A284252 -id:A284252 - cp's OEIS frontend

A284253 a(n) = n / A284252(n).

1, 2, 3, 4, 5, 6, 7, 8, 9, 2, 11, 12, 13, 2, 15, 16, 17, 18, 19, 4, 21, 2, 23, 24, 25, 2, 27, 4, 29, 6, 31, 32, 3, 2, 35, 36, 37, 2, 3, 8, 41, 6, 43, 4, 45, 2, 47, 48, 49, 10, 3, 4, 53, 54, 55, 8, 3, 2, 59, 12, 61, 2, 63, 64, 65, 6, 67, 4, 3, 14, 71, 72, 73, 2, 75, 4, 77, 6, 79, 16, 81, 2, 83, 12, 85, 2, 3, 8, 89, 18, 91, 4, 3, 2, 95, 96, 97, 14, 9, 20, 101, 6

Offset: 1

Antti Karttunen, Mar 24 2017

nonn

Antti Karttunen, Table of n, a(n) for n = 1..10001

Cf. A020639, A284252, A284254, A284255, A284256, A284257, A284258, A284259.

a[n_] := Block[{p = First /@ FactorInteger[n]}, n / SelectFirst[p, # > p[[1]]^2 &, 1]]; Array[a, 120] (* Giovanni Resta, Mar 24 2017 *)

a(n) = if(n<2, return(1), my(f=factor(n)[, 1]); for(i=2, #f, if(f[i]>f[1]^2, return(f[i]))); return(1));
for(n=1, 151, print1(n/a(n),", ")) \\ Indranil Ghosh, after David A. Corneth, Mar 24 2017

from sympy import primefactors
def a(n):
    pf = primefactors(n)
    if pf: min_pf2 = min(pf)**2
    for i in pf:
        if i > min_pf2: return i
    return 1
print([n//a(n) for n in range(1, 151)]) # Indranil Ghosh, Mar 24 2017

Scheme
```
(define (A284253 n) (/ n (A284252 n)))
```

a(n) = n / A284252(n).

A251726 Numbers n > 1 for which gpf(n) < lpf(n)^2, where lpf and gpf (least and greatest prime factor of n) are given by A020639(n) and A006530(n).

Original entry on oeis.org

2, 3, 4, 5, 6, 7, 8, 9, 11, 12, 13, 15, 16, 17, 18, 19, 21, 23, 24, 25, 27, 29, 31, 32, 35, 36, 37, 41, 43, 45, 47, 48, 49, 53, 54, 55, 59, 61, 63, 64, 65, 67, 71, 72, 73, 75, 77, 79, 81, 83, 85, 89, 91, 95, 96, 97, 101, 103, 105, 107, 108, 109, 113, 115, 119, 121, 125, 127, 128, 131, 133, 135, 137, 139, 143, 144

Offset: 1

Antti Karttunen, Dec 17 2014

nonn

Numbers n > 1 for which there exists r <= gpf(n) such that r^k <= lpf(n) and gpf(n) < r^(k+1) for some k >= 0, where lpf and gpf (least and greatest prime factor of n) are given by A020639(n) and A006530(n) (the original, equivalent definition of the sequence).
Numbers n > 1 such that A252375(n) < 1 + A006530(n). Equally, one can substitute A251725 for A252375.
These are numbers n all of whose prime factors "fit between" two consecutive powers of some positive integer which itself is <= the largest prime factor of n.
Conjecture: If any n is in the sequence, then so is A003961(n).
Note: if Legendre's or Brocard's conjecture is true, then the above conjecture is true as well. See my comments at A251728. - Antti Karttunen, Jan 01 2015

			For 35 = 5*7, 7 is less than 5^2, thus 35 is included.
For 90 = 2*3*3*5, 5 is not less than 2^2, thus 90 is NOT included.
For 105 = 3*5*7, 7 is less than 3^2, thus 105 is included.

