A077462 -id:A077462 - 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

A305800 Filter sequence for a(prime) = constant sequences.

Original entry on oeis.org

1, 2, 2, 3, 2, 4, 2, 5, 6, 7, 2, 8, 2, 9, 10, 11, 2, 12, 2, 13, 14, 15, 2, 16, 17, 18, 19, 20, 2, 21, 2, 22, 23, 24, 25, 26, 2, 27, 28, 29, 2, 30, 2, 31, 32, 33, 2, 34, 35, 36, 37, 38, 2, 39, 40, 41, 42, 43, 2, 44, 2, 45, 46, 47, 48, 49, 2, 50, 51, 52, 2, 53, 2, 54, 55, 56, 57, 58, 2, 59, 60, 61, 2, 62, 63, 64, 65, 66, 2, 67, 68, 69, 70, 71, 72, 73, 2, 74, 75, 76, 2, 77, 2, 78, 79, 80, 2, 81, 2, 82, 83, 84, 2, 85, 86, 87, 88, 89, 90, 91, 92, 93, 94, 95, 96

Offset: 1

Antti Karttunen, Jun 14 2018

nonn

Restricted growth sequence transform of A239968.
In the following, A stands for this sequence, A305800, and S -> T (where S and T are sequence A-numbers) indicates that for all i, j: S(i) = S(i) => T(i) = T(j).
For example, the following implications hold:
  A -> A300247 -> A305897 -> A077462 -> A101296,
  A -> A290110 -> A300250 -> A101296.

Antti Karttunen, Table of n, a(n) for n = 1..100000

Cf. A000720, A239968.
Differs from A296073 for the first time at n=125, as a(125) = 96, while A296073(125) = 33.
Cf. also A305900, A305801, A295300, A289626 for other "upper level" filters.

Join[{1},Table[If[PrimeQ[n],2,1+n-PrimePi[n]],{n,2,150}]] (* Harvey P. Dale, Jul 12 2019 *)

A305800(n) = if(1==n,n,if(isprime(n),2,1+n-primepi(n)));

a(1) = 1; for n > 1, a(n) = 2 for prime n, and a(n) = 1+n-A000720(n) for composite n.

A305897 Lexicographically earliest infinite sequence such that a(i) = a(j) => A348717(i) = A348717(j), for all i, j >= 1.

Original entry on oeis.org

1, 2, 2, 3, 2, 4, 2, 5, 3, 6, 2, 7, 2, 8, 4, 9, 2, 10, 2, 11, 6, 12, 2, 13, 3, 14, 5, 15, 2, 16, 2, 17, 8, 18, 4, 19, 2, 20, 12, 21, 2, 22, 2, 23, 7, 24, 2, 25, 3, 26, 14, 27, 2, 28, 6, 29, 18, 30, 2, 31, 2, 32, 11, 33, 8, 34, 2, 35, 20, 36, 2, 37, 2, 38, 10, 39, 4, 40, 2, 41, 9, 42, 2, 43, 12, 44, 24, 45, 2, 46, 6, 47, 30, 48, 14, 49, 2, 50, 15, 51, 2, 52, 2, 53, 16

Offset: 1

Antti Karttunen, Jun 14 2018

nonn

Restricted growth sequence transform of A348717, or equally, of A246277.
Filter sequence for prime factorization patterns, including also information about gaps between prime factors. - Original name, gives the motivation for this sequence. Here the "gaps" refers to differences between the indices of primes present, not the prime gaps as usually understood.
For all i, j:
  A305800(i) = A305800(j) => a(i) = a(j),
  a(i) = a(j) => A077462(i) = A077462(j) => A101296(i) = A101296(j).
  a(i) = a(j) => A243055(i) = A243055(j).

			a(10) = a(21) (= 6) because both have prime exponents [1, 1] and the difference between the prime indices is the same, as 10 = prime(1)*prime(3), and 21 = prime(2)*prime(4).
a(12) != a(18) because the prime exponents [2,1] and [1,2] do not occur in the same order.
a(140) = a(693) (= 71) because both numbers have prime exponents [2, 1, 1] (in this order) and the differences between the indices of the successive prime factors are same: 140 = prime(1)^2 * prime(3) * prime(4), 693 = prime(2)^2 * prime(4) * prime(5).

