A350593 -id:A350593 - cp's OEIS frontend

A350675 Numbers k such that tau(k) + tau(k+1) + tau(k+2) = 10, where tau is the number of divisors function A000005.

6, 11, 13, 17, 21, 37, 57, 157, 177, 381, 501, 541, 717, 877, 1201, 1317, 1381, 1437, 1621, 1821, 2017, 2557, 2577, 2857, 2901, 3061, 3117, 3777, 4281, 4357, 4441, 4677, 4701, 5077, 5097, 5581, 5637, 5701, 5937, 6337, 6637, 6661, 6717, 6997, 7417, 8221, 8781

Offset: 1

Jon E. Schoenfield, Jan 10 2022

nonn

Since tau(k) + tau(k+1) + tau(k+2) = 10 and no three consecutive integers include more than one square, the triple (tau(k), tau(k+1), tau(k+2)) must consist of three even numbers, so it must be one of (2, 2, 6), (2, 4, 4), (2, 6, 2), (4, 2, 4), (4, 4, 2), and (6, 2, 2). Of these, (2, 2, 6) and (6, 2, 2) are impossible. Of the remaining patterns:
(2, 4, 4) requires that k be an odd prime other than 3, followed by two semiprimes, so k is a prime p such that (p+1)/2 and (p+2)/3 are also prime, and such primes are 13, 37, 157, 541, ... (A036570);
(2, 6, 2) requires that (k, k+2) be a twin prime pair whose average has exactly 6 divisors, and is thus either 12 or 18, so k is 11 or 17;
(4, 2, 4) requires that k+1 be an odd prime, with both k and k+2 having exactly 4 divisors, even though one of them is a multiple of 4, so that one is k+2 = 2^3 = 8, so k = 6;
(4, 4, 2) requires that k+2 be an odd prime > 3, preceded by two semiprimes, so k+2 is a prime p such that (p-1)/2 and (p-2)/3 are also prime, so k+2 is in {23, 59, 179, 383, ...} (which is A181841, after its first two terms, 7 and 11), so k is in {A181841(n) - 2} \ {5, 9}, i.e., k is in {21, 57, 177, 381, ...}.
Tau(k) + tau(k+1) + tau(k+2) >= 10 for all sufficiently large k; the only numbers k for which tau(k) + tau(k+1) + tau(k+2) < 10 are 1..5, 7, and 9.

			Each of the patterns (tau(k), ..., tau(k+2)) that appears repeatedly for large k corresponds to one of the two possible orders in which the multipliers m=1..3 can appear among 3 consecutive integers of the form m*prime. E.g., k=37 begins a run of 3 consecutive integers having the form (p, 2*q, 3*r), where p, q, and r are distinct primes > 3; k=57 begins a similar run, but there the 3 consecutive integers have the form (3*p, 2*q, r).
For each of the patterns of tau values that does not occur repeatedly for large k, one or more of the three consecutive integers in k..k+2 has no prime factor > 3; in the table below, each such integer appears in parentheses in the columns on the right.
.
                              factorization as
              # divisors of    m*(prime > 3)
  n  a(n)=k     k  k+1 k+2      k   k+1  k+2
  -  ------    --- --- ---    ---- ---- ----
  1      6      4   2   4      (6)   q   (8)
  2     11      2   6   2       p  (12)   r
  3     13      2   4   4       p   2q   3r
  4     17      2   6   2       p  (18)   r
  5     21      4   4   2      3p   2q    r
  6     37      2   4   4       p   2q   3r
  7     57      4   4   2      3p   2q    r

Jon E. Schoenfield, Table of n, a(n) for n = 1..10000

Cf. A000005, A036570, A181841, A307120, A317670.

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), (this sequence) (N=3), A350686 (N=4), A350699 (N=5), A350769 (N=6), A350773 (N=7), A350854 (N=8).

Position[Plus @@@ Partition[Array[DivisorSigma[0, #] & , 10^4], 3, 1], 10] // Flatten (* Amiram Eldar, Jan 11 2022 *)

isok(k) = numdiv(k) + numdiv(k+1) + numdiv(k+2) == 10; \\ Michel Marcus, Jan 16 2022

{ k : tau(k) + tau(k+1) + tau(k+2) = 10 }.
UNION({6}, {11, 17}, A036570, {A181841(n) - 2} \ {5, 9}).
a(n) = A317670(n) - 1.

