A053347 a(n) = binomial(n+7, 7)*(n+4)/4.

1, 10, 54, 210, 660, 1782, 4290, 9438, 19305, 37180, 68068, 119340, 201552, 329460, 523260, 810084, 1225785, 1817046, 2643850, 3782350, 5328180, 7400250, 10145070, 13741650, 18407025, 24402456, 32040360, 41692024, 53796160

Offset: 0

Author

Barry E. Williams, Jan 06 2000

Keywords

• easy
• nonn

Comments

If a 2-set Y and an (n-3)-set Z are disjoint subsets of an n-set X then a(n-9) is the number of 9-subsets of X intersecting both Y and Z. - Milan Janjic, Sep 08 2007

8-dimensional square numbers, seventh partial sums of binomial transform of [1, 2, 0, 0, 0, ...]. a(n) = sum{i=0,n,C(n+7, i+7)*b(i)}, where b(i) = [1, 2, 0, 0, 0, ...]. - Borislav St. Borisov (b.st.borisov(AT)abv.bg), Mar 05 2009

2*a(n) is number of ways to place 7 queens on an (n+7) X (n+7) chessboard so that they diagonally attack each other exactly 21 times. The maximal possible attack number, p=binomial(k,2)=21 for k=7 queens, is achievable only when all queens are on the same diagonal. In graph-theory representation they thus form the corresponding complete graph. - Antal Pinter, Dec 27 2015

Coefficients in the terminating series identity 1 - 10*n/(n + 9) + 54*n*(n - 1)/((n + 9)*(n + 10)) - 210*n*(n - 1)*(n - 2)/((n + 9)*(n + 10)*(n + 11)) + ... = 0 for n = 1,2,3,.... Cf. A050486. - Peter Bala, Feb 18 2019

References

• Albert H. Beiler, Recreations in the Theory of Numbers, Dover, N.Y., 1964, pp. 194-196.

Links

• Vincenzo Librandi, Table of n, a(n) for n = 0..1000
• Milan Janjic, Two Enumerative Functions
• 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.
• Index entries for linear recurrences with constant coefficients, signature (9,-36,84,-126,126,-84,36,-9,1).
• Index entries for sequences related to Chebyshev polynomials.

Crossrefs

Partial sums of A050486.

Cf. A005585, A040977.

Programs

Magma

[Binomial(n+7,7)+2*Binomial(n+7,8): n in [0..35]]; // Vincenzo Librandi, Jun 09 2013

Maple

A053347:=n->binomial(n+7,7)*(n+4)/4: seq(A053347(n), n=0..50); # Wesley Ivan Hurt, Jul 16 2017

Mathematica

s1=s2=s3=s4=s5=s6=0; lst={}; Do[s1+=n^2; s2+=s1; s3+=s2; s4+=s3; s5+=s4; s6+=s5; AppendTo[lst,s6],{n,0,7!}]; lst (* Vladimir Joseph Stephan Orlovsky, Jan 15 2009 *)
CoefficientList[Series[(1 + x) / (1 - x)^9, {x, 0, 50}], x] (* Vincenzo Librandi, Jun 09 2013 *)
Table[SeriesCoefficient[(1 + x)/(1 - x)^9, {x, 0, n}], {n, 0, 28}] (* or *)
Table[Binomial[n + 7, 7] (n + 4)/4, {n, 0, 28}] (* Michael De Vlieger, Dec 31 2015 *)

PARI

a(n)=binomial(n+7,7)*(n+4)/4 \\ Charles R Greathouse IV, Jun 10 2011

Python

A053347_list, m = [], [2]+[1]*8
for _ in range(10**2):
    A053347_list.append(m[-1])
    print(m[-1])
    for i in range(8):
        m[i+1] += m[i] # Chai Wah Wu, Jan 24 2016

Formula

a(n) = ((-1)^n)*A053120(2*n+8, 8)/2^7 (1/128 of ninth unsigned column of Chebyshev T-triangle, zeros omitted).

G.f.: (1+x)/(1-x)^9.

a(n) = 2*C(n+8, 8) - C(n+7, 7). - Paul Barry, Mar 04 2003

a(n) = A027803(n-3)/35 = C(n+4, n)*C(n+7, 4)/35. - Zerinvary Lajos, May 25 2005

a(n) = C(n+7, 7) + 2*C(n+7, 8). - Borislav St. Borisov (b.st.borisov(AT)abv.bg), Mar 05 2009

a(n) = (n^8 + 32*n^7 + 434*n^6 + 3248*n^5 + 14609*n^4 + 40208*n^3 + 65596*n^2 + 57312*n + 20160)/20160. - Chai Wah Wu, Jan 24 2016

Sum_{n>=0} 1/a(n) = 41503/45 - 280/3*Pi^2. - Jaume Oliver Lafont, Jul 17 2017

Sum_{n>=0} (-1)^n/a(n) = 140*Pi^2/3 - 1379/3. - Amiram Eldar, Jan 25 2022