Antti Karttunen, Table of n, a(n) for n = 1..100000
Index entries for sequences computed from indices in prime factorization

Cf. A077462, A101296, A246277, A300247, A305800, A348717.

Cf. also A286621.

up_to = 65537;
rgs_transform(invec) = { my(om = Map(), outvec = vector(length(invec)), u=1); for(i=1, length(invec), if(mapisdefined(om,invec[i]), my(pp = mapget(om, invec[i])); outvec[i] = outvec[pp] , mapput(om,invec[i],i); outvec[i] = u; u++ )); outvec; };
A246277(n) = if(1==n, 0, my(f = factor(n), k = primepi(f[1,1])-1); for (i=1, #f~, f[i,1] = prime(primepi(f[i,1])-k)); factorback(f)/2);
v305897 = rgs_transform(vector(up_to,n,A246277(n)));
A305897(n) = v305897[n];

Name changed by Antti Karttunen, Apr 30 2022

A329600 Smallest number with the same set of distinct prime exponents as A108951(n).

Original entry on oeis.org

1, 2, 2, 4, 2, 12, 2, 8, 4, 12, 2, 24, 2, 12, 12, 16, 2, 72, 2, 24, 12, 12, 2, 48, 4, 12, 8, 24, 2, 360, 2, 32, 12, 12, 12, 144, 2, 12, 12, 48, 2, 360, 2, 24, 24, 12, 2, 96, 4, 72, 12, 24, 2, 432, 12, 48, 12, 12, 2, 720, 2, 12, 24, 64, 12, 360, 2, 24, 12, 360, 2, 288, 2, 12, 72, 24, 12, 360, 2, 96, 16, 12, 2, 720, 12, 12, 12, 48, 2, 2160, 12, 24, 12, 12, 12

Offset: 1

Antti Karttunen, Nov 17 2019

nonn

Antti Karttunen, Table of n, a(n) for n = 1..20000
Index entries for sequences computed from indices in prime factorization

Cf. A007947, A034386, A071364, A108951, A122111, A181819, A181821, A328400, A329607.

Cf. A077462 (rgs-transform, from its term a(1)=1 onward).

Array[Times @@ MapIndexed[Prime[#2[[1]]]^#1 &, Reverse[Flatten[Cases[FactorInteger[#], {p_, k_} :> Table[PrimePi[p], {k}]]]]] &[Times @@ FactorInteger[#][[All, 1]]] &@ If[# == 1, 1, Times @@ Prime@ FactorInteger[#][[All, -1]]] &[Times @@ Map[#1^#2 & @@ # &, FactorInteger[#] /. {p_, e_} /; e > 0 :> {Times @@ Prime@ Range@ PrimePi@ p, e}]] &, 105] (* Michael De Vlieger, Nov 18 2019, after Gus Wiseman at A181821 *)

A007947(n) = factorback(factorint(n)[, 1]);
A181819(n) = factorback(apply(e->prime(e),(factor(n)[,2])));
A181821(n) = { my(f=factor(n),p=0,m=1); forstep(i=#f~,1,-1,while(f[i,2], f[i,2]--; m *= (p=nextprime(p+1))^primepi(f[i,1]))); (m); };
A328400(n) = A181821(A007947(A181819(n)));
A034386(n) = prod(i=1, primepi(n), prime(i));
A108951(n) = { my(f=factor(n)); prod(i=1, #f~, A034386(f[i, 1])^f[i, 2]) };  \\ From A108951
A329600(n) = A328400(A108951(n));

a(n) = A328400(A108951(n)) = A181821(A007947(A122111(n)))

a(n) = A181821(A329607(n)) = A181821(A122111(A071364(n))).

A319717 Filter sequence combining the largest proper divisor of n (A032742) with modulo 6 residue of the smallest prime factor, A010875(A020639(n)), and a single bit A319710(n) telling whether the smallest prime factor is unitary.

