A319035
Triangular numbers T(k) that have the same number of divisors as their successors T(k+1).
Original entry on oeis.org
6, 10, 15, 66, 153, 406, 435, 561, 861, 903, 1378, 1540, 1770, 2211, 2346, 2556, 2926, 3655, 3916, 4186, 4371, 5151, 5778, 6555, 7626, 9453, 10011, 10296, 11175, 11325, 12720, 14535, 14878, 16110, 16836, 17205, 17391, 17766, 18336, 19306, 19503, 20301, 20706
Offset: 1
T(2) = 6 is a term because 6 = 2 * 3 has 4 divisors (1, 2, 3, 6) and T(3) = 10 = 2 * 5 also has 4 divisors (1, 2, 5, 10).
T(17) = 153 is a term because 153 = 3^2 * 17 has 6 divisors (1, 3, 9, 17, 51, 153) and T(18) = 171 = 3^2 * 19 also has 6 divisors (1, 3, 9, 19, 57, 171).
Cf.
A276542 (indices of these triangular numbers).
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T:=List([1..210],n->n*(n+1)/2);; a:=List(Filtered([1..Length(T)-1],i->Tau(T[i])=Tau(T[i+1])),k->T[k]); # Muniru A Asiru, Dec 06 2018
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t[n_] := n(n+1)/2; aQ[n_] := DivisorSigma[0, t[n]] == DivisorSigma[0, t[n+1]]; t[Select[Range[100], aQ]] (* Amiram Eldar, Dec 06 2018 *)
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lista(nn) = {for (n=1, nn, if (numdiv(t=n*(n+1)/2) == numdiv((n+1)*(n+2)/2), print1(t, ", ")););} \\ Michel Marcus, Dec 06 2018
A319037
a(n) is the length of the longest run of consecutive triangular numbers with n divisors.
Original entry on oeis.org
1, 1, 0, 4, 0, 2, 0, 3, 1, 1, 0, 3, 0, 1, 1, 6, 0, 2, 0, 3, 0, 1, 0, 7, 0, 1, 1, 3, 0, 2, 0, 13, 0, 1, 0, 6, 0, 1, 0, 6, 0, 2, 0, 3, 1, 1, 0
Offset: 1
a(1) = 1 because there is only one triangular number T(k) = k*(k+1)/2 with 1 divisor: T(1) = 1.
a(2) = 1 because there is only one triangular number with 2 divisors: T(2) = 3, the only prime triangular number.
a(3) = 0 because there is no triangular number with 3 divisors.
a(4) = 4 because {6, 10, 15, 21} is the longest run of consecutive triangular numbers with 4 divisors.
a(16) = 6 because T(692993)..T(692998) is a run of 6 consecutive triangular numbers with 16 divisors, and no longer run of such triangular numbers exists.
a(24) = 7 because T(1081135121474335700644)..T(1081135121474335700650) is a run of 7 consecutive triangular numbers with 24 divisors, and no longer run of such triangular numbers exists. - _Jinyuan Wang_, Aug 23 2020
T(1081135121474335700644) is the smallest triangular number that begins a run of 7 consecutive triangular numbers with 24 divisors. - _Jon E. Schoenfield_, May 29 2022
A276553
Numbers n such that n^2 and (n + 1)^2 have the same number of divisors.
Original entry on oeis.org
2, 14, 15, 21, 33, 34, 38, 44, 57, 75, 81, 85, 86, 93, 94, 98, 116, 118, 122, 133, 135, 141, 142, 145, 147, 158, 171, 177, 201, 202, 205, 213, 214, 217, 218, 230, 244, 253, 272, 285, 296, 298, 301, 302, 326, 332, 334, 375, 381, 387, 393, 394, 405, 429, 434, 445
Offset: 1
We see that 14^2 = 196, the divisors of which are 1, 2, 4, 7, 14, 28, 49, 98, 196, and there are nine of them. And we see that 15^2 = 225, the divisors of which are 1, 3, 5, 9, 15, 25, 45, 75, 225, and there are nine of them. Both 14^2 and 15^2 have the same number of divisors, hence 14 is in the sequence.
And we see that 16^2 = 256, the divisors of which are the powers of 2 from 2^0 to 2^8, that's nine divisors. Both 15^2 and 16^2 have the same number of divisors, hence 15 is also in the sequence.
But 16 is not in the sequence, since 17 is prime and 17^2 consequently only has three divisors.
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N:= 1000: # to get all terms <= N
T:= map(t -> numtheory:-tau(t^2), [$1..N+1]):
select(t -> T[t]=T[t+1], [$1..N]); # Robert Israel, Apr 10 2017
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Select[Range[1000], DivisorSigma[0, #^2] == DivisorSigma[0, (# + 1)^2] &]
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k=[]; for(n=1, 1000, a=numdiv(n^2); b=numdiv((n+1)^2); if(a==b, k=concat(k, n))); k
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from sympy.ntheory import divisor_count
print([n for n in range(1, 501) if divisor_count(n**2) == divisor_count((n + 1)**2)]) # Indranil Ghosh, Apr 10 2017
(Scheme, with Antti Karttunen's IntSeq-library) (define A276553 (ZERO-POS 1 1 A284570)) ;; Antti Karttunen, Apr 15 2017
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
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
Showing 1-4 of 4 results.
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