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A053347 a(n) = binomial(n+7, 7)*(n+4)/4. 22

%I

%S 1,10,54,210,660,1782,4290,9438,19305,37180,68068,119340,201552,

%T 329460,523260,810084,1225785,1817046,2643850,3782350,5328180,7400250,

%U 10145070,13741650,18407025,24402456,32040360,41692024,53796160

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

%C 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

%C 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

%C 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

%C 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

%D A. H. Beiler, Recreations in the Theory of Numbers, Dover, N.Y., 1964, pp. 194-196.

%H Vincenzo Librandi, <a href="/A053347/b053347.txt">Table of n, a(n) for n = 0..1000</a>

%H Milan Janjic, <a href="http://www.pmfbl.org/janjic/">Two Enumerative Functions</a>

%H <a href="/index/Rec#order_09">Index entries for linear recurrences with constant coefficients</a>, signature (9,-36,84,-126,126,-84,36,-9,1).

%H <a href="/index/Ch#Cheby">Index entries for sequences related to Chebyshev polynomials.</a>

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

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

%F a(n) = 2*C(n+8, 8) - C(n+7, 7). - _Paul Barry_, Mar 04 2003

%F Equals A027803/35 = C(n+4, n)*C(n+7, 4)/35. - _Zerinvary Lajos_, May 25 2005

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

%F 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

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

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

%t 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 *)

%t CoefficientList[Series[(1 + x) / (1 - x)^9, {x, 0, 50}], x] (* _Vincenzo Librandi_, Jun 09 2013 *)

%t Table[SeriesCoefficient[(1 + x)/(1 - x)^9, {x, 0, n}], {n, 0, 28}] (* or *)

%t Table[Binomial[n + 7, 7] (n + 4)/4, {n, 0, 28}] (* _Michael De Vlieger_, Dec 31 2015 *)

%o (PARI) a(n)=binomial(n+7,7)*(n+4)/4 \\ _Charles R Greathouse IV_, Jun 10 2011

%o (MAGMA) [Binomial(n+7,7)+2*Binomial(n+7,8): n in [0..35]]; // _Vincenzo Librandi_, Jun 09 2013

%o (Python)

%o A053347_list, m = [], [2]+[1]*8

%o for _ in range(10**2):

%o A053347_list.append(m[-1])

%o print(m[-1])

%o for i in range(8):

%o m[i+1] += m[i] # _Chai Wah Wu_, Jan 24 2016

%Y Partial sums of A050486.

%Y Cf. A005585, A040977.

%K easy,nonn

%O 0,2

%A _Barry E. Williams_, Jan 06 2000

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Last modified August 20 06:04 EDT 2019. Contains 326139 sequences. (Running on oeis4.)