A350686 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.

Original entry on oeis.org

12, 17, 19, 20, 26, 31, 211, 716, 1226, 1436, 2306, 2731, 2971, 5636, 8011, 12146, 12721, 16921, 18266, 19441, 24481, 24691, 25796, 28316, 30026, 34651, 35876, 37171, 45986, 49681, 51691, 56036, 58676, 61561, 67531, 77276, 98731, 98996, 104161, 104756, 108571

Offset: 1

Jon E. Schoenfield, Jan 11 2022

nonn

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:
  (2, 4, 4, 6), which first occurs at k = 12721, 16921, 19441, 24481, ... (A163573);
  (2, 6, 4, 4), which first occurs at k = 19, 31, 211, 2731, ...;
  (4, 4, 6, 2), which first occurs at k = 26, 1226, 2306, 12146, ...;
  (6, 4, 4, 2), which first occurs at k = 20, 716, 1436, 5636, ...; ({A247347(n)-3}, other than its first term)
or one of the following, each of which occurs only once:
  (2, 6, 2, 6), which occurs only at k = 17; and
  (6, 2, 4, 4), which occurs only at k = 12.
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.

			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".
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.
.
                                factorization as
                # divisors of     m*(prime > 5)
   n  a(n)=k    k  k+1 k+2 k+3    k  k+1 k+2 k+3   k mod 120
   -  ------   --- --- --- ---   --- --- --- ---   ---------
   1      12    6   2   4   4    (12)  q  2r  3s       12
   2      17    2   6   2   6      p (18)  r  4s       17
   3      19    2   6   4   4      p (20) 3r  2s       19
   4      20    6   4   4   2    (20) 3q  2r   s       20
   5      26    4   4   6   2     2p (27) 4r   s       26
   6      31    2   6   4   4      p (32) 3r  2s       31
   7     211    2   6   4   4      p  4q  3r  2s       91
   8     716    6   4   2   2     4p  3q  2r   s      116
   9    1226    4   4   6   2     2p  3q  4r   s       26
  17   12721    2   4   4   6      p  2q  3r  4s        1

Jon E. Schoenfield, Table of n, a(n) for n = 1..10000

Cf. A000005, A163573, A247347.

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).

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

isok(k) = numdiv(k) + numdiv(k+1) + numdiv(k+2) + numdiv(k+3) == 16; \\ Michel Marcus, Jan 12 2022

from sympy import divisor_count as tau
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

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

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.

Original entry on oeis.org

17, 31, 37, 43, 211, 2305, 2731, 19441, 116131, 174595, 222931, 229945, 232051, 243091, 266401, 334315, 350785, 423481, 495265, 523945, 530545, 535915, 539401, 556705, 600601, 663601, 671035, 689131, 721891, 907195, 908041, 1105105, 1113961, 1289731, 1338241

Offset: 1

Jon E. Schoenfield, Jan 12 2022

nonn

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:
  (2, 4, 4, 6, 4), which first occurs at k = 19441, 266401, 423481, 539401, ... (A204592);
  (2, 6, 4, 4, 4), which first occurs at k = 31, 211, 2731, 116131, ...;
  (4, 4, 4, 6, 2), which first occurs at k = 2305, 229945, 350785, 495265, ...;
  (4, 6, 4, 4, 2), which first occurs at k = 174595, 334315, 535915, 671035, ... ({A247348(n)} - 4);
or one of the following, each of which occurs only once:
  (2, 6, 2, 6, 4), which occurs only at k = 17;
  (2, 4, 4, 8, 2), which occurs only at k = 37;
  (2, 6, 6, 4, 2), which occurs only at k = 43.
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.

