A073483 -id:A073483 - cp's OEIS frontend

A073485 Product of any number of consecutive primes; squarefree numbers with no gaps in their prime factorization.

1, 2, 3, 5, 6, 7, 11, 13, 15, 17, 19, 23, 29, 30, 31, 35, 37, 41, 43, 47, 53, 59, 61, 67, 71, 73, 77, 79, 83, 89, 97, 101, 103, 105, 107, 109, 113, 127, 131, 137, 139, 143, 149, 151, 157, 163, 167, 173, 179, 181, 191, 193, 197, 199, 210, 211, 221, 223, 227, 229, 233

Offset: 1

Reinhard Zumkeller, Aug 03 2002

A073484(a(n)) = 0 and A073483(a(n)) = 1;
See A097889 for composite terms. - Reinhard Zumkeller, Mar 30 2010
A169829 is a subsequence. - Reinhard Zumkeller, May 31 2010
a(A192280(n)) = 1: complement of A193166.
Also fixed points of A053590: a(n) = A053590(a(n)). - Reinhard Zumkeller, May 28 2012
The Heinz numbers of the partitions into distinct consecutive integers. The Heinz number of a partition p = [p_1, p_2, ..., p_r] is defined as Product_{j=1..r} prime(p_j) (concept used by Alois P. Heinz in A215366 as an "encoding" of a partition). Example: (i) 15 (= 3*5) is in the sequence because it is the Heinz number of the partition [2,3]; (ii) 10 (= 2*5) is not in the sequence because it is the Heinz number of the partition [1,3]. - Emeric Deutsch, Oct 02 2015
Except for the term 1, each term can uniquely represented as A002110(k)/A002110(m) for k > m >= 0; 1 = A002110(k)/A002110(k) for all k. - Michel Marcus and Jianing Song, Jun 19 2019

			105 is a term, as 105 = 3*5*7 with consecutive prime factors.

Alois P. Heinz, Table of n, a(n) for n = 1..20000 (first 1000 terms from T. D. Noe)

Complement: A193166.
Intersection of A005117 and A073491.
Subsequence of A277417.
Cf. A053590, A073483, A073484, A073486, A192280.
Cf. A000040, A006094, A002110, A097889, A169829 (subsequences).
Cf. A096334.

a073485 n = a073485_list !! (n-1)
a073485_list = filter ((== 1) . a192280) [1..]
-- Reinhard Zumkeller, May 28 2012, Aug 26 2011

isA073485 := proc(n)
    local plist,p,i ;
    plist := sort(convert(numtheory[factorset](n),list)) ;
    for i from 1 to nops(plist) do
        p := op(i,plist) ;
        if modp(n,p^2) = 0 then
            return false;
        end if;
        if i > 1 then
            if nextprime(op(i-1,plist)) <> p then
                return false;
            end if;
        end if;
    end do:
    true;
end proc:
for n from 1 to 1000 do
    if isA073485(n) then
        printf("%d,",n);
    end if;
end do: # R. J. Mathar, Jan 12 2016
# second Maple program:
q:= proc(n) uses numtheory; n=1 or issqrfree(n) and (s->
      nops(s)=1+pi(max(s))-pi(min(s)))(factorset(n))
    end:
select(q, [$1..288])[];  # Alois P. Heinz, Jan 27 2022

f[n_] := FoldList[ Times, 1, Prime[ Range[n, n + 3]]]; lst = {}; k = 1; While[k < 55, AppendTo[lst, f@k]; k++ ]; Take[ Union@ Flatten@ lst, 65] (* Robert G. Wilson v, Jun 11 2010 *)

list(lim)=my(v=List(primes(primepi(lim))),p,t);for(e=2, log(lim+.5)\log(2), p=1; t=prod(i=1,e-1,prime(i)); forprime(q=prime(e), lim, t*=q/p; if(t>lim,next(2)); listput(v,t); p=nextprime(p+1))); vecsort(Vec(v)) \\ Charles R Greathouse IV, Oct 24 2012

a(n) ~ n log n. - Charles R Greathouse IV, Oct 24 2012

Alternative description added to the name by Antti Karttunen, Oct 29 2016

A073490 Number of prime gaps in factorization of n.

