A000537 -id:A000537 - cp's OEIS frontend

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A163102 a(n) = n^2*(n+1)^2/2.

0, 2, 18, 72, 200, 450, 882, 1568, 2592, 4050, 6050, 8712, 12168, 16562, 22050, 28800, 36992, 46818, 58482, 72200, 88200, 106722, 128018, 152352, 180000, 211250, 246402, 285768, 329672, 378450, 432450, 492032, 557568, 629442, 708050, 793800, 887112, 988418

Offset: 0

Omar E. Pol, Jul 24 2009

Row sums of triangle A163282.
Also, the number of nonattacking placements of 2 rooks on an (n+1) X (n+1) board. - Thomas Zaslavsky, Jun 26 2010
If P_{k}(n) is the n-th k-gonal number, then a(n) = P_{s}(n+1)*P_{t}(n+1) - P_{s+1}(n+1)*P_{t-1}(n+1) for s=t+1. - Bruno Berselli, Sep 05 2014
Subsequence of A000982, see formula. - David James Sycamore, Jul 31 2018
Number of edges in the (n+1) X (n+1) rook complement graph. - Freddy Barrera, May 02 2019
Number of paths from (0,0) to (n+2,n+2) consisting of exactly three forward horizontal steps and three upward vertical steps. - Greg Dresden and Snezhana Tuneska, Aug 24 2023

Seth Chaiken, Christopher R. H. Hanusa, and Thomas Zaslavsky, A q-queens problem, in preparation. - Thomas Zaslavsky, Jun 26 2010

Vincenzo Librandi, Table of n, a(n) for n = 0..1000
Seth Chaiken, Christopher R. H. Hanusa, and Thomas Zaslavsky, A q-Queens Problem. IV. Queens, Bishops, Nightriders (and Rooks), arXiv preprint arXiv:1609.00853 [math.CO], 2016-2020.
A. Umar, Combinatorial Results for Semigroups of Orientation-Preserving Partial Transformations, J. Int. Seq., Vol. 14 (2011), Article 11.7.5.
Eric Weisstein's World of Mathematics, Rook Complement Graph.
Index entries for linear recurrences with constant coefficients, signature (5,-10,10,-5,1).

Column k=2 of A144084.
Cf. A000217, A000537, A000982, A035287.
Cf. A006002, A099903, A163274, A163275, A163276, A163277, A163282.
Cf. A060300, A254371.

List([0..40],n->(n*(n+1))^2/2); # Muniru A Asiru, Aug 02 2018

[n^2*(n+1)^2/2: n in [0..40]]; // Vincenzo Librandi, Mar 26 2012

CoefficientList[Series[2*x*(1+4*x+x^2)/(1-x)^5,{x,0,40}],x] (* Vincenzo Librandi, Mar 26 2012 *)

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

a(n) = 2*A000537(n) = A035287(n+1)/2. - Omar E. Pol, Nov 29 2011
G.f.: 2*x*(1+4*x+x^2)/(1-x)^5. - R. J. Mathar, Nov 30 2011
Let t(n) = A000217(n). Then a(n) = (t(n-1)*(t(n)+t(n+1)) + t(n)*(t(n-1)+t(n+1)) + t(n+1)*(t(n-1)+t(n)))/3. - J. M. Bergot, Jun 21 2012
a(n) = A000982(n*(n+1)). - David James Sycamore, Jul 31 2018
From Amiram Eldar, Nov 02 2021: (Start)
Sum_{n>=1} 1/a(n) = 2*Pi^2/3 - 6.
Sum_{n>=1} (-1)^(n+1)/a(n) = 6 - 8*log(2). (End)
Another identity: ..., a(4) = 200 = 1*(2+4+6+8) + 3*(4+6+8) + 5*(6+8) + 7*(8), a(5) = 450 = 1*(2+4+6+8+10) + 3*(4+6+8+10) + 5*(6+8+10) + 7*(8+10) + 9*(10) = 30+84+120+126+90, and so on. - J. M. Bergot, Aug 25 2022
From Elmo R. Oliveira, Aug 14 2025: (Start)
E.g.f.: x*(2 + x)*(2 + 6*x + x^2)*exp(x)/2.
a(n) = 5*a(n-1) - 10*a(n-2) + 10*a(n-3) - 5*a(n-4) + a(n-5).
a(n) = A254371(n)/4 = A060300(n)/8. (End)

A101097 a(n) = n(n+1)(n+2)(n+3)(n+4)(2 + 4n + n^2)/840.

