A189987 -id:A189987 - cp's OEIS frontend

A101296 n has the a(n)-th distinct prime signature.

1, 2, 2, 3, 2, 4, 2, 5, 3, 4, 2, 6, 2, 4, 4, 7, 2, 6, 2, 6, 4, 4, 2, 8, 3, 4, 5, 6, 2, 9, 2, 10, 4, 4, 4, 11, 2, 4, 4, 8, 2, 9, 2, 6, 6, 4, 2, 12, 3, 6, 4, 6, 2, 8, 4, 8, 4, 4, 2, 13, 2, 4, 6, 14, 4, 9, 2, 6, 4, 9, 2, 15, 2, 4, 6, 6, 4, 9, 2, 12, 7, 4, 2, 13, 4, 4, 4, 8, 2, 13, 4, 6, 4, 4, 4, 16, 2, 6, 6, 11, 2, 9, 2, 8, 9, 4, 2, 15, 2, 9, 4, 12, 2, 9, 4, 6, 6, 4, 4, 17

Offset: 1

David Wasserman, Dec 21 2004

From Antti Karttunen, May 12 2017: (Start)
Restricted growth sequence transform of A046523, the least representative of each prime signature. Thus this partitions the natural numbers to the same equivalence classes as A046523, i.e., for all i, j: a(i) = a(j) <=> A046523(i) = A046523(j), and for that reason satisfies in that respect all the same conditions as A046523. For example, we have, for all i, j: if a(i) = a(j), then:
    A000005(i) = A000005(j), A008683(i) = A008683(j), A286605(i) = A286605(j).
So, this sequence (instead of A046523) can be used for finding sequences where a(n)'s value is dependent only on the prime signature of n, that is, only on the multiset of prime exponents in the factorization of n. (End)
This is also the restricted growth sequence transform of many other sequences, for example, that of A181819. See further comments there. - Antti Karttunen, Apr 30 2022

			From _David A. Corneth_, May 12 2017: (Start)
1 has prime signature (), the first distinct prime signature. Therefore, a(1) = 1.
2 has prime signature (1), the second distinct prime signature after (1). Therefore, a(2) = 2.
3 has prime signature (1), as does 2. Therefore, a(3) = a(2) = 2.
4 has prime signature (2), the third distinct prime signature after () and (1). Therefore, a(4) = 3. (End)
From _Antti Karttunen_, May 12 2017: (Start)
Construction of restricted growth sequences: In this case we start with a(1) = 1 for A046523(1) = 1, and thereafter, for all n > 1, we use the least so far unused natural number k for a(n) if A046523(n) has not been encountered before, otherwise [whenever A046523(n) = A046523(m), for some m < n], we set a(n) = a(m).
For n = 2, A046523(2) = 2, which has not been encountered before (first prime), thus we allot for a(2) the least so far unused number, which is 2, thus a(2) = 2.
For n = 3, A046523(2) = 2, which was already encountered as A046523(1), thus we set a(3) = a(2) = 2.
For n = 4, A046523(4) = 4, not encountered before (first square of prime), thus we allot for a(4) the least so far unused number, which is 3, thus a(4) = 3.
For n = 5, A046523(5) = 2, as for the first time encountered at n = 2, thus we set a(5) = a(2) = 2.
For n = 6, A046523(6) = 6, not encountered before (first semiprime pq with distinct p and q), thus we allot for a(6) the least so far unused number, which is 4, thus a(6) = 4.
For n = 8, A046523(8) = 8, not encountered before (first cube of a prime), thus we allot for a(8) the least so far unused number, which is 5, thus a(8) = 5.
For n = 9, A046523(9) = 4, as for the first time encountered at n = 4, thus a(9) = 3.
(End)
From _David A. Corneth_, May 12 2017: (Start)
(Rough) description of an algorithm of computing the sequence:
Suppose we want to compute a(n) for n in [1..20].
We set up a vector of 20 elements, values 0, and a number m = 1, the minimum number we haven't checked and c = 0, the number of distinct prime signatures we've found so far.
[0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0]
We check the prime signature of m and see that it's (). We increase c with 1 and set all elements up to 20 with prime signature () to 1. In the process, we adjust m. This gives:
[1, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0]. The least number we haven't checked is m = 2. 2 has prime signature (1). We increase c with 1 and set all elements up to 20 with prime signature (1) to 2. In the process, we adjust m. This gives:
[1, 2, 2, 0, 2, 0, 2, 0, 0, 0, 2, 0, 2, 0, 0, 0, 2, 0, 2, 0]
We check the prime signature of m = 4 and see that its prime signature is (2). We increase c with 1 and set all numbers up to 20 with prime signature (2) to 3. This gives:
[1, 2, 2, 3, 2, 0, 2, 0, 3, 0, 2, 0, 2, 0, 0, 0, 2, 0, 2, 0]
Similarily, after m = 6, we get
[1, 2, 2, 3, 2, 4, 2, 0, 3, 4, 2, 0, 2, 4, 4, 0, 2, 0, 2, 0], after m = 8 we get:
[1, 2, 2, 3, 2, 4, 2, 5, 3, 4, 2, 0, 2, 4, 4, 0, 2, 0, 2, 0], after m = 12 we get:
[1, 2, 2, 3, 2, 4, 2, 5, 3, 4, 2, 6, 2, 4, 4, 0, 2, 6, 2, 0], after m = 16 we get:
[1, 2, 2, 3, 2, 4, 2, 5, 3, 4, 2, 6, 2, 4, 4, 7, 2, 6, 2, 0], after m = 20 we get:
[1, 2, 2, 3, 2, 4, 2, 5, 3, 4, 2, 6, 2, 4, 4, 7, 2, 6, 2, 8]. Now, m > 20 so we stop. (End)
The above method is inefficient, because the step "set all elements a(n) up to n = Nmax with prime signature s(n) = S[c] to c" requires factoring all integers up to Nmax (or at least comparing their signature, once computed, with S[c]) again and again. It is much more efficient to run only once over each m = 1..Nmax, compute its prime signature s(m), add it to an ordered list in case it did not occur earlier, together with its "rank" (= new size of the list), and assign that rank to a(m). The list of prime signatures is much shorter than [1..Nmax]. One can also use m'(m) := the smallest n with the prime signature of m (which is faster to compute than to search for the signature) as representative for s(m), and set a(m) := a(m'(m)). Then it is sufficient to have just one counter (number of prime signatures seen so far) as auxiliary variable, in addition to the sequence to be computed. - _M. F. Hasler_, Jul 18 2019

