A052409 -id:A052409 - cp's OEIS frontend

A295924 Number of twice-factorizations of n of type (R,P,R).

1, 1, 1, 3, 1, 1, 1, 4, 3, 1, 1, 1, 1, 1, 1, 8, 1, 1, 1, 1, 1, 1, 1, 1, 3, 1, 4, 1, 1, 1, 1, 8, 1, 1, 1, 3, 1, 1, 1, 1, 1, 1, 1, 1, 1, 1, 1, 1, 3, 1, 1, 1, 1, 1, 1, 1, 1, 1, 1, 1, 1, 1, 1, 17, 1, 1, 1, 1, 1, 1, 1, 1, 1, 1, 1, 1, 1, 1, 1, 1, 8, 1, 1, 1, 1, 1, 1, 1, 1, 1, 1, 1, 1, 1, 1, 1, 1, 1, 1, 3, 1, 1, 1, 1, 1

Offset: 1

Gus Wiseman, Nov 30 2017

nonn

a(n) is the number of ways to choose an integer partition of a divisor of A052409(n).

			The a(16) = 8 twice-factorizations are (2)*(2)*(2)*(2), (2)*(2)*(2*2), (2)*(2*2*2), (2*2)*(2*2), (2*2*2*2), (4)*(4), (4*4), (16).

Antti Karttunen, Table of n, a(n) for n = 1..65537
Index entries for sequences computed from exponents in factorization of n

Cf. A000005, A000041, A001055, A047968, A052409, A052410, A089723, A281113, A284639, A295923, A295931, A295935, A296134.

Table[DivisorSum[GCD@@FactorInteger[n][[All,2]],PartitionsP],{n,100}]

A052409(n) = { my(k=ispower(n)); if(k, k, n>1); }; \\ From A052409
A295924(n) = if(1==n,n,sumdiv(A052409(n),d,numbpart(d))); \\ Antti Karttunen, Jul 29 2018

a(1) = 1; for n > 1, a(n) = Sum_{d|A052409(n)} A000041(d). - Antti Karttunen, Jul 29 2018

More terms from Antti Karttunen, Jul 29 2018

A294337 Number of ways to write 2^n as a finite power-tower a^(b^(c^...)) of positive integers greater than one.

Original entry on oeis.org

1, 2, 2, 4, 2, 4, 2, 6, 4, 4, 2, 7, 2, 4, 4, 10, 2, 7, 2, 7, 4, 4, 2, 10, 4, 4, 6, 7, 2, 8, 2, 12, 4, 4, 4, 12, 2, 4, 4, 10, 2, 8, 2, 7, 7, 4, 2, 15, 4, 7, 4, 7, 2, 10, 4, 10, 4, 4, 2, 13, 2, 4, 7, 16, 4, 8, 2, 7, 4, 8, 2, 16, 2, 4, 7, 7, 4, 8, 2, 15, 10, 4, 2, 13, 4, 4, 4, 10, 2, 13, 4, 7, 4, 4, 4, 18, 2, 7, 7, 12, 2, 8, 2, 10, 8

Offset: 1

Gus Wiseman, Oct 28 2017

nonn

			The a(12) = 7 ways are: 2^12, 4^6, 8^4, 8^(2^2), 16^3, 64^2, 4096.

Antti Karttunen, Table of n, a(n) for n = 1..65537

Cf. A001597, A007916, A052409, A052410, A089723, A164336, A277562, A284639, A288636, A294336, A294338, A294339.

a[n_]:=1+Sum[a[g],{g,Rest[Divisors[GCD@@FactorInteger[n][[All,2]]]]}];
Table[a[2^n],{n,100}]

A052409(n) = { my(k=ispower(n)); if(k, k, n>1); }; \\ From A052409
A294336(n) = if(1==n,n,sumdiv(A052409(n),d,A294336(d)));
A294337(n) = sumdiv(n,d,A294336(d));
\\ Or alternatively, after Mathematica-code as:
A294337(n) = A294336(2^n); \\ Antti Karttunen, Jun 12 2018

a(n) = Sum_{d|n} A294336(d) = A294336(A000079(n)). - Antti Karttunen, Jun 12 2018

More terms from Antti Karttunen, Jun 12 2018

A295931 Number of ways to write n in the form n = (x^y)^z where x, y, and z are positive integers.