Original entry on oeis.org

1, 12, 69, 272, 846, 2232, 5214, 11088, 21879, 40612, 71643, 121056, 197132, 310896, 476748, 713184, 1043613, 1497276, 2110273, 2926704, 3999930, 5393960, 7184970, 9462960, 12333555, 15919956, 20365047, 25833664, 32515032, 40625376, 50410712

Offset: 1

Cecilia Rossiter (cecilia(AT)noticingnumbers.net), Dec 15 2004

Fourth partial sums of cubes (A000578). Partial sums of A101094.

G. C. Greubel, Table of n, a(n) for n = 1..5000
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.
C. Rossiter, Depictions, Explorations and Formulas of the Euler/Pascal Cube. [broken link]
Index entries for linear recurrences with constant coefficients, signature (8,-28,56,-70,56,-28,8,-1).

Cf. A000217, A101102, A101094, A024166, A000537.

A000217:=func; [&+[A000217(k)^2*A000217(n-k+1): k in [1..n]]: n in [1..40]]; // Bruno Berselli, Sep 04 2013

Mathematica

Table[Binomial[n+4,5]*(2+4*n+n^2)/7, {n,0,50}] (* G. C. Greubel, Feb 17 2017 *)

PARI

{A101097(n) = n*(n+1)*(n+2)*(n+3)*(n+4)*(2+4*n+n^2)/840} \\ R. J. Mathar, Dec 06 2011

Sage

[binomial(n+4,5)*(2+4*n+n^2)/7 for n in (1..40)] # G. C. Greubel, Dec 01 2018

Formula

a(n) = n*(n+1)*(n+2)*(n+3)*...*(n+k)*(n*(n+k) + (k-1)*k/6)/((k+3)!/6) for k=4. - Alexander R. Povolotsky, May 17 2008

G.f.: x*(1 + 4*x + x^2)/(1-x)^8. - R. J. Mathar, Jun 13 2008

a(n) = Sum_{k=1..n} A000217(k)^2*A000217(n-k+1). - Bruno Berselli, Sep 04 2013

E.g.f.: x*(840 + 4200*x + 5040*x^2 + 2240*x^3 + 427*x^4 + 35*x^5 + x^6) *exp(x)/840. - G. C. Greubel, Dec 01 2018

Extensions

Edited by Ralf Stephan, Dec 16 2004

A074279 n appears n^2 times.

Original entry on oeis.org

1, 2, 2, 2, 2, 3, 3, 3, 3, 3, 3, 3, 3, 3, 4, 4, 4, 4, 4, 4, 4, 4, 4, 4, 4, 4, 4, 4, 4, 4, 5, 5, 5, 5, 5, 5, 5, 5, 5, 5, 5, 5, 5, 5, 5, 5, 5, 5, 5, 5, 5, 5, 5, 5, 5, 6, 6, 6, 6, 6, 6, 6, 6, 6, 6, 6, 6, 6, 6, 6, 6, 6, 6, 6, 6, 6, 6, 6, 6, 6, 6, 6, 6, 6, 6, 6, 6, 6

Offset: 1

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Author

Jon Perry, Sep 21 2002

Keywords

nonn
easy
tabf
changed

Comments

Since the last occurrence of n comes one before the first occurrence of n+1 and the former is at Sum_{i=0..n} i^2 = A000330(n), we have a(A000330(n)) = a(n*(n+1)*(2n+1)/6) = n and a(1+A000330(n)) = a(1+(n*(n+1)*(2n+1)/6)) = n+1. So the current sequence is, loosely speaking, the inverse function of the square pyramidal sequence A000330. A000330 has many alternative formulas, thus yielding many alternative formulas for the current sequence. - Jonathan Vos Post, Mar 18 2006

