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A091042
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Triangle of even numbered entries of odd numbered rows of Pascal's triangle A007318.
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24
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1, 1, 3, 1, 10, 5, 1, 21, 35, 7, 1, 36, 126, 84, 9, 1, 55, 330, 462, 165, 11, 1, 78, 715, 1716, 1287, 286, 13, 1, 105, 1365, 5005, 6435, 3003, 455, 15, 1, 136, 2380, 12376, 24310, 19448, 6188, 680, 17, 1, 171, 3876, 27132, 75582, 92378, 50388, 11628, 969, 19, 1, 210, 5985, 54264, 203490, 352716, 293930, 116280, 20349, 1330, 21
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OFFSET
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0,3
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COMMENTS
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The row polynomials Pe(n, x) := Sum_{m=0..n} a(n, m)*x^m appear as numerators of the generating functions for the even numbered column sequences of array A034870.
Elements have the same parity as those of Pascal's triangle.
All zeros of polynomial Pe(n, x) are negative. They are -tan^2(Pi/2*n+1), -tan^2(2*Pi/2*n+1), ..., -tan^2(n*Pi/2*n+1). Moreover, for m >= 1, Pe(m, -x^2) is the characteristic polynomial of the linear difference equation with constant coefficients for differences between multiples of 2*m+1 with even and odd digit sum in base 2*m in the interval [0,(2*m)^n). - Vladimir Shevelev and Peter J. C. Moses, May 22 2012
The row polynomial Pe(d, x), multiplied by (2*d)!/d! = A001813(d), gives the numerator polynomial of the o.g.f. of the sequence of the diagonal d, for d >= 0, of the Sheffer triangle Lah[4,1] given in A048854. - Wolfdieter Lang, Oct 12 2017
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REFERENCES
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A. M. Yaglom and I. M. Yaglom, An elementary proof of the Wallis, Leibniz and Euler formulas for pi. Uspekhi Matem. Nauk, VIII (1953), 181-187(in Russian).
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LINKS
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FORMULA
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T(n, m) = binomial(2*n+1, 2*m) = A007318(2*n+1, 2*m), n >= m >= 0, otherwise 0.
E.g.f.: sinh(t)*cosh(sqrt(x)*t) = t + (1 + 3*x)*t^3/3! + (1 + 10*x + 5*x^2)*t^5/5! + (1 + 21*x + 35*x^2 + 7*x^3)*t^7/7! + ....
O.g.f.: A(x,t) = (1 + (x - 1)*t)/( (1 + (x - 1)*t)^2 - 4*t*x ) = 1 + (1 + 3*x)*t + (1 + 10*x + 5*x^2)*t^2 + ...
The function A( x/(x + 4), t*(x + 4)/4 ) = 1 + (1 + x)*t + (1 + 3*x + x^2)*t^2 + ... is the o.g.f. for A085478.
O.g.f. for n-th diagonal: ( Sum_{k = 0..n} binomial(2*n,2*k)*x^k )/(1 - x)^(2*n).
n-th row polynomial R(n,x) = (1/2)*( (1 + sqrt(x))^(2*n+1) - (sqrt(x) - 1)^(2*n+1) ).
T(n, k) = 2*T(n-1,k) + 2*T(n-1,k-1) + 2*T(n-2,k-1) - T(n-2,k) - T(n-2,k-2) with T(0,0)=T(1,0)=1, T(1,1)=3, T(n,k)=0 if k < 0 or if k > n. - Philippe Deléham, Nov 26 2013
Define S(r,N) = Sum_{j = 1..N} j^r. Then the following identity holds for n >= 0: (1/2)*(N^2 + N)^(2*n+1) = T(n,0)*S(2*n+1,N) + T(n,1)*S(2*n+3,N) + ... + T(n,n)* S(4*n+1,N). Some examples are given below. (End)
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EXAMPLE
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Triangle a(n, m) begins:
n\m 0 1 2 3 4 5 6 7 8 9 10 ...
0: 1
1: 1 3
2: 1 10 5
3: 1 21 35 7
4: 1 36 126 84 9
5: 1 55 330 462 165 11
6: 1 78 715 1716 1287 286 13
7: 1 105 1365 5005 6435 3003 455 15
8: 1 136 2380 12376 24310 19448 6188 680 17
9: 1 171 3876 27132 75582 92378 50388 11628 969 19
10: 1 210 5985 54264 203490 352716 293930 116280 20349 1330 21
(1/2)*(N^2 + N) = Sum_{j = 1..N} j.
(1/2)*(N^2 + N)^3 = Sum_{j = 1..N} j^3 + 3*Sum_{j = 1..N} j^5.
(1/2)*(N^2 + N)^5 = Sum_{j = 1..N} j^5 + 10*Sum_{j = 1..N} j^7 + 5*Sum_{j = 1..N} j^9.
(1/2)*(N^2 + N)^7 = Sum_{j = 1..N} j^7 + 21*Sum_{j = 1..N} j^9 + 35*Sum_{j = 1..N} j^11 + 7*Sum_{j = 1..N} j^13. (End)
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MAPLE
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f := (x, t) -> cosh(sqrt(x)*t)*sinh(t); seq(seq(coeff(((2*n+1)!*coeff(series(f(x, t), t, 2*n+2), t, 2*n+1)), x, k), k=0..n), n=0..9); # Peter Luschny, Jul 29 2013
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MATHEMATICA
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T[n_, k_] /; 0 <= k <= n := T[n, k] = 2T[n-1, k] + 2T[n-1, k-1] + 2T[n-2, k-1] - T[n-2, k] - T[n-2, k-2]; T[0, 0] = T[1, 0] = 1; T[1, 1] = 3; T[_, _] = 0;
Table[Binomial[2*n+1, 2*k], {n, 0, 12}, {k, 0, n}]//Flatten (* G. C. Greubel, Aug 01 2019 *)
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PROG
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(PARI) T(n, k) = binomial(2*n+1, 2*k); \\ G. C. Greubel, Aug 01 2019
(Magma) [[Binomial(2*n+1, 2*k): k in [0..n]]: n in [0..12]]; // G. C. Greubel, Aug 01 2019
(Sage) [[binomial(2*n+1, 2*k) for k in (0..n)] for n in (0..12)] # G. C. Greubel, Aug 01 2019
(GAP) Flat(List([0..12], n-> List([0..n], k-> Binomial(2*n+1, 2*k) ))); # G. C. Greubel, Aug 01 2019
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CROSSREFS
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KEYWORD
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AUTHOR
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STATUS
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approved
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