0,2

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

Contribution from Gary W. Adamson (qntmpkt(AT)yahoo.com), Oct 17 2008: (Start)

Received from Herb Conn; Custer, SD:

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)

tax 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. [From Philippe DELEHAM (kolotoko(AT)wanadoo.fr), Nov 12 2008]

A. T. Benjamin and J. J. Quinn, Proofs that really count: the art of combinatorial proof, M.A.A. 2003, id. 136.

L. Carlitz and R. Scoville, Zero-one sequences and Fibonacci numbers, Fibonacci Quarterly, 15 (1977), 246-254.

Eric Weisstein's World of Mathematics, Tangent [From Eric W. Weisstein (eric(AT)weisstein.com), Oct 18 2008]

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 (vladeta(AT)eunet.rs), Mar 20 2005

G.f.=1/[(1-z)^2-tz^2]. - Emeric Deutsch (deutsch(AT)duke.poly.edu), Apr 01 2005

T(n, k) = Sum_{ j = 0, . . ., n} A034839(j, k). - Philippe DELEHAM, 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 (tcjpn(AT)msn.com), Nov 30 2007

T(n,k)=A007318(n,2k)+A007318(n,2k+1). [From Philippe DELEHAM (kolotoko(AT)wanadoo.fr), Nov 12 2008]

Triangle starts:

1;

2;

3,1;

4,4;

5,10,1;

6,20,6;

T(5,2)=6 because only the following binary sequences of length 5 contain 2 subsequences 01: 00101,01001,01010,01011,01101 and 10101.

seq(seq(binomial(n+1, 2*k+1), k=0..floor(n/2)), n=0..14); (Deutsch)

Cf. A007318, A034839.

A034839 [From Gary W. Adamson (qntmpkt(AT)yahoo.com), Oct 17 2008]

Adjacent sequences: A034864 A034865 A034866 this_sequence A034868 A034869 A034870

Sequence in context: A100035 A090244 A096180 this_sequence A055446 A104706 A094137

nonn,tabf,easy

N. J. A. Sloane (njas(AT)research.att.com).

More terms from Emeric Deutsch (deutsch(AT)duke.poly.edu), Apr 01 2005