A038456 -id:A038456 - cp's OEIS frontend

A039833 Smallest of three consecutive squarefree numbers k, k+1, k+2 of the form p*q where p and q are distinct primes.

33, 85, 93, 141, 201, 213, 217, 301, 393, 445, 633, 697, 921, 1041, 1137, 1261, 1345, 1401, 1641, 1761, 1837, 1893, 1941, 1981, 2101, 2181, 2217, 2305, 2361, 2433, 2461, 2517, 2641, 2721, 2733, 3097, 3385, 3601, 3693, 3865, 3901, 3957, 4285, 4413, 4533, 4593, 4881, 5601

Offset: 1

Olivier Gérard

Equivalently: k, k+1 and k+2 all have 4 divisors.
There cannot be four consecutive squarefree numbers as one of them is divisible by 2^2 = 4.
These 3 consecutive squarefree numbers of the form p*q have altogether 6 prime factors always including 2 and 3. E.g., if k = 99985, the six prime factors are {2,3,5,19997,33329,49993}. The middle term is even and not divisible by 3.
Nonsquare terms of A056809. First terms of A056809 absent here are A056809(4)=121=11^2, A056809(14)=841=29^2, A056809(55)=6241=79^2.
Cf. A179502 (Numbers k with the property that k^2, k^2+1 and k^2+2 are all semiprimes). - Zak Seidov, Oct 27 2015
The numbers k, k+1, k+2 have the form 2p-1, 2p, 2p+1 where p is an odd prime.  A195685 gives the sequence of odd primes that generates these maximal runs of three consecutive integers with four positive divisors. - Timothy L. Tiffin, Jul 05 2016
a(n) is always 1 or 9 mod 12. - Charles R Greathouse IV, Mar 19 2022

			33, 34 and 35 all have 4 divisors.
85 is a term as 85 = 17*5, 86 = 43*2, 87 = 29*3.

R. K. Guy, Unsolved Problems in Number Theory, Springer, 1st edition, 1981. See section B18.
David Wells, Curious and interesting numbers, Penguin Books, 1986, p. 114.

Amiram Eldar, Table of n, a(n) for n = 1..10000 (terms 1..1000 from Reinhard Zumkeller)
Roberto Conti, Pierluigi Contucci, and Vitalii Iudelevich, Bounds on tree distribution in number theory, arXiv:2401.03278 [math.NT], 2024. See page 13.

Subsequence of A006881 and A056809 and A263990.

Cf. A038456, A039832, A008683, A007675, A063736, A063838, A070552, A045939, A195685, A179502.

a039833 n = a039833_list !! (n-1)
a039833_list = f a006881_list where
   f (u : vs@(v : w : xs))
     | v == u+1 && w == v+1 = u : f vs
     | otherwise            = f vs
-- Reinhard Zumkeller, Aug 07 2011

lst = {}; Do[z = n^3 + 3*n^2 + 2*n; If[PrimeOmega[z/n] == PrimeOmega[z/(n + 2)] == 4 && PrimeNu[z] == 6, AppendTo[lst, n]], {n, 1, 5601, 2}]; lst (* Arkadiusz Wesolowski, Dec 11 2011 *)
okQ[n_]:=Module[{cl={n,n+1,n+2}},And@@SquareFreeQ/@cl && Union[ DivisorSigma[ 0,cl]]=={4}]; Select[Range[1,6001,2],okQ] (* Harvey P. Dale, Dec 17 2011 *)
SequencePosition[DivisorSigma[0,Range[6000]],{4,4,4}][[All,1]] (* Requires Mathematica version 10 or later *) (* Harvey P. Dale, Aug 17 2017 *)

is(n)=n%4==1 && factor(n)[,2]==[1,1]~ && factor(n+1)[,2]==[1,1]~ && factor(n+2)[,2]==[1,1]~ \\ Charles R Greathouse IV, Aug 29 2016

is(n)=my(t=n%12); if(t==1, isprime((n+2)/3) && isprime((n+1)/2) && factor(n)[,2]==[1,1]~, t==9 && isprime(n/3) && isprime((n+1)/2) && factor(n+2)[,2]==[1,1]~) \\ Charles R Greathouse IV, Mar 19 2022

