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 A139416 a(n) = smallest positive integer k such that d(k) = d(k+2n) = 2n, 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 (list; graph; refs; listen; history; text; internal format)
 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 LINKS 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 Cf. A000005, A137532. A subsequence of A175304. Sequence in context: A085866 A282231 A301522 * A013563 A009788 A102966 Adjacent sequences:  A139413 A139414 A139415 * A139417 A139418 A139419 KEYWORD more,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 STATUS approved

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Last modified April 17 19:51 EDT 2021. Contains 343070 sequences. (Running on oeis4.)