A319037 -id:A319037 - cp's OEIS frontend

A081978 Smallest triangular number with exactly n divisors, or 0 if no such number exists.

1, 3, 0, 6, 0, 28, 0, 66, 36, 496, 0, 276, 0, 8128, 1631432881, 120, 0, 300, 0, 528, 0, 38009927549623740385753, 0, 630, 0, 33550336, 2172602007770041, 8256, 0, 209628, 0, 3570, 0, 8589869056, 0, 2016, 0, 137438691328, 0, 3240, 0, 662976, 0, 2096128, 41616

Offset: 1

Amarnath Murthy, Apr 03 2003

a(p)=0 if p is an odd prime. If n is an odd composite number, then a(n) is a square; see A001110 for numbers that are both triangular and square. - Victoria A Sapko (vsapko(AT)frc.mass.edu), Sep 28 2007
From Jon E. Schoenfield, May 25 2014: (Start)
If n is an odd semiprime, then a triangular number t having exactly n divisors must be of the form t = p^(2r) * q^(2s) = (p^r * q^s)^2, where p and q are distinct primes (p < q) and r and s are positive integers such that (2r+1)*(2s+1) = n.
If t is the k-th triangular number k(k+1)/2, it can be factored as t = u * v where
     u = k    and   v = (k+1)/2   if k is odd, or
     u = k/2  and   v = k+1       if k is even.
Since neither p nor q (each of which is greater than 1) can divide both k and (k+1)/2, or both k/2 and k+1, only four cases need to be considered:
   Case 1:  k is even, q^(2s) = k/2, p^(2r) = k+1
   Case 2:  k is even, p^(2r) = k/2, q^(2s) = k+1
   Case 3:  k is odd,  q^(2s) = k,   p^(2r) = (k+1)/2
   Case 4:  k is odd,  p^(2r) = k,   q^(2s) = (k+1)/2
These yield the following equations:
   Case 1:  2 * q^(2s) + 1 = p^(2r)
   Case 2:  2 * p^(2r) + 1 = q^(2s)
   Case 3:  2 * p^(2r) - 1 = q^(2s)
   Case 4:  2 * q^(2s) - 1 = p^(2r)
Case 1 can be ruled out: since q > p, q is odd, so 2 * q^(2s) + 1 == 3 (mod 16), but p^(2r) cannot be 3 (mod 16).
For a Case 2 solution, since q is odd, 2 * p^(2r) + 1 = q^(2s) == 1 (mod 8), so p^(2r) == 0 (mod 4), so p must be even. Therefore, p = 2, and we must satisfy the equation 2 * 2^(2r) + 1 = q^(2s), whose left-hand side, which is divisible by 3 for every nonnegative integer r, is thus the square of a prime iff it is 3^2. So r=1, q=3, and s=1, yielding t = 2^2 * 3^2 = 36, which is the smallest triangular number with exactly 9 divisors, so a(9)=36.
In Case 3, both p and q must be odd, and p^r must be a number w having the property that 2*w^2 - 1 is square (i.e., (q^s)^2); every such number w is in A001653 (1, 5, 29, 169, 985, ...), and the corresponding value of q^s = sqrt(2*w^2 - 1) is in A002315 (1, 7, 41, 239, 1393, ...). (Note that A001653 and A002315 are defined with offsets of 1 and 0, respectively, so A001653(j) corresponds to A002315(j-1).) However, for odd semiprime n > 9, we need r > 1 and/or s > 1. The only nontrivial power (i.e., number of the form x^m, where both x and m are integers greater than 2) in A001653 is A001653(4) = 169 = 13^2 [Pethö], so the only Case 3 solution with r > 1 is 2 * 13^4 - 1 = 239^2, which yields the 15-divisor triangular number 13^4 * 239^2 = 1631432881 = a(15). A Case 3 solution with r = 1 and s > 1 would require 2 * p^2 - 1 = q^(2s), which is impossible since p < q.
