A342028 -id:A342028 - cp's OEIS frontend

A130091 Numbers having in their canonical prime factorization mutually distinct exponents.

1, 2, 3, 4, 5, 7, 8, 9, 11, 12, 13, 16, 17, 18, 19, 20, 23, 24, 25, 27, 28, 29, 31, 32, 37, 40, 41, 43, 44, 45, 47, 48, 49, 50, 52, 53, 54, 56, 59, 61, 63, 64, 67, 68, 71, 72, 73, 75, 76, 79, 80, 81, 83, 88, 89, 92, 96, 97, 98, 99, 101, 103, 104, 107, 108, 109, 112, 113, 116

Offset: 1

Reinhard Zumkeller, May 06 2007

nonn

This sequence does not contain any number of the form 36n-6 or 36n+6, as such numbers are divisible by 6 but not by 4 or 9. Consequently, this sequence does not contain 24 consecutive integers. The quest for the greatest number of consecutive integers in this sequence has ties to the ABC conjecture (see the MathOverflow link). - Danny Rorabaugh, Sep 23 2015
The Heinz number of an integer partition (y_1,...,y_k) is prime(y_1)*...*prime(y_k), so these are Heinz numbers of integer partitions with distinct multiplicities. The enumeration of these partitions by sum is given by A098859. - Gus Wiseman, May 04 2019
Aktaş and Ram Murty (2017) called these terms "special numbers" ("for lack of a better word"). They prove that the number of terms below x is ~ c*x/log(x), where c > 1 is a constant. - Amiram Eldar, Feb 25 2021
Sequence A005940(1+A328592(n)), n >= 1, sorted into ascending order. - Antti Karttunen, Apr 03 2022

			From _Gus Wiseman_, May 04 2019: (Start)
The sequence of terms together with their prime indices begins:
   1: {}
   2: {1}
   3: {2}
   4: {1,1}
   5: {3}
   7: {4}
   8: {1,1,1}
   9: {2,2}
  11: {5}
  12: {1,1,2}
  13: {6}
  16: {1,1,1,1}
  17: {7}
  18: {1,2,2}
  19: {8}
  20: {1,1,3}
  23: {9}
  24: {1,1,1,2}
  25: {3,3}
  27: {2,2,2}
(End)

Reinhard Zumkeller, Table of n, a(n) for n = 1..10000
Kevser Aktaş and M. Ram Murty, On the number of special numbers, Proceedings - Mathematical Sciences, Vol. 127, No. 3 (2017), pp. 423-430; alternative link.
MathOverflow, Consecutive numbers with mutually distinct exponents in their canonical prime factorization
Carlo Sanna, On the number of distinct exponents in the prime factorization of an integer, arXiv:1902.09224 [math.NT], 2019.
Eric Weisstein's World of Mathematics, Prime Factorization

Complement of A130092. A351564 is the characteristic function.
Subsequence of A351294.
Cf. A000961, A006939, A181818, A304686, A319161, A342028, A342029, A342030, A342031, A342032 (subsequences).
Cf. A005940, A048767, A048768, A056239, A098859, A112798, A118914, A181796, A217605, A325326, A325337, A325368, A327498, A327523, A328592, A336423, A336424, A336569, A336570, A336571, A343012, A343013.

filter:= proc(t) local f;
f:= map2(op,2,ifactors(t)[2]);
nops(f) = nops(convert(f,set));
end proc:
select(filter, [$1..1000]); # Robert Israel, Mar 30 2015

t[n_] := FactorInteger[n][[All, 2]]; Select[Range[400],  Union[t[#]] == Sort[t[#]] &]  (* Clark Kimberling, Mar 12 2015 *)

isok(n) = {nbf = omega(n); f = factor(n); for (i = 1, nbf, for (j = i+1, nbf, if (f[i, 2] == f[j, 2], return (0)););); return (1);} \\ Michel Marcus, Aug 18 2013

isA130091(n) = issquarefree(factorback(apply(e->prime(e), (factor(n)[, 2])))); \\ Antti Karttunen, Apr 03 2022

a(n) < A130092(n) for n<=150, a(n) > A130092(n) for n>150.

A342029 Starts of runs of 3 consecutive numbers that have mutually distinct exponents in their prime factorization (A130091).

