A006558 -id:A006558 - cp's OEIS frontend

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

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

A284597 a(n) is the least number that begins a run of exactly n consecutive numbers with a nondecreasing number of divisors, or -1 if no such number exists.

Original entry on oeis.org

46, 5, 43, 1, 1613, 241, 17011, 12853, 234613, 376741, 78312721, 125938261, 4019167441, 16586155153, 35237422882, 1296230533473, 42301168491121, 61118966262061

Offset: 1

Fred Schneider, Mar 29 2017

The words "begins" and "exactly" in the definition are crucial. The initial values of tau (number of divisors function, A000005) can be partitioned into nondecreasing runs as follows: {1, 2, 2, 3}, {2, 4}, {2, 4}, {3, 4}, {2, 6}, {2, 4, 4, 5}, {2, 6}, {2, 6}, {4, 4}, {2, 8}, {3, 4, 4, 6}, {2, 8}, {2, 6}, {4, 4, 4, 9}, {2, 4, 4, 8}, {2, 8}, {2, 6, 6}, {4}, {2, 10}, ... From this we can see that a(1) = 46 (the first singleton), a(2)=5 (the first pair), a(3)=43 (the first triple), a(4)=1, etc. - Bill McEachen and Giovanni Resta, Apr 26 2017. (see also A303577 and A303578 - N. J. A. Sloane, Apr 29 2018)
Initial values computed with a brute force C++ program.
It seems very likely that one can always find a(n) and that we never need to take a(n) = -1. But this is at present only a conjecture. - N. J. A. Sloane, May 04 2017
Conjecture follows from Dickson's conjecture (see link). - Robert Israel, Mar 30 2020
If a(n) > 1, then A013632(a(n)) >= n. Might be useful to help speed up brute force search. - Chai Wah Wu, May 04 2017
The analog sequence for sigma (sum of divisors) instead of tau (number of divisors) is A285893 (see also A028965). - M. F. Hasler, May 06 2017
a(n) > 3.37*10^14 for n > 18. - Robert Gerbicz, May 14 2017

			241 = 241^1 => 2 divisors
242 = 2^1 * 11^2 => 6 divisors
243 = 3^5 => 6 divisors
244 = 2^2 * 61^1 => 6 divisors
245 = 5^1 * 7^2 => 6 divisors
246 = 2^1 * 3^1 * 41^1 => 8 divisors
247 = 13^1 * 19^1 => 4 divisors
So, 247 breaks the chain. 241 is the lowest number that is the beginning of exactly 6 consecutive numbers with a nondecreasing number of divisors. So it is the 6th term in the sequence.
Note also that a(5) is not 242, even though tau evaluated at 242, 243,..., 246 gives 5 nondecreasing values, because here we deal with full runs and 242 belongs to the run of 6 values starting at 241.

Robert Israel, Dickson's conjecture implies all a(n)>0

Cf. A000005, A000203, A006558, A013632, A028965, A075046, A285893.

See also A286287, A286288, A286289, A303577, A303578.

Function[s, {46}~Join~Map[Function[r, Select[s, Last@ # == r &][[1, 1]]], Range[2, Max[s[[All, -1]] ] ]]]@ Map[{#[[1, 1]], Length@ # + 1} &, DeleteCases[SplitBy[#, #[[-1]] >= 0 &], k_ /; k[[1, -1]] < 0]] &@ MapIndexed[{First@ #2, #1} &, Differences@ Array[DivisorSigma[0, #] &, 10^6]] (* Michael De Vlieger, May 06 2017 *)

genit()={for(n=1,20,q=0;ibgn=1;for(m=ibgn,9E99,mark1=q;q=numdiv(m);if(mark1==0,summ=0;dun=0;mark2=m);if(q>=mark1,summ+=1,dun=1);if(dun>0&&summ==n,print(n," ",mark2);break);if(dun>0&&summ!=n,q=0;m-=1)));} \\ Bill McEachen, Apr 25 2017

