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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.

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%I A319038 #24 Feb 02 2021 22:52:21
%S A319038 1,3,55,0,0,6,28,153,105,66,406,36,496,276,3916,11175,8128,1631432881,
%T A319038 120,1770,17205,106030,457062495,240119995521,300,2556,528,327645,
%U A319038 12607731,38009927549623740385753,630,1540,29646,181503,3181503,18542914232391,584426575442663305723408463937454358857690
%N 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.
%C A319038 The number of terms in the n-th row is A319037(n). Row n has no terms here iff A081978(n) = 0 (i.e., there is no triangular number with n divisors; this is the case for n = 3, 5, 7, 11, 13, 17, 19, 21, 23, 25, 29, 31, 33, 35, 37, 39, 41, 43, 47, 49, 51, 53, 55, 57, 59, 61, ...).
%e A319038 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.)
%e A319038 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
%e A319038 .
%e A319038     1   2   3   4   5   6   7   8   9  10  11  12  13  14  15
%e A319038    /\  /\  /\  /\  /\  /\  /\  /\  /\  /\  /\  /\  /\  /\  /\
%e A319038   /  \/  \/  \/  \/  \/  \/  \/  \/  \/  \/  \/  \/  \/  \/
%e A319038   1   3   5   7   9  11  13  15  17  19  21  23  25  27  29
%e A319038 .
%e A319038 then each triangular number is the product of two connected factors f1 and f2 as follows:
%e A319038 .
%e A319038 f1:      1         2         3         4         5         6
%e A319038         /\        /\        /\        /\        /\        /\
%e A319038        /  \      /  \      /  \      /  \      /  \      /  \
%e A319038       1    3    6   10   15   21   28   36   45   55   66
%e A319038      /      \  /      \  /      \  /      \  /      \  /
%e A319038     /        \/        \/        \/        \/        \/
%e A319038 f2: 1         3         5         7         9        11
%e A319038 .
%e A319038 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
%e A319038 .
%e A319038 f1:       1           2           3           4           5
%e A319038          (1)         (2)         (2)         (3)         (2)
%e A319038          /\          /\          /\          /\          /\
%e A319038         /  \        /  \        /  \        /  \        /  \
%e A319038        1    3      6   10     15   21     28   36     45
%e A319038      (1)    (2)  (4)    (4)  (4)    (4)  (6)    (9)  (6)
%e A319038      /        \  /        \  /        \  /        \  /
%e A319038     /          \/          \/          \/          \/
%e A319038 f2: 1           3           5           7           9
%e A319038    (1)         (2)         (2)         (2)         (3)
%e A319038 .
%e A319038 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:
%e A319038 .
%e A319038      74                75                76
%e A319038 (2*37: tau=4)    (3*5^2: tau=6)   (2^2*19: tau=6)
%e A319038       .                /\                /.
%e A319038        .              /  \              /  .
%e A319038         .            /    \            /    .
%e A319038          .       11175=   11325=   11476=    .
%e A319038        (tau=8)   75*149   75*151   76*151  (tau=36)
%e A319038            .     tau=12   tau=12   tau=12      .
%e A319038             .    /            \    /            .
%e A319038              .  /              \  /              .
%e A319038               ./                \/                .
%e A319038              149               151               153
%e A319038         (prime: tau=2)    (prime: tau=2)   (3^2*17: tau=6)
%e A319038 .
%e A319038 This first run of exactly three consecutive triangular numbers with 12 divisors begins with 11175, so T(12,3) = 11175.
%e A319038 The table begins as follows:
%e A319038 .
%e A319038   row  terms
%e A319038   ---  --------------------------------------------------
%e A319038     1  1;
%e A319038     2  3;
%e A319038     3  (no terms)
%e A319038     4  55, 0, 0, 6;
%e A319038     5  (no terms)
%e A319038     6  28, 153;
%e A319038     7  (no terms)
%e A319038     8  105, 66, 406;
%e A319038     9  36;
%e A319038    10  496;
%e A319038    11  (no terms)
%e A319038    12  276, 3916, 11175;
%e A319038    13  (no terms)
%e A319038    14  8128;
%e A319038    15  1631432881;
%e A319038    16  120, 1770, 17205, 106030, 457062495, 240119995521;
%e A319038    17  (no terms)
%e A319038    18  300, 2556;
%e A319038    19  (no terms)
%e A319038    20  528, 327645, 12607731;
%e A319038    21  (no terms)
%e A319038    22  38009927549623740385753;
%e A319038    23  (no terms)
%e A319038    24  630, 1540, 29646, 181503, 3181503, 18542914232391, 584426575442663305723408463937454358857690
%Y A319038 Cf. A000005, A000217, A081978, A276542, A319035, A319036, A319037.
%K A319038 nonn,tabf,hard
%O A319038 1,2
%A A319038 _Jon E. Schoenfield_, Dec 08 2018
%E A319038 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]