			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.)
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).
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.
.
                                      factorization as
                  # divisors of         m*(prime > 5)
   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
   -  ------   --- --- --- --- ---   --- --- --- --- ---   ---------
   1      17    2   6   2   6   4      p (18)  r (20) 3t       17
   2      31    2   6   4   4   4      p (32) 3r  2s  5t       31
   3      37    2   4   4   8   2      p  2q  3r (40)  t       37
   4      43    2   6   6   4   2      p  4q (45) 2s   t       43
   5     211    2   6   4   4   4      p  4q  3r  2s  5t       91
   6    2305    4   4   4   6   2     5p  2q  3r  4s   t       25
   8   19441    2   4   4   6   4      p  2q  3r  4s  5t        1
  10  174595    4   6   4   4   2     5p  4q  3r  2s   t      115

Jon E. Schoenfield, Table of n, a(n) for n = 1..10000

Cf. A000005, A204592, A247348.

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).

Position[Plus @@@ Partition[Array[DivisorSigma[0, #] &, 10^6], 5, 1], 20] // Flatten (* Amiram Eldar, Jan 13 2022 *)

isok(k) = numdiv(k) + numdiv(k+1) + numdiv(k+2) + numdiv(k+3) + numdiv(k+4) == 20; \\ Michel Marcus, Jan 13 2022

from labmath import divcount
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

{ k : tau(k) + tau(k+1) + tau(k+2) + tau(k+3) + tau(k+4) = 20 }.

A350769 Numbers k such that tau(k) + ... + tau(k+5) = 28, where tau is the number of divisors function A000005.

Original entry on oeis.org

27, 28, 30, 37, 38, 41, 42, 57, 18362, 2914913, 5516281, 6618242, 7224834, 9018353, 9339114, 10780554, 16831081, 17800553, 18164161, 18646202, 20239913, 29743561, 32464433, 32915513, 42464514, 43502033, 45652314, 51755761, 53464314, 62198634, 69899754

Offset: 1

Jon E. Schoenfield, Jan 14 2022

nonn

It can be shown that if tau(k) + ... + tau(k+5) = 28, the sextuple (tau(k), tau(k+1), tau(k+2), tau(k+3), tau(k+4), tau(k+5)) must be one of the following, each of which might plausibly occur infinitely often:
  (2, 4, 4, 6, 4, 8), which first occurs at k = 5516281, 16831081, 18164161, ... (A208455);
  (2, 8, 4, 6, 4, 4), which first occurs at k = 2914913, 9018353, 17800553, ...;
  (4, 4, 6, 4, 8, 2), which first occurs at k = 18362, 6618242, 18646202, ...;
  (8, 4, 6, 4, 4, 2), which first occurs at k = 7224834, 9339114, 10780554, ...;
or one of the following, each of which occurs only once:
  (4, 6, 2, 8, 2, 6),  which occurs only at k = 27;
  (6, 2, 8, 2, 6, 4),  which occurs only at k = 28;
  (8, 2, 6, 4, 4, 4),  which occurs only at k = 30;
  (2, 4, 4, 8, 2, 8),  which occurs only at k = 37;
  (4, 4, 8, 2, 8, 2),  which occurs only at k = 38;
  (2, 8, 2, 6, 6, 4),  which occurs only at k = 41;
  (8, 2, 6, 6, 4, 2),  which occurs only at k = 42;
  (4, 4, 2, 12, 2, 4), which occurs only at k = 57.
The terms of the repeatedly occurring patterns form sequence A071368.
Tau(k) + ... + tau(k+5) >= 28 for all sufficiently large k; the only numbers k for which tau(k) + ... + tau(k+5) < 28 are 1..26, 29, 33, and 34.