Partial sums of A253903. - Jeremy Gardiner, Jan 14 2018

Examples

This can be viewed also as an irregular table consisting of successively larger square matrices: 1; 2, 2; 2, 2; 3, 3, 3; 3, 3, 3; 3, 3, 3; 4, 4, 4, 4; 4, 4, 4, 4; 4, 4, 4, 4; 4, 4, 4, 4; etc. When this is used with any similarly organized sequence, a(n) is the index of the matrix in whose range n is. A121997(n) (= A237451(n)+1) and A238013(n) (= A237452(n)+1) would then yield the index of the column and row within that matrix.

Links

Antti Karttunen, Table of n, a(n) for n = 1..9455 (Table of squares from 1 X 1 to 30 X 30, flattened)

Y.-F. S. Petermann, J.-L. Remy and I. Vardi, Discrete derivatives of sequences, Adv. in Appl. Math. 27 (2001), 562-84.

Crossrefs

Cf. A000217, A000330, A002024, A003881, A006331, A050446, A050447, A000537, A006003, A005900, A064866, A237451, A237452.

Programs

Mathematica

Table[n, {n, 0, 6}, {n^2}] // Flatten (* Arkadiusz Wesolowski, Jan 13 2013 *)

PARI

A074279_vec(N=9)=concat(vector(N,i,vector(i^2,j,i))) \\ Note: This creates a vector; use A074279_vec()[n] to get the n-th term. - M. F. Hasler, Feb 17 2014

PARI

a(n) = my(k=sqrtnint(3*n,3)); k + (6*n > k*(k+1)*(2*k+1)); \\ Kevin Ryde, Sep 03 2025

Python

from sympy import integer_nthroot def A074279(n): return (m:=integer_nthroot(3*n,3)[0])+(6*n>m*(m+1)*((m<<1)+1)) # Chai Wah Wu, Nov 04 2024

Formula

For 1 <= n <= 650, a(n) = floor((3n)^(1/3)+1/2). - Mikael Aaltonen, Jan 05 2015

a(n) = 1 + floor( t(n) + 1 / ( 12 * t(n) ) - 1/2 ), where t(n) = (sqrt(3888*(n-1)^2-1) / (8*3^(3/2)) + 3 * (n-1)/2 ) ^(1/3). - Mikael Aaltonen, Mar 01 2015

a(n) = floor(t + 1/(12*t) + 1/2), where t = (3*n - 1)^(1/3). - Ridouane Oudra, Oct 30 2023

a(n) = m+1 if n > m(m+1)(2m+1)/6 and a(n) = m otherwise where m = floor((3n)^(1/3)). - Chai Wah Wu, Nov 04 2024

Sum_{n>=1} (-1)^(n+1)/a(n) = Pi/4 (A003881). - Amiram Eldar, Jun 30 2025

Extensions

Offset corrected from 0 to 1 by Antti Karttunen, Feb 08 2014

A101102 Fifth partial sums of cubes (A000578).

Original entry on oeis.org

1, 13, 82, 354, 1200, 3432, 8646, 19734, 41613, 82225, 153868, 274924, 472056, 782952, 1259700, 1972884, 3016497, 4513773, 6624046, 9550750, 13550680, 18944640, 26129610, 35592570, 47926125, 63846081, 84211128, 110044792, 142559824, 183185200, 233595912

Offset: 1

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Author

Cecilia Rossiter (cecilia(AT)noticingnumbers.net), Dec 15 2004

Keywords

easy
nonn

Links

Colin Barker, Table of n, a(n) for n = 1..1000

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.

Cecilia Rossiter, Depictions, Explorations and Formulas of the Euler/Pascal Cube. [Archive link]

Cecilia Rossiter, Depictions, Explorations and Formulas of the Euler/Pascal Cube. [Cached copy, May 15 2013]

Index entries for linear recurrences with constant coefficients, signature (9,-36,84,-126,126,-84,36,-9,1).

Crossrefs

Partial sums of A101097.

Cf. A000537, A024166, A101094.

