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

%I #30 Sep 05 2018 02:29:31

%S 1,6,36,1,216,12,1296,108,1,7776,864,18,46656,6480,216,1,279936,46656,

%T 2160,24,1679616,326592,19440,360,1,10077696,2239488,163296,4320,30,

%U 60466176,15116544,1306368,45360,540,1,362797056,100776960,10077696,435456,7560,36

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

%C The numbers in rows of the triangle are along skew diagonals pointing top-left in center-justified triangle given in A013613 ((1+6*x)^n).

%C The coefficients in the expansion of 1/(1-6x-x^2) are given by the sequence generated by the row sums.

%C The row sums are Denominators of continued fraction convergent to sqrt(10), see A005668.

%C If s(n) is the row sum at n, then the ratio s(n)/s(n-1) is approximately 6.162277660..., a metallic mean (see A176398), when n approaches infinity.

%D Shara Lalo and Zagros Lalo, Polynomial Expansion Theorems and Number Triangles, Zana Publishing, 2018, ISBN: 978-1-9995914-0-3, pp. 70, 72, 94.

%H Zagros Lalo, <a href="/A304255/a304255.pdf">Left-justified triangle</a>

%H Zagros Lalo, <a href="/A304255/a304255_1.pdf">Skew diagonals in center-justified triangle of coefficients in expansion of (1+6x)^n</a>

%e Triangle begins:

%e 1;

%e 6;

%e 36, 1;

%e 216, 12;

%e 1296, 108, 1;

%e 7776, 864, 18;

%e 46656, 6480, 216, 1;

%e 279936, 46656, 2160, 24;

%e 1679616, 326592, 19440, 360, 1;

%e 10077696, 2239488, 163296, 4320, 30;

%e 60466176, 15116544, 1306368, 45360, 540, 1;

%e 362797056, 100776960, 10077696, 435456, 7560, 36;

%e 2176782336, 665127936, 75582720, 3919104, 90720, 756, 1;

%e 13060694016, 4353564672, 554273280, 33592320, 979776, 12096, 42;

%e 78364164096, 28298170368, 3990767616, 277136640, 9797760, 163296, 1008, 1;

%e 470184984576, 182849716224, 28298170368, 2217093120, 92378880, 1959552, 18144, 48;

%t t[0, 0] = 1; t[n_, k_] := If[n < 0 || k < 0, 0, 6 t[n - 1, k] + t[n - 2, k - 1]]; Table[t[n, k], {n, 0, 11}, {k, 0, Floor[n/2]}] // Flatten

%o (PARI) T(n, k) = if ((n<0) || (k<0), 0, if ((n==0) && (k==0), 1, 6*T(n-1, k) + T(n-2, k-1)));

%o tabf(nn) = for (n=0, nn, for (k=0, n\2, print1(T(n,k), ", ")); print); \\ _Michel Marcus_, May 26 2018

%Y Row sums give A005668.

%Y Cf. A000400 (column 0), A053469 (column 1), A081136 (column 2), A081144 (column 3).

%Y Cf. A013613.

%Y Cf. A176398.

%K tabf,nonn,easy

%O 0,2

%A _Zagros Lalo_, May 09 2018