			The table below lists each term k with a pattern (tau(k), ..., tau(k+5)) that appears only once (these appear at n = 1..8) as well as each term k that is the smallest one having a pattern that appears repeatedly for large k. (a(12)=6618242 is omitted from the table because it has the same pattern as a(9)=18362.)
Each of the repeatedly occurring patterns corresponds to one of the four possible orders in which the multipliers m=1..6 can appear among 6 consecutive integers of the form m*prime, and thus to a single residue of k modulo 2520; e.g., k=18362 begins a run of 6 consecutive integers having the form (2*p, 3*q, 4*r, 5*s, 6*t, 1*u), where p, q, r, s, t, and u are distinct primes > 6, and all such runs satisfy k == 722 (mod 2520).
For each of the patterns that does not occur repeatedly, one or more of the six consecutive integers in k..k+5 has no prime factor > 6; each such integer appears in parentheses in the "factorization" columns.
.
.                                             factorization as         k
                      # divisors of             m*(prime > 6)         mod
   n    a(n)=k    k  k+1 k+2 k+3 k+4 k+5    k  k+1 k+2 k+3 k+4 k+5   2520
   -  --------   --- --- --- --- --- ---   --- --- --- --- --- ---   ----
   1        27    4   6   2   8   2   6    (27) 4q   r (30)  t (32)    27
   2        28    6   2   8   2   6   4     4p   q (30)  s (32) 3u     28
   3        30    8   2   6   4   4   4    (30)  q (32) 3s  2t  5u     30
   4        37    2   4   4   8   2   8      p  2q  3r (40)  t  6u     37
   5        38    4   4   8   2   8   2     2p  3q (40)  s  6t   u     38
   6        41    2   8   2   6   6   4      p  6q   r  4s (45) 2u     41
   7        42    8   2   6   6   4   2     6p   q  4r (45) 2t   u     42
   8        57    4   4   2  12   2   4     3p  2q   r (60)  t  2u     57
   9     18362    4   4   6   4   8   2     2p  3q  4r  5s  6t   u    722
  10   2914913    2   8   4   6   4   4      p  6q  5r  4s  3t  2u   1793
  11   5516281    2   4   4   6   4   8      p  2q  3r  4s  5t  6u      1
  13   7224834    8   4   6   4   4   2     6p  5q  4r  3s  2t   u   2514

Jon E. Schoenfield, Table of n, a(n) for n = 1..10000

Cf. A000005, A071368, A208455.

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), A350699 (N=5), (this sequence) (N=6), A350773 (N=7), A350854 (N=8).

Position[Plus @@@ Partition[Array[DivisorSigma[0, #] &, 10^7], 6, 1], 28] // Flatten (* Amiram Eldar, Jan 16 2022 *)

from sympy import divisor_count as tau
taulist = [1, 2, 2, 3, 2, 4]
for k in range(1, 10000000):
    if sum(taulist) == 28: print(k, end=", ")
    taulist.append(tau(k+6))
    del taulist[0] # Karl-Heinz Hofmann, Jan 18 2022

{ k : Sum_{j=0..5} tau(k+j) = 28 }.

A350773 Numbers k such that tau(k) + ... + tau(k+6) = 32, where tau is the number of divisors function A000005.

Original entry on oeis.org

18, 26, 27, 28, 53, 73, 2914913, 5516281, 6618241, 9018353, 10780553, 18164161, 20239913, 45652313, 51755761, 62198633, 81235441, 91986833, 158764313, 175472641, 191010953, 197375753, 215206201, 322030801, 322461713, 362007353, 513284401, 668745001, 757892513

Offset: 1

Jon E. Schoenfield, Jan 15 2022

nonn

It can be shown that if tau(k) + ... + tau(k+6) = 32, the septuple (tau(k), tau(k+1), tau(k+2), tau(k+3), tau(k+4), tau(k+5), tau(k+6)) must be one of the following, each of which might plausibly occur infinitely often:
  (2, 4, 4, 6, 4, 8, 4), which first occurs at k = 5516281, 18164161, 51755761, ... (A207825);
  (2, 8, 4, 6, 4, 4, 4), which first occurs at k = 2914913, 9018353, 20239913, ...;
  (4, 4, 4, 6, 4, 8, 2), which first occurs at k = 6618241, 81235441, 215206201, ...;
  (4, 8, 4, 6, 4, 4, 2), which first occurs at k = 10780553, 45652313, 62198633, ...;
or one of the following, each of which occurs only once:
  (6, 2, 6, 4, 4, 2, 8), which occurs only at k = 18;
  (4, 4, 6, 2, 8, 2, 6), which occurs only at k = 26;
  (4, 6, 2, 8, 2, 6, 4), which occurs only at k = 27;
  (6, 2, 8, 2, 6, 4, 4), which occurs only at k = 28;
  (2, 8, 4, 8, 4, 4, 2), which occurs only at k = 53;
  (2, 4, 6, 6, 4, 8, 2), which occurs only at k = 73.
The terms of the repeatedly occurring patterns form sequence A071369.
Tau(k) + ... + tau(k+6) >= 32 for all sufficiently large k; the only numbers k for which tau(k) + ... + tau(k+6) < 32 are 1..17, 19..23, 25, 29, 31, 33, 37, and 41.

