A119479 -id:A119479 - cp's OEIS frontend

A006558 Start of first run of n consecutive integers with same number of divisors.

1, 2, 33, 242, 11605, 28374, 171893, 1043710445721, 2197379769820, 2642166652554075

Offset: 1

The entry 40311 given by Guy and by Wells is incorrect. - Jud McCranie, Jan 20 2002
a(10) <= 2642166652554075, a(11) <= 17707503256664346, a(12) <= 9827470582657267545. - David Wasserman, Feb 22 2008
a(10) > 10^13. - Giovanni Resta, Jul 13 2015
a(12) <= 3842083249515874843. - Hugo van der Sanden, Sep 20 2022
a(13) <= 34169215324203592637988571. - Hugo van der Sanden, Apr 13 2022
a(14) <= 9721439902882994590514319997146. - Hugo van der Sanden, Jun 14 2022
a(15) <= 80215613469168729088982885848674841. - Natalia Makarova, Sep 18 2022
a(16) <= 37981337212463143311694743672867136611416. - Vladimir Letsko, Mar 17 2017
a(17) <= 768369049267672356024049141254832375543516. - Vladimir Letsko, Sep 12 2017
a(18) <= 488900003598703704335810037459507226590256411. - Vladimir Letsko, Jun 03 2022
a(19) <= 5908388043825578351730345292813071711296723319324. - Vladimir Letsko, Apr 09 2022
a(20) <= 17668887847524548413038893976018715843277693308027547. Vladimir Letsko, May 30 2022
Spătaru proves that the longest such run up to N is at most exp(C*sqrt(log N log log N)) for some constant C, hence a(n) >> exp(exp(W((log^2 n)/C))) which is approximately exp(log^2 n/(2 log log n)). - Charles R Greathouse IV, Feb 06 2023

			33 has four divisors (1, 3, 11, and 33), 34 has four divisors (1, 2, 17, and 34), 35 has four divisors (1, 5, 7, and 35).  These are the first three consecutive numbers with the same number of divisors, so a(3)=33.

J.-M. De Koninck, Ces nombres qui nous fascinent, Entry 33, pp 12, Ellipses, Paris 2008.
R. K. Guy, Unsolved Problems in Number Theory, section B18.
N. J. A. Sloane and Simon Plouffe, The Encyclopedia of Integer Sequences, Academic Press, 1995 (includes this sequence).
James J. Tattersall, Elementary Number Theory in Nine Chapters, Cambridge University Press, 1999, page 87.
D. Wells, The Penguin Dictionary of Curious and Interesting Numbers, Penguin Books, NY, 1986, pages 147 and 176.

Pentti Haukkanen, Some computational results concerning the divisor functions d(n) and sigma(n), The Mathematics Student, Vol. 62 Nos. 1-4 (1993) pp. 166-168. See p. 167.
Vladimir A. Letsko, Some new results on consecutive equidivisible integers, arXiv:1510.07081 [math.NT], 2015.
Vladimir A. Letsko and Vasilii Dziubenko On consecutive equidivisible integers (in Russian), Boundaries of knowledge, 2 (45) 2016.
Carlos Rivera, Problem 20: k consecutive numbers with the same number of divisors, The Prime Puzzles and Problems Connection.
Carlos Rivera, Problem 61: problem 20 revisited, The Prime Puzzles and Problems Connection.
Vlad-Titus Spătaru, Runs of consecutive integers having the same number of divisors, arXiv preprint (2023). arXiv:2301.04464 [math.NT]

Cf. A000005, A005237, A005238, A006601, A049051, A019273, A039665.

Cf. A034173, A115158, A119479.

tau = DivisorSigma[0, #]&;
A006558[q_, w_] := Module[{a, k, j, ok, n}, For[j = 0, j <= w, j++, For[n = 1, n <= q, n++, ok = 1; a = tau[n]; For[k = 1, k <= j, k++, If[a != tau[n + k], ok = 0; Break[]]]; If [ok == 1, Print[n]; Break[]]]]];
A006558[2*10^5, 7] (* Jean-François Alcover, Dec 10 2017 *)

isok(n, k)=nb = numdiv(k); for (j=k+1, k+n-1, if (numdiv(j) != nb, return(0));); 1;
a(n) = {k=1; while (!isok(n, k), k++); k;} \\ Michel Marcus, Feb 17 2016

a(8) from Jud McCranie, Jan 20 2002
a(9) conjectured by David Wasserman, Jan 08 2006
a(9) confirmed by Jud McCranie, Jan 14 2006
a(10) by Jud McCranie, Nov 27 2018

A030513 Numbers with 4 divisors.

