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A006975
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Negated coefficients of Chebyshev T polynomials: a(n) = -A053120(n+10, n), n >= 0.
(Formerly M4796)
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11
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1, 11, 72, 364, 1568, 6048, 21504, 71808, 228096, 695552, 2050048, 5870592, 16400384, 44843008, 120324096, 317521920, 825556992, 2118057984, 5369233408, 13463453696, 33426505728, 82239815680, 200655503360, 485826232320
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OFFSET
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0,2
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COMMENTS
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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
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REFERENCES
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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).
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LINKS
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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].
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FORMULA
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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) = 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
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PROG
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(Magma) [2^(n-1)/5*Binomial(n+4, 4)*(n+10): n in [0..25]]; // Brad Clardy, Mar 10 2012
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CROSSREFS
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KEYWORD
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nonn,easy,changed
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AUTHOR
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EXTENSIONS
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STATUS
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approved
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