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A317497
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Triangle read by rows: T(0,0) = 1; T(n,k) = 3 * T(n-1,k) + T(n-3,k-1) for k = 0..floor(n/3); T(n,k)=0 for n or k < 0.
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4
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1, 3, 9, 27, 1, 81, 6, 243, 27, 729, 108, 1, 2187, 405, 9, 6561, 1458, 54, 19683, 5103, 270, 1, 59049, 17496, 1215, 12, 177147, 59049, 5103, 90, 531441, 196830, 20412, 540, 1, 1594323, 649539, 78732, 2835, 15, 4782969, 2125764, 295245, 13608, 135, 14348907, 6908733, 1082565, 61236, 945, 1
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OFFSET
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0,2
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COMMENTS
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The numbers in rows of the triangle are along a "second layer" of skew diagonals pointing top-left in center-justified triangle given in A013610 ((1+3*x)^n) and along a "second layer" of skew diagonals pointing top-right in center-justified triangle given in A027465 ((3+x)^n), see links. (Note: First layer of skew diagonals in triangles of coefficients in expansions of (1+3*x)^n and (3+x)^n are given in A304236 and A304249 respectively.)
The coefficients in the expansion of 1/(1-3x-x^3) are given by the sequence generated by the row sums.
The row sums give A052541.
If s(n) is the row sum at n, then the ratio s(n)/s(n-1) is approximately 3.1038034027355..., when n approaches infinity.
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REFERENCES
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Shara Lalo and Zagros Lalo, Polynomial Expansion Theorems and Number Triangles, Zana Publishing, 2018, ISBN: 978-1-9995914-0-3, pp. 364-366
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LINKS
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Table of n, a(n) for n=0..50.
Zagros Lalo, Second layer skew diagonals in center-justified triangle of coefficients in expansion of (1 + 3x)^n
Zagros Lalo, Second layer skew diagonals in center-justified triangle of coefficients in expansion of (3 + x)^n
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FORMULA
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T(n,k) = 3^(n - 3k) / ((n - 3k)! k!) * (n - 2k)! where n is a nonnegative integer and k = 0..floor(n/3).
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EXAMPLE
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Triangle begins:
1;
3;
9;
27, 1;
81, 6;
243, 27;
729, 108, 1;
2187, 405, 9;
6561, 1458, 54;
19683, 5103, 270, 1;
59049, 17496, 1215, 12;
177147, 59049, 5103, 90;
531441, 196830, 20412, 540, 1;
1594323, 649539, 78732, 2835, 15;
4782969, 2125764, 295245, 13608, 135;
14348907, 6908733, 1082565, 61236, 945, 1;
43046721, 22320522, 3897234, 262440, 5670, 18;
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MATHEMATICA
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t[n_, k_] := t[n, k] = 3^(n - 3k)/((n - 3 k)! k!) (n - 2 k)!; Table[t[n, k], {n, 0, 15}, {k, 0, Floor[n/3]} ] // Flatten
t[0, 0] = 1; t[n_, k_] := t[n, k] = If[n < 0 || k < 0, 0, 3 t[n - 1, k] + t[n - 3, k - 1]]; Table[t[n, k], {n, 0, 15}, {k, 0, Floor[n/3]}] // Flatten
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PROG
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(GAP) Flat(List([0..16], n->List([0..Int(n/3)], k->3^(n-3*k)/(Factorial(n-3*k)*Factorial(k))*Factorial(n-2*k)))); # Muniru A Asiru, Aug 01 2018
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CROSSREFS
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Row sums give A052541.
Cf. A013610, A027465, A317496, A304236, A304249.
Cf. A000244 (column 0), A027471 (column 1), A027472 (column 2), A036216 (column 3), A036217 (column 4).
Sequence in context: A148923 A058143 A126025 * A114181 A036134 A317502
Adjacent sequences: A317494 A317495 A317496 * A317498 A317499 A317500
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KEYWORD
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tabf,nonn,easy
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AUTHOR
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Zagros Lalo, Jul 31 2018
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STATUS
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approved
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