A006601 -id:A006601 - cp's OEIS frontend

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

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.

A332314 Numbers k such that k, k + 1, k + 2 and k + 3 have the same number of divisors in Gaussian integers.

Original entry on oeis.org

263449773, 334047725, 760228973, 862305773, 1965540624, 2136055725, 2362380525, 2477365422, 2515570575, 2613782223, 2939626925, 3181603023, 3814526223, 3987335022, 4610697039, 4771214574, 4981539822, 5018728272, 5035157775, 5186567824, 6189727725, 6329159823, 6569396973

Offset: 1

Amiram Eldar, Feb 09 2020

nonn

			263449773 is a term since 263449773, 263449774, 263449775 and 263449776 each have 72 divisors in Gaussian integers.

Cf. A006601, A062327, A332312, A332313.

gaussNumDiv[n_] := DivisorSigma[0, n, GaussianIntegers -> True]; m = 4; s = gaussNumDiv /@ Range[m]; seq = {}; n = m + 1; While[Length[seq] < 10, If[Length @ Union[s] == 1, AppendTo[seq, n - m + 1]]; n++; s = Join[Rest[s], {gaussNumDiv[n]}]]; seq

A332388 Numbers k such that k, k + 1, k + 2 and k + 3 have the same number of divisors in Eisenstein integers.

Original entry on oeis.org

34193750, 76788050, 78267398, 113004199, 135383873, 148843670, 170293249, 199259222, 311313398, 318128599, 364828550, 368222599, 381026822, 384839047, 420686749, 428129222, 430154150, 432466824, 450050450, 462825847, 492828521, 510703975, 517126773, 518268772

Offset: 1

Amiram Eldar, Feb 10 2020

nonn

			34193750 is a term since 34193750, 34193751, 34193752 and 34193750 each have 24 divisors in Eisenstein integers.

Cf. A006601, A319442, A332314, A332386, A332387.

f[p_, e_] := Switch[Mod[p, 3], 0, 2*e + 1, 1, (e + 1)^2, 2, e + 1]; eisNumDiv[1] = 1; eisNumDiv[n_] := Times @@ f @@@ FactorInteger[n]; m = 4; s = eisNumDiv /@ Range[m]; seq = {}; n = m + 1; While[Length[seq] < 10, If[Length @ Union[s] == 1, AppendTo[seq, n - m + 1]]; n++; s = Join[Rest[s], {eisNumDiv[n]}]]; seq

A338454 Starts of runs of 4 consecutive numbers with the same total binary weight of their divisors (A093653).

Original entry on oeis.org

242, 947767, 1041607, 2545015, 3275463, 8170983, 15720871, 21532430, 23752181, 25135885, 25595913, 27981703, 28226983, 30505142, 30962767, 33364805, 37264493, 49002661, 49766629, 52910454, 53408456, 57917191, 57952016, 58331576, 59230454, 60014053, 60723111, 63378005

Offset: 1

Amiram Eldar, Oct 28 2020

Numbers k such that A093653(k) = A093653(k+1) = A093653(k+2) = A093653(k+3).

			242 is a term since A093653(242) = A093653(243) = A093653(244) = A093653(245) = 18.

Amiram Eldar, Table of n, a(n) for n = 1..10000

Cf. A093653.
Subsequence of A338452 and A338453.
Similar sequences: A006601, A045932, A045940.

f[n_] := DivisorSum[n, DigitCount[#, 2, 1] &]; s = {}; m = 4; fs = f /@ Range[m]; Do[If[Equal @@  fs, AppendTo[s, n - m]]; fs = Rest @ AppendTo[fs, f[n]], {n, m + 1, 10^7}]; s

A347603 Numbers k such that tau(k) = 2*tau(k-1) and sigma(k) = sigma(k-1), where tau(k) and sigma(k) are respectively the number and sum functions of the divisors of k.

Original entry on oeis.org

4365, 74919, 79827, 111507, 347739, 445875, 739557, 2168907, 4481986, 7263945, 7845387, 9309465, 10838247, 12290055, 12673095, 18151479, 22083215, 25645707, 39175955, 62634519, 69076995, 72794967, 80889207, 81166839, 87215967, 94682133, 107522943, 110768835, 119192283

Offset: 1

Claude H. R. Dequatre, Sep 08 2021

Conjecture: the asymptotic density of terms is equal to 0 and this sequence is infinite.

			a(1) = 4365 because the divisors of 4365 are: 1, 3, 5, 9, 15, 45, 97, 291, 485, 873, 1455, 4365; so, tau(4365) = 12 and sigma(4365) = 7644. The divisors of 4364 are: 1, 2, 4, 1091, 2182, 4364; so, tau(4364) = 6 and sigma(4364) = 7644. Thus tau(4365) = 2*tau(4364), sigma(4365) = sigma(4364) and so 4365 is a term.

Kevin P. Thompson, Table of n, a(n) for n = 1..288

Cf. A027750, A000005, A000203.

Similar sequences: A002961, A005237, A005238, A006601, A049051, A347076.