Original entry on oeis.org

6, 8, 10, 14, 15, 21, 22, 26, 27, 33, 34, 35, 38, 39, 46, 51, 55, 57, 58, 62, 65, 69, 74, 77, 82, 85, 86, 87, 91, 93, 94, 95, 106, 111, 115, 118, 119, 122, 123, 125, 129, 133, 134, 141, 142, 143, 145, 146, 155, 158, 159, 161, 166, 177, 178, 183, 185, 187

Offset: 1

Jeff Burch

Essentially the same as A007422.
Numbers which are either the product of two distinct primes (A006881) or the cube of a prime (A030078).
4*a(n) are the solutions to A048272(x) = Sum_{d|x} (-1)^d = 4. - Benoit Cloitre, Apr 14 2002
Since A119479(4)=3, there are never more than 3 consecutive integers in the sequence. Triples of consecutive integers start at 33, 85, 93, 141, 201, ... (A039833). No such triple contains a term of the form p^3. - Ivan Neretin, Feb 08 2016
Numbers that are equal to the product of their proper divisors (A007956) (proof in Sierpiński). - Bernard Schott, Apr 04 2022

Wacław Sierpiński, Elementary Theory of Numbers, Ex. 2 p. 174, Warsaw, 1964.

Cf. A000005, A007422, A007956, A030515, A035533, A048272, A119479.

Equals the disjoint union of A006881 and A030078.

[n: n in [1..200] | DivisorSigma(0, n) eq 4]; // Vincenzo Librandi, Jul 16 2015

Select[Range[200], DivisorSigma[0,#]==4&] (* Harvey P. Dale, Apr 06 2011 *)

is(n)=numdiv(n)==4 \\ Charles R Greathouse IV, May 18 2015

from math import isqrt
from sympy import primepi, integer_nthroot, primerange
def A030513(n):
    def f(x): return int(n+x-primepi(integer_nthroot(x,3)[0])+(t:=primepi(s:=isqrt(x)))+(t*(t-1)>>1)-sum(primepi(x//k) for k in primerange(1, s+1)))
    m, k = n, f(n)
    while m != k:
        m, k = k, f(k)
    return m # Chai Wah Wu, Aug 16 2024

{n : A000005(n) = 4}. - Juri-Stepan Gerasimov, Oct 10 2009

Incorrect comments removed by Charles R Greathouse IV, Mar 18 2010

A030626 Numbers with exactly 8 divisors.

Original entry on oeis.org

24, 30, 40, 42, 54, 56, 66, 70, 78, 88, 102, 104, 105, 110, 114, 128, 130, 135, 136, 138, 152, 154, 165, 170, 174, 182, 184, 186, 189, 190, 195, 222, 230, 231, 232, 238, 246, 248, 250, 255, 258, 266, 273, 282, 285, 286, 290, 296, 297, 310, 318, 322, 328, 344, 345, 351, 354, 357, 366, 370, 374, 375, 376, 385, 399, 402

Offset: 1

Jeff Burch

nonn

Since A119479(8)=7, there are never more than 7 consecutive terms. Runs of 7 consecutive terms start at 171897, 180969, 647385, ... (subsequence of A049053). - Ivan Neretin, Feb 08 2016

Jon E. Schoenfield, Table of n, a(n) for n = 1..10000 (first 1000 terms from R. J. Mathar)
Jérôme Germoni, Nombres à huit diviseurs, Images des Mathématiques, CNRS, 2017 (in French).
Eric Weisstein's World of Mathematics, Divisor Product.

Essentially the same as A111398.

Cf. A000005, A007304, A049053, A065036, A092759, A119479.

