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A104975
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Inverse of a Fredholm-Rueppel triangle.
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4
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1, 0, 1, -1, 0, 1, 0, -1, 0, 1, 1, 0, -1, 0, 1, 0, 1, 0, -1, 0, 1, -2, 0, 1, 0, -1, 0, 1, 0, -2, 0, 1, 0, -1, 0, 1, 3, 0, -2, 0, 1, 0, -1, 0, 1, 0, 3, 0, -2, 0, 1, 0, -1, 0, 1, -4, 0, 3, 0, -2, 0, 1, 0, -1, 0, 1, 0, -4, 0, 3, 0, -2, 0, 1, 0, -1, 0, 1, 6, 0, -4, 0, 3, 0, -2, 0, 1, 0, -1, 0, 1, 0, 6, 0, -4, 0, 3, 0, -2, 0, 1, 0, -1, 0, 1, -10, 0, 6, 0, -4, 0, 3, 0, -2
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OFFSET
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0,22
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COMMENTS
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LINKS
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FORMULA
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Riordan array (x^2/( (Sum_{k>=0} x^(2^k)) - x), x).
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EXAMPLE
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Triangle begins as:
1;
0, 1;
-1, 0, 1;
0, -1, 0, 1;
1, 0, -1, 0, 1;
0, 1, 0, -1, 0, 1;
-2, 0, 1, 0, -1, 0, 1;
0, -2, 0, 1, 0, -1, 0, 1;
3, 0, -2, 0, 1, 0, -1, 0, 1;
0, 3, 0, -2, 0, 1, 0, -1, 0, 1;
-4, 0, 3, 0, -2, 0, 1, 0, -1, 0, 1;
0, -4, 0, 3, 0, -2, 0, 1, 0, -1, 0, 1;
6, 0, -4, 0, 3, 0, -2, 0, 1, 0, -1, 0, 1;
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MATHEMATICA
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t[n_, k_]:= t[n, k]= If[k==n, 1, (1+(-1)^(n-k))/2 Sum[Binomial[k, j]*t[(n-k)/2, j], {j, (n-k)/2}]];
S[n_]:= Sum[(-1)^j*t[n, j], {j, 0, n}]; (* S = A104977 *)
T[n_, k_]:= If[EvenQ[n-k], S[(n-k)/2], 0];
Table[T[n, k], {n, 0, 12}, {k, 0, n}]//Flatten (* G. C. Greubel, Jun 08 2021 *)
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PROG
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(Sage)
@CachedFunction
def t(n, k): return 1 if (k==n) else ((1+(-1)^(n-k))/2)*sum( binomial(k, j)*t((n-k)/2, j) for j in (1..(n-k)//2) )
def S(n): return sum( (-1)^j*t(n, j) for j in (0..n) ) # S = A104977
def T(n, k): return S((n-k)/2) if (mod(n-k, 2)==0) else 0
flatten([[T(n, k) for k in (0..n)] for n in (0..12)]) # G. C. Greubel, Jun 08 2021
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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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