Original entry on oeis.org

1, 1, 1, 3, 1, 1, 1, 3, 3, 1, 1, 1, 1, 1, 1, 6, 1, 1, 1, 1, 1, 1, 1, 1, 3, 1, 3, 1, 1, 1, 1, 3, 1, 1, 1, 3, 1, 1, 1, 1, 1, 1, 1, 1, 1, 1, 1, 1, 3, 1, 1, 1, 1, 1, 1, 1, 1, 1, 1, 1, 1, 1, 1, 9, 1, 1, 1, 1, 1, 1, 1, 1, 1, 1, 1, 1, 1, 1, 1, 1, 6, 1, 1, 1, 1, 1, 1

Offset: 1

Gus Wiseman, Nov 29 2017

nonn

By convention a(1) = 1.

Values can be 1, 3, 6, 9, 10, 15, 18, 21, 27, 28, 30, 36, 45, 54, 60, 63, 84, 90, etc. - Robert G. Wilson v, Dec 10 2017

			The a(256) = 10 ways are:
(2^1)^8    (2^2)^4   (2^4)^2  (2^8)^1
(4^1)^4    (4^2)^2   (4^4)^1
(16^1)^2   (16^2)^1
(256^1)^1

Robert Israel, Table of n, a(n) for n = 1..10000

Cf. A000005, A007425, A052409, A052410, A089723, A093771, A175082, A277562, A281113, A284639, A294786, A294336, A294338, A295920, A295923, A295924, A295935.

f:= proc(n) local m,d,t;
  m:= igcd(seq(t[2],t=ifactors(n)[2]));
  add(numtheory:-tau(d),d=numtheory:-divisors(m))
end proc:
f(1):= 1:
map(f, [$1..100]); # Robert Israel, Dec 19 2017

Table[Sum[DivisorSigma[0,d],{d,Divisors[GCD@@FactorInteger[n][[All,2]]]}],{n,100}]

a(A175082(k)) = 1, a(A093771(k)) = 3.

a(n) = Sum_{d|A052409(n)} A000005(d).

A128164 Least k > 2 such that (n^k - 1)/(n-1) is prime, or 0 if no such prime exists.

Original entry on oeis.org

3, 3, 0, 3, 3, 5, 3, 0, 19, 17, 3, 5, 3, 3, 0, 3, 25667, 19, 3, 3, 5, 5, 3, 0, 7, 3, 5, 5, 5, 7, 0, 3, 13, 313, 0, 13, 3, 349, 5, 3, 1319, 5, 5, 19, 7, 127, 19, 0, 3, 4229, 103, 11, 3, 17, 7, 3, 41, 3, 7, 7, 3, 5, 0, 19, 3, 19, 5, 3, 29, 3, 7, 5, 5, 3, 41, 3, 3, 5, 3, 0, 23, 5, 17, 5, 11, 7, 61, 3, 3

Offset: 2

Alexander Adamchuk, Feb 20 2007

nonn

a(n) = A084740(n) for all n except n = p-1, where p is an odd prime, for which A084740(n) = 2.
All nonzero terms are odd primes.
a(n) = 0 for n = {4,9,16,25,32,36,49,64,81,100,121,125,144,...}, which are the perfect powers with exceptions of the form n^(p^m) where p>2 and (n^(p^(m+1))-1)/(n^(p^m)-1) are prime and m>=1 (in which case a(n^(p^m))=p). - Max Alekseyev, Jan 24 2009
a(n) = 3 for n in A002384, i.e., for n such that n^2 + n + 1 is prime.
a(152) > 20000. - Eric Chen, Jun 01 2015
a(n) is the least number k such that (n^k - 1)/(n-1) is a Brazilian prime, or 0 if no such Brazilian prime exists. - Bernard Schott, Apr 23 2017
These corresponding Brazilian primes are in A285642. - Bernard Schott, Aug 10 2017
a(152) = 270217, see the top PRP link. - Eric Chen, Jun 04 2018
a(184) = 16703, a(200) = 17807, a(210) = 19819, a(306) = 26407, a(311) = 36497, a(326) = 26713, a(331) = 25033; a(185) > 66337, a(269) > 63659, a(281) > 63421, and there are 48 unknown a(n) for n <= 1024. - Eric Chen, Jun 04 2018
Six more terms found: a(522)=20183, a(570)=12907, a(684)=22573, a(731)=15427, a(820)=12043, a(996)=14629. - Michael Stocker, Apr 09 2020

			a(7) = 5 because (7^5 - 1)/6 = 2801 = 11111_7 is prime and (7^k - 1)/6 = 1, 8, 57, 400 for k = 1, 2, 3, 4. - _Bernard Schott_, Apr 23 2017

