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A008311
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Triangle of expansions of powers of x in terms of Chebyshev polynomials T_n (x).
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5
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1, 1, 1, 1, 3, 1, 3, 4, 1, 10, 5, 1, 10, 15, 6, 1, 35, 21, 7, 1, 35, 56, 28, 8, 1, 126, 84, 36, 9, 1, 126, 210, 120, 45, 10, 1, 462, 330, 165, 55, 11, 1, 462, 792, 495, 220, 66, 12, 1, 1716, 1287, 715, 286, 78, 13, 1, 1716, 3003, 2002, 1001, 364, 91, 14, 1, 6435, 5005, 3003
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OFFSET
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0,5
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COMMENTS
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This triangle is the right half of Pascal's triangle (A007318), but with each number along the center of Pascal's triangle (except the 1 at the top) divided by 2. - Benjamin Schak (schak(AT)math.upenn.edu), Dec 02 2005
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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.
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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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EXAMPLE
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Triangle begins:
1;
-, 1;
1, -, 1;
-, 3, -, 1;
3, -, 4, -, 1;
-, 10, -, 5, -, 1;
...
cos(x) = 1*cos(x),
2*cos(x)^2 = 1 + cos(2x),
4*cos(x)^3 = 3*cos(x) + cos(3x),
8*cos(x)^4 = 3 + 4*cos(2x) + cos(4x),
16*cos(x)^5 = 10*cos(x) + 5*cos(3x) + cos(5x), etc. (End)
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MAPLE
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printf("1, ") ; for n from 1 to 20 do for j from n mod 2 to n by 2 do if j = 0 then printf("%d, ", binomial(n, (n-j)/2)/2) ; else printf("%d, ", binomial(n, (n-j)/2)) ; fi ; od ; od ; # R. J. Mathar, May 13 2006
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MATHEMATICA
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row[n_] := If[n == 0, {1}, Table[If[j == 0, Binomial[n, (n - j)/2]/2, Binomial[n, (n - j)/2]], {j, Mod[n, 2], n, 2}]];
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CROSSREFS
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KEYWORD
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nonn,tabf,easy
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AUTHOR
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EXTENSIONS
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STATUS
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approved
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