A350686 - OEIS

%I #43 Mar 21 2023 06:13:10

%S 12,17,19,20,26,31,211,716,1226,1436,2306,2731,2971,5636,8011,12146,

%T 12721,16921,18266,19441,24481,24691,25796,28316,30026,34651,35876,

%U 37171,45986,49681,51691,56036,58676,61561,67531,77276,98731,98996,104161,104756,108571

%N Numbers k such that tau(k) + tau(k+1) + tau(k+2) + tau(k+3) = 16, where tau is the number of divisors function A000005.

%C It can be shown that if tau(k) + tau(k+1) + tau(k+2) + tau(k+3) = 16, the quadruple (tau(k), tau(k+1), tau(k+2), tau(k+3)) must be one of the following, each of which might plausibly occur infinitely often:

%C   (2, 4, 4, 6), which first occurs at k = 12721, 16921, 19441, 24481, ... (A163573);

%C   (2, 6, 4, 4), which first occurs at k = 19, 31, 211, 2731, ...;

%C   (4, 4, 6, 2), which first occurs at k = 26, 1226, 2306, 12146, ...;

%C   (6, 4, 4, 2), which first occurs at k = 20, 716, 1436, 5636, ...; ({A247347(n)-3}, other than its first term)

%C or one of the following, each of which occurs only once:

%C   (2, 6, 2, 6), which occurs only at k = 17; and

%C   (6, 2, 4, 4), which occurs only at k = 12.

%C Tau(k) + tau(k+1) + tau(k+2) + tau(k+3) >= 16 for all sufficiently large k; the only numbers k for which tau(k) + tau(k+1) + tau(k+2) + tau(k+3) < 16 are 1..11, 13, 14, and 16.

%H Jon E. Schoenfield, <a href="/A350686/b350686.txt">Table of n, a(n) for n = 1..10000</a>

%F { k : tau(k) + tau(k+1) + tau(k+2) + tau(k+3) = 16 }.

%e The table below includes all terms k such that at least one of the four numbers k, k+1, k+2, k+3 has no prime factor > 5; each such number appears in parentheses in the columns under "factorization".

%e The table also includes, for each of the patterns (tau(k), tau(k+1), tau(k+2), tau(k+3)) that continues to appear for large k, the smallest such k for which each of the four numbers k, k+1, k+2, k+3 has a prime factor > 5. For each such quadruple, each of the four numbers is the product of a distinct multiplier m from 1..4 and a prime > 5, and each pattern corresponds to a distinct value of k mod 120: the tau patterns (2, 4, 4, 6), (2, 6, 4, 4), (4, 4, 6, 2), and (6, 4, 4, 2) correspond to k mod 120 = 1, 91, 26, and 116, respectively.

%e .

%e                                 factorization as

%e                 # divisors of     m*(prime > 5)

%e    n  a(n)=k    k  k+1 k+2 k+3    k  k+1 k+2 k+3   k mod 120

%e    -  ------   --- --- --- ---   --- --- --- ---   ---------

%e    1      12    6   2   4   4    (12)  q  2r  3s       12

%e    2      17    2   6   2   6      p (18)  r  4s       17

%e    3      19    2   6   4   4      p (20) 3r  2s       19

%e    4      20    6   4   4   2    (20) 3q  2r   s       20

%e    5      26    4   4   6   2     2p (27) 4r   s       26

%e    6      31    2   6   4   4      p (32) 3r  2s       31

%e    7     211    2   6   4   4      p  4q  3r  2s       91

%e    8     716    6   4   2   2     4p  3q  2r   s      116

%e    9    1226    4   4   6   2     2p  3q  4r   s       26

%e   17   12721    2   4   4   6      p  2q  3r  4s        1

%t Position[Plus @@@ Partition[Array[DivisorSigma[0, #] & , 10^5], 4, 1], 16] // Flatten (* _Amiram Eldar_, Jan 12 2022 *)

%o (PARI) isok(k) = numdiv(k) + numdiv(k+1) + numdiv(k+2) + numdiv(k+3) == 16; \\ _Michel Marcus_, Jan 12 2022

%o (Python)

%o from sympy import divisor_count as tau

%o print([k for k in range( 1, 108572) if tau(k) + tau(k+1) + tau(k+2) + tau(k+3) == 16]) # _Karl-Heinz Hofmann_, Jan 12 2022

%Y Cf. A000005, A163573, A247347.

%Y Numbers k such that Sum_{j=0..N-1} tau(k+j) = 2*Sum_{k=1..N} tau(k): A000040 (N=1), A350593 (N=2), A350675 (N=3), (this sequence) (N=4), A350699 (N=5), A350769 (N=6), A350773 (N=7), A350854 (N=8).

%K nonn

%O 1,1

%A _Jon E. Schoenfield_, Jan 11 2022