A006558 -id:A006558 - cp's OEIS frontend

A048892 Start of n consecutive integers with distinct number of divisors.

1, 1, 4, 9, 45, 76, 270, 2204, 3718, 95499, 590890, 16023339, 16475964, 1745175039, 31287652672, 347321438520, 2620400333120, 239919791836864

Offset: 1

Carlos Rivera

			The 5th number of this sequence, 45, means that 45, 46, 47, 48 and 49 all have distinct number of divisors: 6, 4, 2, 10 and 3, respectively.

Carlos Rivera, Problem 20. Divisors (II) K consecutive numbers with the same number of divisors, The Prime Puzzles and Problems Connection. See Problem 20c.

Cf. A000005, A006558.

More terms from Jud McCranie
a(15)-a(16) from Donovan Johnson, Feb 17 2010
a(17)-a(18) from Martin Ehrenstein, Aug 07 2023

A072507 Smallest start of n consecutive integers with n divisors, or 0 if no such number exists.

Original entry on oeis.org

1, 2, 0, 0, 0, 0, 0, 0, 0, 0, 0, 120402988681658048433948, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0

Offset: 1

Amarnath Murthy, Jul 22 2002

a(3) = 0 because only squares of primes have three divisors.
From T. D. Noe, Dec 04 2004: (Start)
"Note that a(n)=0 for odd n > 1 because a number has an odd number of divisors only if it is a square and there are no consecutive positive squares. Also, a(4)=0 because one of four consecutive numbers would be a multiple of 4 and have 4 divisors only if it is 8.
"Similarly, a(6)=0 because one of six consecutive number would be a multiple of 6 and the only multiples of 6 having 6 divisors are 12 and 18. For a(8), one of the eight consecutive numbers must be an odd multiple of 4, which cannot have 8 divisors. Interestingly, the 7 consecutive numbers starting at 171893 have 8 divisors.
"Similarly, for a(10), one of the ten consecutive numbers must be an odd multiple of 4, which would have 3x divisors. It is also easy to verify that a(n)=0 for n=14,16,20,22,26,28,32,34,... It seems likely that a(n)=0 for n>2."  (End)
This sequence is zero for all but finitely many n. If k = floor(log_2(n)), there must be at least one term exactly divisible by 2^j for any j < k; hence the number of divisors must be divisible by j+1, or more generally by lcm_{i<=k} i. The only values of n divisible by this lcm are 1,2,3,4,6,12,24,60 and 120. For example, for n=30, there must be an element divisible by exactly 8, so its number of divisors is divisible by 4. For n = 60, there must by two numbers 8k and 8(k+2) with k odd; then k and k+2 must each have 15 divisors, making them squares. Together with the comments from T. D. Noe, this leaves only 12, 24 and 120 as open questions. - Franklin T. Adams-Watters, Jul 14 2006
If a(120) = k > 0, then a) k+i cannot be 64 (mod 128) since 7 would divide tau(k+i); b) k+i cannot be 120 (mod 144) since then we'd need k+i = 24x^2 with x==2 (mod 3); c) k+i cannot be 168 (mod 288) since then we'd need k+i = 24x^2 with x==3 (mod 4). Hence no possibility (mod 288) exists, and a(120) = 0. - Hugo van der Sanden, Jan 12 2022
a(12) <= 247239052981730986799644. - Hugo van der Sanden, Apr 25 2022

			a(2) = 2 as 2 and 3 are the first (by chance the only) set of two consecutive integers with two divisors.

R. K. Guy, Unsolved Problems in Theory of Numbers, Springer-Verlag, Third Edition, 2004, B12.

Cf. A000005 (number of divisors of n).
Cf. A006558 (start of first run of n consecutive integers with same number of divisors).
Cf. A119479.

More terms from T. D. Noe, Dec 04 2004

a(12) to a(23) from Hugo van der Sanden, Dec 18 2022

A113465 Rectangular array read by antidiagonals: a(n, d) is the smallest number that starts an arithmetic progression with common difference d of n numbers with the same number of divisors.

Original entry on oeis.org

1, 2, 1, 33, 3, 1, 242, 3, 2, 1, 11605, 213, 119, 3, 1, 28374, 213, 3445, 3, 2, 1, 171893, 1383, 15026, 111, 77, 5, 1, 1043710445721, 3091, 74783, 201, 8718, 5, 8, 1

Offset: 1

David Wasserman, Jan 08 2006

First two columns are A006558 and A113466. First four rows are A000012, A065559, A113467 and A113468. a(9, 1) is unknown; the rest of the 9th antidiagonal is 8129,88015,201,8718,5,8,3,1.

			a(4, 3) = 3445 because 3445, 3448, 3451 and 3454 each have 8 divisors.

Cf. A006558, A065559, A113456, A113466, A113467, A113468.

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

A172438 Numbers k such that tau(k^2+1) - tau(k^2) = 1 where the function tau(k) is the number of positive divisors of k.

Original entry on oeis.org

1, 3, 5, 11, 19, 27, 29, 59, 61, 71, 79, 101, 125, 131, 139, 181, 199, 242, 243, 271, 333, 349, 379, 387, 409, 423, 449, 461, 477, 521, 569, 571, 603, 631, 641, 661, 739, 747, 751, 772, 788, 821, 881, 929, 991, 1017, 1031, 1039, 1051, 1058, 1069, 1075, 1083

Offset: 1

Michel Lagneau, Feb 02 2010

nonn

Square roots of perfect squares in A055927. [Juri-Stepan Gerasimov, Apr 06 2011]

			k=1, tau(2) - tau(1) = 2 - 1 = 1.
k=3, tau(10) - tau(9) = 4 - 3 = 1.
k=5, tau(26) - tau(25) = 4 - 3 = 1.
k=387, tau(149770)- tau(149769) = 16 - 15 = 1.