Antti Karttunen, Table of n, a(n) for n = 1..10000
Ratio A251726(n)/A251727(n) (plotted with OEIS-server's Plot2-utility)

Complement: A251727. Subsequences: A251728, A000961 (after 1).
Characteristic function: A252372. Inverse function: A252373.
Gives the positions of zeros in A252459 (after its initial zero), cf. also A284261.
Cf. A252370 (gives the difference between the prime indices of gpf and lpf for each a(n)).
Sequence gives all n > 1 for which A284252(n) (equally: A284254) is 1, and A284256(n) (equally A284258) is 0, and also n > 1 such that A284260(n) = A006530(n).
Cf. A003961, A006530, A020639, A252375, A251725, A253784.
Related permutations: A252757-A252758.

pfQ[n_]:=Module[{f=FactorInteger[n]},f[[-1,1]]Harvey P. Dale, May 01 2015 *)

for(n=2, 150, if(vecmax(factor(n)[,1]) < vecmin(factor(n)[,1])^2, print1(n,", "))) \\ Indranil Ghosh, Mar 24 2017

from sympy import primefactors
print([n for n in range(2, 150) if max(primefactors(n))Indranil Ghosh, Mar 24 2017

Other identities. For all n >= 1:

A252373(a(n)) = n. [A252373 works as an inverse or ranking function for this sequence.]

A new simpler definition found Jan 01 2015 and the original definition moved to the Comments section.

A284254 Largest divisor of n such that all its prime factors are greater than the square of smallest prime factor of n, a(1) = 1.

Original entry on oeis.org

1, 1, 1, 1, 1, 1, 1, 1, 1, 5, 1, 1, 1, 7, 1, 1, 1, 1, 1, 5, 1, 11, 1, 1, 1, 13, 1, 7, 1, 5, 1, 1, 11, 17, 1, 1, 1, 19, 13, 5, 1, 7, 1, 11, 1, 23, 1, 1, 1, 25, 17, 13, 1, 1, 1, 7, 19, 29, 1, 5, 1, 31, 1, 1, 1, 11, 1, 17, 23, 35, 1, 1, 1, 37, 1, 19, 1, 13, 1, 5, 1, 41, 1, 7, 1, 43, 29, 11, 1, 5, 1, 23, 31, 47, 1, 1, 1, 49, 11, 25, 1, 17, 1, 13, 1, 53, 1, 1, 1, 55

Offset: 1

Antti Karttunen, Mar 24 2017

nonn

			For n = 15 = 3*5, no prime factor is larger than 3^2, thus a(15) = 1. In this case the largest divisor satisfying the condition has no prime factors at all.
For n = 50 = 2*5*5, the primes larger than 2^2 are 5 and 5, thus a(50) = 5*5 = 25.

Antti Karttunen, Table of n, a(n) for n = 1..10001

Cf. A020639, A284253, A284255, A284256, A284257, A284258, A284259, A284260.
Cf. A251726 (gives the positions of ones after the initial a(1)=1).
Differs from related A284252 for the first time at n=50, where a(50) = 25, while A284252(50) = 5.

Table[If[n == 1, 1, Function[d, Last[Select[Reverse@ First@ d, Times @@ Boole@ Map[# > Last[d]^2 &, FactorInteger[#][[All, 1]]] == 1 &] /. {} -> {1}]]@ {#, First@ Select[#, PrimeQ]} &@ Divisors@ n], {n, 108}] (* Michael De Vlieger, Mar 24 2017 *)

A(n) = if(n<2, return(1), my(f=factor(n)[, 1]); for(i=2, #f, if(f[i]>f[1]^2, return(f[i]))); return(1));
a(n) = if(A(n)==1, 1, A(n)*a(n/A(n)));
for(n=1, 150, print1(a(n),", ")) \\ Indranil Ghosh, after David A. Corneth, Mar 24 2017

from sympy import primefactors
def A(n):
     for i in primefactors(n):
         if i>min(primefactors(n))**2: return i
     return 1
def a(n): return 1 if A(n) == 1 else A(n)*a(n//A(n))
print([a(n) for n in range(1, 151)]) # Indranil Ghosh, Mar 24 2017

If A284252(n) = 1, then a(n) = 1, otherwise A284252(n) * a(A284253(n)).
Other identities. For all n >= 1:
n / a(n) = A284255(n).
A020639(a(n)) = A284252(n).
A001221(a(n)) = A284258(n).
A001222(a(n)) = A284256(n).