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

Offset: 1

Reinhard Zumkeller, Aug 03 2002

A137723(n) is the smallest number of the first occurring set of exactly n consecutive numbers with at least one prime gap in their factorization: a(A137723(n)+k)>0 for 0<=kA137723(n)-1)=a(A137723(n)+n)=0. - Reinhard Zumkeller, Feb 09 2008

			84 = 2*2*3*7 with one gap between 3 and 7, therefore a(84) = 1;
110 = 2*5*11 with two gaps: between 2 and 5 and between 5 and 11, therefore a(110) = 2.

R. Zumkeller, Table of n, a(n) for n = 1..50000

Cf. A073491, A073492, A073493, A073494, A073495.

Cf. A137721, A137722, A027748, A049084.

a073490 1 = 0
a073490 n = length $ filter (> 1) $ zipWith (-) (tail ips) ips
   where ips = map a049084 $ a027748_row n
-- Reinhard Zumkeller, Jul 04 2012

A073490 := proc(n)
    local a,plist ;
    plist := sort(convert(numtheory[factorset](n),list)) ;
    a := 0 ;
    for i from 2 to nops(plist) do
        if op(i,plist) <> nextprime(op(i-1,plist)) then
            a := a+1 ;
        end if;
    end do:
    a;
end proc:
seq(A073490(n),n=1..110) ; # R. J. Mathar, Oct 27 2019

gaps[n_Integer/;n>0]:=If[n===1, 0, Complement[Prime[PrimePi[Rest[ # ]]-1], # ]&[First/@FactorInteger[n]]]; Table[Length[gaps[n]], {n, 1, 120}] (* Wouter Meeussen, Oct 30 2004 *)
pa[n_, k_] := If[k == NextPrime[n], 0, 1]; Table[Total[pa @@@ Partition[First /@ FactorInteger[n], 2, 1]], {n, 120}] (* Jayanta Basu, Jul 01 2013 *)

from sympy import primefactors, nextprime
def a(n):
    pf = primefactors(n)
    return sum(p2 != nextprime(p1) for p1, p2 in zip(pf[:-1], pf[1:]))
print([a(n) for n in range(1, 121)]) # Michael S. Branicky, Oct 14 2021

a(n) = A073484(A007947(n)).
a(A000040(n))=0; a(A000961(n))=0; a(A006094(n))=0; a(A002110(n))=0; a(A073485(n))=0.
a(A073486(n))>0; a(A073487(n)) = 1; a(A073488(n))=2; a(A073489(n))=3.
a(n)=0 iff A073483(n) = 1.
a(A097889(n)) = 0. - Reinhard Zumkeller, Nov 20 2004
0 <= a(m*n) <= a(m) + a(n) + 1. A137794(n) = 0^a(n). - Reinhard Zumkeller, Feb 11 2008

More terms from Franklin T. Adams-Watters, May 19 2006

A073486 Squarefree numbers having at least one prime gap.

Original entry on oeis.org

10, 14, 21, 22, 26, 33, 34, 38, 39, 42, 46, 51, 55, 57, 58, 62, 65, 66, 69, 70, 74, 78, 82, 85, 86, 87, 91, 93, 94, 95, 102, 106, 110, 111, 114, 115, 118, 119, 122, 123, 129, 130, 133, 134, 138, 141, 142, 145, 146, 154, 155, 158, 159, 161, 165, 166, 170, 174, 177

Offset: 1

Reinhard Zumkeller, Aug 03 2002

nonn

Subsequence of A193166. - Reinhard Zumkeller, Aug 26 2011

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

Cf. A005117, A073487, A073488, A073489, A073485, A073492, A193166.

q[n_] := SequenceCount[FactorInteger[n][[;; , 1]], {p1_, p2_} /; p2 != NextPrime[p1]] > 0; Select[Range[177], SquareFreeQ[#] && q[#] &] (* Amiram Eldar, Apr 10 2021 *)

A073484(a(n)) > 0 and A073483(a(n)) > 1.