Programs

Magma

[Binomial(n+5,6)*(3*n^2+15*n+10)/28: n in [1..30]]; // G. C. Greubel, Dec 01 2018

Mathematica

Table[Binomial[n+5,6]*(3*n^2+15*n+10)/28, {n,1,30}] (* G. C. Greubel, Dec 01 2018 *) Nest[Accumulate,Range[40]^3,5] (* Harvey P. Dale, Feb 06 2023 *)

PARI

a(n)=sum(t=1,n,sum(s=1,t,sum(l=1,s,sum(j=1,l, sum(m=1, j, sum(i=m*(m+1)/2-m+1, m*(m+1)/2,(2*i-1))))))) \\ Alexander R. Povolotsky, May 17 2008

PARI

Vec(-x*(x^2+4*x+1)/(x-1)^9 + O(x^100)) \\ Colin Barker, Apr 23 2015

PARI

a(n) = binomial(n+5,6)*(3*n^2+15*n+10)/28 \\ Charles R Greathouse IV, Apr 23 2015

Sage

[binomial(n+5,6)*(3*n^2+15*n+10)/28 for n in (1..30)] # G. C. Greubel, Dec 01 2018

Formula

a(n) = n*(n+1)*(n+2)*(n+3)*(n+4)*(n+5)*(10 + 3*n*(n+5))/20160.

This sequence could be obtained from the general formula a(n) = n*(n+1)*(n+2)*(n+3)*...*(n+k)*(n*(n+k) + (k-1)*k/6)/((k+3)!/6) at k=5. - Alexander R. Povolotsky, May 17 2008

G.f.: x*(x^2+4*x+1) / (1-x)^9. - Colin Barker, Apr 23 2015

Sum_{n>=1} 1/a(n) = -162*sqrt(21/5)*Pi*tan(sqrt(35/3)*Pi/2) - 136269/100. - Amiram Eldar, Jan 26 2022

Extensions

Edited by Ralf Stephan, Dec 16 2004

A250812 T(n,k)=Number of (n+1)X(k+1) 0..2 arrays with nondecreasing x(i,j)-x(i,j-1) in the i direction and nondecreasing min(x(i,j),x(i-1,j)) in the j direction.

Original entry on oeis.org

36, 100, 129, 225, 379, 432, 441, 873, 1315, 1389, 784, 1731, 3081, 4321, 4356, 1296, 3097, 6171, 10233, 13735, 13449, 2025, 5139, 11116, 20631, 32745, 42769, 41112, 3025, 8049, 18537, 37333, 66291, 102393, 131455, 124869, 4356, 12043, 29145, 62469

Offset: 1

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Author

R. H. Hardin, Nov 27 2014

Keywords

nonn
tabl

Comments

Table starts

......36.....100.....225......441......784.....1296.....2025......3025

.....129.....379.....873.....1731.....3097.....5139.....8049.....12043

.....432....1315....3081.....6171....11116....18537....29145.....43741

....1389....4321...10233....20631....37333....62469....98481....148123

....4356...13735...32745....66291...120304...201741...318585....479845

...13449...42769..102393...207831...377857...634509..1003089...1512163

...41112..131455..315561...641571..1167796..1962717..3104985...4683421

..124869..400681..963513..1961031..3572173..6007149..9507441..14345803

..377676.1214695.2924265..5955891.10854424.18260061.28908345..43630165

.1139169.3669409.8840313.18013431.32839417.55258029.87498129.132077683

Examples

Some solutions for n=4 k=4 ..1..1..0..0..0....1..0..0..0..0....1..1..1..1..0....0..0..0..0..0 ..0..0..1..1..1....0..0..0..0..0....0..0..0..0..0....2..2..2..2..2 ..1..1..2..2..2....2..2..2..2..2....2..2..2..2..2....2..2..2..2..2 ..0..0..1..1..1....1..1..1..1..2....2..2..2..2..2....1..1..1..1..2 ..0..0..1..2..2....0..1..1..1..2....0..0..1..1..2....1..1..1..1..2

Links

R. H. Hardin, Table of n, a(n) for n = 1..241

Crossrefs

Row 1 is A000537(n+2)

