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 A185139 Triangle T(n,k) = Sum_{i=1..n} 2^(i-1)*C(n+2*k-i-1, k-1), 1 <= k <= n. 1

%I

%S 1,3,10,7,25,91,15,56,210,792,31,119,456,1749,6721,63,246,957,3718,

%T 14443,56134,127,501,1969,7722,30251,118456,463828,255,1012,4004,

%U 15808,62322,245480,966416,3803648,511,2035,8086,32071,127024,502588,1987096,7852453,31020445,1023,4082,16263

%N Triangle T(n,k) = Sum_{i=1..n} 2^(i-1)*C(n+2*k-i-1, k-1), 1 <= k <= n.

%C The first term of the m-th row is 2^m-1.

%H G. C. Greubel, <a href="/A185139/b185139.txt">Table of n, a(n) for the first 50 rows, flattened</a>

%H V. Shevelev and P. Moses, <a href="http://arxiv.org/abs/1112.5715">On a sequence of polynomials with hypothetically integer coefficients</a> arXiv:1112.5715 [math.NT], 2011.

%F 2*T_n(k) = T_(n-1)(k+1) + C(n+2*k-1,k).

%F T_n(k) = T_(n-2)(k+1) + C(n+2*k-1,k).

%F T_n(k) = 2*T_(n-1)(k) + C(n+2*k-2,k-1).

%F T_n(k+1) = 4*T_n(k) - (n/k)*C(n+2*k-1,k-1).

%e Triangle begins

%e 1,

%e 3, 10,

%e 7, 25, 91,

%e 15, 56, 210, 792,

%e 31, 119, 456, 1749, 6721,

%e 63, 246, 957, 3718, 14443, 56134,

%e 127, 501, 1969, 7722, 30251, 118456, 463828,

%e 255, 1012, 4004, 15808, 62322, 245480, 966416, 3803648,

%e 511, 2035, 8086, 32071, 127024, 502588, 1987096, 7852453, 31020445,

%e ...

%t Table[Sum[2^(j - 1)*Binomial[n + 2*k - j - 1, k - 1], {j, 1, n}], {n,

%t 1, 50}, {k, 1, n}] // Flatten (* _G. C. Greubel_, Jun 23 2017 *)

%o (PARI) for(n=1,20, for(k=1,n, print1(sum(j=1,n, 2^(j-1)*binomial(n+2*k-j-1,k-1)), ", "))) \\ _G. C. Greubel_, Jun 23 2017

%Y Cf. A174531.

%K nonn,tabl

%O 1,2

%A _Vladimir Shevelev_ and _Peter J. C. Moses_, Feb 04 2012

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Last modified May 29 16:43 EDT 2020. Contains 334704 sequences. (Running on oeis4.)