[n: n in [1..400] | DivisorSigma(0, n) eq 8]; // Vincenzo Librandi, Oct 05 2017

select(numtheory:-tau=8, [$1..1000]); # Robert Israel, Dec 17 2014

Select[Range[400], DivisorSigma[0, #]== 8 &] (* Vincenzo Librandi, Oct 05 2017 *)

Vec(select(x->x==8,vector(500, i, numdiv(i)),1)) \\ Michel Marcus, Dec 17 2014

from sympy import divisor_count
isok = lambda n: divisor_count(n) == 8
print([n for n in range(1, 400) if isok(n)]) # Darío Clavijo, Oct 17 2023

from math import isqrt
from sympy import primepi, primerange, integer_nthroot
def A030626(n):
    def bisection(f,kmin=0,kmax=1):
        while f(kmax) > kmax: kmax <<= 1
        kmin = kmax >> 1
        while kmax-kmin > 1:
            kmid = kmax+kmin>>1
            if f(kmid) <= kmid:
                kmax = kmid
            else:
                kmin = kmid
        return kmax
    def f(x): return int(n+x-sum(primepi(x//(k*m))-b for a,k in enumerate(primerange(integer_nthroot(x,3)[0]+1),1) for b,m in enumerate(primerange(k+1,isqrt(x//k)+1),a+1))-sum(primepi(x//p**3) for p in primerange(integer_nthroot(x,3)[0]+1))+primepi(integer_nthroot(x,4)[0])-primepi(integer_nthroot(x,7)[0]))
    return bisection(f,n,n) # Chai Wah Wu, Feb 21 2025

A000005(a(n))=8. - Juri-Stepan Gerasimov, Oct 10 2009

Equals A065036 (p*q^3) U A007304 (p*q*r) U A092759 (p^7). - Amarnath Murthy, Apr 21 2001

A006601 Numbers k such that k, k+1, k+2 and k+3 have the same number of divisors.

Original entry on oeis.org

242, 3655, 4503, 5943, 6853, 7256, 8392, 9367, 10983, 11605, 11606, 12565, 12855, 12856, 12872, 13255, 13782, 13783, 14312, 16133, 17095, 18469, 19045, 19142, 19143, 19940, 20165, 20965, 21368, 21494, 21495, 21512, 22855, 23989, 26885

Offset: 1

N. J. A. Sloane

M. Abramowitz and I. A. Stegun, eds., Handbook of Mathematical Functions, National Bureau of Standards Applied Math. Series 55, 1964 (and various reprintings), p. 840.
Jean-Marie De Koninck, Ces nombres qui nous fascinent, Entry 242, p. 67, Ellipses, Paris 2008.
Richard K. Guy, Unsolved Problems in Number Theory, 3rd Edition, Springer, 2004, Section B18, pp. 111-113.
N. J. A. Sloane and Simon Plouffe, The Encyclopedia of Integer Sequences, Academic Press, 1995 (includes this sequence).

Amiram Eldar, Table of n, a(n) for n = 1..10000 (terms 1..1000 from T. D. Noe)
M. Abramowitz and I. A. Stegun, eds., Handbook of Mathematical Functions, National Bureau of Standards, Applied Math. Series 55, Tenth Printing, 1972 [alternative scanned copy].
Victor Meally, Letter to N. J. A. Sloane, no date.

Cf. A000005, A006558, A019273, A119479.

Other runs of equidivisor numbers: A005237 (runs of 2), A005238 (runs of 3), A049051 (runs of 5), A049052 (runs of 6), A049053 (runs of 7).

import Data.List (elemIndices)
a006601 n = a006601_list !! (n-1)
a006601_list = map (+ 1) $ elemIndices 0 $
   zipWith3 (((+) .) . (+)) ds (tail ds) (drop 2 ds) where
   ds = map abs $ zipWith (-) (tail a000005_list) a000005_list
-- Reinhard Zumkeller, Jan 18 2014

f[n_]:=Length[Divisors[n]]; lst={};Do[If[f[n]==f[n+1]==f[n+2]==f[n+3],AppendTo[lst,n]],{n,8!}];lst (* Vladimir Joseph Stephan Orlovsky, Dec 14 2009 *)
dsQ[n_]:=Length[Union[DivisorSigma[0,Range[n,n+3]]]]==1; Select[Range[ 30000],dsQ] (* Harvey P. Dale, Nov 23 2011 *)
Flatten[Position[Partition[DivisorSigma[0,Range[27000]],4,1],?(Union[ Differences[ #]]=={0}&),{1},Heads->False]] (* Faster, because the number of divisors for each number is only calculated once *) (* _Harvey P. Dale, Nov 06 2013 *)
SequencePosition[DivisorSigma[0,Range[27000]],{x_,x_,x_,x_}][[All,1]] (* Requires Mathematica version 10 or later *) (* Harvey P. Dale, Jul 03 2017 *)

is(n)=my(t=numdiv(n)); numdiv(n+1)==t && numdiv(n+2)==t && numdiv(n+3)==t \\ Charles R Greathouse IV, Jun 25 2017

More terms from Olivier Gérard

A030515 Numbers with exactly 6 divisors.

