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A054334
1/512 of 11th unsigned column of triangle A053120 (T-Chebyshev, rising powers, zeros omitted).
15
1, 12, 77, 352, 1287, 4004, 11011, 27456, 63206, 136136, 277134, 537472, 999362, 1790712, 3105322, 5230016, 8580495, 13748020, 21559395, 33153120, 50075025, 74397180, 108864405, 157073280, 223689180, 314707536, 437766252, 602516992, 821063892, 1108479152
OFFSET
0,2
COMMENTS
Partial sums of A054333.
If a 2-set Y and an (n-3)-set Z are disjoint subsets of an n-set X then a(n-11) is the number of 11-subsets of X intersecting both Y and Z. - Milan Janjic, Sep 08 2007
10-dimensional square numbers, ninth partial sums of binomial transform of [1,2,0,0,0,...]. a(n)=sum{i=0,n,C(n+9,i+9)*b(i)}, where b(i)=[1,2,0,0,0,...]. [From Borislav St. Borisov (b.st.borisov(AT)abv.bg), Mar 05 2009]
2*a(n) is number of ways to place 9 queens on an (n+9) X (n+9) chessboard so that they diagonally attack each other exactly 36 times. The maximal possible attack number, p=binomial(k,2)=36 for k=9 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
REFERENCES
M. Abramowitz and I. A. Stegun, eds., Handbook of Mathematical Functions, National Bureau of Standards Applied Math. Series 55, 1964 (and various reprintings), p. 795.
Theodore J. Rivlin, Chebyshev polynomials: from approximation theory to algebra and number theory, 2. ed., Wiley, New York, 1990.
LINKS
M. Abramowitz and I. A. Stegun, eds., Handbook of Mathematical Functions, National Bureau of Standards, Applied Math. Series 55, Tenth Printing, 1972 [alternative scanned copy].
Index entries for linear recurrences with constant coefficients, signature (11,-55,165,-330,462,-462,330,-165,55,-11,1).
FORMULA
a(n) = (2*n+10)*binomial(n+9, 9)/10 = ((-1)^n)*A053120(2*n+10, 10)/2^9.
G.f.: (1+x)/(1-x)^11.
a(n) = 2*binomial(n+10, 10) - binomial(n+9, 9). - Paul Barry, Mar 04 2003
a(n) = binomial(n+9,9) + 2*binomial(n+9,10). - Borislav St. Borisov (b.st.borisov(AT)abv.bg), Mar 05 2009
a(n) = binomial(n+9,9)*(n+5)/5. - Antal Pinter, Dec 27 2015
From Amiram Eldar, Jan 26 2022: (Start)
Sum_{n>=0} 1/a(n) = 525*Pi^2 - 1160419/224.
Sum_{n>=0} (-1)^n/a(n) = 525*Pi^2/2 - 82875/32. (End)
MATHEMATICA
Table[(2*n + 10)*Binomial[n + 9, 9]/10, {n, 0, 100}] (* Vladimir Joseph Stephan Orlovsky, Jan 15 2009 *)
PROG
(PARI) vector(40, n, n--; (2*n+10)*binomial(n+9, 9)/10) \\ G. C. Greubel, Dec 02 2018
(Magma) [(2*n+10)*Binomial(n+9, 9)/10: n in [0..40]]; // G. C. Greubel, Dec 02 2018
(Sage) [(2*n+10)*binomial(n+9, 9)/10 for n in range(40)] # G. C. Greubel, Dec 02 2018
(GAP) List([0..30], n -> (2*n+10)*Binomial(n+9, 9)/10); # G. C. Greubel, Dec 02 2018
CROSSREFS
KEYWORD
nonn,easy
STATUS
approved