Formula

Empirical: T(n,k) = (((5/12)*k^4 + (11/3)*k^3 + (157/12)*k^2 + (95/6)*k + 6)*3^n - ((1/2)*k^4 + (7/2)*k^3 + (23/2)*k^2 + (17/2)*k)*2^n + (1/4)*k^4 + 1*k^3 + (9/4)*k^2 - (1/2)*k)/2

Empirical for column k:

k=1: a(n) = 6*a(n-1) -11*a(n-2) +6*a(n-3); a(n) = (39*3^n-24*2^n+3)/2

k=2: a(n) = 6*a(n-1) -11*a(n-2) +6*a(n-3); a(n) = (126*3^n-99*2^n+20)/2

k=3: a(n) = 6*a(n-1) -11*a(n-2) +6*a(n-3); a(n) = (304*3^n-264*2^n+66)/2

k=4: a(n) = 6*a(n-1) -11*a(n-2) +6*a(n-3); a(n) = (620*3^n-570*2^n+162)/2

k=5: a(n) = 6*a(n-1) -11*a(n-2) +6*a(n-3); a(n) = (1131*3^n-1080*2^n+335)/2

k=6: a(n) = 6*a(n-1) -11*a(n-2) +6*a(n-3); a(n) = (1904*3^n-1869*2^n+618)/2

k=7: a(n) = 6*a(n-1) -11*a(n-2) +6*a(n-3); a(n) = (3016*3^n-3024*2^n+1050)/2

Empirical for row n:

n=1: a(n) = (1/4)*n^4 + (5/2)*n^3 + (37/4)*n^2 + 15*n + 9

n=2: a(n) = 1*n^4 + 10*n^3 + 37*n^2 + 54*n + 27

n=3: a(n) = (15/4)*n^4 + 36*n^3 + (527/4)*n^2 + (359/2)*n + 81

n=4: a(n) = 13*n^4 + 121*n^3 + 439*n^2 + 573*n + 243

n=5: a(n) = (171/4)*n^4 + 390*n^3 + (5627/4)*n^2 + (3575/2)*n + 729

n=6: a(n) = 136*n^4 + 1225*n^3 + 4402*n^2 + 5499*n + 2187

n=7: a(n) = (1695/4)*n^4 + 3786*n^3 + (54287/4)*n^2 + (33539/2)*n + 6561

A323541 a(n) = Product_{k=0..n} (k^3 + (n-k)^3).

Original entry on oeis.org

0, 1, 128, 59049, 51380224, 80869140625, 207351578198016, 811509810302822449, 4603095542875667038208, 36344623587588604291790241, 386644580358400000000000000000, 5395532942025804980378907333844441, 96578621213529440721046520779140759552

Offset: 0

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Author

Vaclav Kotesovec, Jan 17 2019

Keywords

nonn

Links

Seiichi Manyama, Table of n, a(n) for n = 0..158

Crossrefs

Cf. A272246, A323496, A323540, A323542, A323543, A323544, A323545, A323546.

Cf. 2*A000537 and A163102 (with sum instead of product).

Programs

Magma

m:=3; [(&*[k^m + (n-k)^m: k in [0..n]]): n in [0..15]]; // G. C. Greubel, Jan 18 2019

Mathematica

Table[Product[k^3+(n-k)^3, {k, 0, n}], {n, 0, 15}]

PARI

m=3; vector(15, n, n--; prod(k=0,n, k^m + (n-k)^m)) \\ G. C. Greubel, Jan 18 2019

Sage

m=3; [product(k^m +(n-k)^m for k in (0..n)) for n in (0..15)] # G. C. Greubel, Jan 18 2019

Formula

a(n) ~ exp(2*(Pi/(3*sqrt(3))-1)*n) * n^(3*n + 3).

A126200 Numbers n such that n^2 is a sum of consecutive cubes larger than 1.