Michel Marcus (terms 1..10000) & Antti Karttunen, Table of n, a(n) for n = 1..100000

Cf. A025487, A046523, A064839 (ordinal transform of this sequence), A181819, and arrays A095904, A179216.
Cf. A000005, A008683.
Sequences that are unions of finite number (>= 2) of equivalence classes determined by the values that this sequence obtains (i.e., sequences mentioned in David A. Corneth's May 12 2017 formula): A001358 (A001248 U A006881, values 3 & 4), A007422 (values 1, 4, 5), A007964 (2, 3, 4, 5), A014612 (5, 6, 9), A030513 (4, 5), A037143 (1, 2, 3, 4), A037144 (1, 2, 3, 4, 5, 6, 9), A080258 (6, 7), A084116 (2, 4, 5), A167171 (2, 4), A217856 (6, 9).
Cf. also A077462, A305897 (stricter variants, with finer partitioning) and A254524, A286603, A286605, A286610, A286619, A286621, A286622, A286626, A286378 for other similarly constructed sequences.

A101296 := proc(n)
    local a046523, a;
    a046523 := A046523(n) ;
    for a from 1 do
        if A025487(a) = a046523 then
            return a;
        elif A025487(a) > a046523 then
            return -1 ;
        end if;
    end do:
end proc: # R. J. Mathar, May 26 2017

With[{nn = 120}, Function[s, Table[Position[Keys@s, k_ /; MemberQ[k, n]][[1, 1]], {n, nn}]]@ Map[#1 -> #2 & @@ # &, Transpose@ {Values@ #, Keys@ #}] &@ PositionIndex@ Table[Times @@ MapIndexed[Prime[First@ #2]^#1 &, Sort[FactorInteger[n][[All, -1]], Greater]] - Boole[n == 1], {n, nn}] ] (* Michael De Vlieger, May 12 2017, Version 10 *)

find(ps, vps) = {for (k=1, #vps, if (vps[k] == ps, return(k)););}
lisps(nn) = {vps = []; for (n=1, nn, ps = vecsort(factor(n)[,2]); ips = find(ps, vps); if (! ips, vps = concat(vps, ps); ips = #vps); print1(ips, ", "););} \\ Michel Marcus, Nov 15 2015; edited by M. F. Hasler, Jul 16 2019

rgs_transform(invec) = { my(occurrences = Map(), outvec = vector(length(invec)), u=1); for(i=1, length(invec), if(mapisdefined(occurrences,invec[i]), my(pp = mapget(occurrences, invec[i])); outvec[i] = outvec[pp] , mapput(occurrences,invec[i],i); outvec[i] = u; u++ )); outvec; };
write_to_bfile(start_offset,vec,bfilename) = { for(n=1, length(vec), write(bfilename, (n+start_offset)-1, " ", vec[n])); }
write_to_bfile(1,rgs_transform(vector(100000,n,A046523(n))),"b101296.txt");
\\ Antti Karttunen, May 12 2017

