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(1, 9) Pascal Triangle read by horizontal rows. Same as A093644, but mirrored and without the additional row/column (1, 9, 9, 9, 9, ...).
5

%I #14 Apr 25 2022 08:03:38

%S 1,1,10,1,11,19,1,12,30,28,1,13,42,58,37,1,14,55,100,95,46,1,15,69,

%T 155,195,141,55,1,16,84,224,350,336,196,64,1,17,100,308,574,686,532,

%U 260,73,1,18,117,408,882,1260,1218,792,333,82

%N (1, 9) Pascal Triangle read by horizontal rows. Same as A093644, but mirrored and without the additional row/column (1, 9, 9, 9, 9, ...).

%C Binomial transform of A017173.

%H G. C. Greubel, <a href="/A172171/b172171.txt">Rows n = 1..50 of the triangle, flattened</a>

%F T(n,k) = T(n-1,k) + 2*T(n-1,k-1) - T(n-2,k-1) - T(n-2,k-2), T(n,1) = 1, T(2,2) = 10, T(n,k) = 0 if k < 1 or if k > n.

%F Sum_{k=0..n} T(n, k) = A139634(n).

%F T(2*n-1, n) = A050489(n).

%e Triangle begins:

%e 1;

%e 1, 10;

%e 1, 11, 19;

%e 1, 12, 30, 28;

%e 1, 13, 42, 58, 37;

%e 1, 14, 55, 100, 95, 46;

%e 1, 15, 69, 155, 195, 141, 55;

%e 1, 16, 84, 224, 350, 336, 196, 64;

%e 1, 17, 100, 308, 574, 686, 532, 260, 73;

%e 1, 18, 117, 408, 882, 1260, 1218, 792, 333, 82;

%e 1, 19, 135, 525, 1290, 2142, 2478, 2010, 1125, 415, 91;

%e 1, 20, 154, 660, 1815, 3432, 4620, 4488, 3135, 1540, 506, 100;

%t T[n_, k_]:= T[n, k]= If[k<1 || k>n, 0, If[k==1, 1, If[n==2 && k==2, 10, T[n-1, k] + 2*T[n-1, k-1] - T[n-2, k-1] - T[n-2, k-2]]]];

%t Table[T[n, k], {n,15}, {k,n}]//Flatten (* _G. C. Greubel_, Apr 24 2022 *)

%o (SageMath)

%o @CachedFunction

%o def T(n,k):

%o if (k<1 or k>n): return 0

%o elif (k==1): return 1

%o elif (n==2 and k==2): return 10

%o else: return T(n-1,k) + 2*T(n-1,k-1) - T(n-2,k-1) - T(n-2,k-2)

%o flatten([[T(n,k) for k in (1..n)] for n in (1..15)]) # _G. C. Greubel_, Apr 24 2022

%Y Cf. A007318, A017173, A050489 (central terms), A093644, A139634 (row sums).

%K nonn,tabl

%O 1,3

%A _Mark Dols_, Jan 28 2010

%E More terms from _Philippe Deléham_, Dec 25 2013