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1, 1, 2, 3, 3, 3, 5, 12, 6, 4, 11, 25, 30, 10, 5, 21, 66, 75, 60, 15, 6, 43, 147, 231, 175, 105, 21, 7, 85, 344, 588, 616, 350, 168, 28, 8, 171, 765, 1548, 1764, 1386, 630, 252, 36, 9
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OFFSET
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1,3
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COMMENTS
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Left border = A001045: (1, 1, 3, 5, 11, 21, 43, 85, ...).
Row sums = (1, 3, 9, 27, ...).
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LINKS
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FORMULA
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Let A007318 (Pascal's triangle) = P. then A131048 = (1/3) * (P^2 - 1/P). Delete right border of zeros.
O.g.f.: 1/(1 - (2*x + 1)*t + (x^2 + x - 2)*t^2) = 1 + (1 + 2*x)*t + (3 + 3*x + 3*x^2)*t^2 + ....
T(n,n-k) = (1/3)*C(n,k)*(2^k - (-1)^k) = C(n,k)*A001045(k).
The row polynomials R(n,x) := Sum_{k = 0..n} T(n,n-k)*x^(n-k) = (1/3)*((x + 2)^n - (x - 1)^n) and have the divisibility property R(n,x) divides R(m,x) in the polynomial ring Z[x] if n divides m.
The polynomials R(n,-x), n >= 2, satisfy a Riemann hypothesis: their zeros lie on the vertical line Re x = 1/2 in the complex plane. Compare with A094440. (End)
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EXAMPLE
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First few rows of the triangle:
1;
1, 2;
3, 3, 3;
5, 12, 6, 4;
11, 25, 30, 10, 5;
21, 66, 75, 60, 15, 6;
43, 147, 231, 175, 105, 21, 7;
...
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CROSSREFS
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KEYWORD
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AUTHOR
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STATUS
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approved
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