A008966(a(n)) * A064911(a(n)) * A008966(a(n)+1) * A064911(a(n)+1) * A008966(a(n)+2) * A064911(a(n)+2) = 1. - Reinhard Zumkeller, Feb 26 2011

Additional comments from Amarnath Murthy, Vladeta Jovovic, Labos Elemer and Benoit Cloitre, May 08 2002

A039832 Numbers k such that k and k+1 both have 4 divisors.

Original entry on oeis.org

14, 21, 26, 33, 34, 38, 57, 85, 86, 93, 94, 118, 122, 133, 141, 142, 145, 158, 177, 201, 202, 205, 213, 214, 217, 218, 253, 298, 301, 302, 326, 334, 381, 393, 394, 445, 446, 453, 481, 501, 514, 526, 537, 542, 553, 565, 622, 633, 634, 694, 697, 698, 706, 717, 745, 766, 778, 793, 802, 817

Offset: 1

N. J. A. Sloane, Felice Russo, Olivier Gérard, Dec 11 1999

nonn

			14 and 15 both have 4 as number of divisors and are consecutive.

David Wells, Curious and interesting numbers, Penguin Books, 1986, p. 91.

Amiram Eldar, Table of n, a(n) for n = 1..10000 (terms 1..1000 from Vincenzo Librandi)

Intersection of A005237 and A030513.

Cf. A038456, A039833.

Flatten[Position[Partition[Table[DivisorSigma[0, n], {n, 1000}], 2, 1], ?(#=={4, 4}&)]] (* _Vincenzo Librandi, Oct 21 2012 *)

isA039832(n) = numdiv(n)==4 && numdiv(n+1)==4 \\ Michael B. Porter, Feb 03 2010

A178034 a(n) = binomial(n*Omega(n),Omega(n)) / n.

Original entry on oeis.org

1, 1, 1, 7, 1, 11, 1, 253, 17, 19, 1, 595, 1, 27, 29, 39711, 1, 1378, 1, 1711, 41, 43, 1, 138415, 49, 51, 3160, 3403, 1, 3916, 1, 25637001, 65, 67, 69, 477191, 1, 75, 77, 657359, 1, 7750, 1, 8515, 8911, 91, 1, 132563501, 97, 11026, 101, 11935, 1, 1633355

Offset: 1

Michel Lagneau, May 17 2010

nonn

Omega(.) = A001222(.) is the number of prime divisors of n (counted with multiplicity).
binomial(nk,k)= n*binomial(nk-1,k-1) ensures that all entries are integers.
Subcases for this sequence:
If n is prime, Omega(n) = 1, and a(n) = binomial (n,1) / n = 1.
If n and n+1 are products of two primes (A070552), then Omega(n) = Omega(n+1) = 2, and binomial(n*Omega(n), Omega(n)) / n = binomial(2*n, 2) / n = 2*n-1 and binomial(2*(n+1), 2) / (n+1) = 2*n+1, and we obtain two consecutive numbers of the form (x, x+2), for example (17,19), (27,29), (41,43),... at n =9, 14...
Chaining this property: If n, n+1, and n+2 are semiprimes (A056809) , we find three consecutive numbers of the form (x, x+2,x+4), for example (65, 67, 69), (169, 171, 173), at n=33, 85.
At places where Omega(n)=3, we find the subsequence A060544, for example a(8) = A060544(8).
At places where Omega(n)=4, we find the subsequence A015219.

			a(8) = binomial(8*Omega(8),Omega(8))/8 = binomial(8*3,3)/8 = 2024/8 = 253.

