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Triangle read by rows: T(0,0) = 1; T(n,k) = 9 T(n-1,k) + T(n-2,k-1) for k = 0..floor(n/2); T(n,k)=0 for n or k < 0.
2

%I #17 Sep 05 2018 02:31:01

%S 1,9,81,1,729,18,6561,243,1,59049,2916,27,531441,32805,486,1,4782969,

%T 354294,7290,36,43046721,3720087,98415,810,1,387420489,38263752,

%U 1240029,14580,45,3486784401,387420489,14880348,229635,1215,1,31381059609,3874204890,172186884,3306744,25515,54

%N Triangle read by rows: T(0,0) = 1; T(n,k) = 9 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 A013616 ((1+9*x)^n) and along skew diagonals pointing top-right in center-justified triangle given in A038291 ((9+x)^n).

%C The coefficients in the expansion of 1/(1-9x-x^2) are given by the sequence generated by the row sums (see A099371).

%C If s(n) is the row sum at n, then the ratio s(n)/s(n-1) is approximately 9.109772228646443655... (a metallic mean), when n approaches infinity; (see A176522: ((9+sqrt(85))/2)).

%D Shara Lalo and Zagros Lalo, Polynomial Expansion Theorems and Number Triangles, Zana Publishing, 2018, ISBN: 978-1-9995914-0-3, pp. 70, 100.

%H Zagros Lalo, <a href="/A317052/a317052.pdf">Left-justified triangle</a>

%H Zagros Lalo, <a href="/A317052/a317052_1.pdf">Skew diagonals in center-justified triangle of coefficients in expansion of (1 + 9x)^n</a>

%H Zagros Lalo, <a href="/A317052/a317052_2.pdf">Skew diagonals in center-justified triangle of coefficients in expansion of (9 + x)^n</a>

%e Triangle begins:

%e 1;

%e 9;

%e 81, 1;

%e 729, 18;

%e 6561, 243, 1;

%e 59049, 2916, 27;

%e 531441, 32805, 486, 1;

%e 4782969, 354294, 7290, 36;

%e 43046721, 3720087, 98415, 810, 1;

%e 387420489, 38263752, 1240029, 14580, 45;

%e 3486784401, 387420489, 14880348, 229635, 1215, 1;

%e 31381059609, 3874204890, 172186884, 3306744, 25515, 54;

%e 282429536481, 38354628411, 1937102445, 44641044, 459270, 1701, 1;

%e 2541865828329, 376572715308, 21308126895, 573956280, 7440174, 40824, 63;

%t t[0, 0] = 1; t[n_, k_] := If[n < 0 || k < 0, 0, 9 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, 9*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_, Jul 20 2018

%Y Row sums give A099371.

%Y Cf. A013616

%Y Cf. A038291

%Y Cf. A176522

%Y Cf. A001019 (column 0), A053540 (column 1), A081139 (column 2), A173187 (column 3), A173000 (column 4).

%K tabf,nonn,easy

%O 0,2

%A _Zagros Lalo_, Jul 20 2018