Original entry on oeis.org

8, 27, 64, 125, 204, 216, 312, 315, 323, 343, 504, 512, 588, 720, 729, 1000, 1331, 1728, 2079, 2170, 2197, 2744, 2940, 3375, 4096, 4472, 4913, 4914, 5187, 5832, 5880, 5984, 6630, 6859, 7497, 8000, 8721, 8778, 9261, 9360, 10296, 10648, 10695, 11024, 12167, 13104

Offset: 1

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Author

Zak Seidov, Mar 11 2007

Keywords

nonn

Comments

Note that all triangular numbers A000217(i) have squares A000217(i)^2=A000537(i), which are sums of consecutive cubes starting with 1. But such decompositions are not counted here. - R. J. Mathar, Nov 02 2007

Also, the positive integers n such that n^2 is the difference of squares of two positive triangular numbers. - Max Alekseyev, Jul 27 2014

Included all cubes > 1.

Examples

204^2=23^3+24^3+25^3, 312^2=14^3+15^3+...24^3+25^3; n^2=sum[i^3, (i=i1...i2)]; {n, i1=initial index of cube, i2=final index of cube}: {8, 4, 4}, {27, 9, 9}, {64, 16, 16}, {125, 25, 25}, {204, 23, 25}, {216, 36, 36}, {312, 14, 25}, {315, 25, 29}, {323, 9, 25}, {343, 49, 49}, {504, 28, 35}, {512, 64, 64}, {588, 14, 34}, {720, 25, 39}, {729, 81, 81}, {1000, 100, 100}, {1331, 121, 121}, {1728, 144, 144}, {2079, 33, 65}, {2170, 96, 100}, {2197, 169, 169}, {2744, 196, 196}.

Links

Donovan Johnson, Table of n, a(n) for n = 1..1000

Crossrefs

Cf. A126203.

Programs

PARI

mc=335241; cb=vector(mc); for(i=2, mc, cb[i]=i^3); v=vector(1000); mx=194104539^2; n=0; for(i=2, mc, s=0; for(j=i, mc, s=s+cb[j]; if(s>mx, next(2)); if(issquare(s,&sr), n++; v[n]=sr))); v=vecsort(v); for(i=1, 1000, write("b126200.txt", i " " v[i])) /* Donovan Johnson, Feb 02 2013 */

Extensions

Better definition from Dean Hickerson, Dec 02 2007

Many terms were missing - thanks to Donovan Johnson for catching this. (Feb 02 2013)

A027602 a(n) = n^3 + (n+1)^3 + (n+2)^3.

Original entry on oeis.org

9, 36, 99, 216, 405, 684, 1071, 1584, 2241, 3060, 4059, 5256, 6669, 8316, 10215, 12384, 14841, 17604, 20691, 24120, 27909, 32076, 36639, 41616, 47025, 52884, 59211, 66024, 73341, 81180, 89559, 98496, 108009, 118116, 128835, 140184

Offset: 0

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Author

Patrick De Geest

Keywords

nonn
easy

Comments

a(3) = 216 = 6^3 (a cube). - Howard Berman (howard_berman(AT)hotmail.com), Nov 07 2008

Pairs [n,a(n)] for n<=10^7 such that a(n) is a perfect power are [0, 9], [1, 36], [3, 216], [23, 41616]. - Joerg Arndt, Jan 25 2011

Sums of three consecutive cubes. - Al Hakanson (hawkuu(AT)gmail.com), May 20 2009

Links

Vincenzo Librandi, Table of n, a(n) for n = 0..750

Patrick De Geest, Palindromic Sums of Cubes of Consecutive Integers

Index entries for linear recurrences with constant coefficients, signature (4,-6,4,-1).

Crossrefs

Cf. A000537, A003215, A006527, A008585, A059100.

Cf. A000578, A005898, A027603, A027604.

