OFFSET
1,2
COMMENTS
EXAMPLE
Row 4 has A319037(4) = 4 terms: 55, 0, 0, 6. T(4,4) = 6 because 6 is the smallest triangular number that begins a run of exactly 4 consecutive triangular numbers with 4 divisors; T(4,1) = 55 because T(10)=55 is the smallest triangular number whose predecessor T(9)=45 and successor T(11)=66 each have a number of divisors other than 4 (so 55 constitutes a "run" of only a single triangular number); and T(4,2) = T(4,3) = 0 because no triangular numbers with 4 divisors are in runs of exactly 2 or 3 successive triangular number with 4 divisors. (In other words, every triangular number with 4 divisors that is not in the run {6, 10, 15, 21} is isolated.)
Every triangular number can be represented as the product of an integer m and one of the two odd integers 2m-1 and 2m+1; graphically, if we represent the integers m and the odd integers 2m-1 and 2m+1 in two rows, with each m connected to 2m-1 and 2m+1 by diagonal lines as
.
1 2 3 4 5 6 7 8 9 10 11 12 13 14 15
/\ /\ /\ /\ /\ /\ /\ /\ /\ /\ /\ /\ /\ /\ /\
/ \/ \/ \/ \/ \/ \/ \/ \/ \/ \/ \/ \/ \/ \/
1 3 5 7 9 11 13 15 17 19 21 23 25 27 29
.
then each triangular number is the product of two connected factors f1 and f2 as follows:
.
f1: 1 2 3 4 5 6
/\ /\ /\ /\ /\ /\
/ \ / \ / \ / \ / \ / \
1 3 6 10 15 21 28 36 45 55 66
/ \ / \ / \ / \ / \ /
/ \/ \/ \/ \/ \/
f2: 1 3 5 7 9 11
.
Writing tau() as the number-of-divisors function, since gcd(m, 2m-1) = gcd(m, 2m+1) = 1, we have tau(f1*f2) = tau(f1)*tau(f2) for each triangular number f1*f2. Showing the number of divisors of each factor f1 and f2 and each triangular number in parentheses, we have
.
f1: 1 2 3 4 5
(1) (2) (2) (3) (2)
/\ /\ /\ /\ /\
/ \ / \ / \ / \ / \
1 3 6 10 15 21 28 36 45
(1) (2) (4) (4) (4) (4) (6) (9) (6)
/ \ / \ / \ / \ /
/ \/ \/ \/ \/
f2: 1 3 5 7 9
(1) (2) (2) (2) (3)
.
A run of consecutive triangular numbers T with a constant tau(T) thus requires a constant tau(f1)=tau1 and a constant tau(f2)=tau2 for all f1 and all f2 that appear as factors in the run. Thus, e.g., a run of 3 consecutive triangular numbers with 12 divisors requires 3 successive connections, hence 4 factors consisting of 2 consecutive integers f1, with tau=tau1, overlapping in the above graphic representation with 2 consecutive odd numbers f2, with tau=tau2, such that tau1*tau2=12. The divisors of 12 are {1, 2, 3, 4, 6, 12}. Neither tau1 nor tau2 can be 1 (only one integer, 1, has tau=1), so neither can be 12/1=12. Neither tau1 nor tau2 can be 3 (every number with 3 divisors is the square of a prime, and no two consecutive integers are squares of primes, nor are any two consecutive odd numbers), so neither can be 12/3=4. Thus tau1 and tau2 must be 2 and 6, in some order. Since tau=2 only for primes, and the only two consecutive integers f1 that are prime are 2 and 3, and the f2 to which they would both connect is 5 (which does not have 6 divisors), tau1 cannot be 2; thus tau1=6, so tau2=2. The two consecutive odd numbers f2 are twin primes, and are not (3,5), so their average is a multiple of 6, so the integer f1 connected to them, being half of that average, is a multiple of 3, and since it has 6 divisors, it must be 3^5 = 243 or a number of the form 3^2*p or 3*p^2 where p is a prime other than 3. As it turns out, the smallest such f1 yielding a run of three consecutive triangular numbers with 12 divisors is 3*5^2 = 75:
.
74 75 76
(2*37: tau=4) (3*5^2: tau=6) (2^2*19: tau=6)
. /\ /.
. / \ / .
. / \ / .
. 11175= 11325= 11476= .
(tau=8) 75*149 75*151 76*151 (tau=36)
. tau=12 tau=12 tau=12 .
. / \ / .
. / \ / .
./ \/ .
149 151 153
(prime: tau=2) (prime: tau=2) (3^2*17: tau=6)
.
This first run of exactly three consecutive triangular numbers with 12 divisors begins with 11175, so T(12,3) = 11175.
The table begins as follows:
.
row terms
--- --------------------------------------------------
1 1;
2 3;
3 (no terms)
4 55, 0, 0, 6;
5 (no terms)
6 28, 153;
7 (no terms)
8 105, 66, 406;
9 36;
10 496;
11 (no terms)
12 276, 3916, 11175;
13 (no terms)
14 8128;
15 1631432881;
16 120, 1770, 17205, 106030, 457062495, 240119995521;
17 (no terms)
18 300, 2556;
19 (no terms)
20 528, 327645, 12607731;
21 (no terms)
22 38009927549623740385753;
23 (no terms)
24 630, 1540, 29646, 181503, 3181503, 18542914232391, 584426575442663305723408463937454358857690
CROSSREFS
KEYWORD
nonn,tabf,hard
AUTHOR
Jon E. Schoenfield, Dec 08 2018
EXTENSIONS
T(22,1), T(24,1)-T(24,7) added to Data, and Comments updated, by Jon E. Schoenfield, Jan 29 2021 [T(24,7) from Jinyuan Wang's Example section entry at A319037]
STATUS
approved