OFFSET
1,1
COMMENTS
Equivalently, T(n,k) is the smallest m such that tau(m+j) = 2*(n+j) for each j in 0..k-1 but not for j = -1 or j = k, where tau is the number-of-divisors function, A000005.
Row 1 is A294528.
No row can have more than 15 terms: if some row n were to have 16 (or more) terms, then T(n,16) would begin a run of 16 consecutive integers, two of which (let's call them x and y, with x < y) would be 4 times an odd number, and thus x and y would each have a number of divisors that is a multiple of 3, but this is impossible since y - x = 8 so tau(y) - tau(x) = 16.
Row 3 has no more than 11 terms.
T(3,7) <= 9392318479793089373.
T(3,8) <= 42121593968048969225.
T(3,9) and T(3,10) seem likely to exist.
T(3,11) cannot exist unless there is a solution to 3*p^4 + 5 = 8*q^4 with p and q prime; if T(3,11) exists, then it exceeds 10^300000.
FORMULA
T(n,k) = min_{j : A000005(m+j) = 2*(n+j) for j = 0..k-1 but not for j = -1 or j = k}.
EXAMPLE
T(2,3) is the smallest m such that tau(m+j) = 2*(2+j) for each j in 0..2 but not for j = -1 or j = 3; i.e., tau(m) = 4, tau(m+1) = 6, and tau(m+2) = 8, but tau(m-1) != 2 and tau(m+3) != 10. The smallest such m is 187:
m = 187 = 11*17 (which has 4 divisors),
m+1 = 188 = 2^2*47 (which has 6 divisors), and
m+2 = 189 = 3^3*7 (which has 8 divisors), but
m-1 = 186 = 2*3*31 (which has 8 divisors, not 2), and
m+3 = 190 = 2*5*19 (which has 8 divisors, not 10).
The first several rows of the table are as follows:
Row n=1: 2, 5, 61, 421, 1524085621; (A294528)
Row n=2: 10, 27, 187, 2545622;
Row n=3: 12, 153, 35557, 363121, 223577456873, 44753756873, ...
Row n=4: 24, 890, 1615, 795056874, 718511874, ...
Row n=5: 48, 1424, 84281875, 1578123, ...
Row n=6: 60, 1215, 53216, ...
Row n=7: 192, 2624, ...
Row n=8: 120, 6699, 31310, ...
Row n=9: 180, 16928, ...
...
CROSSREFS
KEYWORD
nonn,tabf,more
AUTHOR
Jon E. Schoenfield, May 29 2023
STATUS
approved