A045983 -id:A045983 - cp's OEIS frontend

A075042 Duplicate of A045983.

1, 2, 2, 2, 54, 91, 141, 141, 44360, 48919, 218972, 526095, 526095, 526095

Offset: 1

dead

A055927 Numbers k such that k + 1 has one more divisor than k.

1, 3, 9, 15, 25, 63, 121, 195, 255, 361, 483, 729, 841, 1443, 3363, 3481, 3721, 5041, 6241, 10201, 15625, 17161, 18224, 19321, 24963, 31683, 32761, 39601, 58564, 59049, 65535, 73441, 88208, 110889, 121801, 143641, 145923, 149769, 167281

Offset: 1

Labos Elemer, Jul 21 2000

nonn

Numbers k such that d(k+1) - d(k) = 1, where d(k) is A000005(k), the number of divisors.
Numbers k such that A049820(k) = A049820(k+1). - Jaroslav Krizek, Feb 10 2014
Numbers k such that A051950(k+1) = 1. - Danny Rorabaugh, Oct 05 2017

			a(4) = 15, as 15 has 4 and 16 has 5 divisors. a(6) = 63, as 63 and 64 have 6 and 7 divisors respectively.

Giovanni Resta, Table of n, a(n) for n = 1..10000 (first 1000 terms from Donovan Johnson)

Numbers where repetition occurs in A049820.

Cf. A000005, A006073, A045983, A049820, A075044.

Select[ Range[ 200000], DivisorSigma[0, # ] + 1 == DivisorSigma[0, # + 1] &]
Position[Differences[DivisorSigma[0,Range[170000]]],1]//Flatten (* Harvey P. Dale, Jul 06 2025 *)

for(n=1,1000,if(numdiv(n+1)-numdiv(n)==1,print1(n,", "))); /* Joerg Arndt, Apr 09 2011 */

More terms from David W. Wilson, Sep 06 2000, who remarks that every element is of form n^2 or n^2 - 1.

A075046 a(n) = the smallest number k such that the number of divisors of the n numbers from k through k+n-1 are in nondescending order.

Original entry on oeis.org

1, 1, 1, 1, 241, 241, 12853, 12853, 234613, 376741, 78312721, 125938261, 4019167441, 16586155153, 35237422882, 1296230533473, 42301168491121, 61118966262061

Offset: 1

Amarnath Murthy, Sep 03 2002

tau(k) <= tau(k+1) <= ... <= tau(k+n-1).
a(16) > 10^12. - Donovan Johnson, Oct 13 2009
a(17) > 10^13. - Giovanni Resta, Apr 12 2017
a(19) > 2.64*10^15. - Jud McCranie, Mar 27 2019
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

			a(5) = 241 = a(6) as tau(241) = 2 < tau(242) = tau(243) = tau(244) = tau(245) = 6 < tau(246).

Cf. A075028, A075029, A075031, A045983, A006073, A075044, A055927, A075047.

k = 1; Do[ While[t = Table[ DivisorSigma[0, i], {i, k, k + n - 1}]; t != Sort[t], k++ ]; Print[k], {n, 1, 11}]

More terms from Antonio G. Astudillo (afg_astudillo(AT)lycos.com), Apr 19 2003
a(11) from Robert G. Wilson v, Sep 07 2003
a(12)-a(15) from Donovan Johnson, Oct 13 2009
a(16) from Fred Schneider, Mar 29 2017
a(17)-a(18) from Jud McCranie, Mar 27 2019

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
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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
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See Links section.
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a(7) from Rémy Sigrist, Sep 04 2019

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

A323253 a(n) is the smallest number k such that factorizations of n consecutive integers starting at k have the same excess of number of primes counted with multiplicity over number of primes counted without multiplicity (A046660).

Original entry on oeis.org

1, 1, 1, 844, 74849, 671346, 8870025

Offset: 1

Ilya Gutkovskiy, Aug 30 2019

Smallest number k such that n or more consecutive integers starting at k have the same number of proper prime power divisors.
a(8) > 10^9. - Vaclav Kotesovec, Sep 01 2019
a(8) <= 254023231417746. - David A. Corneth, Sep 01 2019
a(8) > 10^13. - Giovanni Resta, Sep 05 2019

			671346 = 2 * 3^2 * 13 * 19 * 151,
671347 = 17^2 * 23 * 101,
671348 = 2^2 * 47 * 3571,
671349 = 3 * 7^2 * 4567,
671350 = 2 * 5^2 * 29 * 463,
671351 = 53^2 * 239.
These the first 6 consecutive numbers with the same number of proper prime power divisors, so a(6) = 671346.

Cf. A001221, A001222, A006558, A045983, A045984, A046660, A246547.