A025487(a(n)) = A046523(n).
Indices of records give A025487. - Michel Marcus, Nov 16 2015
From David A. Corneth, May 12 2017: (Start) [Corresponding characteristic function in brackets]
a(A000012(n)) = 1 (sig.: ()).      [A063524]
a(A000040(n)) = 2 (sig.: (1)).     [A010051]
a(A001248(n)) = 3 (sig.: (2)).     [A302048]
a(A006881(n)) = 4 (sig.: (1,1)).   [A280710]
a(A030078(n)) = 5 (sig.: (3)).
a(A054753(n)) = 6 (sig.: (1,2)).   [A353472]
a(A030514(n)) = 7 (sig.: (4)).
a(A065036(n)) = 8 (sig.: (1,3)).
a(A007304(n)) = 9 (sig.: (1,1,1)). [A354926]
a(A050997(n)) = 10 (sig.: (5)).
a(A085986(n)) = 11 (sig.: (2,2)).
a(A178739(n)) = 12 (sig.: (1,4)).
a(A085987(n)) = 13 (sig.: (1,1,2)).
a(A030516(n)) = 14 (sig.: (6)).
a(A143610(n)) = 15 (sig.: (2,3)).
a(A178740(n)) = 16 (sig.: (1,5)).
a(A189975(n)) = 17 (sig.: (1,1,3)).
a(A092759(n)) = 18 (sig.: (7)).
a(A189988(n)) = 19 (sig.: (2,4)).
a(A179643(n)) = 20 (sig.: (1,2,2)).
a(A189987(n)) = 21 (sig.: (1,6)).
a(A046386(n)) = 22 (sig.: (1,1,1,1)).
a(A162142(n)) = 23 (sig.: (2,2,2)).
a(A179644(n)) = 24 (sig.: (1,1,4)).
a(A179645(n)) = 25 (sig.: (8)).
a(A179646(n)) = 26 (sig.: (2,5)).
a(A163569(n)) = 27 (sig.: (1,2,3)).
a(A179664(n)) = 28 (sig.: (1,7)).
a(A189982(n)) = 29 (sig.: (1,1,1,2)).
a(A179666(n)) = 30 (sig.: (3,4)).
a(A179667(n)) = 31 (sig.: (1,1,5)).
a(A179665(n)) = 32 (sig.: (9)).
a(A189990(n)) = 33 (sig.: (2,6)).
a(A179669(n)) = 34 (sig.: (1,2,4)).
a(A179668(n)) = 35 (sig.: (1,8)).
a(A179670(n)) = 36 (sig.: (1,1,1,3)).
a(A179671(n)) = 37 (sig.: (3,5)).
a(A162143(n)) = 38 (sig.: (2,2,2)).
a(A179672(n)) = 39 (sig.: (1,1,6)).
a(A030629(n)) = 40 (sig.: (10)).
a(A179688(n)) = 41 (sig.: (1,3,3)).
a(A179689(n)) = 42 (sig.: (2,7)).
a(A179690(n)) = 43 (sig.: (1,1,2,2)).
a(A189991(n)) = 44 (sig.: (4,4)).
a(A179691(n)) = 45 (sig.: (1,2,5)).
a(A179692(n)) = 46 (sig.: (1,9)).
a(A179693(n)) = 47 (sig.: (1,1,1,4)).
a(A179694(n)) = 48 (sig.: (3,6)).
a(A179695(n)) = 49 (sig.: (2,2,3)).
a(A179696(n)) = 50 (sig.: (1,1,7)).
(End)

Data section extended to 120 terms by Antti Karttunen, May 12 2017

Minor edits/corrections by M. F. Hasler, Jul 18 2019

A030632 Numbers with 14 divisors.

