A062392 -id:A062392 - cp's OEIS frontend

A271705 Triangle read by rows, T(n,k) = Sum_{j=0..n} C(n,j)*L(j,k), L the unsigned Lah numbers A271703, for n>=0 and 0<=k<=n.

1, 1, 1, 1, 4, 1, 1, 15, 9, 1, 1, 64, 66, 16, 1, 1, 325, 490, 190, 25, 1, 1, 1956, 3915, 2120, 435, 36, 1, 1, 13699, 34251, 23975, 6755, 861, 49, 1, 1, 109600, 328804, 283136, 101990, 17696, 1540, 64, 1, 1, 986409, 3452436, 3534636, 1554966, 342846, 40404, 2556, 81, 1

Offset: 0

Peter Luschny, Apr 14 2016

This is the Sheffer (aka exponential Riordan) matrix T = P*L = A007318*A271703 = (exp(x), x/(1-x)). Note that P = A007318 is Sheffer (exp(t), t) (of the Appell type). The Sheffer a-sequence is [1,1,repeat(0)] and the z-sequence has e.g.f. (x/(1+x))*(1 - exp(-x/(1+x)) given in A288869 / A000027. Because the column k=0 has only entries 1, the z-sequence gives fractional representations of 1. See A288869. - Wolfdieter Lang, Jun 20 2017

			Triangle starts:
  1;
  1,    1;
  1,    4,    1;
  1,   15,    9,    1;
  1,   64,   66,   16,   1;
  1,  325,  490,  190,  25,  1;
  1, 1956, 3915, 2120, 435, 36, 1;
  ...
Recurrence: T(3, 2) = (3/2)*4 + 3*1 = 9. - _Wolfdieter Lang_, Jun 20 2017

G. C. Greubel, Rows n = 0..50 of the triangle, flattened
Marin Knežević, Vedran Krčadinac, and Lucija Relić, Matrix products of binomial coefficients and unsigned Stirling numbers, arXiv:2012.15307 [math.CO], 2020.

Cf. A000290 (diag n, n-1), A062392 (diag n, n-2).
Cf. A007526 (col. 1), A134432 (col. 2).
Cf. A052844 (row sums), A059110 (matrix inverse).
Cf. A007318, A271703, A288869.

B:=Binomial;
A271705:= func< n,k | k eq 0 select 1 else (&+[B(n, j+k)*B(j+k, k)*B(j+k-1, k-1)*Factorial(j): j in [0..n-k]]) >;
[A271705(n,k): k in [0..n], n in [0..12]]; // G. C. Greubel, Jan 09 2022

L := (n,k) -> `if`(k<0 or k>n,0,(n-k)!*binomial(n,n-k)*binomial(n-1,n-k)):
T := (n,k) -> add(L(j,k)*binomial(-j-1,-n-1)*(-1)^(n-j), j=0..n):
seq(seq(T(n,k), k=0..n), n=0..9);

T[n_, k_]:= If[k==0, 1, Sum[((k*j!)/(j+k))*Binomial[n, j+k]*Binomial[j+k, k]^2, {j,0,n-k}]];
Table[T[n, k], {n,0,12}, {k,0,n}]//Flatten (* G. C. Greubel, Jan 09 2022 *)

b=binomial
def A271705(n,k): return 1 if (k==0) else sum(factorial(j-k)*b(n, j)*b(j, k)*b(j-1, k-1) for j in (k..n))
flatten([[A271705(n,k) for k in (0..n)] for n in (0..12)]) # G. C. Greubel, Jan 09 2022

From Wolfdieter Lang, Jun 20 2017: (Start)
T(n, k) = Sum_{m=k..n} A007318(n, m)*A271703(m, k), n >= k >= 0, and 0 for k < m. See also the name.
E.g.f. of column k: exp(x)*(x/(1-x))^k/k! (Sheffer property), k >= 0.
E.g.f. of triangle (or row polynomials in x): exp(z)*exp(x*z/(1-z)).
Recurrence for T(n, k), k >= 1, with T(n, 0) = 1, T(n, k) = 0 if n < k: T(n, k) = (n/k)*T(n-1, k-1) + n*T(n-1, k), n >= 1, k = 1..n. (From the a-sequence with column k=0 as input.) (End)
T(n, k) = Sum_{j=0..n-k} j!*binomial(n, j+k)*binomial(j+k, k)*binomial(j+k-1, k-1) with T(n, 0) = 1. - G. C. Greubel, Jan 09 2022
From Natalia L. Skirrow, Jun 11 2025: (Start)
T(n, k) = C(n, k)*hypergeom([k-n, k], [], -1), which equals C(n, k)*A143409(n-k, k-1) for k>0.
By the saddle point method upon the e.g.f., n-th row polynomial converges with n (for all y) to n^n*exp(2*sqrt(n*y) - n - y/2 + 1)/sqrt(2*sqrt(n/y)); as such, the n-th row's expectation is ~ sqrt(n)-1/4 and the n-th row's variance is ~ (sqrt(n)-1)/2. (End)

