A004006 a(n) = C(n,1) + C(n,2) + C(n,3), or n*(n^2 + 5)/6.

0, 1, 3, 7, 14, 25, 41, 63, 92, 129, 175, 231, 298, 377, 469, 575, 696, 833, 987, 1159, 1350, 1561, 1793, 2047, 2324, 2625, 2951, 3303, 3682, 4089, 4525, 4991, 5488, 6017, 6579, 7175, 7806, 8473, 9177, 9919, 10700, 11521, 12383, 13287, 14234, 15225

Offset

0,3

Comments

3-dimensional analog of centered polygonal numbers.

The Burnside group B(3,n) has order 3^a(n).

Answer to the question: if you have a tall building and 3 plates and you need to find the highest story, a plate thrown from which does not break, what is the number of stories you can handle given n tries? - Leonid Broukhis, Oct 24 2000

Equals row sums of triangle A144329 starting with "1". - Gary W. Adamson, Sep 18 2008

Let A be the Hessenberg matrix of order n, defined by: A[1,j]=A[i,i]:=1, A[i,i-1]=-1, and A[i,j]=0 otherwise. Then, for n>=4, a(n-1)=-coeff(charpoly(A,x),x^(n-3)). - Milan Janjic, Jan 24 2010

From J. M. Bergot, Aug 03 2011: (Start)

If one formed the 3 X 3 square

| n | n+1 | n+2 |

| n+3 | n+4 | n+5 |

| n+6 | n+7 | n+8 |

and found the sum of the horizontal products n*(n + 1)*(n + 2) + (n + 3)*(n + 4)*(n + 5) + (n + 6)*(n + 7)*(n + 8) and added the sum of the vertical products n*(n + 3)*(n + 6) + (n + 1)*(n + 4)*(n + 7) + (n + 2)*(n + 5)(n + 8) one gets 6*n^3 + 72*n^2 + 318*n + 504. This will give 36 times the values of all the terms in this sequence. (End)

a(n) is divisible by n for n congruent to {1,5} mod 6. (see A007310). - Gary Detlefs, Dec 08 2011

From Beimar Naranjo, Feb 22 2024: (Start)

Number of compositions with at most three parts and sum at most n.

Also the number of compositions with at most one part distinct from 1 and with a sum at most n. (End)

References

W. Magnus, A. Karrass and D. Solitar, Combinatorial Group Theory, Wiley, 1966, see p. 380.

Links

Vincenzo Librandi, Table of n, a(n) for n = 0..5000

Michael Boardman, The Egg-Drop Numbers, Mathematics Magazine, 77 (2004), 368-372. [From Parthasarathy Nambi, Sep 30 2009]

F. Javier de Vega, An extension of Furstenberg's theorem of the infinitude of primes, arXiv:2003.13378 [math.NT], 2020.

Oboifeng Dira, A Note on Composition and Recursion, Southeast Asian Bulletin of Mathematics, 2017, Vol. 41 Issue 6, pp. 849-853.

N. Heninger, E. M. Rains and N. J. A. Sloane, On the Integrality of n-th Roots of Generating Functions, arXiv:math/0509316 [math.NT], 2005-2006.

N. Heninger, E. M. Rains and N. J. A. Sloane, On the Integrality of n-th Roots of Generating Functions, J. Combinatorial Theory, Series A, 113 (2006), 1732-1745.

Milan Janjic, Hessenberg Matrices and Integer Sequences , J. Int. Seq. 13 (2010) # 10.7.8.

T. P. Martin, Shells of atoms, Phys. Reports, 273 (1996), 199-241, eq. (11).

Laurent Saloff-Coste, Random walks on finite groups, in Probability on discrete structures, 263-346, Encyclopaedia Math. Sci., 110, Springer, 2004).

Bridget Eileen Tenner, Reduced word manipulation: patterns and enumeration, J. Algebr. Comb. 46, No. 1, 189-217 (2017), w in S_n(231) l(w)=3.

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

Formula

G.f.: x*(1-x+x^2)/(1-x)^4.