A284258 a(n) = number of distinct prime factors of n that are > the square of smallest prime factor of n, a(1) = 0.

Original entry on oeis.org

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

Offset: 1

Antti Karttunen, Mar 24 2017

nonn

			For n = 50, 2*5*5, the prime factor > 2^2 is 5, which is counted only once, thus a(50) = 1.
For n = 70, 2*5*7, the prime factors > 2^2 are 5 and 7, thus a(70) = 2.

Antti Karttunen, Table of n, a(n) for n = 1..10001

Cf. A001221, A020639, A057427, A284252, A284253, A284254, A284255, A284257, A284259, A284261.
Cf. A251726 (gives the positions of zeros after the initial a(1)=0).
Differs from related A284256 for the first time at n=50, where a(50)=1, while A284256(50)=2.

Table[If[n == 1, 0, Count[#, d_ /; d > First[#]^2] &@ FactorInteger[n][[All, 1]]], {n, 120}] (* Michael De Vlieger, Mar 24 2017 *)

A(n) = if(n<2, return(1), my(f=factor(n)[, 1]); for(i=2, #f, if(f[i]>f[1]^2, return(f[i]))); return(1));
a(n) = if(A(n)==1, 1, A(n)*a(n/A(n)));
for(n=1, 150, print1(omega(a(n)),", ")) \\ Indranil Ghosh, after David A. Corneth, Mar 24 2017

from sympy import primefactors
def omega(n): return len(primefactors(n))
def A(n):
    for i in primefactors(n):
        if i>min(primefactors(n))**2: return i
    return 1
def a(n): return 1 if A(n)==1 else A(n)*a(n//A(n))
print([omega(a(n)) for n in range(1, 151)]) # Indranil Ghosh, Mar 24 2017

(define (A284258 n) (A001221 (A284254 n)))

a(n) = Sum_{p|n, p prime and > spf(n)^2} sign(p), where spf(n) (A020639) gives the smallest prime factor of n, and sign(p) = A057427(p) = 1 for all p.
a(n) = A001221(A284254(n)).
a(n) = A001221(n) - A284259(n).
a(n) <= A284256(n).

A284259 a(n) = number of distinct prime factors of n that are < the square of smallest prime factor of n, a(1) = 0.

Original entry on oeis.org

0, 1, 1, 1, 1, 2, 1, 1, 1, 1, 1, 2, 1, 1, 2, 1, 1, 2, 1, 1, 2, 1, 1, 2, 1, 1, 1, 1, 1, 2, 1, 1, 1, 1, 2, 2, 1, 1, 1, 1, 1, 2, 1, 1, 2, 1, 1, 2, 1, 1, 1, 1, 1, 2, 2, 1, 1, 1, 1, 2, 1, 1, 2, 1, 2, 2, 1, 1, 1, 1, 1, 2, 1, 1, 2, 1, 2, 2, 1, 1, 1, 1, 1, 2, 2, 1, 1, 1, 1, 2, 2, 1, 1, 1, 2, 2, 1, 1, 1, 1, 1, 2, 1, 1, 3, 1, 1, 2, 1, 1, 1, 1, 1, 2, 2, 1, 1, 1, 2, 2

Offset: 1

Antti Karttunen, Mar 24 2017

nonn

			For n = 4 = 2*2, the prime factor 2 is less than 2^2, and it is counted only once, thus a(4) = 1.
For n = 45 = 3*3*5, both prime factors 3 and 5 are less than 3^2, thus a(45) = 2.

Antti Karttunen, Table of n, a(n) for n = 1..10001

Cf. A001221, A020639, A057427, A284252, A284253, A284254, A284255, A284257, A284258, A284260.

Cf. A284262 (where obtains first time value n, also positions of records).

