This is a front-end for the Online Encyclopedia of Integer Sequences, made by Christian Perfect. The idea is to provide OEIS entries in non-ancient HTML, and then to think about how they're presented visually. The source code is on GitHub.
%I A350699 #35 Jan 20 2022 02:48:08
%S A350699 17,31,37,43,211,2305,2731,19441,116131,174595,222931,229945,232051,
%T A350699 243091,266401,334315,350785,423481,495265,523945,530545,535915,
%U A350699 539401,556705,600601,663601,671035,689131,721891,907195,908041,1105105,1113961,1289731,1338241
%N A350699 Numbers k such that tau(k) + tau(k+1) + tau(k+2) + tau(k+3) + tau(k+4) = 20, where tau is the number of divisors function A000005.
%C A350699 It can be shown that if tau(k) + ... + tau(k+4) = 20, the quintuple (tau(k), tau(k+1), tau(k+2), tau(k+3), tau(k+4)) must be one of the following, each of which might plausibly occur infinitely often:
%C A350699 (2, 4, 4, 6, 4), which first occurs at k = 19441, 266401, 423481, 539401, ... (A204592);
%C A350699 (2, 6, 4, 4, 4), which first occurs at k = 31, 211, 2731, 116131, ...;
%C A350699 (4, 4, 4, 6, 2), which first occurs at k = 2305, 229945, 350785, 495265, ...;
%C A350699 (4, 6, 4, 4, 2), which first occurs at k = 174595, 334315, 535915, 671035, ... ({A247348(n)} - 4);
%C A350699 or one of the following, each of which occurs only once:
%C A350699 (2, 6, 2, 6, 4), which occurs only at k = 17;
%C A350699 (2, 4, 4, 8, 2), which occurs only at k = 37;
%C A350699 (2, 6, 6, 4, 2), which occurs only at k = 43.
%C A350699 Tau(k) + ... + tau(k+4) >= 20 for all sufficiently large k; the only numbers k for which tau(k) + ... + tau(k+4) < 20 are 1..11, 13, 15, 19, and 25.
%H A350699 Jon E. Schoenfield, <a href="/A350699/b350699.txt">Table of n, a(n) for n = 1..10000</a>
%F A350699 { k : tau(k) + tau(k+1) + tau(k+2) + tau(k+3) + tau(k+4) = 20 }.
%e A350699 The table below lists each term k with a pattern (tau(k), ..., tau(k+4)) that appears only once (these appear at n = 1, 3, and 4), as well as each term k that is the smallest one having a pattern that appears repeatedly for large k (these are at n = 2, 6, 8, and 10). It also includes k = a(5) = 211, which is the smallest k that not only has a pattern that appears repeatedly for large k but also has each of k, ..., k+4 divisible by a prime > 5. (k = a(2) = 31 is a special case in that, while it and k = 211 share the same pattern of tau values, i.e., (2, 6, 4, 4, 4), their prime signatures differ at k+1: both 31+1=32 and 211+1=212 have 6 divisors, but 32 is a 5th power.)
%e A350699 Each of the repeatedly occurring patterns corresponds to one of the four possible orders in which the multipliers m=1..5 can appear among 5 consecutive integers of the form m*prime, and thus to a single residue of k modulo 120; e.g., k=2305 begins a run of 5 consecutive integers having the form (5*p, 2*q, 3*r, 4*s, t), where p, q, r, s, and t are distinct primes > 5, and all such runs satisfy k == 25 (mod 120).
%e A350699 For each of the patterns of tau values that does not occur repeatedly, and also for the special case k = 31, one or more of the five consecutive integers in k..k+4 has no prime factor > 5; each such integer appears in parentheses in the "factorization" columns.
%e A350699 .
%e A350699 factorization as
%e A350699 # divisors of m*(prime > 5)
%e A350699 n a(n)=k k k+1 k+2 k+3 k+4 k k+1 k+2 k+3 k+4 k mod 120
%e A350699 - ------ --- --- --- --- --- --- --- --- --- --- ---------
%e A350699 1 17 2 6 2 6 4 p (18) r (20) 3t 17
%e A350699 2 31 2 6 4 4 4 p (32) 3r 2s 5t 31
%e A350699 3 37 2 4 4 8 2 p 2q 3r (40) t 37
%e A350699 4 43 2 6 6 4 2 p 4q (45) 2s t 43
%e A350699 5 211 2 6 4 4 4 p 4q 3r 2s 5t 91
%e A350699 6 2305 4 4 4 6 2 5p 2q 3r 4s t 25
%e A350699 8 19441 2 4 4 6 4 p 2q 3r 4s 5t 1
%e A350699 10 174595 4 6 4 4 2 5p 4q 3r 2s t 115
%t A350699 Position[Plus @@@ Partition[Array[DivisorSigma[0, #] &, 10^6], 5, 1], 20] // Flatten (* _Amiram Eldar_, Jan 13 2022 *)
%o A350699 (Python) from labmath import divcount
%o A350699 print([k for k in range(1, 1338242) if divcount(k) + divcount(k+1) + divcount(k+2) + divcount(k+3) + divcount(k+4) == 20]) # _Karl-Heinz Hofmann_, Jan 13 2022
%o A350699 (PARI) isok(k) = numdiv(k) + numdiv(k+1) + numdiv(k+2) + numdiv(k+3) + numdiv(k+4) == 20; \\ _Michel Marcus_, Jan 13 2022
%Y A350699 Cf. A000005, A204592, A247348.
%Y A350699 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), A350686 (N=4), (this sequence) (N=5), A350769 (N=6), A350773 (N=7), A350854 (N=8).
%K A350699 nonn
%O A350699 1,1
%A A350699 _Jon E. Schoenfield_, Jan 12 2022