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%I #22 Apr 18 2020 09:21:36
%S 1,13,13,28,44,28,46,90,90,46,67,154,198,154,67,91,239,370,370,239,91,
%T 118,348,627,758,627,348,118,148,484,993,1403,1403,993,484,148,181,
%U 650,1495,2414,2824,2414,1495,650,181,217,849,2163,3927,5256,5256,3927,2163,849,217
%N Triangular array T read by rows (9-diamondization of Pascal's triangle). Step 1: t(n,k) = sum of 9 entries in diamond-shaped subarray of Pascal's triangle having vertices C(n,k), C(n+4,k+2), C(n+2,k), C(n+2,k+2). Step 2: T(n,k) = t(n,k) - t(0,0) + 1.
%H Indranil Ghosh, <a href="/A026907/b026907.txt">Rows 0..125, flattened</a>
%e Triangle starts:
%e 1;
%e 13, 13;
%e 28, 44, 28;
%e 46, 90, 90, 46;
%e 67, 154, 198, 154, 67;
%e 91, 239, 370, 370, 239, 91;
%e ...
%t t[n_, k_]:=Binomial[n + 4, k + 2 ] + Binomial[n + 3, k + 1] + Binomial[n + 3, k + 2] + Binomial[n + 2, k] + Binomial[n + 2, k + 1] + Binomial[n + 2, k + 2] + Binomial[n + 1, k] + Binomial[n + 1, k + 1] + Binomial[n, k] ; T[n_, k_]:=t[n,k] - t[0, 0] + 1; Flatten[Table[T[n, k], {n, 0, 9},{k, 0, n}]] (* _Indranil Ghosh_, Mar 13 2017 *)
%o (PARI) alias(C, binomial);
%o t(n,k) = C(n+4,k+2) + C(n+3,k+1) + C(n+3,k+2) + C(n+2,k) + C(n+2,k+1) + C(n+2,k+2) + C(n+1,k) + C(n+1,k+1) + C(n,k);
%o T(n,k) = t(n,k)-t(0,0)+1;
%o tabl(nn) = for (n=0, nn, for (k=0, n, print1(T(n,k), ", ")); print());
%o \\ _Michel Marcus_, Mar 13 2017
%Y Cf. A007318, A027170.
%K nonn,tabl
%O 0,2
%A _Clark Kimberling_
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