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A139416
a(n) is the smallest positive integer k such that d(k) = d(k+2*n) = 2*n, where d(m) (A000005) is the number of positive divisors of m, or 0 if no such k exists.
1
3, 6, 12, 70, 600281, 60, 1458, 264, 450, 266875, 12670498046853, 480, 3998684814453099, 11968722, 299538, 3640, 49921080474853515591, 1980, 6107691150665283203125, 14960, 575047378296833, 4068173828125, 13936286848094463348388671875, 6552, 5035427051913
OFFSET
1,1
COMMENTS
Does this sequence have a term for every positive integer n, or are there no solutions for some n?
There is no solution for any odd positive integer n to d(k) = d(k+n) = n.
If n is prime, then a(n) exists if and only if there exist either three primes p, q, r such that p^(2*n-1) +- 2*n = q^(n-1)*r or four primes p_1, q_1, p_2, q_2 such that p_1^(n-1)*q_1 + 2*n = p_2^(n-1)*q_2. - Vladimir Shevelev, Jul 14 2015
EXAMPLE
For a(4) we want the smallest integer m such that d(m) = d(m+8) = 8. The positive integers that have 8 divisors each form the sequence: 24, 30, 40, 42, 54, 56, 66, 70, 78, 88, 102, 104, 105, 110, ... (A030626)
The first (not necessarily adjacent) pair of integers with 8 divisors each that is separated by exactly 8 is (70,78). So a(4) is the least element of this pair, which is 70.
Let n=5, a(5) = 600281. According to our comment, 600281 is the smallest number such that there exist either three primes p, q, r such that p^9 +- 10 = q^4*r or four primes p_1, q_1, p_2, q_2 such that p_1^4*q_1 + 10 = p_2^4*q_2. Here p_1 = 11, q_1 = 41, p_2 = 3, q_2 = 7411. - Vladimir Shevelev, Jul 14 2015
MATHEMATICA
f[n_] := Block[{a = {}, d, k, lim = 1000000}, d[x_] := DivisorSigma[0, x]; Do[k = 1; While[Nand[d[k] == d[k + d[k]], d[k] == 2 i] && k <= lim, k++]; If[k > lim, AppendTo[a, 0], AppendTo[a, k]], {i, n}]; a]; f@ 10 (* Michael De Vlieger, Jul 13 2015 *)
PROG
(PARI)
A_simple(n)=local(m=2); n*=2; until(numdiv(m)==n&numdiv(m+n)==n, m++); m
A_try_pair(p, q, n, limit)=
{
/* Helper for A_prime() */
/* Look for solution which is 0 mod p^(n-1) and -n*2 mod q^(n-1) */
local(m = chinese(Mod(0, p^(n-1)), Mod(-n*2, q^(n-1))));
forstep(x=lift(m), limit, component(m, 1),
if(isprime(x\p^(n-1)) & isprime((x+n*2)\q^(n-1)), return(x)));
limit
}
A_try_above_below(m, n)=
{
/* Helper for A_prime() */
/* Function presumes that numdiv(m)==n*2 */
if(numdiv(m-n*2)==n*2, limit=m-n*2,
if(numdiv(m+n*2)==n*2, limit=m,
0))
}
A_prime(n, limit, pairmax=30)=
{
if (n%2==0 || !isprime(n), error("Only works for odd primes"));
if (default(primelimit) < limit\nextprime(pairmax+1)^(n-1),
default(primelimit, limit\nextprime(pairmax+1)^(n-1));
);
/* Evens with numdiv==n*2 are {2^(n*2-1)} u {2*p^(n-1)} u {2^(n-1)*p} */
/* Potential solutions must come from different sets */
/* Try above and below first two sets */
A_try_above_below(2^(n*2-1), n);
forprime(p=3, (limit\2)^(1/(n-1)),
if (A_try_above_below(2*p^(n-1), n), break));
/* Odd numbers with numdiv==n*2 are {p^(n*2-1)} u {p^(n-1)*q} */
/* Try where a(n) and a(n)+n*2 are (small prime)^(n-1)*(big prime) */
forprime(p=3, pairmax, forprime(q=3, pairmax,
if (p!=q, limit = A_try_pair(p, q, n, limit))));
/* Try above and below all other odd numbers with numdiv==n*2 */
forprime(p=pairmax+1, (limit\3)^(1/(n-1)),
forprime(q=3, limit\p^(n-1),
if (p!=q & A_try_above_below(p^(n-1)*q, n), break)));
forprime(p=3, limit^(1/21),
if (A_try_above_below(p^21, n), break));
limit
} /* Martin Fuller, Apr 20 2008 */
CROSSREFS
A subsequence of A175304.
Sequence in context: A085866 A282231 A301522 * A013563 A009788 A102966
KEYWORD
nonn
AUTHOR
Leroy Quet, Apr 20 2008
EXTENSIONS
First 10 terms calculated by M. F. Hasler
a(11)-a(20) from Martin Fuller, Apr 20 2008
a(21)-a(25) from Jinyuan Wang, Sep 24 2021
STATUS
approved