A284597=vector(19);apply(scan(N,s=1,t=numdiv(s))=for(k=s+1,N,t>(t=numdiv(k))||next;k-s>#A284597||A284597[k-s]||printf(" a(%d)=%d,",k-s,s)||A284597[k-s]=s;s=k);done,[10^6]) \\ Finds a(1..10) in ~ 1 sec, but would take 100 times longer to get one more term with scan(10^8). You may extend the search using scan(END,START). - M. F. Hasler, May 06 2017

from sympy import divisor_count
def A284597(n):
    count, starti, s, i = 0,1,0,1
    while True:
        d = divisor_count(i)
        if d < s:
            if count == n:
                return starti
            starti = i
            count = 0
        s = d
        i += 1
        count += 1 # Chai Wah Wu, May 04 2017

a(1), a(2), a(4) corrected by Bill McEachen and Giovanni Resta, Apr 26 2017

a(17)-a(18) from Robert Gerbicz, May 14 2017

A039665 Sets of 4 consecutive numbers with equal number of divisors.

Original entry on oeis.org

242, 243, 244, 245, 3655, 3656, 3657, 3658, 4503, 4504, 4505, 4506, 5943, 5944, 5945, 5946, 6853, 6854, 6855, 6856, 7256, 7257, 7258, 7259, 8392, 8393, 8394, 8395, 9367, 9368, 9369, 9370, 10983, 10984, 10985, 10986, 11605, 11606, 11607, 11608

Offset: 1

Felice Russo

nonn

Taking the first entry in each set gives A006601.

D. Wells, Curious and interesting numbers, Penguin Books, p. 134

Cf. A000005, A005237, A005238, A006601, A049051, A006558, A019273.

More terms from Patrick De Geest, Nov 15 1999

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 *)

A218448 First of a run of 5 consecutive numbers with same prime signature.

Original entry on oeis.org

204323, 3252571, 5205074, 7201674, 20182921, 28387953, 36193650, 43216722, 51049537, 56155074, 57070850, 61961315, 62167075, 65425473, 76647074, 82507473, 92658049, 95943321, 100498849, 107236449, 109751473, 110899321, 112198075, 112477849, 116736323

Offset: 1

M. F. Hasler, Oct 28 2012

nonn

A number n is in this sequence iff n and n+1 is in A175590; also: iff n and n+2 are in A052214 (in which case n+1 is in A052214, too); and also: iff {n,n+1,n+2,n+3} are in A052213.

A034173(6) = A218448(62) = A218448(63)-1 is the least term n such that n+1 is also in the sequence.

M. F. Hasler and Donovan Johnson, Table of n, a(n) for n = 1..1000 (first 140 terms from _M. F. Hasler_)

Cf. A034173, A034174, A006558, A083790, A052213, A052214, A175590.

is_A218448(n)={my(f);!for(i=0,4,f!=(f=vecsort(factor(n+i)[,2])) & i & return)}

f(k)=vecsort(factor(k)[,2]~,,4)
t=f(n=2);while(n<1e8, for(i=n+1, n+4, tt=f(i); if(tt!=t, n=i; t=tt; next(2))); print1(n", "); n++) \\ Charles R Greathouse IV, Oct 28 2012

a(6)-a(8) from Charles R Greathouse IV, Oct 28 2012
a(9)-a(25) from Donovan Johnson, Oct 28 2012
Values up to a(140) computed using b175590.txt from Charles R Greathouse IV  - M. F. Hasler, Oct 28 2012

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.

A318166 a(n) begins the first run of at least n consecutive numbers with the same number of infinitary divisors.

Original entry on oeis.org

1, 2, 2, 2, 32, 141, 141, 141, 42410, 171890, 2648985, 10896843, 10896843, 727940625, 1791416073, 19183907363, 62520703916, 162891847444, 162891847444, 349662337209, 7858045724108

Offset: 1

Amiram Eldar, Aug 20 2018

The infinitary version of A006558.

			a(5) = 32 since the number of infinitary divisors of 32, 33, 34, 35 and 36 is 4, and this is the first run of 5 consecutive numbers.