Original entry on oeis.org

1, 2, 3, 4, 5, 6, 7, 8, 9, 10, 5, 11, 7, 12, 13, 14, 5, 15, 7, 16, 17, 18, 5, 19, 20, 21, 22, 23, 5, 24, 7, 25, 26, 27, 28, 29, 7, 30, 31, 32, 5, 33, 7, 34, 35, 36, 5, 37, 38, 39, 40, 41, 5, 42, 43, 44, 45, 46, 5, 47, 7, 48, 49, 50, 51, 52, 7, 53, 54, 55, 5, 56, 7, 57, 58, 59, 60, 61, 7, 62, 63, 64, 5, 65, 66, 67, 68, 69, 5, 70, 71, 72, 73, 74, 75, 76, 7, 77, 78, 79, 5, 80, 7, 81, 82, 83, 5, 84, 7, 85, 86, 87, 5, 88, 89, 90, 91, 92, 93, 94, 95

Offset: 1

Antti Karttunen, Oct 04 2018

nonn

Restricted growth sequence transform of triple [A010875(A020639(n)), A032742(n), A319710(n)] (with a separate value allotted for a(1)), or equally, of ordered pair [A319716(n), A319710(n)].
In addition to A319716, this filter sequence also records in the value of a(n) also the fact whether the smallest prime factor of n is unitary or not. This information is enough to determine the modulo 6 residues of all the divisors of n, thus sequences like A002324 are essentially functions of this sequence. Moreover, a lot of other information is immediately (and unavoidably) present, for example the exact prime signature of n, including also the relative order of exponents.
Any such filtering sequence can be perceived also in terms of what information it leaves out from a(n) that would be needed to reconstruct whole n from each a(n). If the whole n could be reconstructed from a(n) each time, then sequence a would be injective, and would be useless for filtering, because then it would match with any sequence. In this filter, what is left out is only the exact identity of the smallest prime factor, although its residue class mod 6 is retained. However, when the smallest prime factor is 2 or 3, this can be seen from that residue value, so for any number x in A047229, both A020639(x) and A032742(x) are known, and as x = A020639(x)*A032742(x), it means such numbers must occur in their own singleton equivalence classes.
Likewise, for any n in A283050, even if not divisible by 2 or 3, when we have A319710(n) stored in the triple as 1, this immediately gives away the exact identity of the smallest prime factor, which is equal to A014673(n) = A020639(A032742(n)) in these cases.
Thus there is a substantial subset of N (containing at least the union of A047229 and A283050) which is actually in the "blind sector" of this filter, "where anything goes", as this sequence obtains only unique values in that subdomain.
There is a related filter sequence A319996, which operates by "cleaving n from its high end" (by storing the residue class of the largest prime factor, A006530, instead of the smallest, together with n/A006530(n)), which has its own blind spots, but fortunately, they do not fully coincide with the blind spots of this filter. Naturally, any sequence like A002324 should match both to this sequence and A319996.
For all i, j:
  a(i) = a(j) => A002324(i) = A002324(j),
  a(i) = a(j) => A067029(i) = A067029(j),
  a(i) = a(j) => A071178(i) = A071178(j),
  a(i) = a(j) => A077462(i) = A077462(j) => A101296(i) = A101296(j),
  a(i) = a(j) => A319716(i) = A319716(j) => A319690(i) = A319690(j).

			For n = 65 = 5*13 and 143 = 11*13, the smallest prime factor is of the form 6k+5,  doesn't occur more than once in the factorization, and the largest proper divisor is the same number (13) in both cases, thus a(65) = a(143) (= 51, a running count value allotted by rgs-transform for this equivalence class).
For n = 1805 (5*19^2), 3971 (11*19^2), 6137 (17*19^2), it's like above, but the largest proper divisor is in all three cases 361 = 19^2, thus a(1805) = a(3971) = a(6137) (= 1405).
Note that such nontrivial equivalence classes may only contain numbers that are 5-rough, A007310, with no prime factors 2 or 3, and also, they may not contain numbers from A283050. See the comments section.