Original entry on oeis.org

12, 18, 20, 28, 32, 44, 45, 50, 52, 63, 68, 75, 76, 92, 98, 99, 116, 117, 124, 147, 148, 153, 164, 171, 172, 175, 188, 207, 212, 236, 242, 243, 244, 245, 261, 268, 275, 279, 284, 292, 316, 325, 332, 333, 338, 356, 363, 369, 387, 388, 404, 412, 423, 425, 428

Offset: 1

Jeff Burch

Numbers which are either the 5th power of a prime or the product of a prime and the square of a different prime, i.e., numbers which are in A050997 (5th powers of primes) or A054753. - Henry Bottomley, Apr 25 2000
Also numbers which are the square root of the product of their proper divisors. - Amarnath Murthy, Apr 21 2001
Such numbers are multiplicatively 3-perfect (i.e., the product of divisors of a(n) equals a(n)^3). - Lekraj Beedassy, Jul 13 2005
Since A119479(6)=5, there are never more than 5 consecutive terms. Quintuples of consecutive terms start at 10093613546512321, 14414905793929921, 266667848769941521, ... (A141621). No such quintuple contains a term of the form p^5. - Ivan Neretin, Feb 08 2016

Amarnath Murthy, A note on the Smarandache Divisor sequences, Smarandache Notions Journal, Vol. 11, 1-2-3, Spring 2000.

R. J. Mathar, Table of n, a(n) for n = 1..1000
Amarnath Murthy and Charles Ashbacher, Generalized Partitions and Some New Ideas on Number Theory and Smarandache Sequences, Hexis, Phoenix; USA 2005. See Section 1.4, 1.12.
Eric Weisstein's World of Mathematics, Divisor Product

Cf. A061117.

N:= 1000: # to get all terms <= N
Primes:= select(isprime, {2,seq(i,i=3..floor(N/4))}):
S:= select(`<=`,{seq(p^5, p = Primes),seq(seq(p*q^2, p=Primes minus {q}),q=Primes)},N):
sort(convert(S,list)); # Robert Israel, Feb 10 2016

f[n_]:=Length[Divisors[n]]==6; lst={};Do[If[f[n],AppendTo[lst,n]],{n,6!}];lst (* Vladimir Joseph Stephan Orlovsky, Dec 14 2009 *)
Select[Range[500],DivisorSigma[0,#]==6&] (* Harvey P. Dale, Oct 02 2014 *)

is(n)=numdiv(n)==6 \\ Charles R Greathouse IV, Jan 23 2014

from sympy import divisor_count
def ok(n): return divisor_count(n) == 6
print([k for k in range(429) if ok(k)]) # Michael S. Branicky, Dec 18 2021

from math import isqrt
from sympy import primepi, primerange, integer_nthroot
def A030515(n):
    def bisection(f,kmin=0,kmax=1):
        while f(kmax) > kmax: kmax <<= 1
        kmin = kmax >> 1
        while kmax-kmin > 1:
            kmid = kmax+kmin>>1
            if f(kmid) <= kmid:
                kmax = kmid
            else:
                kmin = kmid
        return kmax
    def f(x): return n+x-sum(primepi(x//p**2) for p in primerange(isqrt(x)+1))+primepi(integer_nthroot(x,3)[0])-primepi(integer_nthroot(x,5)[0])
    return bisection(f,n,n) # Chai Wah Wu, Feb 21 2025

Union of A050997 and A054753. - Lekraj Beedassy, Jul 13 2005

A000005(a(n))=6. - Juri-Stepan Gerasimov, Oct 10 2009

Definition clarified by Jonathan Sondow, Jan 23 2014

A049051 Numbers k such that k through k+4 all have the same number of divisors.