Cf. A006558, A037445, A045983.

idivnum[1] = 1; idivnum[n_] := Times @@ Flatten[2^DigitCount[#, 2, 1] & /@ FactorInteger[n][[All, 2]]]; Seq[n_, q_] := Map[idivnum, Range[n, n + q - 1]]; findConsec[q_, nmin_, nmax_] := Module[{}, s = Seq[1, q]; n = q + 1; found = False;  Do[If[CountDistinct[s] == 1, found = True; Break[]]; s = Rest[AppendTo[s, idivnum[n]]]; n++, {k, nmin, nmax}]; If[found, n - q, 0]]; seq = {1}; nmax = 10000000; Do[n1 = Last[seq]; s1 = findConsec[m, n1, nmax]; If[s1 == 0, Break[]]; AppendTo[seq, s1], {m, 2, 11}];

a(12)-a(21) from Giovanni Resta, Aug 24 2018

A324593 a(n) is the smallest number k such that n consecutive integers starting at k have the same number of odd divisors (A001227).

Original entry on oeis.org

1, 1, 5, 10, 10, 515, 2314, 2314, 1536863, 4053992, 4053992, 18584686, 146237365, 163039279, 4775943486, 13147233734, 86153130379

Offset: 1

Ilya Gutkovskiy, Sep 03 2019

			515 has 4 odd divisors {1, 5, 103, 515}, 516 has 4 odd divisors {1, 3, 43, 129}, 517 has 4 odd divisors {1, 11, 47, 517}, 518 has 4 odd divisors {1, 7, 37, 259}, 519 has 4 odd divisors {1, 3, 173, 519} and 520 has 4 odd divisors {1, 5, 13, 65}. These the first 6 consecutive numbers with the same number of odd divisors, so a(6) = 515.

Rémy Sigrist, C program for A324593

Cf. A001227, A006558, A045983, A045984, A323253, A324594.

C
```
See Links section.
```

a(11)-a(14) from Rémy Sigrist, Sep 04 2019

a(15)-a(17) from Giovanni Resta, Sep 04 2019

A324594 a(n) is the smallest number k such that n consecutive integers starting at k have the same number of nonprime divisors (A033273).

Original entry on oeis.org

1, 1, 1, 19940, 204323, 294590, 310042685, 2587701932494, 2587701932494

Offset: 1

Ilya Gutkovskiy, Sep 03 2019

			19940 has 9 nonprime divisors {1, 4, 10, 20, 1994, 3988, 4985, 9970, 19940}, 19941 has 9 nonprime divisors {1, 51, 69, 289, 391, 867, 1173, 6647, 19941}, 19942 has 9 nonprime divisors {1, 26, 118, 169, 338, 767, 1534, 9971, 19942} and 19943 has 9 nonprime divisors {1, 49, 77, 259, 407, 539, 1813, 2849, 19943}. These the first 4 consecutive numbers with the same number of nonprime divisors, so a(4) = 19940.

Rémy Sigrist, C program for A324594

Cf. A006558, A033273, A045983, A045984, A323253, A324593.

C
```
See Links section.
```

a(7) from Rémy Sigrist, Sep 04 2019

a(8)-a(9) from Giovanni Resta, Sep 04 2019

cp's OEIS Frontend

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

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A284597 a(n) is the least number that begins a run of exactly n consecutive numbers with a nondecreasing number of divisors, or -1 if no such number exists.

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A039665 Sets of 4 consecutive numbers with equal number of divisors.

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

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

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A218448 First of a run of 5 consecutive numbers with same prime signature.

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

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A318166 a(n) begins the first run of at least n consecutive numbers with the same number of infinitary divisors.

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