Max Alekseyev and Eric Chen, Table of n, a(n) for n = 2..184 (terms 2..151 from Max Alekseyev)
Eric Chen, Table of n, a(n) for n = 2..1024 status (updated by Jinyuan Wang)
H. Dubner, Generalized repunit primes, Math. Comp., 61 (1993), 927-930.
Richard Fischer, Generalized repunit primes of the form (B^N-1)/(B-1)
Top PRPs, Search by (152^n-1)/(152-1)
Top PRPs, Search by (b^n-1)/a
Eric Weisstein's World of Mathematics, Repunit

Cf. A084738, A065854, A084740, A084741, A065507, A084742, A066180, A084732, A285642, A085104.
Cf. A002384, A049409, A100330, A162862, A217070-A217089. (numbers b such that (b^p-1)/(b-1) is prime for prime p = 3 to 97)
Cf. A000043, A028491, A004061, A004062, A004063, A004023, A005808, A004064, A016054, A006032, A006033, A006034, A133857, A006035, A127995, A127996, A127997, A204940, A127998, A127999, A128000, A181979, A098438, A128002, A209120, A185073, A128003, A128004, A181987, A128005, A239637, A240765, A294722, A242797, A243279, A267375, A245237, A245442, A173767. (numbers n such that (b^n-1)/(b-1) is prime for b = 2 to 53)
A126589 gives locations of zeros.

Table[Function[m, If[m > 0, k = 3; While[! PrimeQ[(m^k - 1)/(m - 1)], k++]; k, 0]]@ If[Set[e, GCD @@ #[[All, -1]]] > 1, {#, IntegerExponent[n, #]} &@ Power[n, 1/e] /. {{k_, m_} /; Or[Not[PrimePowerQ@ m], Prime@ m, FactorInteger[m][[1, 1]] == 2] :> 0, {k_, m_} /; m > 1 :> n}, n] &@ FactorInteger@ n, {n, 2, 17}] (* Michael De Vlieger, Apr 24 2017 *)

a052409(n) = my(k=ispower(n)); if(k, k, n>1)
a052410(n) = if (ispower(n, , &r), r, n)
is(n) = issquare(n) || (ispower(n) && !ispseudoprime((n^a052410(a052409(n))-1)/(n-1)))
a(n) = if(is(n), 0, forprime(p=3, 2^16, if(ispseudoprime((n^p-1)/(n-1)), return(p)))) \\ Eric Chen, Jun 01 2015, corrected by Eric Chen, Jun 04 2018, after Charles R Greathouse IV in A052409 and Michel Marcus in A052410

a(18) = 25667 found by Henri Lifchitz, Sep 26 2007

A253641 Largest integer b such that n=a^b for some integer a; a(0)=a(1)=1 by convention.

Original entry on oeis.org

1, 1, 1, 1, 2, 1, 1, 1, 3, 2, 1, 1, 1, 1, 1, 1, 4, 1, 1, 1, 1, 1, 1, 1, 1, 2, 1, 3, 1, 1, 1, 1, 5, 1, 1, 1, 2, 1, 1, 1, 1, 1, 1, 1, 1, 1, 1, 1, 1, 2, 1, 1, 1, 1, 1, 1, 1, 1, 1, 1, 1, 1, 1, 1, 6, 1, 1, 1, 1, 1, 1, 1, 1, 1, 1, 1, 1, 1, 1, 1, 1, 4, 1, 1, 1, 1, 1, 1, 1, 1, 1

Offset: 0

M. F. Hasler, Jan 25 2015

A000005(a(n))-1 yields the number of times n is listed in A072103, i.e., the number of ways it can be written differently as perfect power.
For any n, a(n) >= 1 since n = n^1.
Integers n = 0 and n = 1 can be written as n^b with arbitrarily large b; to remain consistent with the preceding formula and the comment, the conventional choice a(n) = 1 seemed the best one.
The same as A052409 if the convention is dropped. - R. J. Mathar, Jan 29 2015

			a(4) = 2 since 4 = 2^2. a(64) = 6 since 64 = 2^6 (although also 64 = 4^3 = 8^2).