Original entry on oeis.org

192, 320, 448, 704, 832, 1088, 1216, 1458, 1472, 1856, 1984, 2368, 2624, 2752, 3008, 3392, 3645, 3776, 3904, 4288, 4544, 4672, 5056, 5103, 5312, 5696, 6208, 6464, 6592, 6848, 6976, 7232, 8019, 8128, 8192, 8384, 8768, 8896, 9477, 9536, 9664, 10048, 10432

Offset: 1

Jeff Burch

nonn

Numbers of the form p^13 (A138031) or p*q^6 (A189987), where p and q are distinct primes. - R. J. Mathar, Mar 01 2010

R. J. Mathar, Table of n, a(n) for n = 1..1000

Cf. A092759.

Select[Range[15000], DivisorSigma[0, #] == 14 &]

is(n)=numdiv(n)==14 \\ Charles R Greathouse IV, Jun 19 2016

from sympy import primepi, primerange, integer_nthroot
def A030632(n):
    def bisection(f,kmin=0,kmax=1):
        while f(kmax) > kmax: kmax <<= 1
        kmin = kmax >> 1
        while kmax-kmin > 1:
            kmid = kmax+kmin>>1
            if f(kmid) <= kmid:
                kmax = kmid
            else:
                kmin = kmid
        return kmax
    def f(x): return n+x-sum(primepi(x//p**6) for p in primerange(integer_nthroot(x,6)[0]+1))+primepi(integer_nthroot(x,7)[0])-primepi(integer_nthroot(x,13)[0])
    return bisection(f,n,n) # Chai Wah Wu, Feb 22 2025

A058061 Number of prime factors (counted with multiplicity) of d(n), the number of divisors of n.

Original entry on oeis.org

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

Offset: 1

Labos Elemer, Nov 23 2000

From Bernard Schott, Mar 24 2020: (Start)
a(n) = 1 iff n = p^(q-1) with p, q primes (A009087).
a(n) = 2 if n=p*q with p, q primes (A006881), or if n=p^2*q with p, q primes (A054753), or if n=p^4*q with p, q primes (A178739), or if n=p^6*q with p, q primes (A189987), or if n=p^2*q^4 with p, q primes (A189988), or if n=p^(m-1) with p prime and m is semiprime in A001358 (not exhaustive). (End)

			For n=120, d(120)=16, a(120)=4.

Michael De Vlieger, Table of n, a(n) for n = 1..10000

Cf. A001222, A000005, A058060, A079057 (partial sums).

Table[PrimeOmega@ DivisorSigma[0, n], {n, 120}] (* Michael De Vlieger, Feb 18 2017 *)

a(n) = bigomega(numdiv(n)); \\ Michel Marcus, Dec 14 2013

a(n) = A001222(A000005(n)).

Additive with a(p^e) = A001222(e+1). - Amiram Eldar, Jan 15 2024

A255231 The number of factorizations n = Product_i b_i^e_i, where all bases b_i are distinct, and all exponents e_i are distinct >=1.

Original entry on oeis.org

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

Offset: 1

Saverio Picozzi, Feb 18 2015

nonn

Not multiplicative: a(48) = a(2^4*3) = 5 <> a(2^4)*a(3) = 4*1 = 4. - R. J. Mathar, Nov 05 2016

			From _R. J. Mathar_, Nov 05 2016: (Start)
a(4)=2: 4^1 = 2^2.
a(8)=2: 8^1 = 2^3.
a(9)=2: 9^1 = 3^2.
a(12)=2: 12^1 = 2^2*3^1.
a(16)=4: 16^1 = 4^2 = 2^2*4^1 = 2^4.
a(18)=2: 18^1 = 2*3^2.
a(20)=2: 20^1 = 2^2*5^1.
a(24)=3: 24^1 = 2^2*6^1 = 2^3*3^1.
a(32)=5: 32^1 = 2^1*4^2 = 2^2*8^1 = 2^3*4^1 = 2^5.
a(36)=4: 36^1 = 6^2 = 3^2*4^1 = 2^2*9^1.
a(48)=5: 48^1 = 3^1*4^2 = 2^2*12^1 = 2^3*6^1 = 2^4*3^1.
a(60)=2 : 60^1 = 2^2*15^1.
a(64)=7: 64^1 = 8^2 = 4^3 = 2^2*16^1 = 2^3*8^1 = 2^4*4^1 = 2^6.
a(72)=6 : 72^1 = 3^2*8^1 = 2^1*6^2 = 2^2*18^1 = 2^3*9^1 = 2^3*3^2.
(End)

R. J. Mathar, Table of n, a(n) for n = 1..419
R. J. Mathar, Factorizations of integers into factors with distinct bases and exponents
Index to sequences related to prime signature

Cf. A000688 (b_i not necessarily distinct).