			The table below lists each term k with a pattern (tau(k), ..., tau(k+6)) that appears only once (these appear at n = 1..6) as well as each term k that is the smallest one having a pattern that appears repeatedly for large k. (a(10)=9018353 is omitted from the table because it has the same pattern as a(7)=2914913.)
Each of the repeatedly occurring patterns corresponds to one of the 4 possible orders in which the multipliers m=1..7 can appear among 7 consecutive integers of the form m*prime, and thus to a single residue of k modulo 2520; e.g., k=2914913 begins a run of 7 consecutive integers having the form (1*p, 6*q, 5*r, 4*s, 3*t, 2*u, 7*v), where p, q, r, s, t, u, and v are distinct primes > 7, and all such runs satisfy k == 1793 (mod 2520).
For each of the patterns that does not occur repeatedly, one or more of the seven consecutive integers in k..k+6 has no prime factor > 7; each such integer appears in parentheses in the columns under "factorization".
.
.                # divisors of          factorization
                  k+j for j =          as m*(prime > 7)
   n    a(n)=k   0 1 2 3 4 5 6    k  k+1 k+2 k+3 k+4 k+5 k+6   k mod 2520
   -  --------   - - - - - - -   --- --- --- --- --- --- ---   ----------
   1        18   6 2 6 4 4 2 8   (18)  q (20)(21) 2t   u (24)       18
   2        26   4 4 6 2 8 2 6    2p (27)(28)  s (30)  u (32)       26
   3        27   4 6 2 8 2 6 4   (27)(28)  r (30)  t (32) 3v        27
   4        28   6 2 8 2 6 4 4   (28)  q (30)  s (32) 3u  2v        28
   5        53   2 8 4 8 4 4 2     p (54) 5r (56) 3t  2u   v        53
   6        73   2 4 6 6 4 8 2     p  2q (75) 4s  7t  6u   v        73
   7   2914913   2 8 4 6 4 4 4     p  6q  5r  4s  3t  2u  7v      1793
   8   5516281   2 4 4 6 4 8 4     p  2q  3r  4s  5t  6u  7v         1
   9   6618241   4 4 4 6 4 8 2    7p  2q  3r  4s  5t  6u   v       721
  11  10780553   4 8 4 6 4 4 2    7p  6q  5r  4s  3t  2u   v      2513

Jon E. Schoenfield, Table of n, a(n) for n = 1..10000

Cf. A000005, A071369, A207825.

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), A350699 (N=5), A350769 (N=6), (this sequence) (N=7), A350854 (N=8).

Position[Plus @@@ Partition[Array[DivisorSigma[0, #] &, 10^7], 7, 1], 32] // Flatten (* Amiram Eldar, Jan 16 2022 *)

from sympy import divisor_count as tau
taulist = [1, 2, 2, 3, 2, 4, 2]
for k in range(2, 10000000):
    taulist.append(tau(k+6))
    del taulist[0]
    if sum(taulist) == 32: print(k, end=", ") # Karl-Heinz Hofmann, Jan 15 2022

{ k : Sum_{j=0..6} tau(k+j) = 32 }.

A350854 Numbers k such that tau(k) + ... + tau(k+7) = 40, where tau is the number of divisors function A000005.

Original entry on oeis.org

38, 39, 41, 51, 55, 67, 82, 10780552, 62198632, 884811061, 1457032501, 3573315892, 7321991041, 7391371681, 8557865812, 11434075381, 16893247141, 21599190901, 22487905441, 28044279892, 28273111012, 37923188932, 50238568801, 59635316161, 77814456292, 86148922852