Antti Karttunen, Table of n, a(n) for n = 1..100000

Cf. A101296, A319707, A319716, A319996.
Cf. also A320004 (analogous sequence for modulo 4 residues).
Differs from A319707 for the first time at n=143, where a(143) = 51, differs from A319716 for the first time at n=121, where a(121) = 95.

up_to = 100000;
rgs_transform(invec) = { my(om = Map(), outvec = vector(length(invec)), u=1); for(i=1, length(invec), if(mapisdefined(om,invec[i]), my(pp = mapget(om, invec[i])); outvec[i] = outvec[pp] , mapput(om,invec[i],i); outvec[i] = u; u++ )); outvec; };
A032742(n) = if(1==n,n,n/vecmin(factor(n)[,1]));
A286476(n) = if(1==n,n,(6*A032742(n) + (n % 6)));
A319710(n) = ((n>1)&&(factor(n)[1,2]>1));
v319717 = rgs_transform(vector(up_to,n,[A286476(n),A319710(n)]));
A319717(n) = v319717[n];

A319994 Let g = A006530(n), the largest prime factor of n. This filter sequence combines (g mod 4), n/g (A052126), and a single bit A319988(n) telling whether the largest prime factor is unitary.

Original entry on oeis.org

1, 2, 3, 4, 5, 6, 3, 7, 8, 9, 3, 10, 5, 6, 11, 12, 5, 13, 3, 14, 15, 6, 3, 16, 17, 9, 18, 10, 5, 19, 3, 20, 15, 9, 21, 22, 5, 6, 11, 23, 5, 24, 3, 10, 25, 6, 3, 26, 27, 28, 11, 14, 5, 29, 21, 16, 15, 9, 3, 30, 5, 6, 31, 32, 33, 24, 3, 14, 15, 34, 3, 35, 5, 9, 36, 10, 37, 19, 3, 38, 39, 9, 3, 40, 33, 6, 11, 16, 5, 41, 42, 10, 15, 6, 21, 43, 5, 44, 31, 45, 5, 19, 3, 23, 46

Offset: 1

Antti Karttunen, Oct 05 2018

nonn

Restricted growth sequence transform of triple [A010873(A006530(n)), A052126(n), A319988(n)], with a separate value allotted for a(1).
Here among the first 100000 terms, only 2331 have a unique value, compared to 69714 in A320004.
For all i, j:
  a(i) = a(j) => A024362(i) = A024362(j),
  a(i) = a(j) => A067029(i) = A067029(j),
  a(i) = a(j) => A071178(i) = A071178(j),
  a(i) = a(j) => A077462(i) = A077462(j) => A101296(i) = A101296(j).

Antti Karttunen, Table of n, a(n) for n = 1..100000

Cf. A006530, A052126, A319988, A320004.

Cf. also A319996 (modulo 6 analog).

up_to = 100000;
rgs_transform(invec) = { my(om = Map(), outvec = vector(length(invec)), u=1); for(i=1, length(invec), if(mapisdefined(om,invec[i]), my(pp = mapget(om, invec[i])); outvec[i] = outvec[pp] , mapput(om,invec[i],i); outvec[i] = u; u++ )); outvec; };
A006530(n) = if(n>1, vecmax(factor(n)[, 1]), 1);
A052126(n) = (n/A006530(n));
A319988(n) = ((n>1)&&(factor(n)[omega(n),2]>1));
A319994aux(n) = if(1==n,0,[A006530(n)%4, A052126(n), A319988(n)]);
v319994 = rgs_transform(vector(up_to,n,A319994aux(n)));
A319994(n) = v319994[n];

A319996 Let g = A006530(n), the largest prime factor of n. This filter sequence combines (g mod 6), n/g (A052126), and a single bit A319988(n) telling whether the largest prime factor is unitary.

Original entry on oeis.org

1, 2, 3, 4, 5, 6, 7, 8, 9, 10, 5, 11, 7, 12, 13, 14, 5, 15, 7, 16, 17, 10, 5, 18, 19, 12, 20, 21, 5, 22, 7, 23, 13, 10, 24, 25, 7, 12, 17, 26, 5, 27, 7, 16, 28, 10, 5, 29, 30, 31, 13, 21, 5, 32, 33, 34, 17, 10, 5, 35, 7, 12, 36, 37, 24, 22, 7, 16, 13, 38, 5, 39, 7, 12, 40, 21, 41, 27, 7, 42, 43, 10, 5, 44, 33, 12, 13, 26, 5, 45, 46, 16, 17, 10, 24, 47, 7, 48, 28, 49, 5, 22, 7, 34