Original entry on oeis.org

1, 2, 3, 7, 11, 16, 17, 18, 23, 27, 43, 47, 48, 52, 71, 79, 96, 97, 107, 135, 147, 151, 162, 171, 191, 241, 242, 243, 331, 351, 359, 367, 387, 423, 431, 486, 507, 539, 547, 567, 575, 576, 599, 603, 639, 907, 927, 1051, 1107, 1123, 1151, 1215, 1249, 1250, 1323

Offset: 1

Amiram Eldar, Feb 25 2021

nonn

			2 is a term since 2, 3 and 4 = 2^2 all have a single exponent in their prime factorization.
4 is not a term since in the run {4, 5, 6} the third member 6 = 2*3 has two equal exponents (1) in its prime factorization.

Amiram Eldar, Table of n, a(n) for n = 1..10000
Kevser Aktaş and M. Ram Murty, On the number of special numbers, Proceedings - Mathematical Sciences, Vol. 127, No. 3 (2017), pp. 423-430; alternative link.
Bernardo Recamán Santos, Consecutive numbers with mutually distinct exponents in their canonical prime factorization, MathOverflow, Mar 30 2015.

Subsequence of A130091 and A342028.

Subsequences: A342030, A342031.

q[n_] := Length[(e = FactorInteger[n][[;; , 2]])] == Length[Union[e]]; v = q /@ Range[3]; seq = {}; Do[If[And @@ v, AppendTo[seq, k - 3]]; v = Join[Rest[v], {q[k]}], {k, 4, 1500}]; seq

A342030 Starts of runs of 4 consecutive numbers that have mutually distinct exponents in their prime factorization (A130091).

Original entry on oeis.org

1, 2, 16, 17, 47, 96, 241, 242, 575, 1249, 2644, 2645, 4049, 4372, 4373, 4799, 9124, 12248, 33749, 72250, 120049, 130436, 281249, 303748, 1431124, 1431125, 1531250, 2101247, 3693761, 4085656, 4910975, 12502348, 12502349, 14268481, 22997761, 25486324, 26693549

Offset: 1

Amiram Eldar, Feb 25 2021

nonn

			2 is a term since 2, 3, 4 = 2^2, and 5 all have a single exponent in their prime factorization.
3 is not a term since in the run {3, 4, 5, 6} the fourth member 6 = 2*3 has two equal exponents (1) in its prime factorization.

Amiram Eldar, Table of n, a(n) for n = 1..196 (terms below 10^11)
Kevser Aktaş and M. Ram Murty, On the number of special numbers, Proceedings - Mathematical Sciences, Vol. 127, No. 3 (2017), pp. 423-430; alternative link.
Bernardo Recamán Santos, Consecutive numbers with mutually distinct exponents in their canonical prime factorization, MathOverflow, Mar 30 2015.

Subsequence of A130091, A342028 and A342029.

A342031 is a subsequence.

q[n_] := Length[(e = FactorInteger[n][[;; , 2]])] == Length[Union[e]]; v = q /@ Range[4]; seq = {}; Do[If[And @@ v, AppendTo[seq, k - 4]]; v = Join[Rest[v], {q[k]}], {k, 5, 10^5}]; seq

A342031 Starts of runs of 5 consecutive numbers that have mutually distinct exponents in their prime factorization (A130091).

Original entry on oeis.org

1, 16, 241, 2644, 4372, 1431124, 12502348, 112753348, 750031648, 2844282247, 5882272324, 6741230497, 8004453748, 87346072024, 130489991521, 218551872247, 245127093748, 460925878624, 804065433748, 1176638279524, 2210511903748, 2404792968748, 2483167488748, 3121595927521

Offset: 1

Amiram Eldar, Feb 25 2021

nonn

Bernardo Recamán Santos (2015) showed that there is no run of more than 23 consecutive numbers, since numbers of the form 36*k - 6 and 36*k + 6 do not have distinct exponents. Pace Nielsen and Adam P. Goucher showed that there can be only finitely many runs of 23 consecutive numbers (see MathOverflow link).

Aktaş and Ram Murty (2017) gave an explicit upper bound to such a run of 23 numbers. They found the first 5 terms of this sequence (and stated that there are a few more known up to 7*10^8), and said that we may conjecture (based on numerical evidence) that there are no 6 consecutive numbers.

