zerinvary Lajos, May 08 2008

CoefficientList[Series[2*x*(2+x)/(1-5*x+10*x^2-10*x^3+ 5*x^4-x^5), {x,0,50}], x] (* Vincenzo Librandi, Apr 28 2012 *)
LinearRecurrence[{5,-10,10,-5,1},{0,4,22,70,170},40] (* Harvey P. Dale, Jul 30 2019 *)

a(n) = sum(k=1,n,(k-1)*k^2) \\ Michel Marcus, Nov 09 2012

a(n) = n*(n-1)*(n+1)*(3*n+2)/12.
G.f.: 2*x^2*(2+x)/(1-5*x+10*x^2-10*x^3+5*x^4-x^5). - Colin Barker, Jan 04 2012
a(n) = 5*a(n-1) - 10*a(n-2) + 10*a(n-3) - 5*a(n-4) + a(n-5). - Vincenzo Librandi, Apr 28 2012
a(n) = A033487(n-1) - A007290(n+1) starting at n=1. - J. M. Bergot, Jun 04 2012
a(n) = Sum_{k=1..n} (k-1)*k^2. - Michel Marcus, Nov 09 2012
a(n) = A000537(n) - A000330(n) = 2*A000914(n-1). - Luciano Ancora, Mar 16 2015
From Amiram Eldar, Jan 10 2022: (Start)
Sum_{n>=2} 1/a(n) = 81*log(3)/5 - 9*sqrt(3)*Pi/5 - 192/25.
Sum_{n>=2} (-1)^n/a(n) = 18*sqrt(3)*Pi/5 - 48*log(2)/5 - 318/25. (End)

A254469 Sixth partial sums of cubes (A000578).

Original entry on oeis.org

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)

A288876 a(n) = binomial(n+4, n)^2. Square of the fifth diagonal sequence of A007318 (Pascal). Fifth diagonal sequence of A008459.

Original entry on oeis.org

1, 25, 225, 1225, 4900, 15876, 44100, 108900, 245025, 511225, 1002001, 1863225, 3312400, 5664400, 9363600, 15023376, 23474025, 35820225, 53509225, 78411025, 112911876, 160022500, 223502500, 308002500, 419225625, 564110001, 751034025, 990046225, 1293121600, 1674446400, 2150733376

Offset: 0

Wolfdieter Lang, Jul 27 2017

This is also the square of the fifth (k = 4) column sequence (without leading zeros) of the Pascal triangle A007318. For the triangle with the squares of the entries of Pascal's triangle see A008459.
For the square of the (d+1)-th diagonal sequence of A007318, PD2(d,n) = binomial(d + n, n)^2, d >= 0, one finds the o.g.f. GPD2(d, x) = Sum_{n>=0} PD2(d,n)*x^n in the following way. Compute the compositional inverse (Lagrange inversion formula) of y(t,x) = x*(1 - t/(1-x)) w.r.t. x, that is x = x(t,y). Then -log(1 - x(t,y)) = Sum_{d=0} y^(d+1)/(d+1)*GPD2(d, x). The r.h.s. can be called the logarithmic generating function (l.g.f.) of the o.g.f.s of the square of the diagonals of Pascal's triangle.
This computation was inspired by an article by P. Bala (see a link in A112007) on the diagonal sequences of special Sheffer triangles (1, f(t)) (Sheffer triangles are there called exponential Riordan triangles, and f is called F). This can be generalized to Sheffer (g, f). For general Riordan triangles R = (G(x), F(x)) a similar analysis can be done. The present entry is then obtained for example of the Pascal triangle P = (1/(1-x), x/(1-x)).
The o.g.f.s for the square of the diagonals of Pascal's triangle turn out to be GPD2(d, x) = P(d,x)/(1 - x)^(2*d+1), with the numerator polynomials given by row n of triangle A008459 (squares of the entries of Pascal's triangle): P(d, x) = Sum_{k=0..d} A008459(d, k)*x^k.

Cf. A007318, A008459.

The squares of the first diagonals are in A000012, A000290(n+1), A000537, A001249 (for d = 0..3).

[Binomial(n+4, n)^2: n in [0..30]]; // Vincenzo Librandi, Aug 02 2017

Table[Binomial[n + 4, n]^2, {n, 0, 30}] (* Michael De Vlieger, Jul 30 2017 *)

a(n) = binomial(n+4, n)^2 \\ Felix Fröhlich, Jul 31 2017

a(n) = binomial(n+4, n)^2, n >= 0.
O.g.f.: (1 + 16*x + 36*x^2 + 16*x^3 + x^4)/(1 - x)^9. (See a comment above and row n=4 of A008459.)
E.g.f: exp(x)*(1 + 24*x + 176*x^2/2! + 624*x^3/3! + 1251*x^4/4!+ 1500*x^5/5!+ 1070*x^6/6! + 420*x^7/7! + 70*x^8/8!), computed from the o.g.f with the formulas (23) - (25) of the W. Lang link given in A060187.
From Amiram Eldar, Sep 20 2022: (Start)
Sum_{n>=0} 1/a(n) = 160*Pi^2/3 - 1576/3.
Sum_{n>=0} (-1)^n/a(n) = 512*log(2)/3 - 352/3. (End)

A093995 n^2 appears n times, triangle read by rows.

Original entry on oeis.org

1, 4, 4, 9, 9, 9, 16, 16, 16, 16, 25, 25, 25, 25, 25, 36, 36, 36, 36, 36, 36, 49, 49, 49, 49, 49, 49, 49, 64, 64, 64, 64, 64, 64, 64, 64, 81, 81, 81, 81, 81, 81, 81, 81, 81, 100, 100, 100, 100, 100, 100, 100, 100, 100, 100, 121, 121, 121, 121, 121, 121, 121, 121, 121, 121, 121

Offset: 1

Reinhard Zumkeller, May 24 2004

Row sums give A000578.

