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A070198
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Smallest nonnegative number m such that m == i (mod i+1) for all 1 <= i <= n.
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9
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0, 1, 5, 11, 59, 59, 419, 839, 2519, 2519, 27719, 27719, 360359, 360359, 360359, 720719, 12252239, 12252239, 232792559, 232792559, 232792559, 232792559, 5354228879, 5354228879, 26771144399, 26771144399, 80313433199, 80313433199
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OFFSET
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0,3
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COMMENTS
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Also, smallest k such that, for 0 <= i < n, i+1 divides k-i.
Suggested by Chinese Remainder Theorem. This sequence can generate others: smallest b(n) such that b(n) == i (mod (i+2)), 1 <= i <= n, gives b(1)=1 and b(n) = a(n+1)-1 for n > 1; smallest c(n) such that c(n) == i (mod (i+3)), 1 <= i <= n, gives c(1)=1, c(2)=17 and c(n) = a(n+2) - 2 for n > 2; smallest d(n) such that c(n) == i (mod (i+4)), 1 <= i <= n, gives d(1)=1, d(2)=26, d(3)=206 and d(n) = a(n+3) - 3 for n > 3, etc.
A070198(n-1) is m such that max(Sum_{i=1..n} m (mod i)) = A000217(n-1).
Example for n = 3:
m\i = 1 2 3 sum
1 0 1 1 2
2 0 0 2 2
3 0 1 0 1
4 0 0 1 1
5 0 1 2 3 <--max remainder sum = 3 = A000217(2)
6 0 0 0 0 first occurs at m = 5 = A070198(2)
(End)
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LINKS
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FORMULA
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a(n) = lcm(1, 2, 3, ..., n+1) - 1 = A003418(n+1) - 1.
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EXAMPLE
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a(3) = 11 because 11 == 1 (mod 2), 11 == 2 (mod 3) and 11 == 3 (mod 4).
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MAPLE
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MATHEMATICA
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f[n_] := ChineseRemainder[ Range[0, n - 1], Range[n]]; Array[f, 28] (* or *)
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PROG
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(Haskell)
a070198 n = a070198_list !! n
a070198_list = map (subtract 1) $ scanl lcm 1 [2..]
(Python)
from math import lcm
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CROSSREFS
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KEYWORD
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easy,nonn
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AUTHOR
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EXTENSIONS
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STATUS
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approved
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