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A304252 Triangle read by rows: T(0,0) = 1; T(n,k) = T(n-1,k) + 6*T(n-2,k-1) for k = 0..floor(n/2); T(n,k)=0 for n or k < 0.

%I A304252 #34 Aug 08 2025 06:41:29
%S A304252 1,1,1,6,1,12,1,18,36,1,24,108,1,30,216,216,1,36,360,864,1,42,540,
%T A304252 2160,1296,1,48,756,4320,6480,1,54,1008,7560,19440,7776,1,60,1296,
%U A304252 12096,45360,46656,1,66,1620,18144,90720,163296,46656,1,72,1980,25920,163296,435456,326592,1,78,2376,35640
%N A304252 Triangle read by rows: T(0,0) = 1; T(n,k) = T(n-1,k) + 6*T(n-2,k-1) for k = 0..floor(n/2); T(n,k)=0 for n or k < 0.
%C A304252 The numbers in rows of the triangle are along skew diagonals pointing top-right in center-justified triangle given in A013613 ((1+6*x)^n).
%C A304252 The coefficients in the expansion of 1/(1-x-6*x^2) are given by the sequence generated by the row sums.
%D A304252 Shara Lalo and Zagros Lalo, Polynomial Expansion Theorems and Number Triangles, Zana Publishing, 2018, ISBN: 978-1-9995914-0-3, pp. 70, 72.
%H A304252 Zagros Lalo, <a href="/A304252/a304252.pdf">Left-justified triangle</a>
%H A304252 Zagros Lalo, <a href="/A304252/a304252_1.pdf">Skew diagonals in center-justified triangle of coefficients in expansion of (1+6x)^n</a>
%F A304252 T(n,k) = 6^k*binomial(n-k,k), n >= 0, 0 <= k <= floor(n/2).
%e A304252 Triangle begins:
%e A304252   1;
%e A304252   1;
%e A304252   1,  6;
%e A304252   1, 12;
%e A304252   1, 18,   36;
%e A304252   1, 24,  108;
%e A304252   1, 30,  216,   216;
%e A304252   1, 36,  360,   864;
%e A304252   1, 42,  540,  2160,   1296;
%e A304252   1, 48,  756,  4320,   6480;
%e A304252   1, 54, 1008,  7560,  19440,    7776;
%e A304252   1, 60, 1296, 12096,  45360,   46656;
%e A304252   1, 66, 1620, 18144,  90720,  163296,   46656;
%e A304252   1, 72, 1980, 25920, 163296,  435456,  326592;
%e A304252   1, 78, 2376, 35640, 272160,  979776, 1306368,  279936;
%e A304252   1, 84, 2808, 47520, 427680, 1959552, 3919104, 2239488;
%t A304252 t[0, 0] = 1; t[n_, k_] := If[n < 0 || k < 0, 0, t[n - 1, k] + 6 t[n - 2, k - 1]]; Table[t[n, k], {n, 0, 13}, {k, 0, Floor[n/2]}] // Flatten (* _Robert G. Wilson v_, May 19 2018 *)
%t A304252 Table[6^k Binomial[n - k, k], {n, 0, 13}, {k, 0, Floor[n/2]}] // Flatten
%o A304252 (PARI) T(n,k) = if ((n<0) || (k<0), 0, if ((n==0) && (k==0), 1, T(n-1,k) + 6*T(n-2,k-1)));
%o A304252 tabf(nn) = for (n=0, nn, for (k=0, n\2, print1(T(n,k), ", ")); print); \\ _Michel Marcus_, May 10 2018
%Y A304252 Row sums give A015441.
%Y A304252 Cf. A013613.
%K A304252 tabf,nonn,easy
%O A304252 0,4
%A A304252 _Zagros Lalo_, May 09 2018