Offset: 1

Antti Karttunen, Oct 05 2018

nonn

Restricted growth sequence transform of triple [A010875(A006530(n)), A052126(n), A319988(n)], with a separate value allotted for a(1).
Many of the same comments as given in A319717 apply also here, except for this filter, the "blind spot" area (where only unique values are possible for a(n)) is different, and contains at least all numbers in A070003. Because presence of 2 or 3 in the prime factorization of n do not force the value of a(n) unique, this is substantially less lax (i.e., more exact) filter than A319717. Here among the first 100000 terms, only 2393 have a unique value, compared to 74355 in A319717.
For all i, j:
  a(i) = a(j) => A002324(i) = A002324(j),
  a(i) = a(j) => A067029(i) = A067029(j),
  a(i) = a(j) => A071178(i) = A071178(j),
  a(i) = a(j) => A077462(i) = A077462(j) => A101296(i) = A101296(j),
  a(i) = a(j) => A319690(i) = A319690(j).

			For n = 15 (3*5) and n = 33 (3*11), the mod 6 residue of the largest prime factor is 5, also in both cases it is unitary (A319988(n) = 1), and the quotient n/A006530(n) is equal, in this case 3. Thus a(15) and a(33) are alloted the same running count (13 in this case) by rgs-transform.
For n = 2275 (5^2 * 7 * 13), n = 3325 (5^2 * 7 * 19), 5425 (5^2 * 7 * 31) and 6475 (5^2 * 7 * 37), the largest prime factor = 1 (mod 6), and A052126(n) = 175, thus these numbers are allotted the same running count (394 in this case) by rgs-transform.

Antti Karttunen, Table of n, a(n) for n = 1..100000

Cf. A006530, A052126, A319988, A319717.
Cf. A007528 (positions of 5's), A002476 (of 7's), A112774 (after its initial term gives the position of 10's in this sequence).
Cf. also A319994 (modulo 4 analog).

up_to = 100000;
rgs_transform(invec) = { my(om = Map(), outvec = vector(length(invec)), u=1); for(i=1, length(invec), if(mapisdefined(om,invec[i]), my(pp = mapget(om, invec[i])); outvec[i] = outvec[pp] , mapput(om,invec[i],i); outvec[i] = u; u++ )); outvec; };
A006530(n) = if(n>1, vecmax(factor(n)[, 1]), 1);
A052126(n) = (n/A006530(n));
A319988(n) = ((n>1)&&(factor(n)[omega(n),2]>1));
A319996aux(n) = if(1==n,0,[A006530(n)%6, A052126(n), A319988(n)]);
v319996 = rgs_transform(vector(up_to,n,A319996aux(n)));
A319996(n) = v319996[n];

A320004 Filter sequence combining the largest proper divisor of n (A032742) with n's residue modulo 4 (A010873), and a single bit (A319710) telling whether the smallest prime factor is unitary.

Original entry on oeis.org

1, 2, 3, 4, 5, 6, 3, 7, 8, 9, 3, 10, 5, 11, 12, 13, 5, 14, 3, 15, 16, 17, 3, 18, 19, 20, 21, 22, 5, 23, 3, 24, 25, 26, 27, 28, 5, 29, 30, 31, 5, 32, 3, 33, 34, 35, 3, 36, 37, 38, 39, 40, 5, 41, 42, 43, 44, 45, 3, 46, 5, 47, 48, 49, 50, 51, 3, 52, 53, 54, 3, 55, 5, 56, 57, 58, 25, 59, 3, 60, 61, 62, 3, 63, 64, 65, 66, 67, 5, 68, 30, 69, 70, 71, 72, 73, 5, 74, 75, 76, 5, 77, 3

Offset: 1

Antti Karttunen, Oct 04 2018

nonn

Restricted growth sequence transform of triple [A010873(A020639(n)), A032742(n), A319710(n)], or equally, of ordered pair [A319714(n), A319710(n)].
Here any nontrivial equivalence classes (that is, when we exclude the singleton classes and two infinite classes of A002144 and A002145), like the example shown, may not contain any even numbers, nor any numbers from A283050. See additional comments in A319717 and A319994.
For all i, j:
  a(i) = a(j) => A024362(i) = A024362(j),
  a(i) = a(j) => A067029(i) = A067029(j),
  a(i) = a(j) => A071178(i) = A071178(j),
  a(i) = a(j) => A077462(i) = A077462(j) => A101296(i) = A101296(j).

