cp's OEIS Frontend

This is a front-end for the Online Encyclopedia of Integer Sequences, made by Christian Perfect. The idea is to provide OEIS entries in non-ancient HTML, and then to think about how they're presented visually. The source code is on GitHub.

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.

Original entry on oeis.org

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

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Author

Rainer Rosenthal, Apr 29 2002

Keywords

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 minimum 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.

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.
		

Crossrefs

Programs

  • 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
  • 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 *)
  • PARI
    a(n)=if(n<5,n,my(r=n,k=n^2-1,t); fordiv(k,p,t=(p+k/p)%n;if(t1, 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