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Search: id:A136397
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| A136397 |
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Triangle read by rows of coefficients of Chebyshev-like polynomials P_{n,5}(x) with 0 omitted (exponents in increasing order). |
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+0
1
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| 1, -5, 2, 10, -11, 4, -10, 25, -24, 8, 5, -30, 61, -52, 16, -1, 20, -85, 146, -112, 32, -7, 70, -231, 344, -240, 64, 1, -34, 225, -608, 800, -512, 128, 9, -138, 681, -1560, 1840, -1088, 256, -1, 52, -501, 1970, -3920, 4192, -2304, 512
(list; graph; listen)
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OFFSET
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5,2
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COMMENT
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If U_n(x), T_n(x) are Chebyshev's polynomials then U_n(x)=P_{n,0}(x), T_n(x)=P_{n,1}(x).
Let n>=5 and k be of the same parity. Consider a set X consisting of (n+k)/2-5 blocks of the size 2 and an additional block of the size 5, then (-1)^((n-k)/2)a(n,k) is the number of n-5-subsets of X intersecting each block of the size 2.
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LINKS
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Milan Janjic, Two enumerative functions.
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FORMULA
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If n>=5 and k are of the same parity then a(n,k)= (-1)^((n-k)/2)*sum((-1)^i*binomial((n+k)/2-5, i)*binomial(n+k-5-2*i, n-5), i=0..(n+k)/2-5), and a(n,k)=0 if n and k are of different parity.
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EXAMPLE
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Rows are (1),(-5,2),(10,-11,4),... since P_{5,5}=x^5, P_{6,5}=-5x^4+2x^6, P_{7,5}=10x^3-11x^5+4x^7,...
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MAPLE
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if modp(n-k, 2)=0 then a[n, k]:=(-1)^((n-k)/2)*sum((-1)^i*binomial((n+k)/2-5, i)*binomial(n+k-5-2*i, n-5), i=0..(n+k)/2-5); end if;
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CROSSREFS
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Cf. A008310, A053117.
Sequence in context: A040024 A036121 A055682 this_sequence A065292 A090812 A050004
Adjacent sequences: A136394 A136395 A136396 this_sequence A136398 A136399 A136400
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KEYWORD
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sign
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AUTHOR
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Milan R. Janjic (agnus(AT)blic.net), Mar 30 2008, revised Apr 05 2008
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