Cf. A001248, A005117, A030078, A030514, A054753, A065036, A085986, A085987, A143610, A178739.

# Count solutions for products if n = dvs_i^exps(i) where i=1..pividx are fixed
Apiv := proc(n,dvs,exps,pividx)
    local dvscnt, expscopy,i,a,expsrt,e ;
    dvscnt := nops(dvs) ;
    a := 0 ;
    if pividx > dvscnt then
        # have exhausted the exponent list: leave of the recursion
        # check that dvs_i^exps(i) is a representation
        if n = mul( op(i,dvs)^op(i,exps),i=1..dvscnt) then
            # construct list of non-0 exponents
            expsrt := [];
            for i from 1 to dvscnt do
                if op(i,exps) > 0 then
                    expsrt := [op(expsrt),op(i,exps)] ;
                end if;
            end do;
            # check that list is duplicate-free
            if nops(expsrt) = nops( convert(expsrt,set)) then
                return 1;
            else
                return 0;
            end if;
        else
            return 0 ;
        end if;
    end if;
    # need a local copy of the list to modify it
    expscopy := [] ;
    for i from 1 to nops(exps) do
        expscopy := [op(expscopy),op(i,exps)] ;
    end do:
    # loop over all exponents assigned to the next base in the list.
    for e from 0 do
        candf := op(pividx,dvs)^e ;
        if modp(n,candf) <> 0 then
            break;
        end if;
        # assign e to the local copy of exponents
        expscopy := subsop(pividx=e,expscopy) ;
        a := a+procname(n,dvs,expscopy,pividx+1) ;
    end do:
    return a;
end proc:
A255231 := proc(n)
    local dvs,dvscnt,exps ;
    if n = 1 then
        return 1;
    end if;
    # candidates for the bases are all divisors except 1
    dvs := convert(numtheory[divisors](n) minus {1},list) ;
    dvscnt := nops(dvs) ;
    # list of exponents starts at all-0 and is
    # increased recursively
    exps := [seq(0,e=1..dvscnt)] ;
    # take any subset of dvs for the bases, i.e. exponents 0 upwards
    Apiv(n,dvs,exps,1) ;
end proc:
seq(A255231(n),n=1..120) ; # R. J. Mathar, Nov 05 2016

a(n)=1 for all n in A005117. a(n)=2 for all n in A001248 and for all n in A054753 and for all n in A085987 and for all n in A030078. a(n)=3 for all n in A065036. a(n)=4 for all n in A085986 and for all n in A030514. a(n)=5 for all n in A178739, all n in A179644 and for all n in A050997. a(n)=6 for all n in A143610, all n in A162142 and all n in A178740. a(n)=7 for all n in A030516. a(n)=9 for all n in A189988 and all n in A189987. a(n)=10 for all n in A092759. a(n) = 11 for all n in A179664. a(n)=12 for all n in A179646. - R. J. Mathar, Nov 05 2016, May 20 2017

Values corrected. Incorrect comments removed. - R. J. Mathar, Nov 05 2016

A337533 1 together with nonsquares whose square part's square root is in the sequence.

Original entry on oeis.org

1, 2, 3, 5, 6, 7, 8, 10, 11, 12, 13, 14, 15, 17, 18, 19, 20, 21, 22, 23, 24, 26, 27, 28, 29, 30, 31, 33, 34, 35, 37, 38, 39, 40, 41, 42, 43, 44, 45, 46, 47, 50, 51, 52, 53, 54, 55, 56, 57, 58, 59, 60, 61, 62, 63, 65, 66, 67, 68

Offset: 1

Peter Munn, Aug 31 2020

The appearance of a number is determined by its prime signature.
Every squarefree number is present, as the square root of the square part of a squarefree number is 1. Other 4th-power-free numbers are present if and only if they are nonsquare.
If the square part of nonsquarefree k is a 4th power, k does not appear.
Every positive integer k is the product of a unique subset S_k of the terms of A050376, which are arranged in array form in A329050 (primes in column 0, squares of primes in column 1, 4th powers of primes in column 2 and so on). k > 1 is in this sequence if and only if the members of S_k occur in consecutive columns of A329050, starting with column 0.
If the qualifying condition in the previous paragraph was based on the rows instead of the columns of A329050, we would get A055932. The self-inverse function defined by A225546 transposes A329050. A225546 also has multiplicative properties such that if we consider A055932 and this sequence as sets, A225546(.) maps the members of either set 1:1 onto the other set.