			16 is a term since 16 = 2^4, 17, 18 = 2*3^2, 19 and 20 = 2^2*5 all have distinct exponents in their prime factorization.

Martin Ehrenstein, Table of n, a(n) for n = 1..43
Kevser Aktaş and M. Ram Murty, On the number of special numbers, Proceedings - Mathematical Sciences, Vol. 127, No. 3 (2017), pp. 423-430; alternative link.
Bernardo Recamán Santos, Consecutive numbers with mutually distinct exponents in their canonical prime factorization, MathOverflow, Mar 30 2015.

Subsequence of A130091, A342028, A342029 and A342030.

q[n_] := Length[(e = FactorInteger[n][[;; , 2]])] == Length[Union[e]]; v = q /@ Range[5]; seq = {}; Do[If[And @@ v, AppendTo[seq, k - 5]]; v = Join[Rest[v], {q[k]}], {k, 6, 1.3*10^6}]; seq

a(15) and beyond from Martin Ehrenstein, Mar 08 2021

A359747 Numbers k such that k*(k+1) has in its canonical prime factorization mutually distinct exponents.

Original entry on oeis.org

1, 3, 4, 7, 8, 16, 24, 27, 31, 48, 63, 71, 72, 107, 108, 124, 127, 199, 242, 243, 256, 400, 431, 432, 499, 512, 576, 647, 783, 863, 967, 971, 1024, 1151, 1152, 1372, 1567, 1600, 1999, 2187, 2311, 2400, 2591, 2592, 2887, 2916, 3087, 3136, 3456, 3887, 3888, 3968, 4000

Offset: 1

Amiram Eldar, Jan 13 2023

nonn

Equivalently, numbers k such that A002378(k) = k*(k+1) is a term of A130091.
Equivalently, numbers k such that the multisets of exponents in the prime factorizations of k and k+1 are disjoint and each have distinct elements.
Either k or k+1 is a powerful number (A001694). Except for k=8, are there terms k such that both k and k+1 are powerful (i.e., terms that are also in A060355)? None of the terms A060355(n) for n = 2..39 is in this sequence.

			3 is a term since 3*4 = 12 = 2^2 * 3^1 has 2 distinct exponents in its prime factorization: 1 and 3.

Amiram Eldar, Table of n, a(n) for n = 1..1000

Subsequence of A130091 and A342028.
A359748 is a subsequence.
Cf. A001694, A002378, A060355.

q[n_] := UnsameQ @@ (FactorInteger[n*(n+1)][[;; , 2]]); Select[Range[4000], q]

is(n) = { my(e = factor(n*(n+1))[, 2]); #Set(e) == #e; }

A368402 Numbers k such that k and k+1 are both in A268375.

Original entry on oeis.org

1, 2, 3, 4, 7, 8, 11, 12, 16, 17, 18, 19, 27, 28, 31, 43, 44, 47, 48, 49, 52, 63, 67, 75, 79, 80, 97, 98, 112, 116, 124, 127, 147, 148, 162, 163, 171, 172, 175, 191, 192, 207, 211, 241, 242, 243, 244, 256, 268, 271, 283, 288, 292, 316, 324, 331, 332, 337, 367

Offset: 1

Amiram Eldar, Dec 23 2023

Analogous to A342028, as A268375 is analogous to A130091.

The Mersenne primes (A000668) are terms.

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

Subsequence of A130091, A268375 and A342028.

Subsequences: A000668, A368403, A368404.

f[e_] := Position[Reverse[IntegerDigits[e, 2]], 1] // Flatten; q[n_] := q[n] = UnsameQ @@ Flatten[f /@ FactorInteger[n][[;; , 2]]]; Select[Range[100], q[#] && q[#+1] &]

isA268375(n) = {my(e = factor(n)[,2], b = 0); for(i=1, #e, b = bitor(b, e[i])); n == 1 || b == vecsum(e);}
lista(kmax) = {my(is1 = 0, is2); for(k = 1, kmax, is2 = isA268375(k); if(is1 && is2, print1(k-1, ", ")); is1 = is2);}

A359748 Numbers k such that k and k+1 are both in A359747.

