A319038 Table read by rows: T(n,k) is the smallest triangular number that begins a run of exactly k consecutive triangular numbers with n divisors, or 0 if no such run exists.
1, 3, 55, 0, 0, 6, 28, 153, 105, 66, 406, 36, 496, 276, 3916, 11175, 8128, 1631432881, 120, 1770, 17205, 106030, 457062495, 240119995521, 300, 2556, 528, 327645, 12607731, 38009927549623740385753, 630, 1540, 29646, 181503, 3181503, 18542914232391, 584426575442663305723408463937454358857690
Offset: 1
Examples
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
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]
Comments