A110450 a(n) = n(n+1)(n^2+n+1)/2.

Original entry on oeis.org

0, 3, 21, 78, 210, 465, 903, 1596, 2628, 4095, 6105, 8778, 12246, 16653, 22155, 28920, 37128, 46971, 58653, 72390, 88410, 106953, 128271, 152628, 180300, 211575, 246753, 286146, 330078, 378885, 432915, 492528, 558096, 630003, 708645, 794430

Offset: 0

Reinhard Zumkeller, Jul 21 2005

This sequence is related to A085461 by 3*A085461(n) = n*a(n) - Sum_{i=0..n-1} a(i) for n>0. - Bruno Berselli, Dec 27 2010

Subsequence of the triangular numbers A000217, see formulas below. - David James Sycamore, Jul 31 2018

G. C. Greubel, Table of n, a(n) for n = 0..5000
Index entries for linear recurrences with constant coefficients, signature (5,-10,10,-5,1).

Cf. A000217, A002378, A061317, A085461, A110449, A014105.

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

[n*(n+1)*(n^2+n+1)/2: n in [0..40]]; // Vincenzo Librandi, Dec 26 2010

A110450:=n->n*(n+1)*(n^2+n+1)/2; seq(A110450(k), k=0..50); # Wesley Ivan Hurt, Sep 27 2013

Table[n (n + 1) (n^2 + n + 1)/2, {n, 0, 100}] (* Wesley Ivan Hurt, Sep 27 2013 *)
CoefficientList[Series[-3 x (x^2 + 2 x + 1)/(x - 1)^5, {x, 0, 36}], x] (* or *)
LinearRecurrence[{5, -10, 10, -5, 1}, {0, 3, 21, 78, 210}, 36] (* Robert G. Wilson v, Jul 31 2018 *)

a(n)=n*(n+1)*(n^2+n+1)/2 \\ Charles R Greathouse IV, Oct 16 2015

a(n) = Sum_{k=0..n} A110449(n,k), sums of rows in triangle A110449.
From Bruno Berselli, Dec 27 2010: (Start)
G.f.: 3*x*(1 + x)^2/(1 - x)^5.
a(n) = A014105(A000217(n)). (End)
a(n) = Sum_{i=1..n*(n+1)} i. - Wesley Ivan Hurt, Sep 27 2013
a(n) = Sum_{i=0..n} i*(2*i^2+1), and these are the partial sums of A061317. - Bruno Berselli, Feb 09 2017
a(n) = t(n,t(n,A000217(n))), where t(n,k) = n*(n+1)/2 + k*n and k=0. - Bruno Berselli, Feb 28 2017
E.g.f.: (x/2)*(6 + 15*x + 8*x^2 + x^3)*exp(x). - G. C. Greubel, Aug 24 2017
a(n) = A000217(n*(n+1)). - David James Sycamore, Jul 31 2018
a(n) = A000217(2*A000217(n)) = A000217(A002378(n)). - Alois P. Heinz, Jul 31 2018
From R. J. Mathar, Mar 23 2021: (Start)
a(n) = A002378(n)+A062392(n).
a(n) = 3*A006325(n+1). (End)
Sum_{n>=1} 1/a(n) = 4 - 2*Pi*tanh(sqrt(3)*Pi/2)/sqrt(3). - Amiram Eldar, May 10 2025

A062393 a(n) = n^5 - (n-1)^5 + (n-2)^5 - ... +(-1)^n*0^5.