Original entry on oeis.org

11605, 12855, 13782, 19142, 21494, 28374, 28375, 40311, 42805, 50582, 55254, 60231, 60663, 79094, 87655, 90181, 90182, 95845, 99655, 103621, 109765, 115591, 120727, 121045, 122151, 122871, 142454, 142806, 152630, 157493, 157494, 171893

Offset: 1

David W. Wilson

nonn

Amiram Eldar, Table of n, a(n) for n = 1..10000 (terms 1..1000 from Donovan Johnson)

Cf. A000005, A006558, A019273, A039665, A119479.

Other runs of equidivisor numbers: A005237 (runs of 2), A005238 (runs of 3), A006601 (runs of 4), A049052 (runs of 6), A049053 (runs of 7).

Flatten[Position[Partition[DivisorSigma[0,Range[200000]],5,1],?(Length[ Union[#]] == 1&),{1},Heads->False]] (* _Harvey P. Dale, May 30 2014 *)
SequencePosition[DivisorSigma[0,Range[200000]],{x_,x_,x_,x_,x_}][[All,1]] (* Requires Mathematica version 10 or later *) (* Harvey P. Dale, Jul 03 2017 *)

A141621 Numbers that begin a run of 5 consecutive integers of the form p^2*q where p and q are distinct primes.

Original entry on oeis.org

10093613546512321, 14414905793929921, 266667848769941521, 562672865058083521, 1579571757660876721, 1841337567664174321, 2737837351207392721, 4456162869973433521, 4683238426747860721, 4993613853242910721, 5037980611623036721, 5174116847290255921

Offset: 1

Matthijs Coster, Aug 23 2008

Old name was "The first number of a series of 5 consecutive numbers with the same signature, i.e., all numbers have the format p^2*q, where p and q are primes. Therefore the number of divisors is the same (6)." [That name could have been confusing in that not every sequence of 5 consecutive integers having the same prime signature has the prime signature p^2*q; e.g., 204323 is the first of 5 consecutive numbers of the form p^2*q*r. - Jon E. Schoenfield, Jun 05 2018]
Each of the five numbers in each such sequence has 6 divisors.
It is easy to prove that any number in this sequence must be congruent to 1 modulo 240. The program below calculates only an element of the sequence. Since the reference A119479 it is the smallest one. If we assume that the first element has the format 7^2*n49, the second number has the format 2*p^2, the third element has the format 3^2*n9 and the fifth element has the format 5^2*n25, then p must be modulo 22050 one out of 1181, 3719, 4219, 9119, 12931, 17831, 18331 or 20869.
It is unclear if these numbers are the smallest ones. - Matthijs Coster, Aug 28 2008 [The terms listed in the Data section are, in fact, the smallest numbers matching the definition. - Jon E. Schoenfield, Jun 05 2018]
The first quintuple not of the aforementioned form starts with 5344962129269790721 = 23^2*prime. - Ivan Neretin, Feb 08 2016
Among the first 200 terms, the frequency with which the squared prime factor p is {7, 17, 23, 31, 41, 47, 73, 127, 193, 1039, 1399} is {171, 10, 6, 4, 3, 1, 1, 1, 1, 1, 1}, respectively. - Jon E. Schoenfield, Jun 09 2018

			a(1) = 10093613546512321, because
10093613546512321 = 7^2 * 205992113194129,
10093613546512322 =   2 * 71040881^2,
10093613546512323 = 3^2 * 1121512616279147,
10093613546512324 = 2^2 * 2523403386628081, and
10093613546512325 = 5^2 * 403744541860493,
so each of the five consecutive integers is of the form p^2*q, and no smaller run of five consecutive integers has this property. [corrected by _Jon E. Schoenfield_, Jun 05 2018]

Ray Chandler, Table of n, a(n) for n = 1..2000 (first 25 terms from Ivan Neretin, through 200 terms from Jon E. Schoenfield)
Richard I. Hess, Problem 1231, Crux Mathematicorum, Vol. 13, No. 4, p. 118, 1987. (takes a long time to download)
Richard I. Hess, Puzzles from Around the World, p. 63, H17.
Carlos Rivera, Problem 20.- Divisors (II) K consecutive numbers with the same number of divisors, The Prime Puzzles & Problems Connection.
StackExchange, Sequence of numbers with prime factorization pq^2
wu :: forums, Same Number of Divisors, Oct 05 2007.
Index to sequences related to prime signature

Cf. A119479, A006558, A005237, A005238, A006601.

