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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.
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%I #28 Oct 05 2021 00:29:43

%S 3,6,12,70,600281,60,1458,264,450,266875,12670498046853,480,

%T 3998684814453099,11968722,299538,3640,49921080474853515591,1980,

%U 6107691150665283203125,14960,575047378296833,4068173828125,13936286848094463348388671875,6552,5035427051913

%N 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.

%C Does this sequence have a term for every positive integer n, or are there no solutions for some n?

%C There is no solution for any odd positive integer n to d(k) = d(k+n) = n.

%C 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

%e 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)

%e 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.

%e 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

%t 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 *)

%o (PARI)

%o A_simple(n)=local(m=2);n*=2;until(numdiv(m)==n&numdiv(m+n)==n,m++);m

%o A_try_pair(p,q,n,limit)=

%o {

%o /* Helper for A_prime() */

%o /* Look for solution which is 0 mod p^(n-1) and -n*2 mod q^(n-1) */

%o local(m = chinese(Mod(0,p^(n-1)), Mod(-n*2,q^(n-1))));

%o forstep(x=lift(m), limit, component(m,1),

%o if(isprime(x\p^(n-1)) & isprime((x+n*2)\q^(n-1)), return(x)));

%o limit

%o }

%o A_try_above_below(m,n)=

%o {

%o /* Helper for A_prime() */

%o /* Function presumes that numdiv(m)==n*2 */

%o if(numdiv(m-n*2)==n*2, limit=m-n*2,

%o if(numdiv(m+n*2)==n*2, limit=m,

%o 0))

%o }

%o A_prime(n,limit,pairmax=30)=

%o {

%o if (n%2==0 || !isprime(n), error("Only works for odd primes"));

%o if (default(primelimit) < limit\nextprime(pairmax+1)^(n-1),

%o default(primelimit, limit\nextprime(pairmax+1)^(n-1));

%o );

%o /* Evens with numdiv==n*2 are {2^(n*2-1)} u {2*p^(n-1)} u {2^(n-1)*p} */

%o /* Potential solutions must come from different sets */

%o /* Try above and below first two sets */

%o A_try_above_below(2^(n*2-1),n);

%o forprime(p=3, (limit\2)^(1/(n-1)),

%o if (A_try_above_below(2*p^(n-1),n), break));

%o /* Odd numbers with numdiv==n*2 are {p^(n*2-1)} u {p^(n-1)*q} */

%o /* Try where a(n) and a(n)+n*2 are (small prime)^(n-1)*(big prime) */

%o forprime(p=3, pairmax, forprime(q=3, pairmax,

%o if (p!=q, limit = A_try_pair(p,q,n,limit))));

%o /* Try above and below all other odd numbers with numdiv==n*2 */

%o forprime(p=pairmax+1, (limit\3)^(1/(n-1)),

%o forprime(q=3, limit\p^(n-1),

%o if (p!=q & A_try_above_below(p^(n-1)*q,n), break)));

%o forprime(p=3, limit^(1/21),

%o if (A_try_above_below(p^21,n), break));

%o limit

%o } /* _Martin Fuller_, Apr 20 2008 */

%Y Cf. A000005, A137532.

%Y A subsequence of A175304.

%K nonn

%O 1,1

%A _Leroy Quet_, Apr 20 2008

%E First 10 terms calculated by _M. F. Hasler_

%E a(11)-a(20) from _Martin Fuller_, Apr 20 2008

%E a(21)-a(25) from _Jinyuan Wang_, Sep 24 2021