Original entry on oeis.org

0, 1, 31, 212, 812, 2313, 5463, 11344, 21424, 37625, 62375, 98676, 150156, 221137, 316687, 442688, 605888, 813969, 1075599, 1400500, 1799500, 2284601, 2869031, 3567312, 4395312, 5370313, 6511063, 7837844, 9372524, 11138625, 13161375

Offset: 0

Henry Bottomley, Jun 21 2001

The general formula for alternating sums of powers is in terms of the Swiss-Knife polynomials P(n,x) A153641 2^(-n-1)(P(n,1)-(-1)^k P(n,2k+1)). Thus a(k) = |2^(-6)(P(5,1)-(-1)^k P(5,2k+1))|. - Peter Luschny, Jul 12 2009

Harry J. Smith, Table of n, a(n) for n = 0..1000
Index entries for linear recurrences with constant coefficients, signature (5,-9,5,5,-9,5,-1).

Cf. A000539, A000584. A062392 for 4th powers, A152725 for 6th powers.

a := n -> (1-(-1)^n+n^2*(n^2*(2*n+5)-5))/4; # Peter Luschny, Jul 12 2009

k=0;lst={k};Do[k=n^5-k;AppendTo[lst, k], {n, 1, 5!}];lst (* Vladimir Joseph Stephan Orlovsky, Dec 11 2008 *)
Table[Total[(Times@@@Partition[Riffle[Range[n,0,-1],{1,-1},{2,-1,2}],2])^5],{n,0,30}] (* or *) LinearRecurrence[ {5,-9,5,5,-9,5,-1},{0,1,31,212,812,2313,5463},40] (* Harvey P. Dale, Feb 01 2013 *)

{ a=0; for (n=0, 1000, write("b062393.txt", n, " ", a=n^5 - a) ) } \\ Harry J. Smith, Aug 07 2009

a(n) = (2*n^5 + 5*n^4 - 5*n^2 + 1 - (-1)^n)/4 = n^5 - a(n-1).
G.f.: x*(x^4 + 26*x^3 + 66*x^2 + 26*x + 1)/((x-1)^6*(x+1)). - Colin Barker, Sep 19 2012
a(0)=0, a(1)=1, a(2)=31, a(3)=212, a(4)=812, a(5)=2313, a(6)=5463, a(n) = 5*a(n-1) - 9*a(n-2) + 5*a(n-3) + 5*a(n-4) - 9*a(n-5) + 5*a(n-6) - a(n-7). - Harvey P. Dale, Feb 01 2013

A086302 a(n) = 4n^4 + 24n^3 + 48n^2 + 36n + 8.

Original entry on oeis.org

8, 120, 528, 1520, 3480, 6888, 12320, 20448, 32040, 47960, 69168, 96720, 131768, 175560, 229440, 294848, 373320, 466488, 576080, 703920, 851928, 1022120, 1216608, 1437600, 1687400, 1968408, 2283120, 2634128, 3024120, 3455880, 3932288, 4456320, 5031048

Offset: 0

Neven Juric (neven.juric(AT)apis-it.hr), Aug 29 2003

Suppose one wishes to find sets of four positive integers (a,b,c,d) such that ab+1, ac+1, ad+1, bc+1, bd+1, cd+1 are perfect squares. Then one may take a = 1, b = x^2 + 2x, c = x^2 + 4x + 3, d = 4x^4 + 24x^3 + 48x^2 + 36x + 8.

			(a,b,c,d) = (1,3,8,120), (1,8,15,528), (1,15,24,1520), (1,24,35,3480), ...

Paolo Xausa, Table of n, a(n) for n = 0..10000
Philip Gibbs, Diophantine quadruples and Cayley's hyperdeterminant, arXiv:math/0107203 [math.NT], 2001.
Eric Weisstein's World of Mathematics, Diophantus Property.
Index entries for linear recurrences with constant coefficients, signature (5,-10,10,-5,1).

Cf. A057769, A062392.

LinearRecurrence[{5, -10, 10, -5, 1}, {8, 120, 528, 1520, 3480}, 50] (* or *)
A086302[n_] := 4 (n + 1) (n + 2) (n^2 + 3 n + 1);
Array[A086302, 50, 0] (* Paolo Xausa, Jan 16 2024 *)

a(n) = A057769(n+1) +  1. - N. J. A. Sloane, Jun 12 2004
G.f.: 8*(1 + 10*x + x^2)/(1 - x)^5. - Colin Barker, Mar 26 2012
a(n) = 4 * (n+1) * (n+2) * (n^2 + 3*n + 1). - Bruno Berselli, Mar 26 2012
a(n) = 8*A062392(n+1). - Bruce J. Nicholson, Jun 05 2017
Sum_{n>=0} 1/a(n) = tan(sqrt(5)*Pi/2)*Pi/(4*sqrt(5)). - Amiram Eldar, Jan 22 2024
E.g.f.: 4*exp(x)*(2 + 28*x + 37*x^2 + 12*x^3 + x^4). - Stefano Spezia, Apr 27 2025

A264854 a(n) = n(n + 1)(11n^2 + 11n - 10)/24.

