Displaying 1-10 of 23 results found.
Members of A143610 for which both neighbors are squarefree.
+20
0
72, 108, 200, 392, 500, 968, 1352, 1372, 4232, 7688, 8788, 13448, 14792, 17672, 19652, 27436, 27848, 35912, 48668, 49928, 55112, 75272, 81608, 84872, 97556, 102152, 119164, 137288, 150152, 154568, 177608, 182408, 197192, 202612, 223112
COMMENTS
Numbers of the form n=p^2*q^3 where p and q are two different primes and where n+1 and n-1 are both in A005117.
MATHEMATICA
f[n_]:=Last/@FactorInteger[n]=={2, 3}||Last/@FactorInteger[n]=={3, 2}; << NumberTheory`NumberTheoryFunctions` lst={}; Do[If[f[n], If[SquareFreeQ[n-1]&&SquareFreeQ[n+1], AppendTo[lst, n]]], {n, 3*9!}]; lst
Number of Abelian groups of order n; number of factorizations of n into prime powers.
(Formerly M0064 N0020)
+10
137
1, 1, 1, 2, 1, 1, 1, 3, 2, 1, 1, 2, 1, 1, 1, 5, 1, 2, 1, 2, 1, 1, 1, 3, 2, 1, 3, 2, 1, 1, 1, 7, 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, 11, 1, 1, 1, 2, 1, 1, 1, 6, 1, 1, 2, 2, 1, 1, 1, 5, 5, 1, 1, 2, 1, 1, 1, 3, 1, 2, 1, 2, 1, 1, 1, 7, 1, 2, 2, 4, 1, 1, 1, 3, 1, 1, 1
COMMENTS
Equivalently, number of Abelian groups with n conjugacy classes. - Michael Somos, Aug 10 2010 a(n) depends only on prime signature of n (cf. A025487). So a(24) = a(375) since 24 = 2^3*3 and 375 = 3*5^3 both have prime signature (3, 1). Also number of rings with n elements that are the direct product of fields; these are the commutative rings with n elements having no nilpotents; likewise the commutative rings where for every element x there is a k > 0 such that x^(k+1) = x. - Franklin T. Adams-Watters, Oct 20 2006 a(n) = 1 if and only if n is from A005117 (squarefree numbers). See the Ahmed Fares comment there, and the formula for n>=2 below. - Wolfdieter Lang, Sep 09 2012 Also, from a theorem of Molnár (see [Molnár]), the number of (non-isomorphic) abelian groups of order 2*n + 1 is equal to the number of non-congruent lattice Z-tilings of R^n by crosses, where a "cross" is a unit cube in R^n for which at each facet is attached another unit cube (Z, R are the integers and reals, respectively). (Cf. [Horak].) - L. Edson Jeffery, Nov 29 2012 Zeta(k*s) is the Dirichlet generating function of the characteristic function of numbers which are k-th powers (k=1 in A000012, k=2 in A010052, k=3 in A010057, see arXiv:1106.4038 Section 3.1). The infinite product over k (here) is the number of representations n=product_i (b_i)^(e_i) where all exponents e_i are distinct and >=1. Examples: a(n=4)=2: 4^1 = 2^2. a(n=8)=3: 8^1 = 2^1*2^2 = 2^3. a(n=9)=2: 9^1 = 3^2. a(n=12)=2: 12^1 = 3*2^2. a(n=16)=5: 16^1 = 2*2^3 = 4^2 = 2^2*4^1 = 2^4. If the e_i are the set {1,2} we get A046951, the number of representations as a product of a number and a square. - R. J. Mathar, Nov 05 2016
REFERENCES
Steven R. Finch, Mathematical Constants, Cambridge, 2003, pp. 274-278.
