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A034867
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Triangle of odd-numbered terms in rows of Pascal's triangle.
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23
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1, 2, 3, 1, 4, 4, 5, 10, 1, 6, 20, 6, 7, 35, 21, 1, 8, 56, 56, 8, 9, 84, 126, 36, 1, 10, 120, 252, 120, 10, 11, 165, 462, 330, 55, 1, 12, 220, 792, 792, 220, 12, 13, 286, 1287, 1716, 715, 78, 1, 14, 364, 2002, 3432, 2002, 364, 14, 15, 455, 3003, 6435, 5005, 1365, 105, 1
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OFFSET
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0,2
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COMMENTS
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Also triangle of numbers of n-sequences of 0,1 with k subsequences of consecutive 01 because this number is C(n+1,2*k+1). - Roger Cuculiere (cuculier(AT)imaginet.fr), Nov 16 2002
Let T = tan x, then
tan x = T
tan 2x = 2T / (1 - T^2)
tan 3x = (3T - T^3) / (1 - 3T^2)
tan 4x = (4T - 4T^3) / (1 - 6T^2 + T^4)
tan 5x = (5T - 10T^3 + T^5) / (1 - 10T^2 + 5T^4)
tan 6x = (6T - 20T^3 + 6T^5) / (1 - 15T^2 + 15T^4 - T^6)
tan 7x = (7T - 35T^3 + 21T^5 - T^7) / (1 - 21T^2 + 35T^4 - 7T^6)
tan 8x = (8T - 56T^3 + 56T^5 - 8T^7) / (1 - 28T^2 + 70T^4 - 28T^6 + T^8)
tan 9x = (9T - 84T^3 + 126T^5 - 36T^7 + T^9) / (1 - 36 T^2 + 126T^4 - 84T^6 + 9T^8)
... To get the next one in the series, (tan 10x), for the numerator add:
9....84....126....36....1 previous numerator +
1....36....126....84....9 previous denominator =
10..120....252...120...10 = new numerator
For the denominator add:
......9.....84...126...36...1 = previous numerator +
1....36....126....84....9.... = previous denominator =
1....45....210...210...45...1 = new denominator
(End)
Triangle, with zeros omitted, given by (2, -1/2, 1/2, 0, 0, 0, 0, 0, 0, 0, ...) DELTA (0, 1/2, -1/2, 0, 0, 0, 0, 0, 0, 0, ...) where DELTA is the operator defined in A084938. - Philippe Deléham, Dec 12 2011
The row polynomials N(n,x) = Sum_{k=0..floor((n-1)/2)} T(n-1,k)*x^k, and D(n,x) = Sum_{k=0..floor(n/2)} A034839(n,k)*x^k, n >= 1, satisfy the recurrences N(n,x) = D(n-1,x) + N(n-1,x), D(n,x) = D(n-1,x) + x*N(n-1,x), with inputs N(1,x) = 1 = D(1,x). This is due to the Pascal triangle A007318 recurrence. Q(n,x) := tan(n*x)/tan(x) satisfies the recurrence Q(n,x) = (1 + Q(n-1,x)/(1 - v(x)*Q(n-1,x)) with input Q(1,x) = 1 and v = v(x) := (tan(x))^2. This recurrence is obtained from the addition theorem for tan(n*x) using n = 1 + (n-1). Therefore Q(n,x) = N(n,-v(x))/D(n,-v(x)). This proves the Gary W. Adamson contribution from above. See also A220673. This calculation was motivated by an e-mail of Thomas Olsen. The Oliver/Prodinger and Ma references resort to HAKEM Al Memo 239, Item 16, for the tan(n*x) formula in terms of tan(x). - Wolfdieter Lang, Jan 17 2013
The infinitesimal generator (infinigen) for the Narayana polynomials A090181/A001263 can be formed from the row polynomials P(n,y) of this entry. The resulting matrix is an instance of a matrix representation of the analytic infinigens presented in A145271 for general sets of binomial Sheffer polynomials and in A001263 and A119900 specifically for the Narayana polynomials. Given the column vector of row polynomials V = (1, P(1,x) = 2x, P(2,y) = 3x + x^2, P(3,y) = 4x + 4x^2, ...), form the lower triangular matrix M(n,k) = V(n-k,n-k), i.e., diagonally multiply the matrix with all ones on the diagonal and below by the components of V. Form the matrix MD by multiplying A132440^Transpose = A218272 = D (representing derivation of o.g.f.s) by M, i.e., MD = M*D. The non-vanishing component of the first row of (MD)^n * V / (n+1)! is the n-th Narayana polynomial. - Tom Copeland, Dec 09 2015
Binomial(n,2k+1) is also the number of permutations avoiding both 132 and 213 with k peaks, i.e., positions with w[i]<w[i+1]>w[i+2]. - Lara Pudwell, Dec 19 2018
Binomial(n,2k+1) is also the number of permutations avoiding both 123 and 132 with k peaks, i.e., positions with w[i]<w[i+1]>w[i+2]. - Lara Pudwell, Dec 19 2018
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REFERENCES
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A. T. Benjamin and J. J. Quinn, Proofs that really count: the art of combinatorial proof, M.A.A. 2003, id. 136.
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LINKS
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K. Oliver and H. Prodinger, The continued fraction expansion of Gauss' hypergeometric function and a new application to the tangent function, Transactions of the Royal Society of South Africa, Vol. 76 (2012), 151-154, [DOI], [PDF]. - From N. J. A. Sloane, Jan 03 2013
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FORMULA
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T(n,k) = C(n+1,2k+1) = Sum_{i=k..n-k} C(i,k) * C(n-i,k).
O.g.f for column k, k>=0: (1/(1-x)^2)*(x/(1-x))^(2*k). See the G.f. of this array given above by Emeric Deutsch. - Wolfdieter Lang, Jan 18 2013
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EXAMPLE
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Triangle starts:
1
2
3 1
4 4
5 10 1
6 20 6
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MAPLE
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seq(seq(binomial(n+1, 2*k+1), k=0..floor(n/2)), n=0..14); # Emeric Deutsch, Apr 01 2005
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MATHEMATICA
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u[1, x_] := 1; v[1, x_] := 1; z = 12;
u[n_, x_] := u[n - 1, x] + x*v[n - 1, x]
v[n_, x_] := u[n - 1, x] + v[n - 1, x]
cu = Table[CoefficientList[u[n, x], x], {n, 1, z}];
TableForm[cu] (* A034839 as a triangle *)
cv = Table[CoefficientList[v[n, x], x], {n, 1, z}];
TableForm[cv] (* A034867 as a triangle *)
Table[Binomial[n+1, 2*k+1], {n, 0, 20}, {k, 0, Floor[n/2]}]//Flatten (* G. C. Greubel, Mar 06 2018 *)
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PROG
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(PARI) for(n=0, 20, for(k=0, floor(n/2), print1(binomial(n+1, 2*k+1), ", "))) \\ G. C. Greubel, Mar 06 2018
(Magma) /* as a triangle */ [[Binomial(n+1, 2*k+1): k in [0..Floor(n/2)]]: n in [0..20]]; // G. C. Greubel, Mar 06 2018
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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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