E.g.f.: x*(1 + x/2 + x^2/6) * exp(x).

a(-n) = -a(n).

a(n) = binomial(n+2,n-1) - binomial(n,n-2). - Zerinvary Lajos, May 11 2006

Euler transform of length 6 sequence [3, 1, 1, 0, 0, -1]. - Michael Somos, May 04 2007

Starting (1, 3, 7, 14, ...) = binomial transform of [1, 2, 2, 1, 0, 0, 0, ...]. - Gary W. Adamson, Apr 24 2008

a(0)=0, a(1)=1, a(2)=3, a(3)=7, a(n) = 4*a(n-1) - 6*a(n-2) + 4*a(n-3) - a(n-4). - Harvey P. Dale, Aug 21 2011

a(n+1) = A000292(n) + n + 1. - Reinhard Zumkeller, Mar 31 2012

a(n) = 2*a(n-1) + (n-1) - a(n-2) with a(0) = 0, a(1) = 1. - Richard R. Forberg, Jan 23 2014

a(n) = Sum_{i=1..n} binomial(n-2i,2). - Wesley Ivan Hurt, Nov 18 2017

a(n) = n + Sum_{k=0..n} k*(n-k). - Gionata Neri, May 19 2018

a(n) = Sum_{k=0..n-1} A000124(k). - Torlach Rush, Aug 05 2018

G.f.: ((1 - x^5)/(1 - x)^5 - 1)/5. - Michael Somos, Dec 29 2019

G.f.: g(f(x)), where g is g.f. of A001477 and f is g.f. of A128834. - Oboifeng Dira, Jun 21 2020

Sum_{n>0} 1/a(n) = 3*(2*gamma + polygamma(0, 1-i*sqrt(5)) + polygamma(0, 1+i*sqrt(5))/5 = 1.6787729555834452106286261834348972248... where i denotes the imaginary unit. - Stefano Spezia, Aug 31 2023

Example

G.f. = x + 3*x^2 + 7*x^3 + 14*x^4 + 25*x^5 + 41*x^6 + 63*x^7 + 92*x^8 + ... - Michael Somos, Dec 29 2019

Maple

A004006 := proc(n) n*(n^2+5)/6 ; end proc:
seq(A004006(n), n=0..10) ; # R. J. Mathar, Jun 05 2011

Mathematica

Table[Total[Table[Binomial[n, i], {i, 3}]], {n, 0, 50}] (* or *) LinearRecurrence[{4, -6, 4, -1}, {0, 1, 3, 7}, 50] (* Harvey P. Dale, Aug 21 2011 *)

Prog

(PARI) {a(n) = n*(n^2 + 5)/6}; /* Michael Somos, May 04 2007 */
(Magma) [n*(n^2+5)/6: n in [0..50]]; // Vincenzo Librandi, May 15 2011
(Haskell)
a004006 n = a000292 n + n + 1  -- Reinhard Zumkeller, Mar 31 2012
(Maxima) A004006(n):=n*(n^2+5)/6$ makelist(A004006(n), n, 0, 30); /* Martin Ettl, Jan 08 2013 */
(Sage) [n*(n^2+5)/6 for n in (0..50)] # G. C. Greubel, Aug 27 2019
(GAP) List([0..50], n-> n*(n^2+5)/6); # G. C. Greubel, Aug 27 2019

Crossrefs

Cf. A051576, A055795, A006552. Differences give A000217 + 1 or A000124.

1/12*t*(n^3-n)+n for t = 2, 4, 6, ... gives A004006, A006527, A006003, A005900, A004068, A000578, A004126, A000447, A004188, A004466, A004467, A007588, A062025, A063521, A063522, A063523.

Cf. A000292, A001477, A007310, A128834, A144329, A228074.

Sequence in context: A171973 A253895 A365641 * A089240 A057524 A293467

Adjacent sequences: A004003 A004004 A004005 * A004007 A004008 A004009

Keyword

nonn,nice,easy

Author

Albert D. Rich (Albert_Rich(AT)msn.com)

Status

approved