Antti Karttunen, Table of n, a(n) for n = 0..10000

Cf. A001597, A072103, A175064, A253642, A052409.

a:=proc(n) if n in {0, 1} then 1 else igcd(map(i->i[2], ifactors(n)[2])[]); fi; end: seq(a(n), n=0..120); # Ridouane Oudra, Jun 10 2025

PARI
```
A253641(n)=max(ispower(n),1)
```

from math import gcd
from sympy import factorint
def A253641(n): return gcd(*factorint(n).values()) if n>1 else 1 # Chai Wah Wu, Aug 13 2024

A302291 a(n) is the period of the binary expansion of n.

Original entry on oeis.org

1, 1, 2, 1, 3, 3, 3, 1, 4, 4, 2, 4, 4, 4, 4, 1, 5, 5, 5, 5, 5, 5, 5, 5, 5, 5, 5, 5, 5, 5, 5, 1, 6, 6, 6, 6, 3, 6, 6, 6, 6, 6, 2, 6, 6, 3, 6, 6, 6, 6, 6, 6, 6, 6, 3, 6, 6, 6, 6, 6, 6, 6, 6, 1, 7, 7, 7, 7, 7, 7, 7, 7, 7, 7, 7, 7, 7, 7, 7, 7, 7, 7, 7, 7, 7, 7, 7

Offset: 0

Rémy Sigrist, Apr 04 2018

Zero is assumed to be represented as 0; otherwise, leading zeros are ignored.

See A302295 for the variant where leading zeros are allowed.

			The first terms, alongside the binary expansion of n with periodic part in parentheses, are:
  n  a(n)    bin(n)
  -- ----    ------
   0    1    (0)
   1    1    (1)
   2    2    (10)
   3    1    (1)(1)
   4    3    (100)
   5    3    (101)
   6    3    (110)
   7    1    (1)(1)(1)
   8    4    (1000)
   9    4    (1001)
  10    2    (10)(10)
  11    4    (1011)
  12    4    (1100)
  13    4    (1101)
  14    4    (1110)
  15    1    (1)(1)(1)(1)
  16    5    (10000)
  17    5    (10001)
  18    5    (10010)
  19    5    (10011)
  20    5    (10100)

Rémy Sigrist, Table of n, a(n) for n = 0..10000
Index entries for sequences related to binary expansion of n

Aperiodic compositions are counted by A000740.
Aperiodic binary words are counted by A027375.
The orderless period of prime indices is A052409.
Numbers whose binary expansion is periodic are A121016.
Periodic compositions are counted by A178472.
Numbers whose prime signature is aperiodic are A329139.
Compositions by number of distinct rotations are A333941.
All of the following pertain to compositions in standard order (A066099):
- Length is A000120.
- Necklaces are A065609.
- Sum is A070939.
- Runs are counted by A124767.
- Rotational symmetries are counted by A138904.
- Strict compositions are A233564.
- Constant compositions are A272919.
- Lyndon compositions are A275692.
- Co-Lyndon compositions are A326774.
- Aperiodic compositions are A328594.
- Rotational period is A333632.
- Co-necklaces are A333764.
- Reversed necklaces are A333943.
Cf. A000031, A001037, A008965, A019536, A020330, A211100, A302295, A328595, A328596, A329312, A329313, A329326.

Table[If[n==0,1,Length[Union[Array[RotateRight[IntegerDigits[n,2],#]&,IntegerLength[n,2]]]]],{n,0,50}] (* Gus Wiseman, Apr 19 2020 *)

a(n) = my (l=max(1, #binary(n))); fordiv (l, w, if (#Set(digits(n, 2^w))<=1, return (w)))

a(n) = A070939(n) / A138904(n).
a(2^n) = n + 1 for any n >= 0.
a(2^n - 1) = 1 for any n >= 0.
a(A020330(n)) = a(n) for any n > 0.

A316112 Number of leaves in the free pure symmetric multifunction (with empty expressions allowed) with e-number n.

