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A276501
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Smallest number k such that k! has at least n terms in its Zeckendorf representation.
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1
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0, 3, 5, 5, 7, 8, 9, 9, 9, 11, 11, 14, 14, 14, 14, 14, 15, 17, 18, 18, 18, 18, 18, 20, 20, 20, 20, 20, 21, 21, 21, 22, 22, 26, 26, 26, 26, 26, 26, 26, 28, 28, 31, 31, 32, 32, 32, 34, 34, 34, 34, 34, 34, 35, 35, 35, 36, 38, 38, 38, 38, 38, 38, 38, 41, 41, 41, 41, 43, 43, 43, 43, 47, 47, 47, 47
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OFFSET
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1,2
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COMMENTS
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Corresponding factorial numbers are 1, 6, 120, 120, 5040, 40320, ...
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LINKS
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EXAMPLE
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a(4) = 5 because Fibonacci(3) + Fibonacci(6) + Fibonacci(8) + Fibonacci(11) = 2 + 8 + 21 + 89 = 120 = 5!.
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PROG
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(PARI) a007895(n) = if(n<4, n>0, my(k=2, s, t); while(fibonacci(k++)<=n, ); while(k && n, t=fibonacci(k); if(t<=n, n-=t; s++); k--); s);
a(n) = {my(k = 0); while(a007895(k!) < n, k++); k; }
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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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