Programs

Magma

[3*n^3 + 9*n^2 + 15*n + 9: n in [0..60]]; // Vincenzo Librandi, Apr 26 2011

Mathematica

f[n_]:=n^3; Table[f[n]+f[n+1]+f[n+2], {n, 0, 100}] (* Vladimir Joseph Stephan Orlovsky, Jan 03 2009 *) Table[3n^3+9n^2+15n+9,{n,0,40}] (* or *) LinearRecurrence[{4,-6,4,-1},{9,36,99,216},40] (* Harvey P. Dale, Nov 27 2024 *)

PARI

a(n)=3*(n^3 + 3*n^2 + 5*n + 3) \\ Charles R Greathouse IV, Jun 11 2015

Python

A027602_list, m = [], [18, 0, 9, 9] for _ in range(10**2): A027602_list.append(m[-1]) for i in range(3): m[i+1] += m[i] # Chai Wah Wu, Dec 15 2015

Sage

[i^3+(i+1)^3+(i+2)^3 for i in range(0,48)] # Zerinvary Lajos, Jul 03 2008

Formula

a(n) = 4*a(n-1) - 6*a(n-2) + 4*a(n-3) - 1*a(n-4) for n>=4.

a(n) = 9*A006527(n+1). - Lekraj Beedassy, Feb 01 2007

a(n) = 3*n^3 + 9*n^2 + 15*n + 9.

G.f.: 9*(1+x^2)/(1-x)^4. - Bruno Berselli, Jan 21 2011

a(n) = A008585(n+1)*A059100(n+1). - Bruno Berselli, Jan 21 2011

E.g.f.: 3*(3 + 9*x + 6*x^2 + x^3)*exp(x). - G. C. Greubel, Aug 24 2022

Sum_{n>=0} 1/a(n) = (2*gamma + polygamma(0, 1-i*sqrt(2)) + polygamma(0, 1+i*sqrt(2)))/12 = 0.161383557127191633050394086192620963436504... where i denotes the imaginary unit. - Stefano Spezia, Aug 31 2023

A058895 a(n) = n^4 - n.

Original entry on oeis.org

0, 0, 14, 78, 252, 620, 1290, 2394, 4088, 6552, 9990, 14630, 20724, 28548, 38402, 50610, 65520, 83504, 104958, 130302, 159980, 194460, 234234, 279818, 331752, 390600, 456950, 531414, 614628, 707252, 809970, 923490, 1048544, 1185888, 1336302, 1500590, 1679580

Offset: 0

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Author

Henry Bottomley, Jan 08 2001

Keywords

easy
nonn

Comments

a(n) is the number of ways to assign 4 different students to n different dorm rooms, each of which can hold at most 3 students. In other words, a(n) is the number of functions f:[4]->[n] with the size of the pre-image set of each element of the codomain at most 3. - Dennis P. Walsh, Mar 21 2013

a(n) are the values of m that yield integer solutions to this family of equations: x = sqrt(m + sqrt(x)), which may also be viewed as an infinitely recursive radical. The real solutions for x at each m = a(n) is n^2, except at n = 1 (m = 0) where x = 0 or 1 is a solution. - Richard R. Forberg, Oct 15 2014

Links

Harry J. Smith, Table of n, a(n) for n = 0..2000

Milan Janjic, Enumerative Formulas for Some Functions on Finite Sets

Index entries for linear recurrences with constant coefficients, signature (5,-10,10,-5,1).

Crossrefs

Cf. A000583, A002061, A002378, A024001, A027444, A027482, A068601, A033562, A185065, A339605.

Programs

Magma

[n^4-n: n in [0..40]]; // Vincenzo Librandi, Feb 23 2012

Maple

seq(n*(n^3-1),n=0..25) ; # R. J. Mathar, Dec 10 2015

Mathematica

Table[n^4 - n, {n, 0, 40}] (* Vladimir Joseph Stephan Orlovsky, Feb 20 2012 *)

PARI

a(n) = { n^4-n } \\ Harry J. Smith, Jun 23 2009

Sage

(2*x^2*(7+4*x+x^2)/(1-x)^5).series(x, 37).coefficients(x, sparse=False) # Stefano Spezia, Jul 09 2021

Formula

a(n) = n*(n-1)*(n^2+n+1) = A000583(n) - n = A002061(n+1) * A002378(n-1) = (n-1) * A027444(n) = -n * A024001(n).

