cp's OEIS Frontend

This is a front-end for the Online Encyclopedia of Integer Sequences, made by Christian Perfect. The idea is to provide OEIS entries in non-ancient HTML, and then to think about how they're presented visually. The source code is on GitHub.

Showing 1-10 of 10 results.

A284261 a(n) = A284258(n) - A284258(A003961(n)).

Original entry on oeis.org

0, 0, 0, 0, 0, 0, 0, 0, 0, 1, 0, 0, 0, 0, 0, 0, 0, 0, 0, 1, 0, 0, 0, 0, 0, 0, 0, 0, 0, 1, 0, 0, 1, 0, 0, 0, 0, 0, 1, 1, 0, 0, 0, 0, 0, 0, 0, 0, 0, 1, 1, 0, 0, 0, 0, 0, 1, 0, 0, 1, 0, 0, 0, 0, 0, 0, 0, 0, 0, 1, 0, 0, 0, 0, 0, 0, 0, 0, 0, 1, 0, 0, 0, 0, 0, 0, 0, 0, 0, 1, 0, 0, 0, 0, 0, 0, 0, 0, 1, 1, 0, 0, 0, 0, 0, 0, 0, 0, 0, 1, 0, 0, 0, 0, 0, 0, 1, 0, 0, 1
Offset: 1

Views

Author

Antti Karttunen, Mar 24 2017

Keywords

Comments

The first 2 occurs at n = 429 = 3*11*13. A003961(429) = 1105 = 5*13*17.
The first 3 occurs at n = 7293 = 3*11*13*17. A003961(7293) = 20995 = 5*13*17*19.

Crossrefs

Programs

  • Mathematica
    Table[If[n == 1, 0, Subtract @@ Map[Count[#, d_ /; d > First[#]^2] &@ FactorInteger[#][[All, 1]] &, {n, Times @@ Map[#1^#2 & @@ # &, FactorInteger[n] /. {p_, e_} /; e > 0 :> {Prime[PrimePi@ p + 1], e}] - Boole[n == 1]}]], {n, 120}] (* Michael De Vlieger, Mar 24 2017 *)
  • Scheme
    (define (A284261 n) (- (A284258 n) (A284258 (A003961 n))))

Formula

a(n) = A284258(n) - A284258(A003961(n)).

Extensions

Secondary offset corrected by Antti Karttunen, Dec 15 2017

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

Views

Author

Antti Karttunen, Dec 17 2014

Keywords

Comments

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

Examples

			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.
		

Crossrefs

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).
Related permutations: A252757-A252758.

Programs

  • Mathematica
    pfQ[n_]:=Module[{f=FactorInteger[n]},f[[-1,1]]Harvey P. Dale, May 01 2015 *)
  • PARI
    for(n=2, 150, if(vecmax(factor(n)[,1]) < vecmin(factor(n)[,1])^2, print1(n,", "))) \\ Indranil Ghosh, Mar 24 2017
    
  • Python
    from sympy import primefactors
    print([n for n in range(2, 150) if max(primefactors(n))Indranil Ghosh, Mar 24 2017

Formula

Other identities. For all n >= 1:
A252373(a(n)) = n. [A252373 works as an inverse or ranking function for this sequence.]

Extensions

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

Views

Author

Antti Karttunen, Mar 24 2017

Keywords

Examples

			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.
		

Crossrefs

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.

Programs

  • Mathematica
    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 *)
  • PARI
    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
    
  • Python
    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

Formula

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).

A284252 a(n) = smallest prime dividing n which is larger than the square of smallest prime dividing n, or 1 if no such prime exists, 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, 5, 17, 13, 1, 1, 1, 7, 19, 29, 1, 5, 1, 31, 1, 1, 1, 11, 1, 17, 23, 5, 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, 7, 11, 5, 1, 17, 1, 13, 1, 53, 1, 1, 1, 5, 37
Offset: 1

Views

Author

Antti Karttunen, Mar 24 2017

Keywords

Examples

			For n=10 = 2*5, the smallest prime divisor > 2^2 is 5, thus a(10) = 5.
For n=15 = 3*5, there are no prime divisors > 3^2, thus a(15) = 1.
For n=165 = 3*5*11, the smallest prime divisor > 3^2 is 11, thus a(165) = 11.
		

Crossrefs

Cf. A251726 (gives the positions of ones after the initial a(1) = 1), A251727 (positions of terms > 1).

Programs

  • Mathematica
    a[n_] := Block[{p = First /@ FactorInteger[n]}, SelectFirst[p, # > p[[1]]^2 &, 1]]; Array[a, 120] (* Giovanni Resta, Mar 24 2017 *)
  • PARI
    a(n) = if(n==1, return(1), my(f=factor(n)[, 1]); s = f[1]; for(i=2,#f, if(f[i]>s^2, return(f[i]))); return(1)) \\ David A. Corneth, Mar 24 2017
    
  • Python
    from sympy import primefactors
    def a(n):
        for i in primefactors(n):
            if i>min(primefactors(n))**2: return i
        return 1
    print([a(n) for n in range(1, 151)]) # Indranil Ghosh, Mar 24 2017
  • Scheme
    (define (A284252 n) (let ((spf1 (A020639 n))) (let loop ((n (/ n spf1))) (let ((spf2 (A020639 n))) (cond ((= 1 spf2) 1) ((> spf2 (* spf1 spf1)) spf2) (else (loop (/ n spf2))))))))
    

Formula

a(n) = A020639(A284254(n)).
a(k) > 1 iff k is in A251727. - David A. Corneth, Mar 25 2017

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

Views

Author

Antti Karttunen, Mar 24 2017

Keywords

Examples

			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.
		

Crossrefs

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

Programs

  • Mathematica
    Table[If[n == 1, 0, Count[#, d_ /; d < First[#]^2] &@ FactorInteger[n][[All, 1]]], {n, 120}] (* Michael De Vlieger, Mar 24 2017 *)
  • PARI
    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
    
  • Python
    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
  • Scheme
    (define (A284259 n) (A001221 (A284255 n)))
    

Formula

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.

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

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, 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

Views

Author

Antti Karttunen, Mar 24 2017

Keywords

Crossrefs

Programs

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

Formula

a(n) = n / A284252(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

Views

Author

Antti Karttunen, Mar 24 2017

Keywords

Examples

			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.
		

Crossrefs

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

Programs

  • Mathematica
    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 *)
  • PARI
    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
    
  • Python
    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)))
    

Formula

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

Views

Author

Antti Karttunen, Mar 24 2017

Keywords

Examples

			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.
		

Crossrefs

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

Programs

  • Mathematica
    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 *)
  • PARI
    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
    
  • Python
    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

Formula

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

Views

Author

Antti Karttunen, Mar 24 2017

Keywords

Examples

			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.
		

Crossrefs

Programs

  • Mathematica
    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 *)
  • PARI
    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
    
  • Python
    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
  • Scheme
    (define (A284257 n) (A001222 (A284255 n)))
    

Formula

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

Views

Author

Antti Karttunen, Mar 24 2017

Keywords

Crossrefs

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

Programs

  • Mathematica
    Table[Last[Function[s, Select[s, # < First[s]^2 &]]@ FactorInteger[n][[All, 1]] /. {} -> {1}], {n, 109}] (* Michael De Vlieger, Mar 24 2017 *)
  • PARI
    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
    
  • Python
    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
  • Scheme
    (define (A284260 n) (A006530 (A284255 n)))
    

Formula

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