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%I #41 Sep 03 2022 17:41:05
%S 5,6,7,10,13,22,37,46,58,61,73,82,106,157,166,178,193,226,262,277,313,
%T 346,358,382,397,421,457,466,478,502,541,562,586,613,661,673,718,733,
%U 757,838,862,877,886,982,997,1018,1093,1153,1186,1201,1213,1237,1282
%N Numbers k such that tau(k) + tau(k+1) = 6, where tau is the number of divisors function A000005.
%C Since tau(k) + tau(k+1) = 6, (tau(k), tau(k+1)) must be (1,5), (2,4), (3,3), (4,2), or (5,1); of these, (1,5) and (5,1) are impossible (tau(m) = 1 only for m=1, but then neither m+1 nor m-1 would have 5 divisors), and (3,3) is also impossible (both k and k+1 would have to be squares of primes), so (tau(k), tau(k+1)) must be either (2,4) or (4,2).
%C For every prime p, tau(p) = 2. For every semiprime s, tau(s) = 4, with the exception of the squares of primes; for p prime, tau(p^2) = 3, since the divisors of p^2 are 1, p, and p^2.
%C The only numbers that have exactly 4 divisors but are not semiprimes are the cubes of primes; for prime p, the divisors of p^3 are 1, p, p^2, and p^3.
%C As a result, this sequence consists of:
%C (1) the primes p such that (p+1)/2 is prime (A005383), with the exception of p=3 (since p+1 = 4 has 3 divisors, not 4),
%C (2) semiprimes of the form prime - 1 (A077065), with the exception of the semiprime 4 (since it does not have 4 divisors), and
%C (3) the special case k = 7, since it is the unique prime p such that p+1 has 4 divisors but is not a semiprime.
%C For all k > 4, tau(k) + tau(k+1) >= 6; for k = 1..4, tau(k) + tau(k+1) = 3, 4, 5, 5.
%H Jon E. Schoenfield, <a href="/A350593/b350593.txt">Table of n, a(n) for n = 1..10000</a>
%F { k : tau(k) + tau(k+1) = 6 }.
%F UNION(A005383 \ {3}, A077065 \ {4}, {7}).
%F a(n) = A164977(n+1) for n>=4. - _Hugo Pfoertner_, Jan 08 2022
%e k tau(k) tau(k+1) tau(k) + tau(k+1)
%e -- ------ -------- -----------------
%e 1 1 2 1 + 2 = 3
%e 2 2 2 2 + 2 = 4
%e 3 2 3 2 + 3 = 5
%e 4 3 2 3 + 2 = 5
%e 5 2 4 2 + 4 = 6 so 5 = a(1)
%e 6 4 2 4 + 2 = 6 so 6 = a(2)
%e 7 2 4 2 + 4 = 6 so 7 = a(3)
%e 8 4 3 4 + 3 = 7
%e 9 3 4 3 + 4 = 7
%e 10 4 2 4 + 2 = 6 so 10 = a(4)
%e 11 2 6 2 + 6 = 8
%e 12 6 2 6 + 2 = 8
%e 13 2 4 2 + 4 = 6 so 13 = a(5)
%t Select[Range[1300], Plus @@ DivisorSigma[0, # + {0, 1}] == 6 &] (* _Amiram Eldar_, Jan 08 2022 *)
%t Position[Total/@Partition[DivisorSigma[0,Range[1300]],2,1],6]//Flatten (* _Harvey P. Dale_, Sep 03 2022 *)
%o (PARI) isok(k) = numdiv(k) + numdiv(k+1) == 6; \\ _Michel Marcus_, Jan 08 2022
%o (Python)
%o from itertools import count, islice
%o from sympy import divisor_count
%o def A350093_gen(): # generator of terms
%o a, b = divisor_count(1), divisor_count(2)
%o for k in count(1):
%o if a + b == 6:
%o yield k
%o a, b = b, divisor_count(k+2)
%o A350093_list = list(islice(A350093_gen(),12)) # _Chai Wah Wu_, Jan 11 2022
%Y Cf. A000005, A005383, A077065, A092405, A164977.
%Y Numbers k such that Sum_{j=0..N-1} tau(k+j) = 2*Sum_{k=1..N} tau(k): A000040 (N=1), (this sequence) (N=2), A350675 (N=3), A350686 (N=4), A350699 (N=5), A350769 (N=6), A350773 (N=7), A350854 (N=8).
%K nonn
%O 1,1
%A _Jon E. Schoenfield_, Jan 08 2022