cp's OEIS Frontend

This is a front-end for the Online Encyclopedia of Integer Sequences, made by Christian Perfect. The idea is to provide OEIS entries in non-ancient HTML, and then to think about how they're presented visually. The source code is on GitHub.

Showing 1-5 of 5 results.

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

Original entry on oeis.org

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

Views

Author

David Wasserman, Dec 21 2004

Keywords

Comments

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

Examples

			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
		

Crossrefs

Cf. A025487, A046523, A064839 (ordinal transform of this sequence), A181819, and arrays A095904, A179216.
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.

Programs

  • Maple
    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
  • 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 *)
  • 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");
    \\ Antti Karttunen, May 12 2017

Formula

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)

Extensions

Data section extended to 120 terms by Antti Karttunen, May 12 2017
Minor edits/corrections by M. F. Hasler, Jul 18 2019

A179696 Numbers with prime signature {7,1,1}, i.e., of form p^7*q*r with p, q and r distinct primes.

Original entry on oeis.org

1920, 2688, 4224, 4480, 4992, 6528, 7040, 7296, 8320, 8832, 9856, 10880, 11136, 11648, 11904, 12160, 14208, 14720, 15232, 15744, 16512, 17024, 18048, 18304, 18560, 19840, 20352, 20608, 21870, 22656, 23424, 23680, 23936, 25728, 25984, 26240, 26752, 27264
Offset: 1

Views

Author

Keywords

Crossrefs

Programs

  • Maple
    a:= proc(n) option remember; local k;
          for k from 1+ `if` (n=1, 1, a(n-1))
            while sort (map (x-> x[2], ifactors(k)[2]), `>`)<>[7, 1, 1]
          do od; k
        end:
    seq (a(n), n=1..40); # Alois P. Heinz, Jan 23 2011
  • Mathematica
    f[n_]:=Sort[Last/@FactorInteger[n]]=={1,1,7}; Select[Range[30000], f]
  • PARI
    list(lim)=my(v=List(),t1,t2);forprime(p=2, (lim\6)^(1/7), t1=p^7;forprime(q=2, lim\t1, if(p==q, next);t2=t1*q;forprime(r=q+1, lim\t2, if(p==r,next);listput(v,t2*r)))); vecsort(Vec(v)) \\ Charles R Greathouse IV, Jul 20 2011
    
  • Python
    from math import isqrt
    from sympy import primerange, primepi, integer_nthroot
    def A179696(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**7)))+(t*(t-1)>>1)-sum(primepi(y//k) for k in primerange(1, s+1)) for r in primerange(integer_nthroot(x,7)[0]+1))+sum(primepi(x//p**8) for p in primerange(integer_nthroot(x,8)[0]+1))-primepi(integer_nthroot(x,9)[0])
        return bisection(f,n,n) # Chai Wah Wu, Mar 27 2025

Extensions

Title edited by Daniel Forgues, Jan 22 2011

A137491 Numbers with 28 divisors.

Original entry on oeis.org

960, 1344, 1728, 2112, 2240, 2496, 3264, 3520, 3648, 4160, 4416, 4928, 5440, 5568, 5824, 5832, 5952, 6080, 7104, 7290, 7360, 7616, 7872, 8000, 8256, 8512, 9024, 9152, 9280, 9920, 10176, 10206, 10304, 11328, 11712, 11840, 11968, 12864, 12992, 13120
Offset: 1

Views

Author

R. J. Mathar, Apr 22 2008

Keywords

Comments

Maple implementation: see A030513.
Numbers of the form p^27 (subset of A122968), p*q^13, p*q*r^6 (A179672) or p^3*q^6 (A179694), where p, q and r are distinct primes. - R. J. Mathar, Mar 01 2010

Crossrefs

Programs

Formula

A000005(a(n)) = 28.

A179702 Numbers of the form p^4*q^5 where p and q are two distinct primes.

