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A069817
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Smallest remainder of p+q modulo n, where p*q=n^2-1, 1 < p < n-1.
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0
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1, 2, 3, 4, 1, 6, 2, 0, 3, 6, 1, 12, 3, 2, 0, 8
(list; graph; refs; listen; history; internal format)
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OFFSET
| 1,2
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COMMENTS
| Factoring n^2-k = p*q for k > 1 gives p+q never divisible by k*a. The remainders p+q mod k*a have minimum > 0 for k > 1. This sequence shows the minima for k=1, where there are zero values, e.g. a(8)=0. The restriction 1 < p < n-1 is due to the fact that (n-1)(n+1) = n^2-1 and (n-1)+(n+1)=2n is zero mod n. Excluding this trivial case as well as 1*(n^2-1) with 1+(n^2-1)=n^2=0 (mod n) gives the more interesting elements.
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LINKS
| IMO 1988 #6
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FORMULA
| If n^2-1 is prime then set a(n) = n. Otherwise check (p+q mod n) for all natural numbers p, q which satisfy: p*q = n^2-1 and 1 < p < n-1 and 1 < q < n-1.
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EXAMPLE
| a(7)=2 since 7^2-1 = 48 = 2*24 = 3*16 = 4*12 where the sums of the factors (mod 7) are: [2+24]=5, [3+16]=5, [4+12]=2.
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CROSSREFS
| Sequence in context: A065331 A066262 A174715 * A071439 A100941 A082119
Adjacent sequences: A069814 A069815 A069816 * A069818 A069819 A069820
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KEYWORD
| easy,nice,nonn
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AUTHOR
| Rainer Rosenthal (r.rosenthal(AT)web.de), Apr 29 2002
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