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0, 0, 0, 0, 0, 1, 1, 1, 1, 1, 3, 3, 3, 3, 3, 6, 6, 6, 6, 6, 10, 10, 10, 10, 10, 16, 17, 18, 19, 20, 23, 24, 25, 26, 27, 32, 33, 34, 35, 36, 43, 44, 45, 46, 47, 56, 57, 58, 59, 60, 73, 76, 79, 82, 85, 91, 94, 97, 100, 103, 112, 115, 118, 121
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graph;
refs;
listen;
history;
text;
internal format)
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OFFSET
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0,11
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COMMENTS
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A combinatorial interpretation is given in the Edgar link.
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LINKS
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FORMULA
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Let b(0) = 1 and b(n) = b(n-1) + b(floor(n/5)) and let c(n) = Product_{i=0..k}(n_i+1) where n = Sum_{i=0..k}n_i*5^i is the base 5 representation of n. Then a(n) = (1/5)*(b(n) - c(n)).
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MATHEMATICA
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b[0] = 1; b[n_] := b[n] = b[n-1] + b[Floor[n/5]];
c[n_] := If[OddQ[n], 2 Count[Table[Binomial[n, k], {k, 0, (n-1)/2}], c_ /; !Divisible[c, 5]], 2 Count[Table[Binomial[n, k], {k, 0, (n-2)/2}], c_ /; !Divisible[c, 5]] + Boole[!Divisible[Binomial[n, n/2], 5]]];
a[n_] := (b[n] - c[n])/5;
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PROG
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(Sage)
def b(n):
A=[1]
for i in [1..n]:
A.append(A[i-1] + A[i//5])
return A[n]
print([(b(n)-prod(x+1 for x in n.digits(5)))/5 for n in [0..63]])
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CROSSREFS
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KEYWORD
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nonn
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AUTHOR
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STATUS
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approved
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