D. S. Mitrinovic et al., Handbook of Number Theory, Kluwer, Section XIII.12, p. 468.
J. S. Rose, A Course on Group Theory, Camb. Univ. Press, 1978, see p. 7.
N. J. A. Sloane, A Handbook of Integer Sequences, Academic Press, 1973 (includes this sequence).
N. J. A. Sloane and Simon Plouffe, The Encyclopedia of Integer Sequences, Academic Press, 1995 (includes this sequence).
A. Speiser, Die Theorie der Gruppen von endlicher Ordnung, 4. Auflage, Birkhäuser, 1956.
FORMULA
Multiplicative with a(p^k) = number of partitions of k = A000041(k); a(mn) = a(m)a(n) if (m, n) = 1. a(n) = Product_{j = 1..N(n)} A000041(e(j)), n >= 2, if n = Product_{j = 1..N(n)} prime(j)^e(j), N(n) = A001221(n). See the Richert reference, quoting A. Speiser's book on finite groups (in German, p. 51 in words). - Wolfdieter Lang, Jul 23 2011 In terms of the cycle index of the symmetric group: Product_{q=1..m} [z^{v_q}] Z(S_v) 1/(1-z) where v is the maximum exponent of any prime in the prime factorization of n, v_q are the exponents of the prime factors, and Z(S_v) is the cycle index of the symmetric group on v elements. - Marko Riedel, Oct 03 2014 Dirichlet g.f.: Sum_{n >= 1} a(n)/n^s = Product_{k >= 1} zeta(ks) [Kendall]. - Álvar Ibeas, Nov 05 2014 Asymptotic mean: lim_{n->oo} (1/n) * Sum_{k=1..n} a(k) = A021002. (End)
EXAMPLE
a(1) = 1 since the trivial group {e} is the only group of order 1, and it is Abelian; alternatively, since the only factorization of 1 into prime powers is the empty product.
a(p) = 1 for any prime p, since the only factorization into prime powers is p = p^1, and (in view of Lagrange's theorem) there is only one group of prime order p; it is isomorphic to (Z/pZ,+) and thus Abelian.
a(8) = 3 because 8 = 2^3, hence a(8) = pa(3) = A000041(3) = 3 from the partitions (3), (2, 1) and (1, 1, 1), leading to the 3 factorizations of 8: 8, 4*2 and 2*2*2. a(36) = 4 because 36 = 2^2*3^2, hence a(36) = pa(2)*pa(2) = 4 from the partitions (2) and (1, 1), leading to the 4 factorizations of 36: 2^2*3^2, 2^2*3^1*3^1, 2^1*2^1*3^2 and 2^1*2^1*3^1*3^1.
(End)
MAPLE
with(combinat): readlib(ifactors): for n from 1 to 120 do ans := 1: for i from 1 to nops(ifactors(n)[2]) do ans := ans*numbpart(ifactors(n)[2][i][2]) od: printf(`%d, `, ans): od: # James A. Sellers, Dec 07 2000
MATHEMATICA
f[n_] := Times @@ PartitionsP /@ Last /@ FactorInteger@n; Array[f, 107] (* Robert G. Wilson v, Sep 22 2006 *)
PROG
(Sage)
def a(n):
F=factor(n)
return prod([number_of_partitions(F[i][1]) for i in range(len(F))])
(Haskell)
a000688 = product . map a000041 . a124010_row
(Python)
from sympy import factorint, npartitions
from math import prod
def A000688(n): return prod(map(npartitions, factorint(n).values())) # Chai Wah Wu, Jan 14 2022
CROSSREFS
Cf. A000001, A021002, A060689, A000041, A000961, A001055, A005361, A034382, A046054, A046055, A046056, A046101, A050360, A055653, A057521, A101872 (bisection), A101876 (quadrisection), A124010, A050361, A051532, A129667 (Dirichlet inverse).
n has the a(n)-th distinct prime signature.