Cf. A054753, A074172, A178032, A308683.

## Warning: this program appears to be incorrect [Joerg Arndt, Feb 29 2016]
for m in range(5000):
    p = 22050*m+17831
    if is_prime(p):
        n = 2*p^2-2
        n4 = n/4+1
        if is_prime(n4):
            n49 = floor((n+1)/49)
            if (49*n49 == n+1) and is_prime(n49):
                n9 = floor((n+3)/9)
                if (9*n9 == n+3) and is_prime(n9):
                    n25 = floor((n+5)/25)
                    if (25*n25 == n+5) and is_prime(n25):
                        print(n+1, n49, p, n9, n4, n25)

Two more terms Matthijs Coster, Aug 28 2008
Missing terms added and extended by Ivan Neretin, Feb 08 2016
New name from Jon E. Schoenfield, Jun 05 2018

A049053 Numbers k such that k through k+6 all have the same number of divisors.

Original entry on oeis.org

171893, 180965, 647381, 1039493, 1071829, 1450261, 1563653, 1713413, 2129029, 2384101, 4704581, 4773301, 5440853, 5775365, 6627061, 6644405, 6697253, 8556661, 8833429, 10531253, 12101509, 12238453, 12307141, 13416661, 13970405

Offset: 1

David W. Wilson

nonn

Allan Swett found that the first term not congruent to 5 mod 16 is 67073285. - Ralf Stephan, Nov 15 2004

Since A119479(n) < 7 for n < 8, no term has fewer than 8 divisors; the first that has more is a(30)=17476613. - Ivan Neretin, Feb 05 2016

Amiram Eldar, Table of n, a(n) for n = 1..10000 (terms 1..1000 from Donovan Johnson)
Jean-Marie De Koninck, Those Fascinating Numbers, Amer. Math. Soc., 2009, page 244.

Cf. A000005, A006558, A019273, A119479.

Other runs of equidivisor numbers: A005237 (runs of 2), A005238 (runs of 3), A006601 (runs of 4), A049051 (runs of 5), A049052 (runs of 6).

isok(n) = {my(nb = numdiv(n)); for (k=1, 6, if (numdiv(n+k) != nb, return (0));); 1;} \\ Michel Marcus, Feb 06 2016

A049052 Numbers k such that k through k+5 all have the same number of divisors.

Original entry on oeis.org

28374, 90181, 157493, 171893, 171894, 180965, 180966, 210133, 298694, 346502, 369061, 376742, 610310, 647381, 647382, 707286, 729542, 769862, 1039493, 1039494, 1071829, 1071830, 1243541, 1302005, 1449605, 1450261, 1450262

Offset: 1

David W. Wilson

nonn

Amiram Eldar, Table of n, a(n) for n = 1..10000 (terms 1..1000 from Donovan Johnson)

Cf. A000005, A006558, A019273, A119479.

Other runs of equidivisor numbers: A005237 (runs of 2), A005238 (runs of 3), A006601 (runs of 4), A049051 (runs of 5), A049053 (runs of 7).

SequencePosition[DivisorSigma[0,Range[1451000]],{x_,x_,x_,x_,x_,x_}][[All,1]] (* Requires Mathematica version 10 or later *) (* Harvey P. Dale, Nov 03 2020 *)

A292580 T(n,k) is the start of the first run of exactly k consecutive integers having exactly 2n divisors. Table read by rows.

Original entry on oeis.org

5, 2, 6, 14, 33, 12, 44, 603, 242, 10093613546512321, 24, 104, 230, 3655, 11605, 28374, 171893, 48, 2511, 7939375, 60, 735, 1274, 19940, 204323, 368431323, 155385466971, 18652995711772, 15724736975643, 2973879756088065948, 9887353188984012120346

Offset: 1

Jon E. Schoenfield, Sep 19 2017

The number of terms in row n is A119479(2n).
Düntsch and Eggleton (1989) has typos for T(3,5) and T(10,3) (called D(6,5) and D(20,3) in their notation). Letsko (2015) and Letsko (2017) both have a wrong value for T(7,3).
The first value required to extend the data is T(6,13) <= 586683019466361719763403545; the first unknown value that may exist is T(12,19). See the a-file for other known values and upper bounds up to T(50,7).