Original entry on oeis.org

0, 1, 14, 61, 175, 400, 791, 1414, 2346, 3675, 5500, 7931, 11089, 15106, 20125, 26300, 33796, 42789, 53466, 66025, 80675, 97636, 117139, 139426, 164750, 193375, 225576, 261639, 301861, 346550, 396025, 450616, 510664, 576521, 648550, 727125, 812631, 905464, 1006031

Offset: 0

Ilya Gutkovskiy, Nov 26 2015

Partial sums of centered 11-gonal (or hendecagonal) pyramidal numbers.

OEIS Wiki, Figurate numbers
Eric Weisstein's World of Mathematics, Pyramidal Number
Index entries for linear recurrences with constant coefficients, signature (5,-10,10,-5,1).

Cf. A004467.

Cf. similar sequences provided by the partial sums of centered k-gonal pyramidal numbers: A006522 (k=1), A006007 (k=2), A002817 (k=3), A006325 (k=4), A006322 (k=5), A000537 (k=6), A006323 (k=7), A006324 (k=8), A236770 (k=9), A264853 (k=10), this sequence (k=11), A062392 (k=12), A264888 (k=13).

[n*(n+1)*(11*n^2+11*n-10)/24: n in [0..50]]; // Vincenzo Librandi, Nov 27 2015

Table[n (n + 1) (11 n^2 + 11 n - 10)/24, {n, 0, 50}]

a(n)=n*(n+1)*(11*n^2+11*n-10)/24 \\ Charles R Greathouse IV, Jul 26 2016

G.f.: x*(1 + 9*x + x^2)/(1 - x)^5.
a(n) = Sum_{k = 0..n} A004467(k).
a(n) = 5*a(n-1) - 10*a(n-2) + 10*a(n-3) - 5*a(n-4) + a(n-5). - Vincenzo Librandi, Nov 27 2015

A181475 a(n) = 3n^4 + 6n^3 - 3*n + 1.

Original entry on oeis.org

1, 7, 91, 397, 1141, 2611, 5167, 9241, 15337, 24031, 35971, 51877, 72541, 98827, 131671, 172081, 221137, 279991, 349867, 432061, 527941, 638947, 766591, 912457, 1078201, 1265551, 1476307, 1712341, 1975597, 2268091, 2591911, 2949217, 3342241, 3773287, 4244731

Offset: 0

Bruno Berselli, Oct 25 2010 - Oct 29 2010

If gcd(n,7) = gcd(n+1,7) = gcd(2*n+1,7) = 1 then a(n) == 0 (mod 7) (E. Picutti, see References).

Ettore Picutti, Sul numero e la sua storia, Feltrinelli Economica, 1977, p. 208.

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

Subsequence of A003215.

Magma
```
[3*n^4+6*n^3-3*n+1: n in [0..31]];
```

Table[3 n^4 + 6 n^3 - 3 n + 1, {n, 0, 40}] (* Vincenzo Librandi, Mar 26 2013 *)
LinearRecurrence[{5,-10,10,-5,1},{1,7,91,397,1141},40] (* Harvey P. Dale, Jul 12 2022 *)

G.f.: (1 + 2*x + 66*x^2 + 2*x^3 + x^4)/(1-x)^5.
a(n) = a(-n-1) = 5*a(n-1) - 10*a(n-2) + 10*a(n-3) - 5*a(n-4) + a(n-5).
a(n) = 4*a(n-1) - 6*a(n-2) + 4*a(n-3) - a(n-4) + 6*12.
a(n) = 3*a(n-1) - 3*a(n-2) + a(n-3) + 6*A008594(n-1).
a(n) = 2*a(n-1) - a(n-2) + 6*A003154(n).
a(n) = a(n-1) + 6*A007588(n).
a(n) = 1 + 6*A062392(n).
a(n) = 7*A000540(n)/A000330(n) = A154105(A000096(n-1)) for n > 0.
Sum_{i=0..n} a(i) = (3*n^5 + 15*n^4 + 20*n^3 - 3*n + 5)/5.
a(n) = 7*(3*n^2 + 3*n - 1)*(Sum_{k=1..n} k^6)/(5*Sum_{k=1..n} k^4), n > 0. - Gary Detlefs, Oct 18 2011

Formula, program and crossref added by Bruno Berselli, Aug 22 2011

A101384 a(n) = n(n-1)^3(n^2-n-1)/2.