Do[find = 0; k = 0; While[find == 0, k++; If[Length[Union[Table[PrimeOmega[j] - PrimeNu[j], {j, k, k + n - 1}]]] == 1, find = 1; Print[k]]], {n, 1, 5}] (* Vaclav Kotesovec, Sep 01 2019 *)
(* faster program *) fak = Table[f = FactorInteger[j]; Total[Transpose[f][[2]]] - Length[f], {j, 1, 10000000}]; m = Max[fak]; Table[Min[Table[SequencePosition[fak, ConstantArray[j, n]], {j, 0, m}]], {n, 1, 7}] (* Vaclav Kotesovec, Sep 01 2019 *)

excess(n) = bigomega(n) - omega(n);
score(n) = my(t=excess(n)); for(k=1, oo, if(excess(n+k) != t, return(k)));
upto(nn) = my(n=1); for(k=1, nn, while(score(k) >= n, print1(k, ", "); n++)); \\ Daniel Suteu, Sep 01 2019

a(7) from Daniel Suteu and Vaclav Kotesovec, Sep 01 2019

A075044 a(0) = 1; a(n) = the smallest number k such that n numbers from k to k+n-1 have n distinct prime divisors, or 0 if no such number exists.

Original entry on oeis.org

1, 2, 14, 644, 134043, 129963314, 626804494291

Offset: 0

Amarnath Murthy, Sep 03 2002

a(7) <= 45164156742722455667280. - Giorgos Kalogeropoulos, Apr 10 2025

Cf. A006049, A045983, A006073.

Do[k = 1; While[Length /@ FactorInteger /@ Range[k, k+n-1] != Table[n, {n}], k++ ]; Print[k], {n, 0, 5}] (* Ryan Propper, Oct 01 2005 *)

Corrected and extended by Ryan Propper, Oct 01 2005

Offset corrected and a(6) from Donovan Johnson, Aug 03 2009

A304463 a(n) begins the first run of at least n consecutive numbers with the same number of bi-unitary divisors.

Original entry on oeis.org

1, 2, 2, 2, 91, 6850, 6850, 10281, 108771, 171890, 3760204, 3760204, 727940626, 5704384304, 13264434091, 13264434091, 63719307522, 287480681209, 607635436331

Offset: 1

Amiram Eldar, Aug 20 2018

The bi-unitary version of A006558.

a(20) > 5*10^12. - Giovanni Resta, Aug 23 2018

			a(5) = 91 since the number of bi-unitary divisors of 91, 92, 93, 94 and 95 is 4, and this is the first run of 5 consecutive numbers.

Cf. A006558, A045983 (equivalent for unitary divisors), A286324.

bdivnum[1] = 1; bdivnum[n_] := Times @@ ((# + Mod[#, 2]) & /@ Last /@ FactorInteger[n]); Seq[n_, q_] := Map[bdivnum, 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, bdivnum[n]]]; n++, {k, nmin, nmax}]; If[found, n - q, 0]]; seq = {1}; nmax = 100000000; Do[n1 = Last[seq]; s1 = findConsec[m, n1, nmax]; If[s1 == 0, Break[]]; AppendTo[seq, s1], {m, 2, 13}];

a(14)-a(19) from Giovanni Resta, Aug 23 2018

A338628 a(n) is the smallest number k such that n consecutive integers starting at k have the same number of square divisors (A046951).

Original entry on oeis.org

1, 1, 1, 844, 3624, 22020, 671346, 8870024, 264459172, 463239475, 1407472722, 108494875170, 12385053656370, 145065154350545

Offset: 1

Ilya Gutkovskiy, Nov 04 2020

			844 has 2 square divisors {1, 4}, 845 has 2 square divisors {1, 169}, 846 has 2 square divisors {1, 9} and 847 has 2 square divisors {1, 121}. These are the first 4 consecutive numbers with the same number of square divisors, so a(4) = 844.

Index entries for sequences related to divisors of numbers

Cf. A006558, A045983, A045984, A046951, A324593, A324594.

Do[find = 0; k = 0; While[find == 0, k++; If[Length[Union[Table[Length[Select[Divisors[j], IntegerQ[Sqrt[#]] &]], {j, k, k + n - 1}]]] == 1, find = 1; Print[k]]], {n, 1, 7}]

isok(n, k) = #Set(apply(x->sumdiv(x, d, issquare(d)), vector(n, i, k+i-1))) == 1;
a(n) = my(k=1); while(! isok(n, k), k++); k; \\ Michel Marcus, Nov 05 2020

a(8)-a(11) from Amiram Eldar, Nov 04 2020

a(12)-a(14) from Martin Ehrenstein, Jul 19 2023

cp's OEIS Frontend

A075042 Duplicate of A045983.

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A055927 Numbers k such that k + 1 has one more divisor than k.

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A075046 a(n) = the smallest number k such that the number of divisors of the n numbers from k through k+n-1 are in nondescending order.

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

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

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A323253 a(n) is the smallest number k such that factorizations of n consecutive integers starting at k have the same excess of number of primes counted with multiplicity over number of primes counted without multiplicity (A046660).

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A075044 a(0) = 1; a(n) = the smallest number k such that n numbers from k to k+n-1 have n distinct prime divisors, or 0 if no such number exists.

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A304463 a(n) begins the first run of at least n consecutive numbers with the same number of bi-unitary divisors.

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