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%I #72 Jul 12 2020 03:33:57
%S 1,1,3,1,8,7,1,17,31,15,1,34,96,94,31,1,67,258,382,253,63,1,132,645,
%T 1280,1275,636,127,1,261,1545,3845,5115,3831,1531,255,1,518,3598,
%U 10766,17920,17906,10738,3578,511,1,1031,8212,28700,57358,71666,57316,28652,8185,1023
%N Triangle of scaled 1-tiered binomial coefficients, T(n,k) = 2^(n+1)*(n-k,k)_1 (n >= 0, 0 <= k <= n), where (N,M)_1 is the 1-tiered binomial coefficient.
%H Michael E. Hoffman, <a href="https://www.usna.edu/Users/math/meh/mpi18.pdf">(Poly)logarithmic Integrals and Multiple Zeta Values</a>, Number Theory Talk, Max-Planck-Institut für Mathematik, Bonn, 20 June 2018. See Slide 29.
%H Michael E. Hoffman and Markus Kuba, <a href="https://arxiv.org/abs/1906.08347">Logarithmic integrals, zeta values, and tiered binomial coefficients</a>, arXiv:1906.08347 [math.CO], 2019-2020. See Example 4 on pp. 12-13.
%F (N,M)_1 = ((N-1,M)_1 + (N,M-1)_1 + (N,M-1)_0)/2 for N, M >= 0 and N + M > 0 with initial conditions: (N,-1)_i = 0 = (-1,M)_i for i = 0, 1; (0,0)_1 = 1/2; and (N,M)_0 = binomial(N+M, M) for N, M >= 0. (This is the recurrence for the square array presentation of the regular triangle.) [Edited by _Petros Hadjicostas_, Jul 06 2020]
%F (N,M)_1 = binomial(N+M-1, M-1) - (binomial(N+M, N)/2 - binomial(N+M-1, N-1))/2^(N+M) for N,M >= 0 and N + M > 0 with (0,0)_1 = 1/2. - _Petros Hadjicostas_, Jul 06 2020
%F G.f. for (N,M)_1: (1-x)/((1-x-y)*(2-x-y)). - _Jean-François Alcover_, Jun 21 2019
%F Scaled coefficients satisfy T(n,0) = 1 for n >= 0 and T(n,k) = T(n-1,k) + T(n-1,k-1) + 2^n*C(n-1,k-1) for n >= k+1 >= 1. - _Charlie Neder_, Jun 21 2019 [Corrected by _Petros Hadjicostas_, Jul 06 2020]
%F From _Petros Hadjicostas_, Jul 07 2020: (Start)
%F (N,M)_1 + (M,N)_1 = (N,M)_0 = binomial(N+M, N) for N, M >= 0.
%F (n-k,k)_1 + (k, n-k)_1 = binomial(n,k) for n >= k >= 0.
%F T(n,k) + T(n,n-k) = 2^(n+1)*binomial(n,k) = 2*A038208(n,k) for n >= k >= 0.
%F T(n,k) = 2^(n + 1)*binomial(n-1, k-1) + 2*binomial(n-1,k) - binomial(n,k) for n >= k >= 0 and (n,k) <> (0,0) with T(0,0) = 1.
%F G.f. for T(n,k): (1 - 2*x)/((1 - 2*x*(1 + y))*(1 - x*(1 + y))). (End)
%e From _Petros Hadjicostas_, Jul 07 2020: (Start)
%e Square array for (N,M)_1 of 1-tiered binomial coefficients (N, M >= 0):
%e 1/2, 3/4, 7/8, 15/16, 31/32, 63/64, 127/128, ...
%e 1/4, 1, 31/16, 47/16, 253/64, 159/32, 1531/256, ...
%e 1/8, 17/16, 3, 191/32, 1275/128, 3831/256, 5369/256, ...
%e 1/16, 17/16, 129/32, 10, 5115/256, 8953/256, 14329/256, ...
%e 1/32, 67/64, 645/128, 3845/256, 35, 35833/512, 129003/1024, ...
%e ... (End)
%e Triangle (n-k,k)_1 of 1-tiered binomial coefficients (n >= 0 and k = 0..n):
%e 1/2,
%e 1/4, 3/4,
%e 1/8, 1, 7/8,
%e 1/16, 17/16, 31/16, 15/16,
%e 1/32, 17/16, 3, 47/16, 31/32,
%e ...
%e Scaled triangle T(n,k) after multiplying each row by 2^(n+1):
%e 1,
%e 1, 3,
%e 1, 8, 7,
%e 1, 17, 31, 15,
%e 1, 34, 96, 94, 31,
%e ...
%t rows = 10;
%t cc = CoefficientList[# + O[y]^rows, y]& /@ CoefficientList[(1-x)/((1-x-y)* (2-x-y)) + O[x]^rows, x];
%t T[n_, m_, 1] := cc[[n-m+1, m+1]];
%t Table[2^(n+1) Table[T[n, m, 1], {m, 0, n}], {n, 0, rows-1}] (* _Jean-François Alcover_, Jun 21 2019 *)
%o (PARI) T(n,m) = if ((n==0) && (m==0), 1/2, binomial(n+m-1, m-1) - (binomial(n+m,n)/2 - binomial(n+m-1,n-1))/2^(n+m));
%o TT(n, k) = T(n-k, k);
%o tabls(nn) = for (n=0, nn, for (k=0, n, print1(2^(n+1)*TT(n, k), ", ")));
%Y Cf. A007318 (Pascal triangle: 0-tiered binomial coefficient), A038208, A145071 (column k = 1).
%K nonn,tabl
%O 0,3
%A _Michel Marcus_, Jun 21 2019
%E Name edited by _Petros Hadjicostas_, Jul 07 2020
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