A006975 Negated coefficients of Chebyshev T polynomials: a(n) = -A053120(n+10, n), n >= 0.

1, 11, 72, 364, 1568, 6048, 21504, 71808, 228096, 695552, 2050048, 5870592, 16400384, 44843008, 120324096, 317521920, 825556992, 2118057984, 5369233408, 13463453696, 33426505728, 82239815680, 200655503360, 485826232320

Offset: 0

Author

Simon Plouffe

Keywords

• nonn
• easy

Comments

Binomial transform of A069038. - Paul Barry, Feb 19 2003

If X_1, X_2, ..., X_n are 2-blocks of a (2n+1)-set X then, for n>=4, a(n-4) is the number of (n+5)-subsets of X intersecting each X_i, (i=1,2,...,n). - Milan Janjic, Nov 18 2007

The 5th corrector line for transforming 2^n offset 0 with a leading 1 into the Fibonacci sequence. - Al Hakanson (hawkuu(AT)gmail.com), Jun 01 2009

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.
• N. J. A. Sloane and Simon Plouffe, The Encyclopedia of Integer Sequences, Academic Press, 1995 (includes this sequence).

Links

• Paolo Xausa, Table of n, a(n) for n = 0..1000
• Milan Janjic, Two Enumerative Functions
• 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 sequences related to Chebyshev polynomials.
• Index entries for linear recurrences with constant coefficients, signature (12,-60,160,-240,192,-64).

Crossrefs

First differences of A054849.

Cf. A039991, A053120, A069038.

Programs

Magma

[2^(n-1)/5*Binomial(n+4,4)*(n+10): n in [0..25]]; // Brad Clardy, Mar 10 2012

Mathematica

Table[2^(n-1)/5*Binomial[n+4, 4]*(n+10), {n, 0, 30}] (* Paolo Xausa, Jun 26 2024 *)

Formula

G.f.: (1-x)/(1-2*x)^6. a(n) = 2^(n-1)*binomial(n+4, 4)*(n+10)/5, for n >= 0. [a(n) from Mar 06 2000 rewritten. See the Brad Clardy formula below, and a comment in A053120 on subdiagonals. - Wolfdieter Lang, Jan 03 2020]

a(n) = 2^(n-4)*(n+1)(n+2)(n+3)(n+4)(n+10)/15. - Paul Barry, Feb 19 2003

a(n) = Sum_{k=0..floor((n+10)/2)} C(n+10, 2k)*C(k, 5). - Paul Barry, May 15 2003

a(n) = -A039991(n+10, 10). - N. J. A. Sloane, May 16 2003

a(n) = binomial transform of b(n)= (2*n^5 + 10*n^4 + 30*n^3 + 50*n^2 + 43*n + 15) / 15 offset 0. a(3) = 364. - Al Hakanson (hawkuu(AT)gmail.com), Jun 01 2009

a(n) = 2^(n-1)/5*binomial(n+4,4)*(n+10). - Brad Clardy, Mar 10 2012

E.g.f.: (1/15)*exp(2*x)*(15+135*x+240*x^2+140*x^3+30*x^4+2*x^5). - Stefano Spezia, Jan 03 2020

Extensions

Name clarified by Wolfdieter Lang, Nov 26 2019