			For n = 33 (3*11) and n = 77 (7*11), the modulo 4 residue of the smallest prime factor is 3, and the largest proper divisors (A032742) is also equal 11, and the smallest prime factor is unitary. Thus a(33) = a(77) (= 25, a running count value allotted by rgs-transform).

Antti Karttunen, Table of n, a(n) for n = 1..100000

Cf. A319704, A319714, A319994.
Cf. also A319717 (analogous sequence for modulo 6 residues).
Cf. A002145 (positions of 3's), A002144 (positions of 5's).
Differs from A319704 for the first time at n=77, and from A319714 for the first time at n=49.

up_to = 100000;
rgs_transform(invec) = { my(om = Map(), outvec = vector(length(invec)), u=1); for(i=1, length(invec), if(mapisdefined(om,invec[i]), my(pp = mapget(om, invec[i])); outvec[i] = outvec[pp] , mapput(om,invec[i],i); outvec[i] = u; u++ )); outvec; };
A032742(n) = if(1==n,n,n/vecmin(factor(n)[,1]));
A286474(n) = if(1==n,n,(4*A032742(n) + (n % 4)));
A319710(n) = ((n>1)&&(factor(n)[1,2]>1));
v320004 = rgs_transform(vector(up_to,n,[A286474(n),A319710(n)]));
A320004(n) = v320004[n];

0 = 1, 1 = p, 2 = p^2, 3 = p.q, 4 = p^3, 5 = p^2.q, 6 = p^4, 7 = p.q^2, 8 = p^3.q, 9 = p.q.r, 10 = p^5, 11 = p^2.q^2, 12 = p^4.q (p

			Primes are given 1. The next prime factorization pattern is 4=p^2, so a(4)=2 and similarly a(6)=3.

Antti Karttunen, Table of n, a(n) for n = 1..20000

Cf. A037916, A077462.

primetemplate(n)=local(f,fl,res,eres); f=factor(n); fl=length(f[,1]); res=""; for (i=1,fl,res=concat(res,f[,2][i])); eres=eval(res); if (v[eres]==0,v[eres]=vc; vc++); eres vc=1; v=vector(10000); for (j=2,50,print1(v[primetemplate(j)]","))

A079616(n) = (A077462(n)-1); \\ Antti Karttunen, Jun 13 2018, uses code in A077462.

a(n) = A077462(n) - 1. - David Wasserman, Dec 21 2004

More terms from David Wasserman, Dec 21 2004

Term a(1) = 0 prepended, superfluous 1 after that removed, extended up to a(105) - Antti Karttunen, Jun 13 2018

cp's OEIS Frontend

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

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A305800 Filter sequence for a(prime) = constant sequences.

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A305897 Lexicographically earliest infinite sequence such that a(i) = a(j) => A348717(i) = A348717(j), for all i, j >= 1.

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A329600 Smallest number with the same set of distinct prime exponents as A108951(n).

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A319717 Filter sequence combining the largest proper divisor of n (A032742) with modulo 6 residue of the smallest prime factor, A010875(A020639(n)), and a single bit A319710(n) telling whether the smallest prime factor is unitary.

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A319994 Let g = A006530(n), the largest prime factor of n. This filter sequence combines (g mod 4), n/g (A052126), and a single bit A319988(n) telling whether the largest prime factor is unitary.

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A319996 Let g = A006530(n), the largest prime factor of n. This filter sequence combines (g mod 6), n/g (A052126), and a single bit A319988(n) telling whether the largest prime factor is unitary.

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A320004 Filter sequence combining the largest proper divisor of n (A032742) with n's residue modulo 4 (A010873), and a single bit (A319710) telling whether the smallest prime factor is unitary.

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