			4 is square and not 1, so 4 is not in the sequence.
12 = 3 * 2^2 is nonsquare, and has square part 4, whose square root (2) is in the sequence. So 12 is in the sequence.
32 = 2 * 4^2 is nonsquare, but has square part 16, whose square root (4) is not in the sequence. So 32 is not in the sequence.

Robert Israel, Table of n, a(n) for n = 1..10000
Eric Weisstein's World of Mathematics, Square part
Index to sequences related to prime signature

Complement of A337534.
Closed under A000188(.).
A209229, A267116 are used in a formula defining this sequence.
Subsequences: A005117, A036537, A179646, A179666, A179669, A189987, A191554, A191555, A298207, A317090.
Subsequence of A164514.
A007913, A008833, A008835, A335324 give the squarefree, square and comparably related parts of a number.
Cf. A050376, A329050.
Related to A055932 via A225546.

S:= {1}:
for n from 2 to 100 do
  if not issqr(n) then
    F:= ifactors(n)[2];
    s:= mul(t[1]^floor(t[2]/2),t=F);
    if member(s,S) then S:= S union {n} fi
  fi
od:
sort(convert(S,list)); # Robert Israel, Jan 07 2025

pow2Q[n_] := n == 2^IntegerExponent[n, 2]; Select[Range[100], # == 1 || pow2Q[1 + BitOr @@ (FactorInteger[#][[;; , 2]])] &] (* Amiram Eldar, Sep 18 2020 *)

Numbers m such that A209229(A267116(m) + 1) = 1.
If A008835(a(n)) > 1 then A335324(a(n)) > 1.
If A008833(a(n)) > 1 then A007913(a(n)) > 1.

A319240 Positions of zeros in A316441, the list of coefficients in the expansion of Product_{n > 1} 1/(1 + 1/n^s).

Original entry on oeis.org

4, 6, 9, 10, 12, 14, 15, 18, 20, 21, 22, 25, 26, 28, 33, 34, 35, 38, 39, 44, 45, 46, 48, 49, 50, 51, 52, 55, 57, 58, 62, 63, 65, 68, 69, 72, 74, 75, 76, 77, 80, 82, 85, 86, 87, 91, 92, 93, 94, 95, 98, 99, 106, 108, 111, 112, 115, 116, 117, 118, 119, 121, 122

Offset: 1

Gus Wiseman, Sep 15 2018

From Tian Vlasic, Dec 31 2021: (Start)
Numbers that have an equal number of even and odd-length unordered factorizations.
There are infinitely many terms since p^2 is a term for prime p.
Out of all numbers of the form p^k with p prime (listed in A000961), only the numbers of the form p^2 (A001248) are terms.
Out of all numbers of the form p*q^k, p and q prime, only the numbers of the form p*q (A006881), p*q^2 (A054753), p*q^4 (A178739) and p*q^6 (A189987) are terms.
Similar methods can be applied to all prime signatures. (End)

			12 = 2*6 = 3*4 = 2*2*3 has an equal number of even-length factorizations and odd-length factorizations (2). - _Tian Vlasic_, Dec 09 2021

David A. Corneth, Table of n, a(n) for n = 1..10000

Complement of A319239.

Cf. A001055, A025487, A045778, A114592, A162247, A190938, A281116, A281118, A303386, A316441, A319238.

facs[n_]:=If[n<=1,{{}},Join@@Table[Map[Prepend[#,d]&,Select[facs[n/d],Min@@#>=d&]],{d,Rest[Divisors[n]]}]];
Join@@Position[Table[Sum[(-1)^Length[f],{f,facs[n]}],{n,100}],0]

A382292 Numbers k such that A382290(k) = 1.