Finally, in Case 4, both p and q must be odd, q^s must be in A001653, and p^r must be the corresponding term in A002315. However, using the only nontrivial power in A001653 (i.e., 169 = 13^2) as q^s would not yield a valid solution here because it would mean p = 239 and q = 13 (contradicting p < q). Thus, if a Case 4 solution exists for odd semiprime n > 9, we must have s = 1 and r > 1, so n = (2r+1)*(2s+1) = (2r+1)*3, where 2r+1 is prime. Such a solution requires an index j satisfying two conditions: (1) A001653(j) = q^1 = q is prime, and (2) the corresponding term A002315(j-1) = p^r is a nontrivial prime power. There are no nontrivial powers (whether of primes or composites) among the terms in A002315 below 10^5000. Moreover, the terms in A001653 are the odd-indexed terms from A000129 (Pell numbers), so condition (1) requires that j satisfy A000129(2j-1) = q. A096650 lists the indices of all prime or probable prime Pell numbers up to 100000. A check of the value A002315(j-1) corresponding to each prime or probable prime among the odd-indexed Pell numbers A000129(2j-1) up to j=50000 determined that none were nontrivial powers, so if any Case 4 solution with n > 9 exists, it will yield a triangular number t = p^(2r) * q^2 = (2 * q^2 - 1) * q^2, where q >= A000129(100001) = 3.16...*10^38277, so t > 10^153110. Since there are no nontrivial powers at all in A002315 below 10^5000, and since prime Pell numbers above A000129(50000) seem so scarce, it seems extremely unlikely that any such solution exists.
Thus, a(n) = 0 for every odd semiprime n that is not divisible by 3, and assuming that no Case 4 solution for odd semiprime n > 9 exists, the only nonzero a(n) where n is an odd semiprime greater than 9 is a(15) = 13^4 * 239^2 = 1631432881.
If j is prime and n=2j, then a(n) (if nonzero) must be of the form p^r * q, where p and q are distinct primes, r = j-1, and q is one of 3 functions of p^r:
    q = f1(p^r) = 2*p^r - 1
    q = f2(p^r) = 2*p^r + 1
    q = f3(p^r) = (p^r - 1)/2
Of these, q = f1(p^r) for all but two cases among n < 1000:
    at n=362, q = f2(p^r), with p=3;
    at n=514, q = f3(p^r), with p=331.
Conjecture: a(2j) > 0 for all j > 1. (This conjecture holds at least through n = 2j = 1000. The largest a(n) for even n <= 1000 is a(898) = 20599^448 * (2 * 20599^448 - 1) = 3.21...*10^3865.) (End)
For more known terms, and information about unknown terms, see Links. - Jon E. Schoenfield, May 26 2014
If d(k*(k+1)/2) = 21, note that 2*q^2 = p^6 + 1 = (p^2 + 1)*(p^4 - p^2 + 1) has no prime solutions, so then k = p^2 and k+1 = 2*q^6, where p and q are distinct primes. We can prove 2*x^3 - y^2 = 1 has only one positive solution (1, 1) which shows that p^2 + 1 = 2*q^6 has no prime solutions. In the ring of Gaussian integers, x^3 = (1+y*i)*(1-y*i)/((1+i)*(1-i)) and (1+y*i)/(1+i) is coprime to (1-y*i)/(1-i), thus (1+y*i)/(1+i) = (u+v*i)^3 and (1-y*i)/(1-i) = (u-v*i)^3 for some integers u and v. Note that 1+y*i = (1+i)*(u+v*i)^3 = (u+v)*(u^2+v^2-4*u*v) + (u-v)*(u^2+v^2+4*u*v)*i, we have (u+v)*(u^2+v^2-4*u*v) = 1. Therefore, u = 1 and v = 0 if u > v, which means y = (u-v)*(u^2+v^2+4*u*v) = 1. This implies that a(21) = 0. - Jinyuan Wang, Aug 22 2020
a(n) is a perfect number for all n such that n/2 is in A000043. - J. Lowell, Mar 16 2024

			a(2)=3 because the smallest triangular number with 2 divisors is T(2)=3.

