A189988 -id:A189988 - 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

A179646 Product of the 5th power of a prime and different distinct prime of the 2nd power (p^5*q^2).

Original entry on oeis.org

288, 800, 972, 1568, 3872, 5408, 6075, 9248, 11552, 11907, 12500, 16928, 26912, 28125, 29403, 30752, 41067, 43808, 53792, 59168, 67228, 70227, 70688, 87723, 89888, 111392, 119072, 128547, 143648, 151263, 153125, 161312, 170528, 199712

Offset: 1

Vladimir Joseph Stephan Orlovsky, Jul 21 2010

nonn

288=2^5*3^2, 800=2^5*5^2,..

T. D. Noe, Table of n, a(n) for n = 1..1000
Will Nicholes, Prime Signatures
Index to sequences related to prime signature

Cf. A030636, A046308, A007774.

Cf. A085548, A085965, A085967.

f[n_]:=Sort[Last/@FactorInteger[n]]=={2,5}; Select[Range[200000], f]

list(lim)=my(v=List(),t);forprime(p=2,(lim\4)^(1/5),t=p^5;forprime(q=2,sqrt(lim\t),if(p==q,next);listput(v,t*q^2)));vecsort(Vec(v)) \\ Charles R Greathouse IV, Jul 20 2011

from math import isqrt
from sympy import primepi, primerange, integer_nthroot
def A189988(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(isqrt(x//p**4)) for p in primerange(integer_nthroot(x,4)[0]+1))+primepi(integer_nthroot(x,6)[0])
    return bisection(f,n,n) # Chai Wah Wu, Feb 21 2025

Sum_{n>=1} 1/a(n) = P(2)*P(5) - P(7) = A085548 * A085965 - A085967 = 0.007886..., where P is the prime zeta function. - Amiram Eldar, Jul 06 2020

A030633 Numbers with 15 divisors.

Original entry on oeis.org

144, 324, 400, 784, 1936, 2025, 2500, 2704, 3969, 4624, 5625, 5776, 8464, 9604, 9801, 13456, 13689, 15376, 16384, 21609, 21904, 23409, 26896, 29241, 29584, 30625, 35344, 42849, 44944, 55696, 58564, 59536, 60025, 68121, 71824, 75625

Offset: 1

Jeff Burch

nonn

Numbers of the form p^14 (subset of A010802) or p^2*q^4 (A189988) 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. A000005, A010802, A030630, A030631, A030632, A189988.

Select[Range[300000],DivisorSigma[0,#]==15&] (* Vladimir Joseph Stephan Orlovsky, May 05 2011 *)

is(n)=numdiv(n)==15 \\ Charles R Greathouse IV, Jun 19 2016

from math import isqrt
from sympy import primepi, primerange, integer_nthroot
def A030633(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(isqrt(x//p**4)) for p in primerange(integer_nthroot(x,4)[0]+1))+primepi(integer_nthroot(x,6)[0])-primepi(integer_nthroot(x,14)[0])
    return bisection(f,n,n) # Chai Wah Wu, Feb 22 2025

From Amiram Eldar, Jul 03 2022: (Start)
A000005(a(n)) = 15.
Sum_{n>=1} 1/a(n) = P(2)*P(4) - P(6) + P(14) = 0.0178111..., where P is the prime zeta function. (End)

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

A217584 Numbers k such that d(k^2)/d(k) is an integer, where d(k) is the number of divisors of k.

