A086460 -id:A086460 - cp's OEIS frontend

A086461 Symmetric version of square array A086460.

1, 1, 1, 1, 1, 1, 1, 2, 2, 1, 1, 3, 4, 3, 1, 1, 4, 6, 6, 4, 1, 1, 5, 8, 9, 8, 5, 1, 1, 6, 10, 12, 12, 10, 6, 1, 1, 7, 12, 15, 16, 15, 12, 7, 1, 1, 8, 14, 18, 20, 20, 18, 14, 8, 1, 1, 9, 16, 21, 24, 25, 24, 21, 16, 9, 1, 1, 10, 18, 24, 28, 30, 30, 28, 24, 18, 10, 1, 1, 11, 20, 27, 32, 35, 36

Offset: 0

Paul Barry, Jul 21 2003

Rows include A028310, A004277, A008486, A008574, A008706, A008458. Main diagonal is n^2+0^n (A000290, preceded by extra 1).

			Rows begin
  1 1 1 1 1 ...
  1 1 2 3 4 ...
  1 2 4 6 8 ...
  1 3 6 9 12 ...
  1 4 8 12 16 ...
As a triangle:
  {1},
  {1, 1},
  {1, 1, 1},
  {1, 2, 2, 1},
  {1, 3, 4, 3, 1},
  {1, 4, 6, 6, 4, 1},
  {1, 5, 8, 9, 8, 5, 1},
  {1, 6, 10, 12, 12, 10, 6, 1},
  {1, 7, 12, 15, 16, 15, 12, 7, 1},
  {1, 8, 14, 18, 20, 20, 18, 14, 8, 1},
  {1, 9, 16, 21, 24, 25, 24, 21, 16, 9, 1}

t[n_, m_] = If[ n == 0 || n == m || m == 0, 1, n - m]*If[n == m || n == 0 || m == 0, 1, m]; Table[t[n, m], {n, 0, 11}, {m, 0, n}] // Flatten (* Roger L. Bagula, Sep 06 2008 *)

T(0, k)=T(n, 0)=1, T(n, k)=nk+0^n, n, k>0

Alternatively, triangle read by rows with formula t(n,m)=If[n == 0 || n == m || m == 0, 1, n - m]*If[n == m || n == 0 || m == 0, 1, m]. - Roger L. Bagula, Sep 06 2008

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

Original entry on oeis.org

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

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

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

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).

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

Cf. A144257, A086283.

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

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

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

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

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 *)

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

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

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

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

cp's OEIS Frontend

A086461 Symmetric version of square array A086460.

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

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cp's OEIS Frontend

A086461 Symmetric version of square array A086460.

Views

Author

Keywords

Comments

Examples

Programs

Mathematica

Formula

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

Views

Author

Keywords

Comments

Links

Crossrefs

Programs

GAP

Haskell

Magma

Maple

Mathematica

PARI

Python

Sage

Formula

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