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%I #23 Apr 28 2022 03:26:02
%S 1,9,11,9,20,11,9,29,31,11,9,38,60,42,11,9,47,98,102,53,11,9,56,145,
%T 200,155,64,11,9,65,201,345,355,219,75,11,9,74,266,546,700,574,294,86,
%U 11,9,83,340,812,1246,1274,868,380,97,11,9,92,423,1152,2058,2520,2142,1248,477,108,11
%N (9,11) Pascal triangle.
%C Sums of NW-SE diagonals give A022114 (apart from first two terms).
%C Triangle T(n,k), read by rows, given by (9,-8,0,0,0,0,0,0,0,...) DELTA (11,-10,0,0,0,0,0,0,0,...) where DELTA is the operator defined in A084938. - _Philippe Deléham_, Oct 09 2011
%C Row n: Expansion of (9+11x)*(1+x)^(n-1), n > 0. - _Philippe Deléham_, Oct 09 2011
%H G. C. Greubel, <a href="/A172185/b172185.txt">Rows n = 0..50 of the triangle, flattened</a>
%F T(n,k) = T(n-1,k-1) + T(n-1,k) with T(0,0)=1, T(1,0)=9, T(1,1)=11. - _Philippe Deléham_, Oct 09 2011
%F G.f.: (1+8*x+10*y*x)/(1-x-y*x). - _Philippe Deléham_, Apr 13 2012
%F From _G. C. Greubel_, Apr 28 2022: (Start)
%F T(n, k) = 9*binomial(n, k) + 2*binomial(n-1, k-1) with T(0, 0) = 1.
%F Sum_{k=0..n} T(n, k) = 10*2^n - 9*[n=0]. (End)
%e Triangle begins:
%e 1;
%e 9, 11;
%e 9, 20, 11;
%e 9, 29, 31, 11;
%e 9, 38, 60, 42, 11;
%e 9, 47, 98, 102, 53, 11;
%e 9, 56, 145, 200, 155, 64, 11;
%e 9, 65, 201, 345, 355, 219, 75, 11;
%e 9, 74, 266, 546, 700, 574, 294, 86, 11;
%e 9, 83, 340, 812, 1246, 1274, 868, 380, 97, 11;
%e 9, 92, 423, 1152, 2058, 2520, 2142, 1248, 477, 108, 11;
%t T[n_, k_]:= If[n==0, 1, (9 + 2*k/n)*Binomial[n, k]]
%t Table[T[n, k], {n,0,12}, {k,0,n}]//Flatten (* _G. C. Greubel_, Apr 28 2022 *)
%o (SageMath)
%o def A172185(n,k): return 9*binomial(n,k) +2*binomial(n-1,k-1) -8*bool(n==0)
%o flatten([[A172185(n,k) for k in (0..n)] for n in (0..12)]) # _G. C. Greubel_, Apr 28 2022
%Y Cf. A022114, A093644, A172179.
%K nonn,tabl
%O 0,2
%A _Mark Dols_, Jan 28 2010
%E Corrected and extended by _Philippe Deléham_, Oct 09 2011
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