A069817 Smallest remainder of p+q modulo n, where p*q = n^2 - 1, 1 < p < n-1, or n if no such factorization exists.
1, 2, 3, 4, 1, 6, 2, 0, 3, 6, 1, 12, 3, 2, 0, 8, 2, 18, 1, 4, 0, 10, 2, 0, 5, 6, 3, 12, 1, 30, 2, 8, 12, 2, 0, 12, 7, 10, 3, 16, 1, 42, 3, 4, 15, 14, 2, 0, 2, 14, 3, 20, 3, 6, 0, 0, 15, 22, 4, 60, 11, 2, 0, 8, 5, 18, 5, 20, 3, 26, 1, 72, 13, 10, 15
Offset: 1
Examples
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.
Links
- Charles R Greathouse IV, Table of n, a(n) for n = 1..10000
- Art of Problem Solving, 1988 IMO Problems/Problem 6
Programs
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Haskell
a069817 1 = 1 a069817 n = if null ms then n else minimum $ map (`mod` n) ms where ms = zipWith (+) ds $ map (div n') ds ds = takeWhile (< n - 1) $ tail $ a027750_row n' n' = n ^ 2 - 1 -- Reinhard Zumkeller, May 16 2013
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Mathematica
a[n_] := (r = Reduce[1 < p < n - 1 && n^2 - 1 == p*q, {p, q}, Integers]; factors = If[r === False, n, {p, q} /. {ToRules[r]}]; Mod[#[[1]] + #[[2]], n] & /@ factors // Min); Table[ a[n] , {n, 1, 16}] (* Jean-François Alcover, Apr 04 2013 *)
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PARI
a(n)=if(n<5,n,my(r=n,k=n^2-1,t); fordiv(k,p,t=(p+k/p)%n;if(t
1, if(p+1==n,return(r),r=t)))) \\ Charles R Greathouse IV, Apr 04 2013
Formula
If n-1 and n+1 are 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.
Extensions
a(17)-a(75) from Charles R Greathouse IV, Apr 04 2013
Comments