Select[Range[2, 10^6], DivisorSigma[0, #] == 2*DivisorSigma[0, # - 1] && DivisorSigma[1, #] == DivisorSigma[1, # - 1] &] (* Amiram Eldar, Sep 08 2021 *)

for(k=2,100000000,if(numdiv(k)==2*numdiv(k-1) && sigma(k)==sigma(k-1),print1(k", ")))

from sympy import divisor_count as tau, divisor_sigma as sigma
print([k for k in range(2, 10**6) if tau(k) == 2*tau(k-1) and sigma(k) == sigma(k-1)]) # Karl-Heinz Hofmann, Jan 15 2022

A355712 Starts of runs of 4 consecutive numbers with the same number of 5-smooth divisors.

Original entry on oeis.org

28374, 133623, 136374, 187623, 190374, 298374, 349623, 352374, 457623, 460374, 511623, 619623, 622374, 673623, 676374, 781623, 838374, 943623, 946374, 997623, 1000374, 1108374, 1159623, 1162374, 1267623, 1270374, 1321623, 1429623, 1432374, 1483623, 1486374, 1591623

Offset: 1

Amiram Eldar, Jul 15 2022

nonn

Numbers k such that A355583(k) = A355583(k+1) = A355583(k+2) = A355583(k+3).

Are there runs of 5 consecutive numbers with the same number of 5-smooth divisors? There are no such runs below 10^10.

			28374 is a term since A355583(28374) = A355583(28375) = A355583(28376) = A355583(28377) = 4.

Amiram Eldar, Table of n, a(n) for n = 1..10000

Cf. A355583.
Subsequence of A355710 and A355711.
Similar sequences: A006601, A332314, A332388.

f[n_] := Times @@ (1 + IntegerExponent[n, {2, 3, 5}]); s = {}; m = 4; fs = f /@ Range[m]; Do[If[Equal @@ fs, AppendTo[s, n - m]]; fs = Rest @ AppendTo[fs, f[n]], {n, m + 1, 10^6}]; s

s(n) = (valuation(n, 2) + 1) * (valuation(n, 3) + 1) * (valuation(n, 5) + 1);
s1 = s(1); s2 = s(2); s3 = s(3); for(k = 4, 1.6e6, s4 = s(k); if(s1 == s2 && s2 == s3 && s3 == s4, print1(k-3,", ")); s1 = s2; s2 = s3; s3 = s4);

A356766 Least number k such that k and k+2 both have exactly 2n divisors, or -1 if no such number exists.

Original entry on oeis.org

3, 6, 18, 40, 127251, 198, 26890623, 918, 17298, 6640, 25269208984375, 3400, 3900566650390623, 640062, 8418573, 18088, 1164385682220458984373, 41650, 69528379848480224609373, 128464, 34084859373, 12164094, 150509919493198394775390625, 90270, 418514293125, 64505245696

Offset: 1

Jean-Marc Rebert, Aug 26 2022

nonn

			For n=1, numdiv(3) = numdiv(5) = 2 = 2*1, and no number < 3 satisfies this, hence a(1) = 3.

Numbers k such that k and k+2 both have exactly m divisors: A001359 (m=2), A356742 (m=4), A356743 (m=6), A356744 (m=8).

Cf. A000005 (d(n)), A003680, A005238, A006558, A006601, A062832, A067888, A067889, A075036.

a={}; n=1; nmax=10; For[k=1, n<=nmax, k++, If[DivisorSigma[0, k] == DivisorSigma[0, k+2] == 2n, AppendTo[a, k]; k=1; n++]]; a (* Stefano Spezia, Aug 26 2022 *)
Flatten[Table[SequencePosition[DivisorSigma[0,Range[27*10^6]],{2n,,2n},1],{n,10}],1][[;;,1]] (* The program generates the first 10 terms of the sequence. To generate more, increase the Range constant but the program will take a long time to run. *) (* _Harvey P. Dale, Jul 01 2023 *)

a(n)=for(k=1,+oo,if(numdiv(k)==2*n&&numdiv(k+2)==2*n,return(k)))

More terms from Jinyuan Wang, Aug 28 2022

cp's OEIS Frontend

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

Views

Author

Keywords

Links

Crossrefs

Programs

Mathematica

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

Views

Author

Keywords

Comments

Examples

Links

Crossrefs

Formula

Extensions

A332314 Numbers k such that k, k + 1, k + 2 and k + 3 have the same number of divisors in Gaussian integers.

Views

Author

Keywords

Examples

Crossrefs

Programs

Mathematica

A332388 Numbers k such that k, k + 1, k + 2 and k + 3 have the same number of divisors in Eisenstein integers.

Views

Author

Keywords

Examples

Crossrefs

Programs

Mathematica

A338454 Starts of runs of 4 consecutive numbers with the same total binary weight of their divisors (A093653).

Views

Author

Keywords

Comments

Examples

Links

Crossrefs

Programs

Mathematica

A347603 Numbers k such that tau(k) = 2*tau(k-1) and sigma(k) = sigma(k-1), where tau(k) and sigma(k) are respectively the number and sum functions of the divisors of k.

Views

Author

Keywords

Comments

Examples

Links

Crossrefs

Programs

Mathematica

PARI

Python

A355712 Starts of runs of 4 consecutive numbers with the same number of 5-smooth divisors.

Views

Author

Keywords

Comments

Examples

Links

Crossrefs

Programs

Mathematica

PARI

A356766 Least number k such that k and k+2 both have exactly 2n divisors, or -1 if no such number exists.

Views

Author

Keywords

Examples

Crossrefs

Programs

Mathematica

PARI

Extensions