a(n) = 2*A027482(n). - Zerinvary Lajos, Jan 28 2008

a(n) = floor(n^7/(n^3+1)). - Gary Detlefs, Feb 11 2010

a(n)^3 = (a(n)/n)^4 + (a(n)/n)^3. - Vincenzo Librandi, Feb 23 2012

a(n)^3 + A068601(n)^3 + A033562(n)^3 = A185065(n)^3, for n > 0. - Vincenzo Librandi, Mar 13 2012

G.f.: 2*x^2*(7 + 4*x + x^2)/(1 - x)^5. - Colin Barker, Apr 23 2012

a(n) = 14*C(n,2) + 36*C(n,3) + 24*C(n,4). - Dennis P. Walsh, Mar 21 2013

Sum_{n>=2} (-1)^n/a(n) = (Pi/3)*sech(Pi*sqrt(3)/2) + 4*log(2)/3 - 1 = 0.06147271494... . - Amiram Eldar, Jul 04 2020

Sum_{n>=2} 1/a(n) = A339605. - R. J. Mathar, Jan 08 2021

E.g.f.: exp(x)*x^2*(7 + 6*x + x^2). - Stefano Spezia, Jul 09 2021

a(n) = 12*A000332(n+2) + 2*A000537(n-1). - Yasser Arath Chavez Reyes, Apr 05 2024

A223918 T(n,k)=Number of nXk 0..2 arrays with rows and antidiagonals unimodal and columns nondecreasing.

Original entry on oeis.org

3, 9, 6, 22, 36, 10, 46, 158, 100, 15, 86, 548, 666, 225, 21, 148, 1600, 3311, 2111, 441, 28, 239, 4102, 13347, 14123, 5548, 784, 36, 367, 9503, 45988, 74040, 48182, 12752, 1296, 45, 541, 20299, 140236, 323394, 319156, 139925, 26494, 2025, 55, 771, 40570

Offset: 1

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Author

R. H. Hardin Mar 29 2013

Keywords

nonn
tabl

Comments

Table starts

..3....9.....22......46.......86.......148........239.........367..........541

..6...36....158.....548.....1600......4102.......9503.......20299........40570

.10..100....666....3311....13347.....45988.....140236......387671.......988447

.15..225...2111...14123....74040....323394....1226491.....4157102.....12856218

.21..441...5548...48182...319156...1721356....7906972....31947798....116289938

.28..784..12752..139925..1147712...7526024...41334135...196691651....831996762

.36.1296..26494..359344..3588729..28247478..183720802..1022160764...4994941017

.45.2025..50863..837243.10031780..93683295..715114500..4620831232..25961964085

.55.3025..91634.1802306.25574524.280338123.2489822632.18558764134.119279565646

.66.4356.156682.3633256.60357024.769012348.7881608430.67311575302.492221386729

Examples

Some solutions for n=3 k=4 ..0..2..1..0....1..2..2..0....0..0..1..0....1..0..0..0....1..0..0..0 ..0..2..1..1....1..2..2..0....0..0..1..1....1..1..1..1....1..0..0..0 ..1..2..1..1....1..2..2..0....0..1..2..1....2..2..2..2....2..2..2..0

Links

R. H. Hardin, Table of n, a(n) for n = 1..480

Crossrefs

Column 1 is A000217(n+1)

Column 2 is A000537(n+1)

Row 1 is A223718

Formula

Empirical: columns k=1..7 are polynomials of degree 2*k for n>0,0,0,1,2,3,4

Empirical: rows n=1..7 are polynomials of degree 4*n

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cp's OEIS Frontend

A163102 a(n) = n^2*(n+1)^2/2.

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A101097 a(n) = n*(n+1)*(n+2)*(n+3)*(n+4)*(2 + 4*n + n^2)/840.

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A074279 n appears n^2 times.

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A101102 Fifth partial sums of cubes (A000578).

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A250812 T(n,k)=Number of (n+1)X(k+1) 0..2 arrays with nondecreasing x(i,j)-x(i,j-1) in the i direction and nondecreasing min(x(i,j),x(i-1,j)) in the j direction.

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A323541 a(n) = Product_{k=0..n} (k^3 + (n-k)^3).

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A101097 a(n) = n(n+1)(n+2)(n+3)(n+4)(2 + 4n + n^2)/840.