Attila Pethö, The Pell sequence contains only trivial perfect powers, Kossuth Lajos University, Department of Computer Science, H-4010 Debrecen, P.O. Box 12, Hungary, Dedicated to V.T. Sós and A. Hajnal, May 18 2008.
Project Euler, Highly divisible triangular number, Problem 12.
Jon E. Schoenfield, Table of n, a(n) for known terms in n = 1..100, with notes on unknown terms

Greedy inverse of A063440.

Cf. A000129, A001110, A001653, A002315, A081979, A242585, A319037.

More terms from Victoria A Sapko (vsapko(AT)frc.mass.edu), Sep 28 2007
a(14) corrected and a(20) added by Jon E. Schoenfield, May 11 2014
a(21)-a(24) from Jinyuan Wang, Aug 22 2020
a(25)-a(45) from Jon E. Schoenfield, Jan 28 2021

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.

Original entry on oeis.org

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

Jon E. Schoenfield, Dec 08 2018

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, ...).

			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

Cf. A000005, A000217, A081978, A276542, A319035, A319036, A319037.

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]

A323743 Table read by rows: row n lists the numbers k for which there exist only finitely many runs of n consecutive integers whose number-of-divisors function sums to k.

Original entry on oeis.org

1, 3, 4, 5, 5, 7, 8, 9, 8, 9, 11, 12, 13, 14, 15, 10, 13, 15, 17, 18, 19, 14, 15, 17, 18, 19, 21, 22, 23, 24, 25, 26, 27, 16, 19, 20, 21, 22, 23, 25, 26, 27, 29, 30, 31, 20, 22, 24, 27, 29, 30, 31, 33, 34, 35, 36, 37, 38, 39

Offset: 1

Jon E. Schoenfield, Apr 02 2019

Row n lists the numbers k such that
  0 < |{m : Sum_j={m..m+n-1} tau(j) = k}| < infinity
where tau(j) = A000005(j) is the number of divisors of j.

			There is only one number with exactly 1 divisor (namely, k=1), but there are infinitely many numbers with j divisors for every j >= 2, so row 1 consists only of the single term 1.
The sequence of values tau(k) for k >= 1 is A000005, which begins 1, 2, 2, 3, 2, 4, 2, 4, 3, 4, ..., from which the sums of two consecutive terms are 1+2=3, 2+2=4, 2+3=5, 3+2=5, 2+4=6, 4+2=6, 2+4=6, 4+3=7, 3+4=7, ...; no number j < 3 appears as such a sum, every j >= 6 appears infinitely many times as such a sum, and each j in {3,4,5} appears as such a sum only finitely many times, so row 2 is {3, 4, 5}.
Row 3 does not contain 6 as a term because there exists no run of 3 consecutive numbers whose sum of tau values is exactly 6.
The first six rows of the table are as follows:
  row 1: {1};
  row 2: {3, 4, 5};
  row 3: {5, 7, 8, 9};
  row 4: {8, 9, 11, 12, 13, 14, 15};
  row 5: {10, 13, 15, 17, 18, 19};
  row 6: {14, 15, 17, 18, 19, 21, 22, 23, 24, 25, 26, 27}.

Cf. A000005, A005237, A006558, A048892, A072507, A100366, A119479, A141621, A284596, A284597, A292580, A319037, A319045, A319046.

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A081978 Smallest triangular number with exactly n divisors, or 0 if no such number exists.

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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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A323743 Table read by rows: row n lists the numbers k for which there exist only finitely many runs of n consecutive integers whose number-of-divisors function sums to k.

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A375640 The smallest triangular number that begins a run of at least n consecutive triangular numbers with the same number of divisors.

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