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A127749
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Inverse of number triangle A(n,k) = 1/(2n+1) if k <= n <= 2k, 0 otherwise.
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2
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1, 0, 3, 0, -3, 5, 0, 3, -5, 7, 0, 0, 0, -7, 9, 0, -3, 5, 0, -9, 11, 0, 0, 0, 0, 0, -11, 13, 0, 3, -5, 7, 0, 0, -13, 15, 0, 0, 0, 0, 0, 0, 0, -15, 17, 0, 0, 0, -7, 9, 0, 0, 0, -17, 19, 0, 0, 0, 0, 0, 0, 0, 0, 0, -19, 21, 0, -3, 5
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OFFSET
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0,3
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COMMENTS
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Conjectures: row sums modulo 2 are the Fredholm-Rueppel sequence A036987; row sums of triangle modulo 2 are A111982. Row sums are A127750.
The first conjecture is equivalent to the row sums conjecture in A111967. - R. J. Mathar, Apr 21 2021
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LINKS
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FORMULA
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EXAMPLE
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Triangle begins
1;
0, 3;
0, -3, 5;
0, 3, -5, 7;
0, 0, 0, -7, 9;
0, -3, 5, 0, -9, 11;
0, 0, 0, 0, 0, -11, 13;
0, 3, -5, 7, 0, 0, -13, 15;
0, 0, 0, 0, 0, 0, 0, -15, 17;
0, 0, 0, -7, 9, 0, 0, 0, -17, 19;
0, 0, 0, 0, 0, 0, 0, 0, 0, -19, 21;
0, -3, 5, 0, -9, 11, 0, 0, 0, 0, -21, 23;
0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, -23, 25;
Inverse of triangle
1;
0, 1/3;
0, 1/5, 1/5;
0, 0, 1/7, 1/7;
0, 0, 1/9, 1/9, 1/9;
0, 0, 0, 1/11, 1/11, 1/11;
0, 0, 0, 1/13, 1/13, 1/13, 1/13;
0, 0, 0, 0, 1/15, 1/15, 1/15, 1/15;
0, 0, 0, 0, 1/17, 1/17, 1/17, 1/17, 1/17;
0, 0, 0, 0, 0, 1/19, 1/19, 1/19, 1/19, 1/19;
0, 0, 0, 0, 0, 1/21, 1/21, 1/21, 1/21, 1/21, 1/21;
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MAPLE
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option remember ;
if k > n then
0 ;
elif k = n then
2*n+1 ;
else
-(2*k+1)*add( procname(n, i)/(2*i+1), i=k+1..min(n, 2*k)) ;
end if;
end proc:
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MATHEMATICA
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nmax = 10;
A[n_, k_] := If[k <= n <= 2k, 1/(2n+1), 0];
invA = Inverse[Table[A[n, k], {n, 0, nmax}, {k, 0, nmax}]];
T[n_, k_] := invA[[n+1, k+1]];
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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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