Original entry on oeis.org

1, 144, 324, 400, 784, 1936, 2025, 2500, 2704, 3600, 3969, 4624, 5625, 5776, 7056, 8100, 8464, 9604, 9801, 13456, 13689, 15376, 15876, 17424, 19600, 21609, 21904, 22500, 23409, 24336, 26896, 29241, 29584, 30625, 35344, 39204, 41616, 42849, 44944, 48400, 51984

Offset: 1

Michel Marcus, Oct 07 2012

nonn

The ratio d(k^2)/d(k) is: 1 for the number 1, 3 for numbers of the form p^4*q^2, 5 for numbers of the form p^4*q^2*r^2 (p, q, r being different primes).
Primes can't be in the sequence. A prime p has two divisors, while p^2 has three divisors: 1, p, p^2. - Alonso del Arte, Oct 07 2012
All the terms are squares since d(m) is odd if and only if m is a square, so d(k^2) is odd and since d(k)|d(k^2), d(k) is also odd, so k is a square. The ratio d(k^2)/d(k) can take values other than 1, 3, and 5: 1587600 is the least term with a ratio 9, and 192099600 is the least term with a ratio 15. - Amiram Eldar, May 23 2020
From Bernard Schott, May 29 2020 and Nov 22 2020: (Start)
This sequence comes from the 3rd problem, proposed by Belarus, during the 39th International Mathematical Olympiad in 1998 at Taipei (Taiwan) [see the link IMO].
If the prime signature of k is (u_1, u_2, ... , u_q) then d(k^2)/d(k) = Product_{i=1..q} (2*u_i+1)/(u_i+1). Two results:
1) If k is a term such that d(k^2)/d(k) = m, then all numbers that have the same prime signature of k are also terms and give the same ratio (see examples below).
2) The set of the integer values of the ratio d(k^2)/d(k) is exactly the set of all positive odd integers (see Marcin E. Kuczma reference).
Some examples:
For numbers with prime signature = (4, 2) (A189988), the ratio is 3 and the smallest such integer is 144 = 2^4 * 3^2.
For numbers with prime signature = (4, 2, 2) (A179746), the ratio is 5 and the smallest such integer is 3600 = 2^4 * 3^2 * 5^2.
For numbers with prime signature = (4, 4, 2, 2) the ratio is 9 and the smallest such integer is 1587600 = 2^4 * 3^4 * 5^2 * 7^2.
For numbers with prime signature = (8, 4, 4, 2, 2) the ratio is 17 and the smallest such integer is 76839840000 = 2^8 * 3^4 * 5^4 * 7^2 * 11^2 (found by David A. Corneth with other prime signatures). (End)

			d(1^2)/d(1) = d(1)/d(1) = 1 an integer, so 1 belongs to the sequence.
144^2 has 45 divisors: 1, 2, 3, 4, 6, 8, 9, 12, ..., 20736, while 144 has 15 divisors: 1, 2, 3, 4, 6, 8, 9, 12, ..., 144; 45/15 = 3 and so 144 is in the sequence.

Marcin E. Kuczma, International Mathematical Olympiads, 1986-1999, The Mathematical Association of America, 2003, pages 134-135.

Amiram Eldar, Table of n, a(n) for n = 1..10000 (terms 1..100 from Paolo P. Lava)
The IMO Compendium, Problem 3, 39th IMO 1998.
Kin Y. Li, Problem 3, Mathematical Excalibur, Vol. 4, No. 3, Jan.-Mar. 1999.
Index to sequences related to Olympiads.

Cf. A000005, A003593, A025487, A048691.
Subsequences: A189988 (d(k^2)/d(k) = 3), A179746 (d(k^2)/d(k) = 5).
Cf. A339055 (values taken by d(a(n)^2)/d(a(n))), A339056 (smallest k such that d(k^2)/d(k) = n-th odd).

Select[Range[1000], IntegerQ[DivisorSigma[0, #^2]/DivisorSigma[0, #]] &] (* Alonso del Arte, Oct 07 2012 *)
Select[Range[228]^2, Divisible[DivisorSigma[0, #^2], DivisorSigma[0, #]] &] (* Amiram Eldar, May 23 2020 *)

dn2dn(n)= {for (i=1, n, if (denominator(numdiv(i^2)/numdiv(i))==1, print1(i,", ");););}

A175752 Numbers with 45 divisors.