Original entry on oeis.org

3, 7, 71, 107, 242, 431, 1151, 2591, 3887, 21599, 49391, 76831, 79999, 107647, 139967, 179999, 197567, 268911, 345599, 346111, 401407, 438047, 472391, 995327, 1031047, 1143071, 1249999, 1254527, 1349999, 1438207, 1685447, 2056751, 2411207, 2829887, 3269807, 4464071

Offset: 1

Amiram Eldar, Jan 13 2023

nonn

Are there 3 terms in A359747 that are consecutive integers?

			7 is a term since 7*8 = 56 = 2^3 * 3^1 has 2 distinct exponents in its prime factorization (1 and 3) and 8*9 = 72 = 2^3 * 3^2 also has 2 distinct exponents in its prime factorization (2 and 3).

Amiram Eldar, Table of n, a(n) for n = 1..100

Subsequence of A130091, A342028, A342029 and A359748.

q[n_] := UnsameQ @@ (FactorInteger[n*(n+1)][[;; , 2]]); Select[Range[10^5], q[#] && q[#+1] &]

is(n) = { my(e1 = factor(n*(n+1))[, 2], e2 = factor((n+1)*(n+2))[, 2]); #Set(e1) == #e1 && #Set(e2) == #e2; }

A342032 Numbers with mutually distinct exponents in their prime factorization (A130091) with a record gap to the next term of A130091.

Original entry on oeis.org

1, 5, 13, 32, 200, 212, 1759, 2313, 5351, 16144, 51071, 128056, 159233, 630737, 1555349, 1627984, 2666309, 6838261, 12243457, 14619901, 25282087, 65891668, 78971281, 121377079, 543433039, 684779072, 1675445647, 2078471579, 2228572121, 11135788439, 42801667036

Offset: 1

Amiram Eldar, Feb 25 2021

nonn

This sequence is infinite since the asymptotic density of A130091 is 0.

The corresponding values of the record gaps are 1, 2, 3, 5, 7, 11, 13, 20, 22, 29, 33, 40, 51, 55, 59, 67, 72, 82, 84, 87, 100, 121, 126, 132, 138, 147, 149, 150, 195, 209, 211, ...

			a(1) = 1 since both 1 and 1+1 = 2 are in A130091.
a(2) = 5 since 5 and 5+2 = 7 are in A130091 and 6 is not.
a(3) = 13 since 13 and 13+3 = 16 are in A130091 and 14 and 15 are not.

Cf. A130091, A342028, A342029, A342030, A342031.

q[n_] := Length[(e = FactorInteger[n][[;; , 2]])] == Length[Union[e]]; seq = {}; m = 1; dm = 0; Do[If[q[n], d = n - m; If[d > dm, dm = d; AppendTo[seq, m]]; m = n], {n, 1, 10^6}]; seq

cp's OEIS Frontend

A130091 Numbers having in their canonical prime factorization mutually distinct exponents.

Views

Author

Keywords

Comments

Examples

Links

Crossrefs

Programs

Maple

Mathematica

PARI

PARI

Formula

A342029 Starts of runs of 3 consecutive numbers that have mutually distinct exponents in their prime factorization (A130091).

Views

Author

Keywords

Examples

Links

Crossrefs

Programs

Mathematica

A342030 Starts of runs of 4 consecutive numbers that have mutually distinct exponents in their prime factorization (A130091).

Views

Author

Keywords

Examples

Links

Crossrefs

Programs

Mathematica

A342031 Starts of runs of 5 consecutive numbers that have mutually distinct exponents in their prime factorization (A130091).

Views

Author

Keywords

Comments

Examples

Links

Crossrefs

Programs

Mathematica

Extensions

A359747 Numbers k such that k*(k+1) has in its canonical prime factorization mutually distinct exponents.

Views

Author

Keywords

Comments

Examples

Links

Crossrefs

Programs

Mathematica

PARI

A368402 Numbers k such that k and k+1 are both in A268375.

Views

Author

Keywords

Comments

Links

Crossrefs

Programs

Mathematica

PARI

A359748 Numbers k such that k and k+1 are both in A359747.

Views

Author

Keywords

Comments

Examples

Links

Crossrefs

Programs

Mathematica

PARI

A342032 Numbers with mutually distinct exponents in their prime factorization (A130091) with a record gap to the next term of A130091.

Views

Author

Keywords