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.
T. M. Apostol, Introduction to Analytic Number Theory, Springer-Verlag, 1976, page 38.
D. S. Mitrinovic et al., Handbook of Number Theory, Kluwer, Chap. II. (For inequalities, etc.)

Marius A. Burtea, Table of n, a(n) for n = 1..5587
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].
G. E. Andrews, Some debts I owe, Séminaire Lotharingien de Combinatoire, Paper B42a, Issue 42, 2000; see (7.1).
P. A. MacMahon, Divisors of numbers and their continuations in the theory of partitions, Proc. London Math. Soc., 19 (1919), 75-113.
S. Ramanujan, On The Number Of Divisors Of A Number

Cf. A055927, A000005, A002183, A002182 (records), A005237, A006558.

Cf. A048691, A193432.

[m:m in [1..1100]| #Divisors(m^2+1) - #Divisors(m^2) eq 1]; // Marius A. Burtea, Jul 12 2019

with(numtheory): for n from 1 to 100000 do; if tau(n^2+1)-tau(n^2)= 1 then print(n); else fi ; od;

dsQ[n_]:=Module[{n2=n^2},DivisorSigma[0,n2+1]-DivisorSigma[0,n2]==1]; Select[Range[1200],dsQ] (* Harvey P. Dale, May 05 2011 *)
Select[Sqrt[#]&/@Flatten[Position[Partition[DivisorSigma[0,Range[1200000]],2,1],?(#[[2]]-#[[1]]==1&),1,Heads->False]],IntegerQ] (* _Harvey P. Dale, Apr 09 2022 *)

A218455 First of a run of 6 consecutive numbers with same prime signature.

Original entry on oeis.org

380480345, 2713001274, 6282718946, 7209536449, 9809067073, 10684724346, 12008728850, 14824913049, 17231547073, 17552118546, 17659180314, 18036555273, 20473171322, 21507097001, 23676804346, 24742649321, 25401767522, 25694056449, 27656894273, 28259097818

Offset: 1

M. F. Hasler, Oct 29 2012

nonn

A number n is in this sequence iff n and n+1 is in A218448; see the comment there for other characterizations in terms of membership in A175590 or A052214 or A052213.

Donovan Johnson, Table of n, a(n) for n = 1..1000

Cf. A034173, A034174, A006558, A083790, A052213, A052214, A175590, A218448, A218865.

is_A218455(n)={my(s(n)=vecsort(factor(n)[,2]),t=s(n));!for(m=n+1,n+5, t!=s(m) & return)}

a(2)-a(20) from Donovan Johnson, Oct 29 2012

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

A358044 a(n) is the smallest number k such that n consecutive integers starting at k have the same number of triangular divisors (A007862).

Original entry on oeis.org

1, 1, 55, 5402, 2515069

Offset: 1

Ilya Gutkovskiy, Oct 26 2022

Any subsequent terms are > 10^10. - Lucas A. Brown, Jan 06 2023

			55 has 2 triangular divisors {1, 55}, 56 has 2 triangular divisors {1, 28} and 57 has 2 triangular divisors {1, 3}. These are the first 3 consecutive numbers with the same number of triangular divisors, so a(3) = 55.

Index entries for sequences related to divisors of numbers
Lucas A. Brown, Python program.

Cf. A000217, A006558, A007862, A045983, A045984, A324593, A324594, A338628.

A173156 Numbers n such that max(tau(n),tau(n+1),tau(n+2),tau(n+3))- min(tau(n),tau(n+1),tau(n+2),tau(n+3)) = 1.

Original entry on oeis.org

2, 20164, 155236, 293761, 293762, 643204, 1435204, 1444802, 5216653, 6120676, 8421601, 8421602, 14047501, 15194404, 15984004, 17606413, 19114383, 22829284, 25786083, 25989602, 35259843, 35259844, 36264484, 41499364, 42876301, 44382241, 50523662, 50523663

Offset: 1

Michel Lagneau, Feb 11 2010

nonn

			For n = 20164, max(tau(20164),tau(20165),tau(20166),tau(20167)) - min(tau(20164),tau(20165),tau(20166),tau(20167)) = max(9,8,8,8) - min(9,8,8,8) = 1.

Giovanni Resta, Table of n, a(n) for n = 1..1000

Cf. A000005, A006558.

with(numtheory):for n from 200000 to 1500000 do;if max(tau(n),tau(n+1),tau(n+2),tau(n+3))- min(tau(n),tau(n+1),tau(n+2),tau(n+3))= 1 then print(n); else fi ; od;

Position[Partition[DivisorSigma[0,Range[5053*10^4]],4,1],?(Max[#]-Min[#] == 1&)]// Flatten (* _Harvey P. Dale, Jan 23 2023 *)

a(13)-a(28) from Giovanni Resta, Jun 12 2016

cp's OEIS Frontend

A048892 Start of n consecutive integers with distinct number of divisors.

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A072507 Smallest start of n consecutive integers with n divisors, or 0 if no such number exists.

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A113465 Rectangular array read by antidiagonals: a(n, d) is the smallest number that starts an arithmetic progression with common difference d of n numbers with the same number of divisors.

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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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A172438 Numbers k such that tau(k^2+1) - tau(k^2) = 1 where the function tau(k) is the number of positive divisors of k.

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A218455 First of a run of 6 consecutive numbers with same prime signature.

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

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