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A034839 Triangular array formed by taking every other term of each row of Pascal's triangle. 37
1, 1, 1, 1, 1, 3, 1, 6, 1, 1, 10, 5, 1, 15, 15, 1, 1, 21, 35, 7, 1, 28, 70, 28, 1, 1, 36, 126, 84, 9, 1, 45, 210, 210, 45, 1, 1, 55, 330, 462, 165, 11, 1, 66, 495, 924, 495, 66, 1, 1, 78, 715, 1716, 1287, 286, 13 (list; graph; refs; listen; history; text; internal format)
OFFSET
0,6
COMMENTS
Number of compositions of n having k parts greater than 1. Example: T(5,2)=5 because we have 3+2, 2+3, 2+2+1, 2+1+2 and 1+2+2. Number of binary words of length n-1 having k runs of consecutive 1's. Example: T(5,2)=5 because we have 1010, 1001, 0101, 1101 and 1011. - Emeric Deutsch, Mar 30 2005
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)
Triangle, with zeros omitted, given by (1, 0, 1, 0, 0, 0, 0, 0, 0, 0, ...) DELTA (0, 1, -1, 0, 0, 0, 0, 0, 0, 0, ...) where DELTA is the operator defined in A084938. - Philippe Deléham, Dec 12 2011
The row (1,66,495,924,495,66,1) plays a role in expansions of powers of the Dedekind eta function. See the Chan link, p. 534. - Tom Copeland, Dec 12 2016
Binomial(n,2k) is also the number of permutations avoiding both 123 and 132 with k ascents, i.e., positions with w[i]<w[i+1]. - Lara Pudwell, Dec 19 2018
Coefficients in expansion of ((x-1)^n+(x+1)^n)/2 or ((x-i)^n+(x+i)^n)/2 with alternating sign. - Eugeniy Sokol, Sep 20 2020
Number of permutations of length n avoiding simultaneously the patterns 213 and 312 with the maximum number of non-overlapping descents equal k (equivalently, with the maximum number of non-overlapping ascents equal k). An ascent (resp., descent) in a permutation a(1)a(2)...a(n) is position i such that a(i) < a(i+1) (resp., a(i) > a(i+1)). - Tian Han, Nov 16 2023
LINKS
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.
H. Chan, S. Cooper, and P. Toh, The 26th power of Dedekind's eta function Advances in Mathematics, 207 (2006) 532-543.
C. Corsani, D. Merlini, and R. Sprugnoli, Left-inversion of combinatorial sums, Discrete Mathematics, 180 (1998) 107-122.
Tian Han and Sergey Kitaev, Joint distributions of statistics over permutations avoiding two patterns of length 3, arXiv:2311.02974 [math.CO], 2023.
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
E.g.f.: exp(x)*cosh(x*sqrt(y)). - Vladeta Jovovic, Mar 20 2005
From Emeric Deutsch, Mar 30 2005: (Start)
T(n, k) = binomial(n, 2*k), for n >= 0 and k = 0, 1, ..., floor(n/2).
G.f.: (1-z)/((1-z)^2 - t*z^2). (End)
O.g.f. for column no. k is (1/(1-x))*(x/(1-x))^(2*k), k >= 0 [from the g.f. given in the preceding formula]. - Wolfdieter Lang, Jan 18 2013
From Peter Bala, Jul 14 2015: (Start)
Stretched Riordan array ( 1/(1 - x ), x^2/(1 - x)^2 ) in the terminology of Corsani et al.
Denote this array by P. Then P * A007318 = A201701.
P * transpose(P) is A119326 read as a square array.
Let Q denote the array ( (-1)^k*binomial(2*n,k) )n,k>=0. Q is a signed version of A034870. Then Q*P = the identity matrix, that is, Q is a left-inverse array of P (see Corsani et al., p. 111).
P * A034870 = A080928. (End)
Even rows are A086645. An aerated version of this array is A099174 with each diagonal divided by the first element of the diagonal, the double factorials A001147. - Tom Copeland, Dec 12 2015
EXAMPLE
Triangluar array T(n, k) begins:
1
1
1 1
1 3
1 6 1
1 10 5
1 15 15 1 ...
- Philippe Deléham, Dec 12 2011
MAPLE
for n from 0 to 13 do seq(binomial(n, 2*k), k=0..floor(n/2)) od; # yields sequence in triangular form; # Emeric Deutsch, Mar 30 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, k], {n, 0, 13}, {k, 0, Floor[n, 2], 2}] // Flatten (* Michael De Vlieger, Dec 13 2016 *)
PROG
(PARI) for(n=0, 15, for(k=0, floor(n/2), print1(binomial(n, 2*k), ", "))) \\ G. C. Greubel, Feb 23 2018
(Magma) /* As a triangle */ [[Binomial(n, 2*k):k in [0..Floor(n/2)]] : n in [0..10]]; // G. C. Greubel, Feb 23 2018
CROSSREFS
Cf. A008619 (row lengths), A086645.
Sequence in context: A127096 A130541 A128489 * A089732 A158905 A098076
KEYWORD
nonn,easy,tabf
AUTHOR
STATUS
approved

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Last modified July 17 08:59 EDT 2024. Contains 374363 sequences. (Running on oeis4.)