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A105150
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Approximation to leading digit of n-th Fibonacci number.
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0
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0, 1, 1, 2, 3, 5, 8, 1, 2, 3, 5, 8, 1, 2, 3, 5, 8, 1, 2, 3, 5, 8, 1, 2, 3, 5, 8, 1, 2, 3, 5, 8, 1, 2, 3, 5, 8, 1, 2, 3, 5, 8, 1, 2, 3, 5, 8, 1, 2, 3, 5, 8, 1, 2, 3, 5, 8, 1, 2, 3, 5, 8, 1, 2, 3, 5, 8, 1, 2, 3, 5, 8, 1, 2, 3, 5, 8, 1, 2, 3, 5, 8, 1, 2, 3, 5, 8, 1, 2, 3, 5, 8, 1, 2, 3, 5, 8, 1, 2, 3, 5, 8, 1, 2, 3
(list; graph; refs; listen; history; internal format)
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OFFSET
| 0,4
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COMMENTS
| a(0) = 1, a(1) = a(2) = 1 and for n > 2:
a(n) = floor(w/10) + (w mod 10)*0^floor(w/10) where
w = (x+y)*0^(z-1) + y + z*0^floor((11-z)/10),
x = a(n-3), y = a(n-2) and z = a(n-1);
a(n) = A008963(n) = A000030(A000045(n) for n<=14.
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EXAMPLE
| n=11, x=2, y=3, z=5: w = (2+3) * 0^(5-1)+3+5*0^[(11-5)/10] = 5*0^4+3+5*0^0 = 0+3+5*1 = 8, a(11) = [8/10] + (8 mod 10) * 0^[8/10] = 0 + 8*0^0 = 8;
n=12, x=3, y=5, z=8: w = (3+5) * 0^(8-1)+5+8*0^[(11-8)/10] = 8*0^7+5+8*0^0 = 0+5+8*1 = 13, a(12) = [13/10] + (13 mod 10) * 0^[13/10] = 1 + 3*0^1 = 1;
n=13, x=5, y=8, z=1: w = (5+8) * 0^(1-1)+8+1*0^[(11-1)/10] = 13*0^0+8+1*0^1 = 13*1+8+1*0 = 21, a(13) = [21/10] + (21 mod 10) * 0^[21/10] = 2 + 1*0^2 = 2.
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CROSSREFS
| Sequence in context: A136740 A105994 A120496 * A008963 A031324 A093086
Adjacent sequences: A105147 A105148 A105149 * A105151 A105152 A105153
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KEYWORD
| nonn,base
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AUTHOR
| Reinhard Zumkeller (reinhard.zumkeller(AT)gmail.com), Apr 10 2005
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