+10
109
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
COMMENTS
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: 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
FORMULA
From David A. Corneth, May 12 2017: (Start) [Corresponding characteristic function in brackets] a( A085987(n)) = 13 (sig.: (1,1,2)). a( A189975(n)) = 17 (sig.: (1,1,3)). a( A179643(n)) = 20 (sig.: (1,2,2)). 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( A163569(n)) = 27 (sig.: (1,2,3)). a( A189982(n)) = 29 (sig.: (1,1,1,2)). a( A179667(n)) = 31 (sig.: (1,1,5)). a( A179669(n)) = 34 (sig.: (1,2,4)). a( A179670(n)) = 36 (sig.: (1,1,1,3)). a( A162143(n)) = 38 (sig.: (2,2,2)). a( A179672(n)) = 39 (sig.: (1,1,6)). a( A179688(n)) = 41 (sig.: (1,3,3)). a( A179690(n)) = 43 (sig.: (1,1,2,2)). a( A179691(n)) = 45 (sig.: (1,2,5)). a( A179693(n)) = 47 (sig.: (1,1,1,4)). a( A179695(n)) = 49 (sig.: (2,2,3)). a( A179696(n)) = 50 (sig.: (1,1,7)). (End)
EXAMPLE
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)
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)
(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
MAPLE
local a046523, a;
for a from 1 do
return a;
return -1 ;
end if;
end do:
MATHEMATICA
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 *)
PROG
(PARI) 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 (PARI)
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");
CROSSREFS
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.
Numbers with 12 divisors.
+10
15
60, 72, 84, 90, 96, 108, 126, 132, 140, 150, 156, 160, 198, 200, 204, 220, 224, 228, 234, 260, 276, 294, 306, 308, 315, 340, 342, 348, 350, 352, 364, 372, 380, 392, 414, 416, 444, 460, 476, 486, 490, 492, 495, 500, 516, 522, 525, 532, 544, 550
MATHEMATICA
Select[Range[600], DivisorSigma[0, #]==12&] (* Harvey P. Dale, Jun 01 2016 *)
PROG
(PARI) for(n=1, 1e3, if(numdiv(n)==12, print1(n, ", "))) \\ Altug Alkan, Nov 11 2015
Product of the 5th power of a prime ( A050997) and a different prime (p^5*q).
+10
11
96, 160, 224, 352, 416, 486, 544, 608, 736, 928, 992, 1184, 1215, 1312, 1376, 1504, 1696, 1701, 1888, 1952, 2144, 2272, 2336, 2528, 2656, 2673, 2848, 3104, 3159, 3232, 3296, 3424, 3488, 3616, 4064, 4131, 4192, 4384, 4448, 4617, 4768, 4832, 5024, 5216
MATHEMATICA
With[{nn=50}, Take[Union[Flatten[{#[[1]]^5 #[[2]], #[[1]]#[[2]]^5}&/@Subsets[ Prime[ Range[nn]], {2}]]], nn]] (* Harvey P. Dale, Mar 18 2013 *)
PROG
(PARI) list(lim)=my(v=List(), t); forprime(p=2, (lim\2)^(1/5), t=p^5; forprime(q=2, lim\t, if(p==q, next); listput(v, t*q))); vecsort(Vec(v)) \\ Altug Alkan, Nov 11 2015 (PARI) isok(n)=my(f=factor(n)[, 2]); f==[5, 1]~||f==[1, 5]~
for(n=1, 1e4, if(isok(n), print1(n, ", "))) \\ Altug Alkan, Nov 11 2015
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.
+10
9
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
COMMENTS
Not multiplicative: a(48) = a(2^4*3) = 5 <> a(2^4)*a(3) = 4*1 = 4. - R. J. Mathar, Nov 05 2016
EXAMPLE
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)
MAPLE
# 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:
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:
CROSSREFS
Cf. A000688 (b_i not necessarily distinct).
EXTENSIONS
Values corrected. Incorrect comments removed. - R. J. Mathar, Nov 05 2016
If n = Product (p_j^k_j) then a(n) = Product (k_j + 2), with a(1) = 1.