Table[If[n == 1, 0, Count[#, d_ /; d < First[#]^2] &@ FactorInteger[n][[All, 1]]], {n, 120}] (* Michael De Vlieger, Mar 24 2017 *)

A(n) = if(n<2, return(1), my(f=factor(n)[, 1]); for(i=2, #f, if(f[i]>f[1]^2, return(f[i]))); return(1));
a(n) = if(A(n)==1, 1, A(n)*a(n/A(n)));
for(n=1, 150, print1(omega(n/a(n)),", ")) \\ Indranil Ghosh, after David A. Corneth, Mar 24 2017

from sympy import primefactors
def omega(n): return len(primefactors(n))
def A(n):
    for i in primefactors(n):
        if i>min(primefactors(n))**2: return i
    return 1
def a(n): return 1 if A(n)==1 else A(n)*a(n//A(n))
print([omega(n//a(n)) for n in range(1, 151)]) # Indranil Ghosh, Mar 24 2017

(define (A284259 n) (A001221 (A284255 n)))

a(n) = Sum_{p|n, p prime and < spf(n)^2} sign(p), where spf(n) (A020639) gives the smallest prime factor of n, and sign(p) = A057427(p) = 1 for all p.
a(n) = A001221(A284255(n)).
a(n) = A001221(n) - A284258(n).
a(n) <= A284257(n).
a(A284262(n)) = n.

A284255 Largest divisor of n such that all its prime factors are less than the square of the smallest prime factor of n, a(1) = 1.

Original entry on oeis.org

1, 2, 3, 4, 5, 6, 7, 8, 9, 2, 11, 12, 13, 2, 15, 16, 17, 18, 19, 4, 21, 2, 23, 24, 25, 2, 27, 4, 29, 6, 31, 32, 3, 2, 35, 36, 37, 2, 3, 8, 41, 6, 43, 4, 45, 2, 47, 48, 49, 2, 3, 4, 53, 54, 55, 8, 3, 2, 59, 12, 61, 2, 63, 64, 65, 6, 67, 4, 3, 2, 71, 72, 73, 2, 75, 4, 77, 6, 79, 16, 81, 2, 83, 12, 85, 2, 3, 8, 89, 18, 91, 4, 3, 2, 95, 96, 97, 2, 9, 4, 101, 6

Offset: 1

Antti Karttunen, Mar 24 2017

nonn

			For n = 50 = 2*5*5, only prime less than 2^2 is 2, thus a(50) = 2.
For n = 90 = 2*3*3*5, the primes less than 2^2 are 2, 3 and 3, thus a(90) = 2*3*3 = 18.

Antti Karttunen, Table of n, a(n) for n = 1..10001

Cf. A001221, A001222, A006530, A020639, A284252, A284254, A284256, A284257, A284258, A284259, A284260.

Differs from A284253 for the first time at n=50, where a(50) = 2, while A284253(50) = 10.

Table[If[n == 1, 1, Function[d, First[Select[Reverse@ First@ d, Times @@ Boole@ Map[# < Last[d]^2 &, FactorInteger[#][[All, 1]]] == 1 &] /. {} -> {1}]]@ {#, First@ Select[#, PrimeQ]} &@ Divisors@ n], {n, 102}] (* Michael De Vlieger, Mar 24 2017 *)

A(n) = if(n<2, return(1), my(f=factor(n)[, 1]); for(i=2, #f, if(f[i]>f[1]^2, return(f[i]))); return(1));
a(n) = if(A(n)==1, 1, A(n)*a(n/A(n)));
for(n=1, 150, print1(n/a(n),", ")) \\ Indranil Ghosh, after David A. Corneth, Mar 24 2017

from sympy import primefactors
def A(n):
    pf = primefactors(n)
    if pf: min_pf2 = min(pf)**2
    for i in pf:
        if i > min_pf2: return i
    return 1
def a(n): return 1 if A(n)==1 else A(n)*a(n//A(n))
print([n//a(n) for n in range(1, 151)]) # Indranil Ghosh, Mar 24 2017

Scheme
```
(define (A284255 n) (/ n (A284254 n)))
```

a(n) = n / A284254(n).
Other identities. For all n >= 1:
A006530(a(n)) = A284260(n).
A020639(a(n)) = A020639(n).
A001221(a(n)) = A284259(n).
A001222(a(n)) = A284257(n).

A284256 a(n) = number of prime factors of n that are > the square of smallest prime factor of n (counted with multiplicity), a(1) = 0.