Original entry on oeis.org

0, 0, 1, 60, 594, 3040, 10875, 30996, 75460, 163584, 324405, 599500, 1046166, 1740960, 2783599, 4301220, 6453000, 9435136, 13486185, 18892764, 25995610, 35196000, 46962531, 61838260, 80448204, 103507200, 131828125, 166330476, 208049310, 258144544, 317910615

Offset: 0

N. J. A. Sloane, Jan 15 2005

T. A. Gulliver, Sequences from Cubes of Integers, Int. Math. Journal, 4 (2003), 439-445.

Vincenzo Librandi, Table of n, a(n) for n = 0..1000
Index entries for linear recurrences with constant coefficients, signature (7,-21,35,-35,21,-7,1).

Cf. A062392.

[n*(n-1)^3*(n^2-n-1)/2: n in [0..40]]; // Vincenzo Librandi, Jun 15 2011

A101384:= n-> n*(n-1)^3*(n^2 -n -1)/2: seq(A101384(n), n=0..35); # G. C. Greubel, Mar 11 2021

Table[n*(n-1)^3*(n^2 -n -1)/2, {n, 0, 35}] (* G. C. Greubel, Mar 11 2021 *)

[n*(n-1)^3*(n^2 -n -1)/2 for n in (0..35)] # G. C. Greubel, Mar 11 2021

G.f.: x^2*(1 + 53*x + 195*x^2 + 107*x^3 + 4*x^4)/(1 - x)^7. - Ilya Gutkovskiy, Feb 24 2017

E.g.f.: x^2*(1 + 19*x + 30*x^2 + 11*x^3 + x^4)*exp(x)/2. - G. C. Greubel, Mar 11 2021

A261032 a(n) = (-1)^n(n^8 + 4n^7 - 14n^5 + 28n^3 - 17*n)/2.

Original entry on oeis.org

0, -1, 255, -6306, 59230, -331395, 1348221, -4416580, 12360636, -30686085, 69313915, -145044966, 284936730, -530793991, 944995065, -1617895560, 2677071736, -4298685705, 6721274871, -10262288170, 15337711830, -22485147531, 32390726005, -45920259276, 64155054900, -88432835725

Offset: 0

Ilya Gutkovskiy, Nov 18 2015

Alternating sum of eighth powers (A001016).

For n>0, a(n) is divisible by A000217(n).

			a(0) = 0^8 = 0,
a(1) = 0^8 -1^8 = -1,
a(2) = 0^8 -1^8 + 2^8 = 255,
a(3) = 0^8 -1^8 + 2^8 - 3^8 = -6306,
a(4) = 0^8 -1^8 + 2^8 - 3^8 + 4^8 = 59230,
a(5) = 0^8 -1^8 + 2^8 - 3^8 + 4^8 - 5^8 = -331395, etc.

Robert Israel, Table of n, a(n) for n = 0..10000
Index entries for linear recurrences with constant coefficients, signature (-9,-36,-84,-126,-126,-84,-36,-9,-1 ).

Cf. A000217, A001016, A000542, A089594, A232599, A062392, A062393, A152725, A152726.

[(-1)^n*(n^8+4*n^7-14*n^5+28*n^3-17*n)/2: n in [0..30]]; // Vincenzo Librandi, Nov 20 2015

seq((-1)^n*(n^8 + 4*n^7 - 14*n^5 + 28*n^3 - 17*n)/2, n = 0 .. 100); # Robert Israel, Nov 18 2015

Table[(1/2) (-1)^n n (n + 1) (n^6 + 3 n^5 - 3 n^4 - 11 n^3 + 11 n^2 + 17 n - 17), {n, 0, 25}]

vector(100, n, n--; (-1)^n*(n^8+4*n^7-14*n^5+28*n^3-17*n)/2) \\ Altug Alkan, Nov 18 2015

[(-1)^n*(n^8 +4*n^7 -14*n^5 +28*n^3 -17*n)/2 for n in (0..40)] # G. C. Greubel, Apr 02 2021

