A050407 a(n) = n*(n^2 - 6*n + 11)/6.

0, 1, 1, 1, 2, 5, 11, 21, 36, 57, 85, 121, 166, 221, 287, 365, 456, 561, 681, 817, 970, 1141, 1331, 1541, 1772, 2025, 2301, 2601, 2926, 3277, 3655, 4061, 4496, 4961, 5457, 5985, 6546, 7141, 7771, 8437, 9140, 9881, 10661, 11481, 12342, 13245, 14191

Offset: 0

Author

Klaus Strassburger (strass(AT)ddfi.uni-duesseldorf.de), Dec 22 1999

Keywords

• easy
• nonn

Comments

Number of invertible shuffles of n-2 cards. - Adam C. McDougall (mcdougal(AT)stolaf.edu) and David Molnar (molnar(AT)stolaf.edu), Apr 09 2002

If Y is a 3-subset of an n-set X then, for n>=3, a(n-2) is the number of (n-3)-subsets of X which have neither one element nor two elements in common with Y. - Milan Janjic, Dec 28 2007

Let A be the Hessenberg n X n matrix defined by: A[1,j]=j mod 2, 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

Starting with offset 3: (1, 2, 5, 11, 21, ...) = triangle A144257 * [1,2,3,...]. - Gary W. Adamson, Feb 18 2010

(1 + 2x + 5x^2 + 11x^3 + ...) = (1 + 2x + 3x^2 + ...)*(1 + 2x^2 + 3x^3 + ...). - Gary W. Adamson, Jul 26 2010

Starting (1, 2, 5, 11, ...) = binomial transform of [1, 1, 2, 1, 0, 0, 0, ...]. - Gary W. Adamson, Aug 25 2010

For n > 1: abs(abs(a(n+2) - a(n+1)) - abs(a(n+1) - a(n))) = n - 1; see also A086283. - Reinhard Zumkeller, Oct 17 2014

For n > 0, a(n) is the number of valid hook configurations of permutations of [n-1] that avoid the patterns 132 and 321. - Colin Defant, Apr 28 2019

For n >= 1, a(n+2) is the number of Grassmannian permutations that avoid a pattern, sigma, where sigma is a pattern of size 4 with exactly one descent. - Jessica A. Tomasko, Nov 15 2022

The number of bigrassmannian permutations in S_n is a(n+2) = binomial(n+1, 3) + 1. - Joshua Swanson, Jan 08 2024

Links

• Vincenzo Librandi, Table of n, a(n) for n = 0..1000
• Kassie Archer and Aaron Geary, Descents in powers of permutations, arXiv:2406.09369 [math.CO], 2024.
• Colin Defant, Motzkin intervals and valid hook configurations, arXiv preprint arXiv:1904.10451 [math.CO], 2019.
• Robert DiSario, Problem 10931, Amer. Math. Monthly, 109 (No. 3, 2002), 298.
• J. B. Gil and J. Tomasko, Restricted Grassmannian permutations, ECA 2:4 (2022) Article S4PP6.
• Nurul Hilda Syani Putri, Mashadi, and Sri Gemawati, Sequences from heptagonal pyramid corners of integer, International Mathematical Forum, Vol. 13, 2018, no. 4, 193-200.
• Luis Manuel Rivera, Integer sequences and k-commuting permutations, arXiv preprint arXiv:1406.3081 [math.CO], 2014.
• Amit Kumar Singh, Akash Kumar and Thambipillai Srikanthan, Accelerating Throughput-aware Run-time Mapping for Heterogeneous MPSoCs, ACM Transactions on Design Automation of Electronic Systems, 2012. - From _N. J. A. Sloane_, Dec 25 2012
• Index entries for linear recurrences with constant coefficients, signature (4,-6,4,-1).

Crossrefs

Apart from initial terms, one more than the tetrahedral numbers A000292.

Cf. A144257, A086283.

Programs

GAP

List([0..50], n-> n*(n^2-6*n+11)/6); # G. C. Greubel, Oct 30 2019

Haskell

a050407 n = n * (n ^ 2 - 6 * n + 11) `div` 6
-- Reinhard Zumkeller, Oct 17 2014

Magma

I:=[0, 1, 1, 1]; [n le 4 select I[n] else 4*Self(n-1)-6*Self(n-2)+4*Self(n-3)-Self(n-4): n in [1..50]]; // Vincenzo Librandi, Jun 22 2012

Maple

seq(binomial(n-1, 3) + 1, n = 0..46); # Zerinvary Lajos, Jul 24 2006

Mathematica

Table[n*(n^2-6*n+11)/6, {n,0,50}] (* Vladimir Joseph Stephan Orlovsky, Dec 17 2008 *)
LinearRecurrence[{4,-6,4,-1},{0,1,1,1},50] (* Vincenzo Librandi, Jun 22 2012 *)
Join[{0,1,1},Nest[Accumulate,Range[0,50],2]+1] (* Harvey P. Dale, Sep 23 2017 *)

PARI

a(n)=n*(n^2-6*n+11)/6 \\ Charles R Greathouse IV, Oct 07 2015

Python

for n in range(0,50): print(n*(n**2 - 6*n + 11)/6, end=', ') # Stefano Spezia, Jan 05 2019

Sage

[n*(n^2-6*n+11)/6 for n in (0..50)] # G. C. Greubel, Oct 30 2019

Formula

From Paul Barry, Jul 21 2003: (Start)

Diagonal sums of square array A086460 (starting 1, 1, 2, ...).

a(n+2) = 1 + n*(n+1)*(n-1)/6 = Sum_{k=0..n} (0^k + (n-k)*k). (End)

a(n) = binomial(n-1,3) + binomial(n-1,0), n>=0. - Zerinvary Lajos, Jul 24 2006

G.f.: x*(1-3*x+3*x^2)/(1-x)^4. - Colin Barker, May 06 2012

a(n) = 4*a(n-1) - 6*a(n-2) + 4*a(n-3) - a(n-4). - Vincenzo Librandi, Jun 22 2012

a(n) = A000292(n-3) + 1, n > 2. - Ivan N. Ianakiev, Apr 27 2014

E.g.f.: x*(6 - 3*x + x^2)*exp(x)/6. - G. C. Greubel, Oct 30 2019

a(n+2) = 1 + Sum_{i=3..4} binomial(n, i-1) for n >= 1. - Jessica A. Tomasko, Nov 15 2022