+10
9
1, 3, 3, 4, 3, 9, 3, 5, 4, 9, 3, 12, 3, 9, 9, 6, 3, 12, 3, 12, 9, 9, 3, 15, 4, 9, 5, 12, 3, 27, 3, 7, 9, 9, 9, 16, 3, 9, 9, 15, 3, 27, 3, 12, 12, 9, 3, 18, 4, 12, 9, 12, 3, 15, 9, 15, 9, 9, 3, 36, 3, 9, 12, 8, 9, 27, 3, 12, 9, 27, 3, 20, 3, 9, 12, 12, 9, 27, 3, 18
COMMENTS
Inverse Moebius transform of A056671. a(n) depends only on the prime signature of n (see formulas). - Bernard Schott, May 03 2021
FORMULA
a(n) = Sum_{d|n, gcd(d, n/d) = 1} tau(d).
a(p^k) = k+2 for p prime, or signature [k].
a( A006881(n)) = 9 for signature [1, 1]. a( A054753(n)) = 12 for signature [2, 1]. a( A065036(n)) = 15 for signature [3, 1]. a( A085986(n)) = 16 for signature [2, 2]. a( A178739(n)) = 18 for signature [4, 1]. a( A143610(n)) = 20 for signature [3, 2]. a( A007304(n)) = 27 for signature [1, 1, 1]. (End) Dirichlet g.f.: zeta(s)^2 * Product_{primes p} (1 + 1/p^s - 1/p^(2*s)). - Vaclav Kotesovec, Feb 11 2023
MATHEMATICA
a[1] = 1; a[n_] := Times @@ ((#[[2]] + 2) & /@ FactorInteger[n]); Table[a[n], {n, 80}]
a[n_] := Sum[If[GCD[d, n/d] == 1, DivisorSigma[0, d], 0], {d, Divisors[n]}]; Table[a[n], {n, 80}]
PROG
(PARI) a(n) = sumdiv(n, d, if(gcd(d, n/d)==1, numdiv(d))) \\ Andrew Howroyd, Apr 15 2021 (PARI) for(n=1, 100, print1(direuler(p=2, n, (1 + X - X^2)/(1-X)^2)[n], ", ")) \\ Vaclav Kotesovec, Feb 11 2023
CROSSREFS
Cf. A000005, A001221, A005361, A007425, A025847, A034444, A056671, A064549, A074816, A107758, A107759, A348018, A363194.
Numbers of the form a^2*b^3, where a >= 2 and b >= 2.
+10
5
32, 72, 108, 128, 200, 243, 256, 288, 392, 432, 500, 512, 576, 648, 675, 800, 864, 968, 972, 1024, 1125, 1152, 1323, 1352, 1372, 1568, 1600, 1728, 1800, 1944, 2000, 2048, 2187, 2304, 2312, 2592, 2700, 2888, 2916, 3087, 3125, 3136, 3200, 3267, 3456, 3528, 3872, 3888, 4000, 4096, 4232, 4500, 4563, 4608
FORMULA
Sum_{n>=1} 1/a(n) = 1 + ((zeta(2)-1)*(zeta(3)-1)-1)/zeta(6) - P(6) = 0.12806919584708298724..., where P(s) is the prime zeta function. - Amiram Eldar, Feb 07 2023
MATHEMATICA
With[{max = 5000}, Union[Table[i^2*j^3, {j, 2, max^(1/3)}, {i, 2, Sqrt[max/j^3]}] // Flatten]] (* Amiram Eldar, Feb 07 2023 *)
PROG
(PARI) list(lim)=my(v=List()); for(b=2, sqrtnint(lim\4, 3), for(a=2, sqrtint(lim\b^3), listput(v, a^2*b^3))); Set(v) \\ Charles R Greathouse IV, Jan 03 2014 (Python)
from math import isqrt
from sympy import mobius, integer_nthroot, primepi
def squarefreepi(n): return int(sum(mobius(k)*(n//k**2) for k in range(1, isqrt(n)+1)))
def bisection(f, kmin=0, kmax=1):
while f(kmax) > kmax: kmax <<= 1
while kmax-kmin > 1:
kmid = kmax+kmin>>1
if f(kmid) <= kmid:
kmax = kmid
else:
kmin = kmid
return kmax
def f(x):
j, b = isqrt(x), integer_nthroot(x, 6)[0]
l, c = 0, n+x-1+primepi(b)+sum(mobius(k)*(j//k**3) for k in range(1, b+1))
while j>1:
k2 = integer_nthroot(x//j**2, 3)[0]+1
w = squarefreepi(k2-1)
c -= j*(w-l)
l, j = w, isqrt(x//k2**3)
return c+l
Numbers of ordered pairs of divisors d < e of n such that gcd(d, e) > 1.
