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A136398
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Triangle read by rows of coefficients of Chebyshev-like polynomials P_{n,6}(x) with 0 omitted (exponents in increasing order).
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1
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1, -6, 2, 15, -13, 4, -20, 36, -28, 8, 15, -55, 85, -60, 16, -6, 50, -146, 198, -128, 32, 1, -27, 155, -377, 456, -272, 64, 8, -104, 456, -952, 1040, -576, 128, -1, 43, -363, 1289, -2360, 2352, -1216, 256, -10, 190, -1182, 3530, -5760, 5280, -2560, 512
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OFFSET
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6,2
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COMMENTS
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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>=6 and k be of the same parity. Consider a set X consisting of (n+k)/2-6 blocks of the size 2 and an additional block of the size 6, then (-1)^((n-k)/2)a(n,k) is the number of n-6-subsets of X intersecting each block of the size 2.
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LINKS
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FORMULA
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If n>=6 and k are of the same parity then a(n,k)= (-1)^((n-k)/2)*sum((-1)^i*binomial((n+k)/2-6, i)*binomial(n+k-6-2*i, n-6), i=0..(n+k)/2-6) 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),(-6,2),(15,-13,4),... since P_{6,6}=x^6, P_{7,6}=-6x^5+2x^7, P_{8,6}=15x^4-13x^6+4x^8,...
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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)/2-6, i)*binomial(n+k-6-2*i, n-6), i=0..(n+k)/2-6); end if;
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MATHEMATICA
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DeleteCases[#, 0] &@ Flatten@ Table[(-1)^((n - k)/2) * Sum[(-1)^i * Binomial[(n + k)/2 - 6, i] Binomial[n + k - 6 - 2 i, n - 6], {i, 0, (n + k)/2 - 6}], {n, 6, 15}, {k, 0 + Boole[OddQ@ n], n, 2}] (* Michael De Vlieger, Jul 05 2019 *)
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CROSSREFS
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KEYWORD
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sign,tabf
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AUTHOR
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STATUS
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approved
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