A193314 The smallest k such that the product k*(k+1) is divisible by the first n primes and no others.

1, 2, 5, 14, 384, 1715, 714, 633555

Offset: 1

Author

Robert G. Wilson v, Aug 17 2011

Keywords

• nonn

Comments

a(9)-a(21) do not exist. It seems unlikely that a(n) exists for larger n. [Charles R Greathouse IV, Aug 18 2011]

If a term beyond a(8) exists, it is larger than 2.29*10^25. - Giovanni Resta, Nov 30 2019

Examples

			n  smallest k   k*(k+1) prime factorization
1  1            2
2  2            2*3
3  5            2*3*5
4  14           2*3*5*7
5  384          2^7*3*5*7*11
6  1715         2^2*3*7^3*11*13
7  714          2*3*5*7*11*13*17
8  633555       2^2*3^3*5*7*11^3*13*17*19^2
		

Links

• Carlos Rivera, Puzzle 358. Ruth-Aaron pairs revisited, The Prime Puzzles & Problems Connection.

Crossrefs

Cf. A006145, A039945.

Cf. A002110, A002378, A006530, A118478.

Programs

Haskell

a193314 n = head [k | k <- [1..], let kk' = a002378 k,
                      mod kk' (a002110 n) == 0, a006530 kk' == a000040 n]
-- Reinhard Zumkeller, Jun 14 2015

Mathematica

f[n_] := Block[{k = 1, p = Fold[ Times, 1, Prime@ Range@ n], tst = Prime@ Range@ n},While[ First@ Transpose@ FactorInteger[ k*p]!=tst || IntegerQ@ Sqrt[ 4k*p+1], k++]; Floor@ Sqrt[k*p]]; Array[f, 8]
(* the search for a(9), I also used *) lst = {}; p = Prime@ Range@ 9; Do[ q = {a, b, c, d, e, f, g, h, i}; If[ IntegerQ[ Sqrt[4Times @@ (p^q) + 1]], r = Floor@ Sqrt@ Times @@ (p^q); Print@ r; AppendTo[lst, r]], {i, 9}, {h, 9}, {g, 9}, {f, 10}, {e, 11}, {d, 14}, {c, 16}, {b, 24}, {a, 8}]

PARI

a(n)={
  my(v=[Mod(0,1)],u,P=1,t,g,k);
  forprime(p=2,prime(n),
    P*=p;
    u=List();
    for(i=1,#v,
      listput(u,chinese(v[i],Mod(-1,p)));
      listput(u,chinese(v[i],Mod(0,p)))
    );
    v=0;v=Vec(u)
  );
  v=vecsort(lift(v));
  while(1,
    for(i=1,#v,
      t=(v[i]+k)*(v[i]+k+1)/P;
      if(!t,next);
      while((g=gcd(P,t))>1, t/=g);
        if (t==1, return(v[i]+k))
    );
    k += P
  )
}; \\ Charles R Greathouse IV, Aug 18 2011