%I A034867
%S A034867 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,
%T A034867 120,10,11,165,462,330,55,1,12,220,792,792,220,12,13,286,1287,1716,715,
%U A034867 78,1,14,364,2002,3432,2002,364,14,15,455,3003,6435,5005,1365,105,1
%N A034867 Triangle of odd-numbered terms in rows of Pascal's triangle.
%C A034867 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
%C A034867 Contribution from Gary W. Adamson (qntmpkt(AT)yahoo.com), Oct 17 2008: 
               (Start)
%C A034867 Received from Herb Conn; Custer, SD:
%C A034867 Let T = tan x, then
%C A034867 tan x = T
%C A034867 tan 2x = 2T / (1 - T^2)
%C A034867 tan 3x = (3T - T^3) / (1 - 3T^2)
%C A034867 tan 4x = (4T - 4T^3) / (1 - 6T^2 + T^4)
%C A034867 tan 5x = (5T - 10T^3 + T^5) / (1 - 10T^2 + 5T^4)
%C A034867 tan 6x = (6T - 20T^3 + 6T^5) / (1 - 15T^2 + 15T^4 - T^6)
%C A034867 tan 7x = (7T - 35T^3 + 21T^5 - T^7) / (1 - 21T^2 + 35T^4 - 7T^6)
%C A034867 tax 8x = (8T - 56T^3 + 56T^5 - 8T^7) /
%C A034867 (1 - 28T^2 + 70T^4 - 28T^6 + T^8)
%C A034867 tan 9x = (9T - 84T^3 + 126T^5 - 36T^7 + T^9) /
%C A034867 (1 - 36 T^2 + 126T^4 - 84T^6 + 9T^8)
%C A034867 ... To get the next one in the series, (tan 10x), for the numerator add:
%C A034867 9....84....126....36....1 previous numerator +
%C A034867 1....36....126....84....9 previous denominator =
%C A034867 10..120....252...120...10 = new numerator
%C A034867 For the denominator add:
%C A034867 ......9.....84...126...36...1 = previous numerator +
%C A034867 1....36....126....84....9.... = previous denominator =
%C A034867 1....45....210...210...45...1 = new denominator
%C A034867 ...where numerators = A034867, denominators = A034839
%C A034867 (End)
%C A034867 Column k is the sum of columns 2k and 2k+1 of A007318. [From Philippe 
               DELEHAM (kolotoko(AT)wanadoo.fr), Nov 12 2008]
%D A034867 A. T. Benjamin and J. J. Quinn, Proofs that really count: the art of 
               combinatorial proof, M.A.A. 2003, id. 136.
%D A034867 L. Carlitz and R. Scoville, Zero-one sequences and Fibonacci numbers, 
               Fibonacci Quarterly, 15 (1977), 246-254.
%H A034867 Eric Weisstein's World of Mathematics, <a href="http://mathworld.wolfram.com/
               Tangent.html">Tangent</a> [From Eric W. Weisstein (eric(AT)weisstein.com), 
               Oct 18 2008]
%F A034867 T(n, k) = C(n+1, 2k+1) = Sum[i=k..n-k, C(i, k)*C(n-i, k) ].
%F A034867 E.g.f.: 1+(exp(x)*sinh(x*sqrt(y)))/sqrt(y). - Vladeta Jovovic (vladeta(AT)eunet.rs), 
               Mar 20 2005
%F A034867 G.f.=1/[(1-z)^2-tz^2]. - Emeric Deutsch (deutsch(AT)duke.poly.edu), Apr 
               01 2005
%F A034867 T(n, k) = Sum_{ j = 0, . . ., n} A034839(j, k). - Philippe DELEHAM, May 
               18 2005
%F A034867 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 (tcjpn(AT)msn.com), Nov 
               30 2007
%F A034867 T(n,k)=A007318(n,2k)+A007318(n,2k+1). [From Philippe DELEHAM (kolotoko(AT)wanadoo.fr), 
               Nov 12 2008]
%e A034867 Triangle starts:
%e A034867 1;
%e A034867 2;
%e A034867 3,1;
%e A034867 4,4;
%e A034867 5,10,1;
%e A034867 6,20,6;
%e A034867 T(5,2)=6 because only the following binary sequences of length 5 contain 
               2 subsequences 01: 00101,01001,01010,01011,01101 and 10101.
%p A034867 seq(seq(binomial(n+1,2*k+1),k=0..floor(n/2)),n=0..14); (Deutsch)
%Y A034867 Cf. A007318, A034839.
%Y A034867 A034839 [From Gary W. Adamson (qntmpkt(AT)yahoo.com), Oct 17 2008]
%Y A034867 Sequence in context: A100035 A090244 A096180 this_sequence A055446 A104706 
               A094137
%Y A034867 Adjacent sequences: A034864 A034865 A034866 this_sequence A034868 A034869 
               A034870
%K A034867 nonn,tabf,easy
%O A034867 0,2
%A A034867 N. J. A. Sloane (njas(AT)research.att.com).
%E A034867 More terms from Emeric Deutsch (deutsch(AT)duke.poly.edu), Apr 01 2005