A054334 1/512 of 11th unsigned column of triangle A053120 (T-Chebyshev, rising powers, zeros omitted).

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

Author

Wolfdieter Lang

Keywords

• nonn
• easy

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

• T. D. Noe, Table of n, a(n) for n = 0..1000
• 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].
• 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 (11,-55,165,-330,462,-462,330,-165,55,-11,1).
• Index entries for sequences related to Chebyshev polynomials.

Crossrefs

Cf. A053120, A054333.

Cf. A005585, A040977, A050486, A053347, A054333.

Programs

GAP

List([0..30], 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

Mathematica

Table[(2*n + 10)*Binomial[n + 9, 9]/10, {n, 0, 100}] (* Vladimir Joseph Stephan Orlovsky, Jan 15 2009 *)
LinearRecurrence[{11,-55,165,-330,462,-462,330,-165,55,-11,1},{1,12,77,352,1287,4004,11011,27456,63206,136136,277134},30] (* Harvey P. Dale, May 11 2025 *)

PARI

vector(40, n, n--; (2*n+10)*binomial(n+9, 9)/10) \\ 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

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)