Original entry on oeis.org

3600, 7056, 8100, 15876, 17424, 19600, 20736, 22500, 24336, 39204, 41616, 48400, 51984, 54756, 67600, 76176, 86436, 93636, 94864, 99225, 104976, 115600, 116964, 121104, 122500, 132496, 138384, 144400, 147456, 160000, 171396, 197136

Offset: 1

Jaroslav Krizek, Aug 27 2010

nonn

Numbers of the forms p^44, p^14*q^2, p^8*q^4 (squares of A189988) and p^4*q^2*r^2 (A179746), where p, q, and r are distinct primes.

T. D. Noe, Table of n, a(n) for n = 1..1000
OEIS Wiki, Index entries for number of divisors

Cf. A000005, A139574, A175751.

Select[Range[400000],DivisorSigma[0,#]==45&] (* Vladimir Joseph Stephan Orlovsky, May 06 2011 *)

is(n)=numdiv(n)==45 \\ Charles R Greathouse IV, Jun 19 2016

A000005(a(n)) = 45.

Sum_{n>=1} 1/a(n) = (P(2)^2*P(4) - P(4)^2)/2 - P(2)*P(6) + P(8) + P(2)*P(14) - P(16) + P(4)*P(8) - P(12) + P(44) = 0.00133023..., where P is the prime zeta function. - Amiram Eldar, Jul 03 2022

A179746 Numbers of the form p^4q^2r^2 where p, q, and r are distinct primes.

Original entry on oeis.org

3600, 7056, 8100, 15876, 17424, 19600, 22500, 24336, 39204, 41616, 48400, 51984, 54756, 67600, 76176, 86436, 93636, 94864, 99225, 115600, 116964, 121104, 122500, 132496, 138384, 144400, 171396, 197136, 211600, 226576, 240100, 242064, 245025

Offset: 1

Vladimir Joseph Stephan Orlovsky, Jul 25 2010

nonn

Numbers k such that tau(k^2)/tau(k) = 5 where tau(n) is the number of divisors of n (A000005). - Bernard Schott, Nov 27 2020

T. D. Noe, Table of n, a(n) for n = 1..1000
Will Nicholes, List of Prime Signatures
Index to sequences related to prime signature

Subsequence of A217584.

Cf. A189988 (tau(k^2)/tau(k) = 3).

f[n_]:=Sort[Last/@FactorInteger[n]]=={2,2,4}; Select[Range[200000],f]

list(lim)=my(v=List(),t1,t2);forprime(p=2, (lim\36)^(1/4), t1=p^4;forprime(q=2, sqrt(lim\t1), if(p==q, next);t2=t1*q^2;forprime(r=q+1, sqrt(lim\t2), if(p==r,next);listput(v,t2*r^2)))); vecsort(Vec(v)) \\ Charles R Greathouse IV, Jul 24 2011

from math import isqrt
from sympy import primepi, primerange, integer_nthroot
def A179746(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((t:=primepi(s:=isqrt(y:=x//r**2)))+(t*(t-1)>>1)-sum(primepi(y//k) for k in primerange(1, s+1)) for r in primerange(isqrt(x)+1))+sum(primepi(x//p**3) for p in primerange(integer_nthroot(x,3)[0]+1))-primepi(integer_nthroot(x,4)[0])
    return bisection(f,n,n)**2 # Chai Wah Wu, Mar 27 2025

Sum_{n>=1} 1/a(n) = (P(2)^2*P(4) - P(4)^2)/2 - P(2)*P(6) + P(8) = 0.00125114..., where P is the prime zeta function. - Amiram Eldar, Jul 03 2022

a(n) = A085987(n)^2. - R. J. Mathar, May 05 2023

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]

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

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A179646 Product of the 5th power of a prime and different distinct prime of the 2nd power (p^5*q^2).

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A030633 Numbers with 15 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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A217584 Numbers k such that d(k^2)/d(k) is an integer, where d(k) is the number of divisors of k.

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A175752 Numbers with 45 divisors.

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A179746 Numbers of the form p^4q^2r^2 where p, q, and r are distinct primes.