Offset: 1

Jon E. Schoenfield, Jan 19 2022

nonn

It can be shown that if tau(k) + ... + tau(k+7) = 40, the octuple (tau(k), tau(k+1), tau(k+2), tau(k+3), tau(k+4), tau(k+5), tau(k+6), tau(k+7)) must be one of the following, each of which might plausibly occur infinitely often:
  (2, 4, 4, 6, 4, 8, 4, 8), which first occurs at k = 7321991041, 7391371681, 22487905441, ...;
  (2, 4, 4, 8, 4, 8, 4, 6), which first occurs at k = 884811061, 1457032501, 11434075381, ...;
  (6, 4, 8, 4, 8, 4, 4, 2), which first occurs at k = 3573315892, 8557865812, 28044279892, ...;
  (8, 4, 8, 4, 6, 4, 4, 2), which first occurs at k = 10780552, 62198632, 139738178152, ...;
or one of the following, each of which occurs only once:
  (4,  4,  8,  2,  8,  2,  6,  6), which occurs only at k = 38;
  (4,  8,  2,  8,  2,  6,  6,  4), which occurs only at k = 39;
  (2,  8,  2,  6,  6,  4,  2, 10), which occurs only at k = 41;
  (4,  6,  2,  8,  4,  8,  4,  4), which occurs only at k = 51;
  (4,  8,  4,  4,  2, 12,  2,  4), which occurs only at k = 55;
  (2,  6,  4,  8,  2, 12,  2,  4), which occurs only at k = 67;
  (4,  2, 12,  4,  4,  4,  8,  2), which occurs only at k = 82.
The terms of the repeatedly occurring patterns form sequence A071370.
Tau(k) + ... + tau(k+7) >= 40 for all sufficiently large k; the only numbers k for which tau(k) + ... + tau(k+7) < 40 are 1..34, 36, 37, 40, 43, 46, 52, and 61.

			The table below lists each term k that is the smallest one having a pattern (tau(k), ..., tau(k+7)) that appears repeatedly for large k. Each such pattern corresponds to one of the 4 possible orders in which the multipliers m=1..8 can appear among 8 consecutive integers of the form m*prime, and thus to a single residue of k modulo 2520; e.g., k=884811061 begins a run of 8 consecutive integers having the form (p, 2*q, 3*r, 8*s, 5*t, 6*u, 7*v, 4*w), where p, q, r, s, t, u, v, and w are distinct primes > 8, and all such runs satisfy k == 1261 (mod 2520).
.
.                   # divisors of    factorization of k+j as
                     k+j for j =      m*(prime > 8) for j =
   n      a(n)=k   0 1 2 3 4 5 6 7    0  1  2  3  4  5  6  7   k mod 2520
   -  ----------   - - - - - - - -   -- -- -- -- -- -- -- --   ----------
   8    10780552   8 4 8 4 6 4 4 2   8p 7q 6r 5s 4t 3u 2v  w      2512
  10   884811061   2 4 4 8 4 8 4 6    p 2q 3r 8s 5t 6u 7v 4w      1261
  12  3573315892   6 4 8 4 8 4 4 2   4p 7q 6r 5s 8t 3u 2v  w      1252
  13  7321991041   2 4 4 6 4 8 4 8    p 2q 3r 4s 5t 6u 7v 8w         1

Cf. A000005, A071370.

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), A350699 (N=5), A350769 (N=6), A350773 (N=7), (this sequence) (N=8).

Position[Plus @@@ Partition[Array[DivisorSigma[0, #] &, 100], 8, 1], 40] // Flatten (* Amiram Eldar, Jan 19 2022 *)

from sympy import divisor_count as tau
taulist = [1, 2, 2, 3, 2, 4, 2, 4]
for k in range(1, 10000000):
    if sum(taulist) == 40: print(k, end=", ")
    taulist.append(tau(k+8))
    del taulist[0] # Karl-Heinz Hofmann, Jan 21 2022

{ k : Sum_{j=0..7} tau(k+j) = 40 }.

cp's OEIS Frontend

A350675 Numbers k such that tau(k) + tau(k+1) + tau(k+2) = 10, where tau is the number of divisors function A000005.

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A350686 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.

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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.

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A350769 Numbers k such that tau(k) + ... + tau(k+5) = 28, where tau is the number of divisors function A000005.

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A350773 Numbers k such that tau(k) + ... + tau(k+6) = 32, where tau is the number of divisors function A000005.

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A350854 Numbers k such that tau(k) + ... + tau(k+7) = 40, where tau is the number of divisors function A000005.

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