Triangle sums give A000537.

			First few rows of the triangle are:
   1;
   4,  4;
   9,  9,  9;
  16, 16, 16, 16;
  25, 25, 25, 25, 25;
  36, 36, 36, 36, 36, 36;
  49, 49, 49, 49, 49, 49, 49;
  ...

Reinhard Zumkeller, Rows n = 1..120 of triangle, flattened

Cf. A000290, A000537, A000578, A016742, A016754, A127733, A199332, A265645.

a093995 n k = a093995_tabl !! (n-1) !! (k-1)
a093995_row n = a093995_tabl !! (n-1)
a093995_tabl = zipWith replicate [1..] $ tail a000290_list
a093995_list = concat a093995_tabl
-- Reinhard Zumkeller, Nov 11 2012, Mar 18 2011, Oct 17 2010

[n^2: k in [1..n], n in [1..13]]; // G. C. Greubel, Dec 27 2021

seq(seq(n^2, i=1..n), n=1..20); # Ridouane Oudra, Jun 18 2019

Flatten[Table[Table[n^2,{n}],{n,11}]]  (* Harvey P. Dale, Feb 18 2011 *)
Table[PadRight[{},n,n^2],{n,12}]//Flatten (* Harvey P. Dale, Jun 28 2023 *)

from math import isqrt
def A093995(n): return ((m:=isqrt(k:=n<<1))+(k>m*(m+1)))**2 # Chai Wah Wu, Nov 07 2024

flatten([[n^2 for k in (1..n)] for n in (1..13)]) # G. C. Greubel, Dec 27 2021

T(n, k) = n^2, 1<=k<=n.
a(n) = floor(sqrt(2*n - 1) + 1/2)^2. - Ridouane Oudra, Jun 18 2019
From G. C. Greubel, Dec 27 2021: (Start)
T(n, n-k) = T(n, k).
Sum_{k=1..floor(n/2)} T(n, k) = [n=1] + A265645(n+1).
Sum_{k=1..floor(n/2)} T(n-k, k) = (1/48)*n*(n-1)*(7*(2*n-1) + 3*(-1)^n).
T(2*n-1, n) = A016754(n).
T(2*n, n) = A016742(n). (End)

Edited by N. J. A. Sloane, Jul 03 2008 at the suggestion of R. J. Mathar

Definition clarified by N. J. A. Sloane, Nov 09 2024

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

A107891 a(n) = (n+1)(n+2)^2(n+3)^2(n+4)(3n^2 + 15n + 20)/2880.

Original entry on oeis.org

1, 19, 155, 805, 3136, 9996, 27468, 67320, 150645, 313027, 611611, 1134497, 2012920, 3436720, 5673648, 9093096, 14194881, 21643755, 32310355, 47319349, 68105576, 96479020, 134699500, 185562000, 252493605, 339663051, 452103939

Offset: 0

Emeric Deutsch, Jun 12 2005

Kekulé numbers for certain benzenoids.

Partial sums of A114239. First differences of A047819. - Peter Bala, Sep 21 2007

S. J. Cyvin and I. Gutman, Kekulé structures in benzenoid hydrocarbons, Lecture Notes in Chemistry, No. 46, Springer, New York, 1988 (see pp. 167, 187 and p. 105 eq. (iii) for k=2 and m=5).

Shawn A. Broyles, Table of n, a(n) for n = 0..1000
Paolo Aluffi, Degrees of projections of rank loci, arXiv:1408.1702 [math.AG], 2014. ["After compiling the results of many explicit computations, we noticed that many of the numbers d_{n,r,S} appear in the existing literature in contexts far removed from the enumerative geometry of rank conditions; we owe this surprising (to us) observation to perusal of [Slo14]."]
Index entries for linear recurrences with constant coefficients, signature (9,-36,84,-126,126,-84,36,-9,1).

Cf. A005585, A006542, A047819, A114239, A114242.

a:=n->(1/2880)*(n+1)*(n+2)^2*(n+3)^2*(n+4)*(3*n^2+15*n+20): seq(a(n),n=0..32);

Table[((1+n) (2+n)^2 (3+n)^2 (4+n) (20+3 n (5+n)))/2880,{n,0,40}] (* or *) LinearRecurrence[{9,-36,84,-126,126,-84,36,-9,1},{1,19,155,805,3136,9996,27468,67320,150645},40] (* Harvey P. Dale, Dec 10 2021 *)

a(n-2) = (1/8) * Sum_{1 <= x_1, x_2 <= n} (x_1*x_2)^2*(det V(x_1,x_2))^2 = 1/8*sum {1 <= i,j <= n} (i*j*(i-j))^2, where V(x_1,x_2) is the Vandermonde matrix of order 2. - Peter Bala, Sep 21 2007
G.f.: (1+10*x+20*x^2+10*x^3+x^4)/(1-x)^9. - Colin Barker, Feb 08 2012
a(n) = (A000330(n+2)*A000538(n+2) - (A000537(n+2))^2)/4. - J. M. Bergot, Sep 17 2013
Sum_{n>=0} 1/a(n) = 17095/4 - 240*Pi^2 - 162*sqrt(15)*Pi*tanh(sqrt(5/3)*Pi/2). - Amiram Eldar, May 29 2022

cp's OEIS Frontend

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

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A107891 a(n) = (n+1)(n+2)^2(n+3)^2(n+4)(3n^2 + 15n + 20)/2880.