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Triangle T(n, k, q) = ceiling(binomial(n, k)/f(n, q)) with T(0, 0) = 1, f(n, q) = 1 + tanh((n-1)/q), and q = 4, read by rows.
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%I #6 Sep 08 2022 08:45:51

%S 1,1,1,1,2,1,1,3,3,1,1,3,4,3,1,1,3,6,6,3,1,1,4,9,11,9,4,1,1,4,12,19,

%T 19,12,4,1,1,5,15,29,37,29,15,5,1,1,5,19,43,65,65,43,19,5,1,1,6,23,61,

%U 107,128,107,61,23,6,1,1,6,28,84,167,233,233,167,84,28,6,1

%N Triangle T(n, k, q) = ceiling(binomial(n, k)/f(n, q)) with T(0, 0) = 1, f(n, q) = 1 + tanh((n-1)/q), and q = 4, read by rows.

%H G. C. Greubel, <a href="/A174447/b174447.txt">Rows n = 0..50 of the triangle, flattened</a>

%F T(n, k, q) = ceiling(binomial(n, k)/f(n, q)) with T(0, 0) = 1, f(n, q) = 1 + tanh((n-1)/q), and q = 4.

%e Triangle begins as:

%e 1;

%e 1, 1;

%e 1, 2, 1;

%e 1, 3, 3, 1;

%e 1, 3, 4, 3, 1;

%e 1, 3, 6, 6, 3, 1;

%e 1, 4, 9, 11, 9, 4, 1;

%e 1, 4, 12, 19, 19, 12, 4, 1;

%e 1, 5, 15, 29, 37, 29, 15, 5, 1;

%e 1, 5, 19, 43, 65, 65, 43, 19, 5, 1;

%e 1, 6, 23, 61, 107, 128, 107, 61, 23, 6, 1;

%t f[n_, q_]= 1 + Tanh[(n-1)/q];

%t T[n_, k_, q_]= If[n==0, 1, Ceiling[Binomial[n, k]/f[n, q]]];

%t Table[T[n,k,4], {n,0,12}, {k,0,n}]//Flatten

%o (Magma)

%o T:= func< n,k,q | n eq 0 select 1 else Ceiling(Binomial(n,k)/(1 + Tanh((n-1)/q))) >;

%o [T(n,k,4): k in [0..n], n in [0..12]]; // _G. C. Greubel_, Aug 05 2021

%o (Sage)

%o def T(n, k, q): return 1 if (n==0) else ceil(binomial(n, k)/(1 + tanh((n-1)/q)))

%o flatten([[T(n,k,4) for k in (0..n)] for n in (0..12)]) # _G. C. Greubel_, Aug 05 2021

%Y Cf. A174446 (q=1), this sequence (q=4), A174448 (q=12).

%K nonn,tabl

%O 0,5

%A _Roger L. Bagula_, Mar 20 2010

%E Edited by _G. C. Greubel_, Aug 05 2021