G.f.: -x*(1 - 246*x + 4047*x^2 - 11572*x^3 + 4047*x^4 - 246*x^5 + x^6)/(1 + x)^9.
a(n) = Sum_{k = 0..n} (-1)^k*k^8.
a(n) = (-1)^n*n*(n + 1)*(n^6 + 3*n^5 - 3*n^4 - 11*n^3 + 11*n^2 + 17*n - 17)/2.
Sum_{n>0} 1/a(n) = -0.9962225712723456482...
Sum_{j=0..9} binomial(9,j)*a(n-j) = 0. - Robert Israel, Nov 18 2015
E.g.f.: (x/2)*(-2 +253*x -1848*x^2 +2961*x^3 -1596*x^4 +350*x^5 -32*x^6 +x^7)*exp(-x). - G. C. Greubel, Apr 02 2021

A384329 Table read by rows: row n is the unique primitive Pythagorean triple (a,b,c) such that (a-b+c)/2 = A000217(n) and its long leg and hypotenuse are consecutive natural numbers, n >= 0.

Original entry on oeis.org

-1, 0, 1, 1, 0, 1, 5, 12, 13, 11, 60, 61, 19, 180, 181, 29, 420, 421, 41, 840, 841, 55, 1512, 1513, 71, 2520, 2521, 89, 3960, 3961, 109, 5940, 5941, 131, 8580, 8581, 155, 12012, 12013, 181, 16380, 16381, 209, 21840, 21841, 239, 28560, 28561, 271, 36720, 36721, 305, 46512, 46513, 341, 58140, 58141

Offset: 0

Miguel-Ángel Pérez García-Ortega, May 26 2025

Row n = 0 and n = 1 are included by convention and correspond to the Pythagorean triples (-1)^2 + 0^2 = 1^2 and 1^2 + 0^2 = 1^2.

			  n=0:     -1,     0,     1;
  n=1:      1,     0,     1;
  n=2:      5,    12,    13;
  n=3:     11,    60,    61;
  ...

José Miguel Blanco Casado and Miguel-Ángel Pérez García-Ortega, El Libro de las Ternas Pitagóricas

Cf. A000217, A165900 (short leg), A062392 (semiperimeter), A384498 (sum of the legs).

a=Table[(n(n+1))/2,{n,0,18}];Apply[Join,Map[{2#-1,2#^2-2#,2#^2-2#+1}&,a]]

row(n) = (2*T(n) - 1, 2*T(n)*(T(n) - 1), 2*T(n)*(T(n) - 1) + 1) where T(n) = A000217(n).

cp's OEIS Frontend

A271705 Triangle read by rows, T(n,k) = Sum_{j=0..n} C(n,j)*L(j,k), L the unsigned Lah numbers A271703, for n>=0 and 0<=k<=n.

Views

Author

Keywords

Comments

Examples

Links

Crossrefs

Programs

Magma

Maple

Mathematica

Sage

Formula

A110450 a(n) = n*(n+1)*(n^2+n+1)/2.

Views

Author

Keywords

Comments

Links

Crossrefs

Programs

GAP

Magma

Maple

Mathematica

PARI

Formula

A062393 a(n) = n^5 - (n-1)^5 + (n-2)^5 - ... +(-1)^n*0^5.

Views

Author

Keywords

Comments

Links

Crossrefs

Programs

Maple

Mathematica

PARI

Formula

A086302 a(n) = 4*n^4 + 24*n^3 + 48*n^2 + 36*n + 8.

Views

Author

Keywords

Comments

Examples

Links

Crossrefs

Programs

Mathematica

Formula

A264854 a(n) = n*(n + 1)*(11*n^2 + 11*n - 10)/24.

Views

Author

Keywords

Comments

Links

Crossrefs

Programs

Magma

Mathematica

PARI

Formula

A181475 a(n) = 3*n^4 + 6*n^3 - 3*n + 1.

Views

Author

Keywords

Comments

References

Links

Crossrefs

Programs

Magma

Mathematica

Formula

Extensions

A101384 a(n) = n*(n-1)^3*(n^2-n-1)/2.

Views

Author

A110450 a(n) = n(n+1)(n^2+n+1)/2.

A086302 a(n) = 4n^4 + 24n^3 + 48n^2 + 36n + 8.

A264854 a(n) = n(n + 1)(11n^2 + 11n - 10)/24.

A181475 a(n) = 3n^4 + 6n^3 - 3*n + 1.

A101384 a(n) = n(n-1)^3(n^2-n-1)/2.

A261032 a(n) = (-1)^n(n^8 + 4n^7 - 14n^5 + 28n^3 - 17*n)/2.