Original entry on oeis.org

1, 1, 1, 2, 1, 2, 1, 2, 2, 2, 1, 2, 2, 2, 1, 3, 2, 2, 2, 1, 3, 2, 2, 2, 2, 1, 2, 3, 2, 2, 2, 2, 2, 1, 2, 3, 3, 2, 2, 2, 2, 2, 1, 2, 3, 3, 2, 2, 2, 2, 2, 2, 1, 2, 3, 3, 2, 2, 2, 2, 2, 2, 1, 3, 2, 3, 3, 2, 2, 2, 2, 2, 2, 1, 3, 2, 3, 3, 2, 2, 3, 2, 2, 2, 2, 1, 3

Offset: 1

Gus Wiseman, Aug 18 2018

nonn

If n = 1 let e(n) be the leaf symbol "o". Given a positive integer n > 1 we construct a unique free pure symmetric multifunction e(n) with one atom by expressing n as a power of a number that is not a perfect power to a product of prime numbers: n = rad(x)^(prime(y_1) * ... * prime(y_k)) where rad = A007916. Then e(n) = e(x)[e(y_1), ..., e(y_k)]. For example, e(21025) = o[o[o]][o] because 21025 = rad(rad(1)^prime(rad(1)^prime(1)))^prime(1).

			e(21025) = o[o[o]][o] has 4 leaves so a(21025) = 4.

Cf. A007916, A052409, A052410, A109129, A277576, A277996, A300626, A316112, A317056, A317658, A317765, A317994.

nn=1000;
radQ[n_]:=If[n==1,False,GCD@@FactorInteger[n][[All,2]]==1];
rad[n_]:=rad[n]=If[n==0,1,NestWhile[#+1&,rad[n-1]+1,Not[radQ[#]]&]];
Clear[radPi];Set@@@Array[radPi[rad[#]]==#&,nn];
a[n_]:=If[n==1,1,With[{g=GCD@@FactorInteger[n][[All,2]]},a[radPi[Power[n,1/g]]]+Sum[a[PrimePi[pr[[1]]]]*pr[[2]],{pr,If[g==1,{},FactorInteger[g]]}]]];
Table[a[n],{n,100}]

a(rad(x)^(prime(y_1) * ... * prime(y_k))) = a(x) + a(y_1) + ... + a(y_k) where rad = A007916.

A317056 Depth of the free pure symmetric multifunction (with empty expressions allowed) with e-number n.

Original entry on oeis.org

0, 1, 2, 1, 3, 2, 4, 2, 2, 3, 5, 3, 3, 4, 6, 1, 4, 4, 5, 7, 2, 5, 5, 6, 3, 8, 2, 3, 6, 6, 7, 3, 4, 9, 3, 2, 4, 7, 7, 8, 4, 5, 10, 4, 3, 5, 8, 8, 4, 9, 5, 6, 11, 5, 4, 6, 9, 9, 5, 10, 6, 7, 12, 2, 6, 5, 7, 10, 10, 6, 11, 7, 8, 13, 3, 7, 6, 8, 11, 11, 2, 7, 12

Offset: 1

Gus Wiseman, Aug 18 2018

nonn

If n = 1 let e(n) be the leaf symbol "o". Given a positive integer n > 1 we construct a unique free pure symmetric multifunction e(n) with one atom by expressing n as a power of a number that is not a perfect power to a product of prime numbers: n = rad(x)^(prime(y_1) * ... * prime(y_k)) where rad = A007916. Then e(n) = e(x)[e(y_1), ..., e(y_k)]. For example, e(21025) = o[o[o]][o] because 21025 = rad(rad(1)^prime(rad(1)^prime(1)))^prime(1).

			e(21025) = o[o[o]][o] has depth 3 so a(21025) = 3.

Cf. A007916, A052409, A052410, A109082, A277576, A277996, A300626, A316112, A317056, A317658, A317765, A317994.

nn=1000;
radQ[n_]:=If[n===1,False,GCD@@FactorInteger[n][[All,2]]===1];
rad[n_]:=rad[n]=If[n===0,1,NestWhile[#+1&,rad[n-1]+1,Not[radQ[#]]&]];
Clear[radPi];Set@@@Array[radPi[rad[#]]==#&,nn];
exp[n_]:=If[n===1,"o",With[{g=GCD@@FactorInteger[n][[All,2]]},Apply[exp[radPi[Power[n,1/g]]],exp/@Flatten[Cases[FactorInteger[g],{p_?PrimeQ,k_}:>ConstantArray[PrimePi[p],k]]]]]];
Table[Max@@Length/@Position[exp[n],_],{n,200}]

A317994 Number of inequivalent leaf-colorings of the free pure symmetric multifunction with e-number n.