+10
5
0, 0, 0, 1, 0, 2, 0, 3, 1, 2, 0, 8, 0, 2, 2, 6, 0, 8, 0, 8, 2, 2, 0, 18, 1, 2, 3, 8, 0, 15, 0, 10, 2, 2, 2, 24, 0, 2, 2, 18, 0, 15, 0, 8, 8, 2, 0, 32, 1, 8, 2, 8, 0, 18, 2, 18, 2, 2, 0, 44, 0, 2, 8, 15, 2, 15, 0, 8, 2, 15, 0, 49, 0, 2, 8, 8, 2, 15, 0, 32, 6, 2
COMMENTS
Number of elements in the set {(x, y): x|n, y|n, x < y, gcd(x, y) > 1}.
Every element of the sequence is repeated indefinitely, for instance:
a(n)=0 if n prime;
a(n)=1 if n = p^2 for p prime ( A001248); a(n)=2 if n is a squarefree semiprime ( A006881); a(n)=3 if n = p^3 for p prime ( A030078); a(n)=6 if n = p^4 for p prime ( A030514); a(n)=8 if n is a number which is the product of a prime and the square of a different prime ( A054753); a(n)=10 if n = p^5 for p prime ( A050997); a(n)=15 if n is in the set { A007304} union {64} = {30, 42, 64, 66, 70,...} = {Sphenic numbers} union {64}; a(n)=18 if n is the product of the cube of a prime ( A030078) and a different prime (see A065036); a(n)=21 if n = p^7 for p prime ( A092759); a(n)=24 if n is square of a squarefree semiprime ( A085986); a(n)=32 if n is the product of the 4th power of a prime ( A030514) and a different prime (see A178739); a(n)=36 if n = p^9 for p prime ( A179665); a(n)=44 if n is the product of exactly four primes, three of which are distinct ( A085987); a(n)=45 if n is a number with 11 divisors ( A030629); a(n)=49 if n is of the form p^2*q^3, where p,q are distinct primes ( A143610); a(n)=50 if n is the product of the 5th power of a prime ( A050997) and a different prime (see A178740); a(n)=55 if n if n = p^11 for p prime( A079395); a(n)=72 if n is a number with 14 divisors ( A030632); a(n)=80 if n is the product of four distinct primes ( A046386); a(n)=83 if n is a number with 15 divisors ( A030633); a(n)=89 if n is a number with prime factorization pqr^3 ( A189975); a(n)=96 if n is a number that are the cube of a product of two distinct primes ( A162142); a(n)=98 if n is the product of the 7th power of a prime and a distinct prime (p^7*q) ( A179664); a(n)=116 if n is the product of exactly 2 distinct squares of primes and a different prime (p^2*q^2*r) ( A179643); a(n)=126 if n is the product of the 5th power of a prime and different distinct prime of the 2nd power (p^5*q^2) ( A179646); a(n)=128 if n is the product of the 8th power of a prime and a distinct prime (p^8*q) ( A179668); a(n)=150 if n is the product of the 4th power of a prime and 2 different distinct primes (p^4*q*r) ( A179644); a(n)=159 if n is the product of the 4th power of a prime and a distinct prime of power 3 (p^4*q^3) ( A179666). It is possible to continue with a(n) = 162, 178, 209, 224, 227, 238, 239, 260, 289, 309, 320, 333,...
