%I M4796 #66 Jun 27 2024 04:45:47
%S 1,11,72,364,1568,6048,21504,71808,228096,695552,2050048,5870592,
%T 16400384,44843008,120324096,317521920,825556992,2118057984,
%U 5369233408,13463453696,33426505728,82239815680,200655503360,485826232320
%N Negated coefficients of Chebyshev T polynomials: a(n) = -A053120(n+10, n), n >= 0.
%C Binomial transform of A069038. - _Paul Barry_, Feb 19 2003
%C 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
%C 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
%D 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.
%D N. J. A. Sloane and Simon Plouffe, The Encyclopedia of Integer Sequences, Academic Press, 1995 (includes this sequence).
%H Paolo Xausa, <a href="/A006975/b006975.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 M. Abramowitz and I. A. Stegun, eds., <a href="http://www.convertit.com/Go/ConvertIt/Reference/AMS55.ASP">Handbook of Mathematical Functions</a>, National Bureau of Standards, Applied Math. Series 55, Tenth Printing, 1972 [alternative scanned copy].
%H <a href="/index/Ch#Cheby">Index entries for sequences related to Chebyshev polynomials.</a>
%H <a href="/index/Rec#order_06">Index entries for linear recurrences with constant coefficients</a>, signature (12,-60,160,-240,192,-64).
%F 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]
%F a(n) = 2^(n-4)*(n+1)(n+2)(n+3)(n+4)(n+10)/15. - _Paul Barry_, Feb 19 2003
%F a(n) = Sum_{k=0..floor((n+10)/2)} C(n+10, 2k)*C(k, 5). - _Paul Barry_, May 15 2003
%F a(n) = -A039991(n+10, 10). - _N. J. A. Sloane_, May 16 2003
%F 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
%F a(n) = 2^(n-1)/5*binomial(n+4,4)*(n+10). - _Brad Clardy_, Mar 10 2012
%F 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
%t Table[2^(n-1)/5*Binomial[n+4, 4]*(n+10), {n, 0, 30}] (* _Paolo Xausa_, Jun 26 2024 *)
%o (Magma) [2^(n-1)/5*Binomial(n+4,4)*(n+10): n in [0..25]]; // _Brad Clardy_, Mar 10 2012
%Y First differences of A054849.
%Y Cf. A039991, A053120, A069038.
%K nonn,easy
%O 0,2
%A _Simon Plouffe_
%E Name clarified by _Wolfdieter Lang_, Nov 26 2019