Original entry on oeis.org

1, 1, 1, 2, 1, 2, 1, 2, 2, 2, 1, 2, 2, 2, 1, 4, 2, 2, 2, 1, 4, 2, 2, 2, 2, 1, 2, 4, 2, 2, 2, 2, 2, 1, 2, 5, 4, 2, 2, 2, 2, 2, 1, 2, 5, 4, 2, 2, 2, 2, 2, 2, 1, 2, 5, 4, 2, 2, 2, 2, 2, 2, 1, 5, 2, 5, 4, 2, 2, 2, 2, 2, 2, 1, 5, 2, 5, 4, 2, 2, 4, 2, 2, 2, 2, 1, 5

Offset: 1

Gus Wiseman, Aug 18 2018

nonn

If n = 1 let e(n) be the leaf symbol "o". Given a positive integer n > 1 we construct a unique free pure symmetric multifunction (with empty expressions allowed) e(n) with one atom by expressing n as a power of a number that is not a perfect power to a product of prime numbers: n = rad(x)^(prime(y_1) * ... * prime(y_k)) where rad = A007916. Then e(n) = e(x)[e(y_1), ..., e(y_k)]. For example, e(21025) = o[o[o]][o] because 21025 = rad(rad(1)^prime(rad(1)^prime(1)))^prime(1).

			Inequivalent representatives of the a(441) = 11 colorings of the expression e(441) = o[o,o][o] are the following.
  1[1,1][1]
  1[1,1][2]
  1[1,2][1]
  1[1,2][2]
  1[1,2][3]
  1[2,2][1]
  1[2,2][2]
  1[2,2][3]
  1[2,3][1]
  1[2,3][2]
  1[2,3][4]

Cf. A007916, A052409, A052410, A277576, A277996, A300626, A316112, A317056, A317658, A317765.

A303709 Number of periodic factorizations of n using elements of A007916 (numbers that are not perfect powers).

Original entry on oeis.org

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

Offset: 1

Gus Wiseman, Apr 29 2018

nonn

A periodic factorization of n is a finite multiset of positive integers greater than 1 whose product is n and whose multiplicities have a common divisor greater than 1. Note that a factorization of a number that is not a perfect power (A007916) is always aperiodic (A303386), so the indices of nonzero entries of this sequence all lie at perfect powers (A001597).

			The a(900) = 5 periodic factorizations are (2*2*3*3*5*5), (2*2*15*15), (3*3*10*10), (5*5*6*6), (30*30).

Antti Karttunen, Table of n, a(n) for n = 1..100000

Cf. A000740, A000837, A001055, A001597, A007716, A007916, A052409, A052410, A281116, A303547, A303552, A303553, A303386, A303707, A303708, A303710.

radQ[n_]:=Or[n===1,GCD@@FactorInteger[n][[All,2]]===1];
facsr[n_]:=If[n<=1,{{}},Join@@Table[Map[Prepend[#,d]&,Select[facsr[n/d],Min@@#>=d&]],{d,Select[Rest[Divisors[n]],radQ]}]];
Table[Length[Select[facsr[n],GCD@@Length/@Split[#]!=1&]],{n,200}]

gcd_of_multiplicities(lista) = { my(u=length(lista)); if(u<2, u, my(g=0, pe = lista[1], j=1); for(i=2,u,if(lista[i]==pe, j++, g = gcd(j,g); j=1; pe = lista[i])); gcd(g,j)); }; \\ the supplied lista (newfacs) should be monotonic
A303709(n, m=n, facs=List([])) = if(1==n, (1!=gcd_of_multiplicities(facs)), my(s=0, newfacs); fordiv(n, d, if((d>1)&&(d<=m)&&!ispower(d), newfacs = List(facs); listput(newfacs,d); s += A303709(n/d, d, newfacs))); (s)); \\ Antti Karttunen, Dec 06 2018

a(n) <= A303553(n) <= A001055(n). - Antti Karttunen, Dec 06 2018

Changed a(1) to 1 by Gus Wiseman, Dec 06 2018

cp's OEIS Frontend

A295924 Number of twice-factorizations of n of type (R,P,R).

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A294337 Number of ways to write 2^n as a finite power-tower a^(b^(c^...)) of positive integers greater than one.

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A295931 Number of ways to write n in the form n = (x^y)^z where x, y, and z are positive integers.

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A128164 Least k > 2 such that (n^k - 1)/(n-1) is prime, or 0 if no such prime exists.

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A253641 Largest integer b such that n=a^b for some integer a; a(0)=a(1)=1 by convention.

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A302291 a(n) is the period of the binary expansion of n.

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A316112 Number of leaves in the free pure symmetric multifunction (with empty expressions allowed) with e-number n.

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