Harvey P. Dale, Table of n, a(n) for n = 1..1000

Cf. A001358, A038456

A178034 := proc(n)
        local o ;
        o := numtheory[bigomega](n) ;
        binomial(n*o,o)/n ;
end proc: # R. J. Mathar, Jul 08 2012

bon[n_]:=Module[{o=PrimeOmega[n]},Binomial[n*o,o]/n]; Array[bon,60] (* Harvey P. Dale, Jul 22 2014 *)

a(n)=my(b=bigomega(n));binomial(n*b,b)/n \\ Charles R Greathouse IV, Oct 25 2012

A346549 Runs (of length > 1) of consecutive squarefree semiprimes.

Original entry on oeis.org

14, 15, 21, 22, 33, 34, 35, 38, 39, 57, 58, 85, 86, 87, 93, 94, 95, 118, 119, 122, 123, 133, 134, 141, 142, 143, 145, 146, 158, 159, 177, 178, 201, 202, 203, 205, 206, 213, 214, 215, 217, 218, 219, 253, 254, 298, 299, 301, 302, 303, 326, 327, 334, 335, 381, 382, 393, 394, 395, 445, 446, 447, 453, 454, 481, 482

Offset: 1

Ositadima Chukwu, Sep 16 2021

nonn

Runs of length bigger than 3 are impossible as one of four consecutive numbers is divisible by 4.
The existence of consecutive pairs of such numbers is connected with primes of the form (p*q +- 1)/2 where p and q are odd primes.
From Michael S. Branicky, Sep 21 2021: (Start)
This only differs from the supersequence A038456 in the terms 26, 27 (present there but not here).
Proof. Numbers with 4 divisors are of the form p*q for p, q prime, p != q or of the form r^3 for r prime.  For two such numbers to be consecutive terms of A038456 (but not terms here) requires p*q + 1 = r^3 or p*q = r^3 + 1 for primes p, q, r, p != q.  There is no solution to either form with p, q distinct primes for r = 2. Thus, r^3 is odd and p*q must be even, so wlog p = 2. Thus, we need to solve for case 1: 2*q + 1 = r^3 for q, r prime. But r^3 - 1 = (r - 1)*(r^2 + r + 1), so r = 3 is the only prime solution producing the factor 2, leading to the pair 26, 27. Likewise, case 2: r^3 + 1 = (r + 1)*(r^2 - r + 1) has no solution with prime r producing the required factor 2. (End)

			14 and 15 are consecutive and both have prime signature {1, 1}

Wikipedia, Prime signature

Intersection of A006881 and A260143.

Subsequence of A038456.

s = Union@ Flatten@ Table[Prime[m] Prime[n], {n, Log2[#/3]}, {m, n + 1, PrimePi[#/Prime[n]]}] &[482]; s[[Flatten@ Map[Append[#, Last[#] + 1] &, Position[Differences[s], 1]]]] (* Michael De Vlieger, Oct 28 2021 *)
With[{sp=If[SquareFreeQ[#]&&PrimeOmega[#]==2,1,0]&/@Range[ 500]},DeleteDuplicates[ Flatten[SequencePosition[sp,{1,1}]]]] (* Harvey P. Dale, Jul 29 2022 *)

consecutive(n)=my(f=vecsort(factor(n)[, 2])); f==vecsort(factor(n-1)[, 2]) || f==vecsort(factor(n+1)[, 2]) \\  based on A260143
squarefree_semiprime(n)=(bigomega(n)==2&&omega(n)==2) \\ based on A006881
for(n=1, 500, if(squarefree_semiprime(n) && consecutive(n), print1(n, ", ")))

from sympy import factorint
def aupto(limit):
    aset, prevsig = set(), [1]
    for k in range(3, limit+2):
        sig = sorted(factorint(k).values())
        if sig == prevsig == [1, 1]: aset.update([k - 1, k])
        prevsig = sig
    return sorted(aset)
print(aupto(482)) # Michael S. Branicky, Sep 20 2021

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A039833 Smallest of three consecutive squarefree numbers k, k+1, k+2 of the form p*q where p and q are distinct primes.

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A039832 Numbers k such that k and k+1 both have 4 divisors.

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A178034 a(n) = binomial(n*Omega(n),Omega(n)) / n.

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A346549 Runs (of length > 1) of consecutive squarefree semiprimes.

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