Original entry on oeis.org

8, 24, 27, 32, 40, 54, 56, 64, 72, 88, 96, 104, 108, 120, 125, 135, 136, 152, 160, 168, 184, 189, 192, 200, 224, 232, 243, 248, 250, 264, 270, 280, 288, 296, 297, 312, 320, 328, 343, 344, 351, 352, 360, 375, 376, 378, 392, 408, 416, 424, 432, 440, 448, 456, 459, 472, 480, 486, 488, 500

Offset: 1

Amiram Eldar, Mar 21 2025

First differs from A374590 and A375432 at n = 25: A374590(25) = A375432(25) = 216 is not a term of this sequence.
Numbers k such that A382291(k) = 2, i.e., numbers whose number of infinitary divisors is twice the number of their unitary divisors.
Numbers whose prime factorization has a single exponent that is a sum of two distinct powers of 2 (A018900) and all the other exponents, if they exist, are powers of 2. Equivalently, numbers of the form p^e * m, where p is a prime, e is a term in A018900, and m is a term in A138302 that is coprime to p.
If k is a term then k^2 is also a term. If m is a term in A138302 that is coprime to k then k * m is also a term. The primitive terms, i.e., the terms that cannot be generated from smaller terms using these rules, are the numbers of the form p^(2^i+1), where p is prime and i >= 1.
Analogous to A060687, which is the sequence of numbers k with prime excess A046660(k) = 2.
The asymptotic density of this sequence is A271727 * Sum_{p prime} (((1 - 1/p)/f(1/p)) * Sum_{k>=1} 1/p^A018900(k)) = 0.11919967112489084407..., where f(x) = 1 - x^3 + Sum_{k>=2} (x^(2^k)-x^(2^k+1)).

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

Cf. A018900, A046660, A060687, A271727, A374590, A375432, A382290, A382291.

Subsequences (numbers of the form): A030078 (p^3), A050997 (p^5), A030516 (p^6), A179665 (p^9), A030629 (p^10), A030631 (p^12), A065036 (p^3*q), A178740 (p^5*q), A189987 (p^6*q), A179692 (p^9*q), A143610 (p^2*q^3), A179646 (p^5*q^2), A189990 (p^2*q^6), A179702 (p^4*q^5), A179666 (p^4*q^3), A190464 (p^4*q^6), A163569 (p^3*q^2*r), A189975 (p*q*r^3), A190115 (p^2*q^3*r^4), A381315, A048109.

f[p_, e_] := DigitCount[e, 2, 1] - 1; q[1] = False; q[n_] := Plus @@ f @@@ FactorInteger[n] == 1; Select[Range[500], q]

isok(k) = vecsum(apply(x -> hammingweight(x) - 1, factor(k)[, 2])) == 1;

A275345 Characteristic polynomials of a square matrix based on A051731 where A051731(1,N)=1 and A051731(N,N)=0 and where N=size of matrix, analogous to the Redheffer matrix.

Original entry on oeis.org

1, 1, -1, -1, -1, 1, -1, 0, 2, -1, 0, 0, 2, -3, 1, -1, 2, 1, -5, 4, -1, 1, -3, 5, -8, 9, -5, 1, -1, 4, -4, -5, 15, -14, 6, -1, 0, -1, 6, -17, 29, -31, 20, -7, 1, 0, 0, 2, -13, 36, -55, 50, -27, 8, -1, 1, -7, 23, -50, 84, -112, 112, -78, 35, -9, 1