A192280(a(n)) = 0. - Reinhard Zumkeller, Aug 26 2011

A073484 Number of gaps in factors of the n-th squarefree number.

Original entry on oeis.org

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

Offset: 1

Reinhard Zumkeller, Aug 03 2002

nonn

			The 69th squarefree number is 110=2*5*11, therefore a(69)=2, as there are two gaps: between 2 and 5 and between 5 and 11.

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

Cf. A005117, A073483.

gaps[n_] := SequenceCount[FactorInteger[n][[;; , 1]], {p1_, p2_} /; p2 != NextPrime[p1], Overlaps -> True]; gaps /@ Select[Range[200], SquareFreeQ] (* Amiram Eldar, Apr 10 2021 *)

a(A000040(n))=0; a(A006094(n))=0; a(A002110(n))=0; a(A073485(n))=0.
a(A073486(n))>0; a(A073487(n))=1; a(A073488(n))=2; a(A073489(n))=3.
a(n)=0 iff A073483(n)=1.

A117214 a(n) = (A117213(n))/(n-th squarefree positive integer).

Original entry on oeis.org

1, 1, 2, 6, 1, 30, 3, 210, 2310, 15, 2, 30030, 510510, 10, 105, 9699690, 1155, 223092870, 1, 6469693230, 70, 15015, 6, 200560490130, 255255, 770, 7420738134810, 5, 304250263527210, 4849845, 13082761331670030, 10010

Offset: 1

Leroy Quet, Mar 03 2006

nonn

Product of all primes up to greatest prime factor of n-th squarefree number that do not divide the n-th squarefree number. - Franklin T. Adams-Watters, Oct 09 2006

a(n) = least k such that k*A005117(n) is a primorial number. Every term is squarefree. Let m be any squarefree number, and let P be the smallest primorial such that m|P. Then a(P/m) = m, and for any primorial number Q > P, a(Q/m) = m. Since there are infinitely many Q > P it follows that every squarefree number appears in this sequence infinitely many times. - David James Sycamore, Jul 04 2024

			10 is the 7th squarefree integer. And 2*3*5 = 30 is the smallest primorial number divisible by 10 = 2*5. So a(7) = 30/10 = 3.

Reinhard Zumkeller, Table of n, a(n) for n = 1..1000

Cf. A117213.
Cf. A006530, A005117, A073482, A073483.
Cf.A000720, A002110, A006530,

a117214 n = product $
   filter ((> 0) . (mod m)) $ takeWhile (< a006530 m) a000040_list
   where m = a005117 n
-- Reinhard Zumkeller, Jan 14 2012

Product[Prime@ i, {i, PrimePi@ FactorInteger[#][[-1, 1]]}]/# & /@ Select[Range@ 52, SquareFreeQ] (* Michael De Vlieger, Sep 30 2017 *)

a(n) = A002110(A000720(A005117(n))))/A005117(n). a(A002110(n)) = 1 for all n >= 0. a(A000040(n) = A002110(n-1), n > 1. - David James Sycamore, Jul 04 2024

More terms from Franklin T. Adams-Watters, Oct 09 2006

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A073485 Product of any number of consecutive primes; squarefree numbers with no gaps in their prime factorization.

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A073490 Number of prime gaps in factorization of n.

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A073486 Squarefree numbers having at least one prime gap.

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A073484 Number of gaps in factors of the n-th squarefree number.

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A117214 a(n) = (A117213(n))/(n-th squarefree positive integer).

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