Original entry on oeis.org

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

Offset: 1

Antti Karttunen, Mar 24 2017

nonn

			For n = 10 = 2*5, there is a single prime factor 5 that is > 2^2, thus a(10) = 1.
For n = 15 = 3*5, there are no prime factors larger than 3^2, thus a(15) = 0.
For n = 50 = 2*5*5, the prime factors larger than 2^2 are 5*5, thus a(50) = 2.

Antti Karttunen, Table of n, a(n) for n = 1..10001

Cf. A001222, A020639, A284252, A284253, A284254, A284257, A284258, A284259, A284260.

Cf. A251726 (gives the positions of zeros after the initial a(1)=0).

Table[If[n == 1, 0, Count[#, d_ /; d > First[#]^2] &@ Flatten@ Map[ConstantArray[#1, #2] & @@ # &, FactorInteger@ n]], {n, 120}] (* Michael De Vlieger, Mar 24 2017 *)

A(n) = if(n<2, return(1), my(f=factor(n)[, 1]); for(i=2, #f, if(f[i]>f[1]^2, return(f[i]))); return(1));
a(n) = if(A(n)==1, 0, 1 + a(n/A(n)));
for(n=1, 150, print1(a(n),", ")) \\ Indranil Ghosh, after David A. Corneth, Mar 24 2017

from sympy import primefactors
def A(n):
    pf = primefactors(n)
    if pf: min_pf2 = min(pf)**2
    for i in pf:
        if i > min_pf2: return i
    return 1
def a(n): return 0 if A(n)==1 else 1 + a(n//A(n))
print([a(n) for n in range(1, 151)]) # Indranil Ghosh, Mar 24 2017

If A284252(n) = 1, a(n) = 0, otherwise a(n) = 1 + a(A284253(n)).
a(n) = A001222(A284254(n)).
a(n) = A001222(n) - A284257(n).

A284257 a(n) = number of prime factors of n that are < the square of smallest prime factor of n (counted with multiplicity), a(1) = 0.

Original entry on oeis.org

0, 1, 1, 2, 1, 2, 1, 3, 2, 1, 1, 3, 1, 1, 2, 4, 1, 3, 1, 2, 2, 1, 1, 4, 2, 1, 3, 2, 1, 2, 1, 5, 1, 1, 2, 4, 1, 1, 1, 3, 1, 2, 1, 2, 3, 1, 1, 5, 2, 1, 1, 2, 1, 4, 2, 3, 1, 1, 1, 3, 1, 1, 3, 6, 2, 2, 1, 2, 1, 1, 1, 5, 1, 1, 3, 2, 2, 2, 1, 4, 4, 1, 1, 3, 2, 1, 1, 3, 1, 3, 2, 2, 1, 1, 2, 6, 1, 1, 2, 2, 1, 2, 1, 3, 3, 1, 1, 5, 1, 1, 1, 4, 1, 2, 2, 2, 2, 1, 2, 4

Offset: 1

Antti Karttunen, Mar 24 2017

nonn

			For n = 45 = 3*3*5, all prime factors 3, 3 and 5 are less than 3^2, thus a(45) = 3.
For n = 120 = 2*2*2*3*5, the prime factors less than 2^2 are 2*2*2*3, thus a(120) = 4.

Antti Karttunen, Table of n, a(n) for n = 1..10001

Cf. A001222, A020639, A284252, A284253, A284254, A284256, A284258, A284259.

Table[If[n == 1, 0, Count[#, d_ /; d < First[#]^2] &@ Flatten@ Map[ConstantArray[#1, #2] & @@ # &, FactorInteger@ n]], {n, 120}] (* Michael De Vlieger, Mar 24 2017 *)

A(n) = if(n<2, return(1), my(f=factor(n)[, 1]); for(i=2, #f, if(f[i]>f[1]^2, return(f[i]))); return(1));
a(n) = if(A(n)==1, 1, A(n)*a(n/A(n)));
for(n=1, 150, print1(bigomega(n/a(n)),", ")) \\ Indranil Ghosh, Mar 24 2017, after David A. Corneth

from sympy import primefactors
def Omega(n): return 0 if n==1 else Omega(n//min(primefactors(n))) + 1
def A(n):
    pf = primefactors(n)
    if pf: min_pf2 = min(pf)**2
    for i in pf:
        if i > min_pf2: return i
    return 1
def a(n): return 1 if A(n)==1 else A(n)*a(n//A(n))
print([Omega(n//a(n)) for n in range(1, 151)]) # Indranil Ghosh, Mar 24 2017

(define (A284257 n) (A001222 (A284255 n)))

a(n) = A001222(A284255(n)).

a(n) = A001222(n) - A284256(n).

