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A193314
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The smallest k such that the product k*(k+1) is divisible by the first n primes and no others.
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2
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OFFSET
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1,2
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COMMENTS
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If a term beyond a(8) exists, it is larger than 2.29*10^25. - Giovanni Resta, Nov 30 2019
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LINKS
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EXAMPLE
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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
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MATHEMATICA
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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}]
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PROG
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(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
)
(Haskell)
a193314 n = head [k | k <- [1..], let kk' = a002378 k,
mod kk' (a002110 n) == 0, a006530 kk' == a000040 n]
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CROSSREFS
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KEYWORD
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nonn
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AUTHOR
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STATUS
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approved
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