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Search: id:A119467
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| A119467 |
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A masked Pascal triangle. |
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+0 5
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| 1, 0, 1, 1, 0, 1, 0, 3, 0, 1, 1, 0, 6, 0, 1, 0, 5, 0, 10, 0, 1, 1, 0, 15, 0, 15, 0, 1, 0, 7, 0, 35, 0, 21, 0, 1, 1, 0, 28, 0, 70, 0, 28, 0, 1, 0, 9, 0, 84, 0, 126, 0, 36, 0, 1, 1, 0, 45, 0, 210, 0, 210, 0, 45, 0, 1
(list; table; graph; listen)
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OFFSET
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0,8
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COMMENT
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Row sums are A011782. Diagonal sums are F(n+1)*(1+(-1)^n)/2 (aerated version of A001519). Product by Pascal's triangle A007318 is A119468. Schur product of (1/(1-x),x/(1-x)) and (1/(1-x^2),x).
Exponential Riordan array (cosh(x),x). Inverse is (sech(x),x) or A119879. - Paul Barry (pbarry(AT)wit.ie), May 26 2006
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FORMULA
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G.f.: (1-xy)/(1-2xy-x^2+x^2*y^2); T(n,k)=C(n,k)*(1+(-1)^(n-k))/2; Column k has g.f. (1/(1-x^2)(x/(1-x^2))^k*sum{j=0..k+1, C(k+1,j)*sin((j+1)*pi/2)^2*x^j};
Column k has e.g.f. cosh(x)*x^k/k! - Paul Barry (pbarry(AT)wit.ie), May 26 2006
Let Pascal's triangle, A007318 = P; then this triangle = (1/2) * (P + 1/P). Also A131047 = (1/2) * (P - 1/P). - Gary W. Adamson (qntmpkt(AT)yahoo.com), Jun 12 2007
Equals A007318 - A131047 since the zeros of the triangle are masks for the terms of A131047. Thus A119467 + A131047 = Pascal's triangle. - Gary W. Adamson (qntmpkt(AT)yahoo.com), Jun 12 2007
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EXAMPLE
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Triangle begins
1,
0, 1,
1, 0, 1,
0, 3, 0, 1,
1, 0, 6, 0, 1,
0, 5, 0, 10, 0, 1,
1, 0, 15, 0, 15, 0, 1,
0, 7, 0, 35, 0, 21, 0, 1,
1, 0, 28, 0, 70, 0, 28, 0, 1,
0, 9, 0, 84, 0, 126, 0, 36, 0, 1,
1, 0, 45, 0, 210, 0, 210, 0, 45, 0, 1
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CROSSREFS
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Cf. A131047.
Cf. A131047.
Adjacent sequences: A119464 A119465 A119466 this_sequence A119468 A119469 A119470
Sequence in context: A132884 A094675 A112743 this_sequence A110235 A036856 A036855
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KEYWORD
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easy,nonn,tabl
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AUTHOR
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Paul Barry (pbarry(AT)wit.ie), May 21 2006
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