Offset: 0

Mats Granvik, Jul 24 2016

From Mats Granvik, Sep 30 2017: (Start)
Conjecture: The largest absolute value of the eigenvalues of these characteristic polynomials appear to have the same prime signature in the factorization of the matrix sizes N.
In other words: Let b(N) equal the sequence of the largest absolute values of the eigenvalues of the characteristic polynomials of the matrices of size N. b(N) is then a sequence of truncated eigenvalues starting:
b(N=1..infinity)
= 1.00000, 1.61803, 1.61803, 2.00000, 1.61803, 2.20557, 1.61803, 2.32472, 2.00000, 2.20557, 1.61803, 2.67170, 1.61803, 2.20557, 2.20557, 2.61803, 1.61803, 2.67170, 1.61803, 2.67170, 2.20557, 2.20557, 1.61803, 3.08032, 2.00000, 2.20557, 2.32472, 2.67170, 1.61803, 2.93796, 1.61803, 2.89055, 2.20557, 2.20557, 2.20557, 3.21878, 1.61803, 2.20557, 2.20557, 3.08032, 1.61803, 2.93796, 1.61803, 2.67170, 2.67170, 2.20557, 1.61803, 3.45341, 2.00000, 2.67170, 2.20557, 2.67170, 1.61803, 3.08032, 2.20557, 3.08032, 2.20557, 2.20557, 1.61803, 3.53392, 1.61803, 2.20557, 2.67170, ...
It then appears that for n = 1,2,3,4,5,...,infinity we have the table:
Prime signature: b(Axxxxxx(n)) = Largest abs(eigenvalue):
p^0      : b(1)          = 1.0000000000000000000000000000...
p        : b(A000040(n)) = 1.6180339887498949025257388711...
p^2      : b(A001248(n)) = 2.0000000000000000000000000000...
p*q      : b(A006881(n)) = 2.2055694304005917238953315973...
p^3      : b(A030078(n)) = 2.3247179572447480566665944934...
p^2*q    : b(A054753(n)) = 2.6716998816571604358216518448...
p^4      : b(A030514(n)) = 2.6180339887498917939012699207...
p^3*q    : b(A065036(n)) = 3.0803227214906021558249449299...
p*q*r    : b(A007304(n)) = 2.9379558827528557962693867011...
p^5      : b(A050997(n)) = 2.8905508875432590620846440288...
p^2*q^2  : b(A085986(n)) = 3.2187765853016649941764626419...
p^4*q    : b(A178739(n)) = 3.4534111136673804054453285061...
p^2*q*r  : b(A085987(n)) = 3.5339198574905377192578725953...
p^6      : b(A030516(n)) = 3.1478990357047909043330946587...
p^3*q^2  : b(A143610(n)) = 3.7022736187975437971431347250...
p^5*q    : b(A178740(n)) = 3.8016448153137023524550386355...
p^3*q*r  : b(A189975(n)) = 4.0600260453688532535920785448...
p^7      : b(A092759(n)) = 3.3935083220984414431597997463...
p^4*q^2  : b(A189988(n)) = 4.1453038440113498808159420150...
p^2*q^2*r: b(A179643(n)) = 4.2413382309993874486053755390...
p^6*q    : b(A189987(n)) = 4.1311805192254587026923218218...
p*q*r*s  : b(A046386(n)) = 3.8825338629275134572083061357...
...
b(Axxxxxx(1)) in the sequences above, is given by A025487.
(End)
First column in the coefficients of the characteristic polynomials is the Möbius function A008683.
Row sums of coefficients start: 0, -1, 0, 0, 0, 0, 0, 0, 0, ...
Third diagonal is a signed version of A000096.
Most of the eigenvalues are equal to 1. The number of eigenvalues equal to 1 are given by A075795 for n>1.
The first three of the eigenvalues above can be calculated as nested radicals. The fourth eigenvalue 2.205569430400590... minus 1 = 1.205569430400590... is also a nested radical.

			{
{ 1},
{ 1, -1},
{-1, -1,  1},
{-1,  0,  2,  -1},
{ 0,  0,  2,  -3,  1},
{-1,  2,  1,  -5,  4,   -1},
{ 1, -3,  5,  -8,  9,   -5,   1},
{-1,  4, -4,  -5, 15,  -14,   6,  -1},
{ 0, -1,  6, -17, 29,  -31,  20,  -7,  1},
{ 0,  0,  2, -13, 36,  -55,  50, -27,  8, -1},
{ 1, -7, 23, -50, 84, -112, 112, -78, 35, -9, 1}
}

Mats Granvik, Mathematica program to verify that eigenvalues determine prime signature
OEIS Wiki, Prime signatures
Eric Weisstein, Prime signature
Index to sequences related to prime signature

Cf. A051731, A008683, A000040, A001248, A006881, A030078, A030514, A054753, A000096, A001622, A272874, A075795.

Cf. A025487. - Mats Granvik, Sep 30 2017

Clear[x, AA, nn, s]; Monitor[AA = Flatten[Table[A = Table[Table[If[Mod[n, k] == 0, 1, 0], {k, 1, nn}], {n, 1, nn}]; MatrixForm[A]; a = A[[1, nn]]; A[[1, nn]] = A[[nn, nn]]; A[[nn, nn]] = a; CoefficientList[CharacteristicPolynomial[A, x], x], {nn, 1, 10}]], nn]

cp's OEIS Frontend

A101296 n has the a(n)-th distinct prime signature.

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A030632 Numbers with 14 divisors.

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A058061 Number of prime factors (counted with multiplicity) of d(n), the number of divisors of n.

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A255231 The number of factorizations n = Product_i b_i^e_i, where all bases b_i are distinct, and all exponents e_i are distinct >=1.

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A337533 1 together with nonsquares whose square part's square root is in the sequence.

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A319240 Positions of zeros in A316441, the list of coefficients in the expansion of Product_{n > 1} 1/(1 + 1/n^s).

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A382292 Numbers k such that A382290(k) = 1.

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