			T(1,1) = 5 because 5 is the start of the first "run" of exactly 1 integer having exactly 2*1=2 divisors (5 is the first prime p such that both p-1 and p+1 are nonprime);
T(1,2) = 2 because 2 is the start of the first run of exactly 2 consecutive integers having exactly 2*1=2 divisors (2 and 3 are the only consecutive integers that are prime);
T(3,4) = 242 because the first run of exactly 4 consecutive integers having exactly 2*3=6 divisors is 242 = 2*11^2, 243 = 3^5, 244 = 2^2*61, 245 = 5*7^2.
Table begins:
   n  T(n,1), T(n,2), ...
  ==  ========================================================
   1  5, 2;
   2  6, 14, 33;
   3  12, 44, 603, 242, 10093613546512321;
   4  24, 104, 230, 3655, 11605, 28374, 171893;
   5  48, 2511, 7939375;
   6  60, 735, 1274, 19940, 204323, 368431323, 155385466971, 18652995711772, 15724736975643, 2973879756088065948, 9887353188984012120346, 120402988681658048433948, T(6,13), ...;
   7  192, 29888, 76571890623;
   8  120, 2295, 8294, 153543, 178086, 5852870, 17476613;
   9  180, 6075, 959075, 66251139635486389922, T(9,5);
  10  240, 5264, 248750, 31805261872, 1428502133048749, 8384279951009420621, 189725682777797295066519373;
  11  3072, 2200933376, 104228508212890623;
  12  360, 5984, 72224, 2919123, 15537948, 973277147, 33815574876, 1043710445721, 2197379769820, 2642166652554075, 17707503256664346, T(12,12), ...;
  13  12288, 689278976, 1489106237081787109375;
  14  960, 156735, 23513890624, 4094170438109373, 55644509293039461218749, 4230767238315793911295500109374, 273404501868270838132985214432619890621;
  15  720, 180224, 145705879375, 10868740069638250502059754282498, T(15,5);
  16  840, 21735, 318680, 6800934, 57645182, 1194435205, 14492398389;
  ...

Hugo van der Sanden, Table of n, a(n) for n = 1..32
Ivo Düntsch and Roger B. Eggleton, Equidivisible consecutive integers, 1989.
Vladimir A. Letsko, Some new results on consecutive equidivisible integers, arXiv:1510.07081 [math.NT], 2015.
Vladimir A. Letsko, Table of a(n) for all even n such that exact value of a(n) is proved, 2017.
Carlos Rivera, Problem 20: k consecutive numbers with the same number of divisors, The Prime Puzzles and Problems Connection.
Hugo van der Sanden, calculation of T(6,11).
Hugo van der Sanden, calculation of T(6,12).
Hugo van der Sanden, A-file with known values and bounds up to T(50,7)

Cf. A000005, A006558, A072507, A119479.
Cf. A005237, A005238, A006601, A049051, A049052, A049053.
Cf. A003680, A075036, A075040.

T(n,2) = A075036(n). - Jon E. Schoenfield, Sep 23 2017

a(1)-a(25) from Düntsch and Eggleton (1989) with corrections by Jon E. Schoenfield, Sep 19 2017
a(26)-a(27) from Giovanni Resta, Sep 20 2017
a(28)-a(29) from Hugo van der Sanden, Jan 12 2022
a(30) from Hugo van der Sanden, Sep 03 2022
a(31) added by Hugo van der Sanden, Dec 05 2022; see "calculation of T(6,11)" link for a list of the people involved.
a(32) added by Hugo van der Sanden, Dec 18 2022; see "calculation of T(6,12)" link for a list of the people involved.

cp's OEIS Frontend

A006558 Start of first run of n consecutive integers with same number of divisors.

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A030513 Numbers with 4 divisors.

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Magma

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A030626 Numbers with exactly 8 divisors.

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A006601 Numbers k such that k, k+1, k+2 and k+3 have the same number of divisors.

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A030515 Numbers with exactly 6 divisors.

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A049051 Numbers k such that k through k+4 all have the same number of divisors.

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