A284260 Greatest prime dividing n which is less than A020639(n)^2, where A020639(n) is the smallest prime dividing n, a(1) = 1.

Original entry on oeis.org

1, 2, 3, 2, 5, 3, 7, 2, 3, 2, 11, 3, 13, 2, 5, 2, 17, 3, 19, 2, 7, 2, 23, 3, 5, 2, 3, 2, 29, 3, 31, 2, 3, 2, 7, 3, 37, 2, 3, 2, 41, 3, 43, 2, 5, 2, 47, 3, 7, 2, 3, 2, 53, 3, 11, 2, 3, 2, 59, 3, 61, 2, 7, 2, 13, 3, 67, 2, 3, 2, 71, 3, 73, 2, 5, 2, 11, 3, 79, 2, 3, 2, 83, 3, 17, 2, 3, 2, 89, 3, 13, 2, 3, 2, 19, 3, 97, 2, 3, 2, 101, 3, 103, 2, 7, 2, 107, 3, 109

Offset: 1

Antti Karttunen, Mar 24 2017

nonn

Antti Karttunen, Table of n, a(n) for n = 1..10001

Cf. A006530, A020639, A284252, A284253, A284254, A284255, A284256, A284257, A284258, A284259.

Cf. A251726 (gives n > 1 such that a(n) = A006530(n)).

Table[Last[Function[s, Select[s, # < First[s]^2 &]]@ FactorInteger[n][[All, 1]] /. {} -> {1}], {n, 109}] (* Michael De Vlieger, Mar 24 2017 *)

A(n) = if(n<2, return(1), my(f=factor(n)[, 1]); for(i=2, #f, if(f[i]>f[1]^2, return(f[i]))); return(1));
a(n) = if(A(n)==1, 1, A(n)*a(n/A(n)));
gpf(n) = if(n>1, vecmax(factor(n)[,1]),1);
for(n=1, 150, print1(gpf(n/a(n)),", ")) \\ Indranil Ghosh, Mar 24 2017, after David A. Corneth

from sympy import primefactors
def A(n):
    for i in primefactors(n):
        if i>min(primefactors(n))**2: return i
    return 1
def a(n): return 1 if A(n)==1 else A(n)*a(n//A(n))
def gpf(n): return 1 if n<2 else max(primefactors(n))
print([gpf(n//a(n)) for n in range(1, 151)]) # Indranil Ghosh, Mar 24 2017

(define (A284260 n) (A006530 (A284255 n)))

a(n) = A006530(A284255(n)).

cp's OEIS Frontend

A284253 a(n) = n / A284252(n).

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A251726 Numbers n > 1 for which gpf(n) < lpf(n)^2, where lpf and gpf (least and greatest prime factor of n) are given by A020639(n) and A006530(n).

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A284254 Largest divisor of n such that all its prime factors are greater than the square of smallest prime factor of n, a(1) = 1.

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A284258 a(n) = number of distinct prime factors of n that are > the square of smallest prime factor of n, a(1) = 0.

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Scheme

Formula

A284259 a(n) = number of distinct prime factors of n that are < the square of smallest prime factor of n, a(1) = 0.

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Programs

Mathematica

PARI

Python

Scheme

Formula

A284255 Largest divisor of n such that all its prime factors are less than the square of the smallest prime factor of n, a(1) = 1.

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Programs

Mathematica

PARI

Python

Scheme

Formula

A284256 a(n) = number of prime factors of n that are > the square of smallest prime factor of n (counted with multiplicity), a(1) = 0.

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