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 A034867 Triangle of odd-numbered terms in rows of Pascal's triangle. 22
 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 (list; graph; refs; listen; history; text; internal format)
 OFFSET 0,2 COMMENTS 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 From Gary W. Adamson, Oct 17 2008: (Start) Received from Herb Conn: 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 ...where numerators = A034867, denominators = A034839 (End) Column k is the sum of columns 2k and 2k+1 of A007318. - Philippe Deléham, Nov 12 2008 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 The diagonals of this entry are A078812 (also shifted A128908 and unsigned A053122, which are embedded in A030528, A102426, A098925, A109466, A092865). Equivalently, the antidiagonals of A078812 are the rows of A034867. - Tom Copeland, Dec 12 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+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+2]. - Lara Pudwell, Dec 19 2018 REFERENCES A. T. Benjamin and J. J. Quinn, Proofs that really count: the art of combinatorial proof, M.A.A. 2003, id. 136. LINKS G. C. Greubel, Table of n, a(n) for the first 100 rows, flattened M. Bukata, R. Kulwicki, N. Lewandowski, L. Pudwell, J. Roth, and T. Wheeland, Distributions of Statistics over Pattern-Avoiding Permutations, arXiv preprint arXiv:1812.07112 [math.CO], 2018. L. Carlitz and R. Scoville, Zero-one sequences and Fibonacci numbers, Fibonacci Quarterly, 15 (1977), 246-254. S.-M. Ma, On some binomial coefficients related to the evaluation of tan(nx), arXiv preprint arXiv:1205.0735 [math.CO], 2012. - From N. J. A. Sloane, Oct 13 2012 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 Eric Weisstein's World of Mathematics, Tangent [From Eric W. Weisstein, Oct 18 2008] FORMULA T(n,k) = C(n+1,2k+1) = Sum_{i=k..n-k} C(i,k) * C(n-i,k). E.g.f.: 1+(exp(x)*sinh(x*sqrt(y)))/sqrt(y). - Vladeta Jovovic, Mar 20 2005 G.f.: 1/((1-z)^2-t*z^2). - Emeric Deutsch, Apr 01 2005 T(n,k) = Sum_{j = 0..n} A034839(j,k). - Philippe Deléham, May 18 2005 Pell(n+1) = A000129(n+1) = Sum_{k=0..n} T(n,k) * 2^k = (1/n!) Sum_{k=0..n} A131980(n,k) * 2^k. - Tom Copeland, Nov 30 2007 T(n,k) = A007318(n,2k) + A007318(n,2k+1). - Philippe Deléham, Nov 12 2008 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 T(n,k) = (x^(2*k+1))*((1+x)^n-(1-x)^n)/2. - L. Edson Jeffery, Jan 15 2014 EXAMPLE Triangle starts:   1   2   3  1   4  4   5 10 1   6 20 6 - Philippe Deléham, Dec 12 2011 MAPLE seq(seq(binomial(n+1, 2*k+1), k=0..floor(n/2)), n=0..14); # Emeric Deutsch, Apr 01 2005 MATHEMATICA 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 *) (* Clark Kimberling, Feb 18 2012 *) Table[Binomial[n+1, 2*k+1], {n, 0, 20}, {k, 0, Floor[n/2]}]//Flatten (* G. C. Greubel, Mar 06 2018 *) PROG (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 CROSSREFS Cf. A000129, A007318, A034839, A034867, A084938, A131980, A220673. Cf. A001263, A090181, A119900, A132440, A145271. Cf. A030528, A053122, A078812, A092865, A098925, A102426, A109466, A128908, A218272. Sequence in context: A324336 A324752 A298637 * A323893 A329721 A193790 Adjacent sequences:  A034864 A034865 A034866 * A034868 A034869 A034870 KEYWORD nonn,tabf,easy AUTHOR EXTENSIONS More terms from Emeric Deutsch, Apr 01 2005 STATUS approved

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Last modified February 26 11:03 EST 2020. Contains 332279 sequences. (Running on oeis4.)