FORMULA
a(n) = Sum_{d1|n, d2|n, d1<d2} (1-[gcd(d1,d2) = 1]), where [ ] is the Iverson bracket. - Wesley Ivan Hurt, Jan 01 2021
EXAMPLE
a(12) = 8 because the divisors of 12 are {1, 2, 3, 4, 6, 12} and GCD(d_i, d_j)>1 for the 8 following pairs of divisors: (2,4), (2,6), (2,12), (3,6), (3,12), (4,6), (4,12) and (6,12).
MAPLE
with(numtheory):nn:=100:
for n from 1 to nn do:
x:=divisors(n):n0:=nops(x):it:=0:
for i from 1 to n0 do:
for j from i+1 to n0 do:
if gcd(x[i], x[j])>1
then
it:=it+1:
else
fi:
od:
od:
printf(`%d, `, it):
od:
MATHEMATICA
Table[Sum[Sum[(1 - KroneckerDelta[GCD[i, k], 1]) (1 - Ceiling[n/k] + Floor[n/k]) (1 - Ceiling[n/i] + Floor[n/i]), {i, k - 1}], {k, n}], {n, 100}] (* Wesley Ivan Hurt, Jan 01 2021 *)
PROG
(PARI) a(n)=my(d=divisors(n)); sum(i=2, #d, sum(j=1, i-1, gcd(d[i], d[j])>1)) \\ Charles R Greathouse IV, Aug 03 2016 (PARI) a(n)=my(f=factor(n)[, 2], t=prod(i=1, #f, f[i]+1)); t*(t-1)/2 - (prod(i=1, #f, 2*f[i]+1)+1)/2 \\ Charles R Greathouse IV, Aug 03 2016
CROSSREFS
Cf. A001248, A006881, A007304, A030078, A030514, A030632, A046386, A050997, A054753, A063647, A065036, A066446, A079395, A085986, A085987, A092759, A143610, A162142, A178739, A178740, A179644, A179646, A179664, A189975.
Number of strict factorizations of n using elements of A007916 (numbers that are not perfect powers).
+10
5
1, 1, 1, 0, 1, 2, 1, 0, 0, 2, 1, 2, 1, 2, 2, 0, 1, 2, 1, 2, 2, 2, 1, 2, 0, 2, 0, 2, 1, 5, 1, 0, 2, 2, 2, 3, 1, 2, 2, 2, 1, 5, 1, 2, 2, 2, 1, 2, 0, 2, 2, 2, 1, 2, 2, 2, 2, 2, 1, 7, 1, 2, 2, 0, 2, 5, 1, 2, 2, 5, 1, 4, 1, 2, 2, 2, 2, 5, 1, 2, 0, 2, 1, 7, 2, 2, 2
EXAMPLE
The a(72) = 4 factorizations are (2*3*12), (3*24), (6*12), (72). Missing from this list and not strict are (2*2*2*3*3), (2*2*3*6), (2*6*6), (2*2*18), while missing from the list and using perfect powers are (2*36), (2*4*9), (3*4*6), (4*18), (8*9).
MATHEMATICA
radQ[n_]:=Or[n==1, GCD@@FactorInteger[n][[All, 2]]==1];
facssr[n_]:=If[n<=1, {{}}, Join@@Table[Map[Prepend[#, d]&, Select[facssr[n/d], Min@@#>d&]], {d, Select[Rest[Divisors[n]], radQ]}]];
Table[Length[facssr[n]], {n, 100}]
CROSSREFS
Cf. A001055, A001597, A007916, A025147, A045778, A052410, A303707, A323054, A323087, A323088, A323089.
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