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A207644
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a(n) = 1 + (n-1) + (n-2)*[(n-3)/2] + (n-3)*[(n-4)/2]*[(n-5)/3] + (n-4)*[(n-5)/2]*[(n-6)/3]*[(n-7)/4] +... where [x] = floor(x), with summation extending over the initial [n/2+1] products only.
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6
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1, 1, 2, 3, 4, 8, 10, 17, 30, 42, 55, 116, 172, 220, 391, 683, 1024, 1616, 2050, 3675, 6520, 9504, 12505, 22421, 35572, 56918, 85701, 138110, 202765, 326231, 503632, 860497, 1376870, 1927446, 2818531, 4892966, 7784671, 11432772, 17287295, 30423457, 46453786, 71810414
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OFFSET
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0,3
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COMMENTS
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Radius of convergence of g.f. A(x) is near 0.637..., with A(1/phi) = 23.059250143... where phi = (sqrt(5)+1)/2.
Compare the definition of a(n) with the binomial sum:
Fibonacci(n) = 1 + (n-1) + (n-2)*((n-3)/2) + (n-3)*((n-4)/2)*((n-5)/3) + (n-4)*((n-5)/2)*((n-6)/3)*((n-7)/4) +...
with summation extending over the initial [n/2+1] products only.
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LINKS
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FORMULA
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a(n) = 1 + Sum_{k=1..[n/2]} Product_{j=1..k} floor( (n-k-j+1) / j ).
Equals the antidiagonal sums of triangle A207645.
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EXAMPLE
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a(3) = 1 + 2 = 3;
a(4) = 1 + 3 + 2*[1/2] = 4;
a(5) = 1 + 4 + 3*[2/2] = 8;
a(6) = 1 + 5 + 4*[3/2] + 3*[2/2]*[1/3] = 10;
a(7) = 1 + 6 + 5*[4/2] + 4*[3/2]*[2/3] = 17;
a(8) = 1 + 7 + 6*[5/2] + 5*[4/2]*[3/3] + 4*[3/2]*[2/3]*[1/4] = 30;
a(9) = 1 + 8 + 7*[6/2] + 6*[5/2]*[4/3] + 5*[4/2]*[3/3]*[2/4] = 42; ...
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MATHEMATICA
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a[n_] := 1 + Sum[ Product[ Floor[ (n-k-j+1)/j ], {j, 1, k}], {k, 1, n/2}]; Table[a[n], {n, 0, 41}] (* Jean-François Alcover, Mar 06 2013 *)
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PROG
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(PARI) {a(n)=1+sum(k=1, n\2, prod(j=1, k, floor((n-k-j+1)/j)))}
for(n=0, 60, print1(a(n), ", "))
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CROSSREFS
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KEYWORD
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nonn,nice
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AUTHOR
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STATUS
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approved
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