Original entry on oeis.org

2592, 3888, 20000, 50000, 76832, 151875, 253125, 268912, 468512, 583443, 913952, 1361367, 2576816, 2672672, 3557763, 4170272, 5940688, 6940323, 7503125, 8954912, 10504375, 13045131, 20295603, 22632992, 22717712, 29552672, 30074733
Offset: 1

Views

Author

Keywords

Comments

Subsequence of A046312 and of A137493. - R. J. Mathar, Jul 27 2010

Crossrefs

Programs

  • Mathematica
    fQ[n_] := Sort[Last /@ FactorInteger @n] == {4, 5}; Select[ Range@ 31668000, fQ] (* fixed by Robert G. Wilson v, Aug 26 2010 *)
    lst = {}; Do[ If[p != q, AppendTo[lst, Prime@p^4*Prime@q^5]], {p, 12}, {q, 10}]; Take[ Sort@ Flatten@ lst, 27] (* Robert G. Wilson v, Aug 26 2010 *)
    Take[Union[First[#]^4 Last[#]^5&/@Flatten[Permutations/@Subsets[ Prime[ Range[30]],{2}],1]],30] (* Harvey P. Dale, Jan 01 2012 *)
  • PARI
    list(lim)=my(v=List(),t);forprime(p=2, (lim\16)^(1/5), t=p^5;forprime(q=2, (lim\t)^(1/4), if(p==q, next);listput(v,t*q^4))); vecsort(Vec(v)) \\ Charles R Greathouse IV, Jul 20 2011
    
  • Python
    from sympy import primepi, integer_nthroot, primerange
    def A179702(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(integer_nthroot(x//p**5,4)[0]) for p in primerange(integer_nthroot(x,5)[0]+1))+primepi(integer_nthroot(x,9)[0])
        return bisection(f,n,n) # Chai Wah Wu, Mar 27 2025

Formula

Sum_{n>=1} 1/a(n) = P(4)*P(5) - P(9) = A085964 * A085965 - A085969 = 0.000748..., where P is the prime zeta function. - Amiram Eldar, Jul 06 2020

Extensions

Edited and extended by Ray Chandler and R. J. Mathar, Jul 26 2010

A355462 Powerful numbers divisible by exactly 2 distinct primes.

Original entry on oeis.org

36, 72, 100, 108, 144, 196, 200, 216, 225, 288, 324, 392, 400, 432, 441, 484, 500, 576, 648, 675, 676, 784, 800, 864, 968, 972, 1000, 1089, 1125, 1152, 1156, 1225, 1296, 1323, 1352, 1372, 1444, 1521, 1568, 1600, 1728, 1936, 1944, 2000, 2025, 2116, 2304, 2312, 2500
Offset: 1

Views

Author

Amiram Eldar, Jul 03 2022

Keywords

Comments

First differs from A286708 at n = 25.
Number of the form p^i * q^j, where p != q are primes and i,j > 1.
Numbers k such that A001221(k) = 2 and A051904(k) >= 2.
The possible values of the number of the divisors (A000005) of terms in this sequence is any composite number that is not 8 or twice a prime (A264828 \ {1, 8}).
675 = 3^3*5^2 and 676 = 2^2*13^2 are 2 consecutive integers in this sequence. There are no other such pairs below 10^22 (the lesser members of such pairs are terms of A060355).

Examples

			36 is a term since 36 = 2^2 * 3^2.
		

Crossrefs

Programs

  • Mathematica
    Select[Range[2500], Length[(e = FactorInteger[#][[;; , 2]])] == 2 && Min[e] > 1 &]
  • PARI
    is(n) = {my(f=factor(n)); #f~ == 2 && vecmin(f[,2]) > 1};

Formula

Sum_{n>=1} 1/a(n) = ((Sum_{p prime} (1/(p*(p-1))))^2 - Sum_{p prime} (1/(p^2*(p-1)^2)))/2 = 0.1583860791... .
Showing 1-5 of 5 results.