A284003 a(n) = A007913(A283477(n)) = A019565(A006068(n)).

1, 2, 6, 3, 30, 15, 5, 10, 210, 105, 35, 70, 7, 14, 42, 21, 2310, 1155, 385, 770, 77, 154, 462, 231, 11, 22, 66, 33, 330, 165, 55, 110, 30030, 15015, 5005, 10010, 1001, 2002, 6006, 3003, 143, 286, 858, 429, 4290, 2145, 715, 1430, 13, 26, 78, 39, 390, 195, 65, 130, 2730, 1365, 455, 910, 91, 182, 546, 273, 510510, 255255, 85085, 170170, 17017

Offset: 0

Author

Antti Karttunen, Mar 18 2017

Keywords

• nonn
• look

Comments

A squarefree analog of A302783. Each term is either a divisor or a multiple of the next one. In contrast to A302033 at each step the previous term can be multiplied (or divided), not just by a single prime, but possibly by a product of several distinct ones, A019565(A000975(k)). E.g., a(3) = 3, a(4) = 2*5*a(3) = 30. - Antti Karttunen, Apr 17 2018

Links

• Antti Karttunen, Table of n, a(n) for n = 0..8191
• Index entries for sequences related to binary expansion of n

Crossrefs

A permutation of A005117.

Cf. A000975, A001222, A006068, A007913, A019565, A046523, A048675, A064707, A209281, A283477, A284004.

Cf. also A277811, A283475, A302033, A302783.

Programs

Mathematica

Table[Apply[Times, FactorInteger[#] /. {p_, e_} /; e > 0 :> Times @@ (p^Mod[e, 2])] &[Times @@ Map[#1^#2 & @@ # &, FactorInteger[#] /. {p_, e_} /; e == 1 :> {Times @@ Prime@ Range@ PrimePi@ p, e}] &[Times @@ Prime@ Flatten@ Position[#, 1] &@ Reverse@ IntegerDigits[n, 2]]], {n, 0, 52}] (* Michael De Vlieger, Mar 18 2017 *)

PARI

A007913(n) = core(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 Charles R Greathouse IV, Jun 28 2015
A019565(n) = {my(j,v); factorback(Mat(vector(if(n, #n=vecextract(binary(n), "-1..1")), j, [prime(j), n[j]])~))}; \\ This function from M. F. Hasler
A283477(n) = A108951(A019565(n));
A284003(n) = A007913(A283477(n));

PARI

A006068(n)= { my(s=1, ns); while(1, ns = n >> s; if(0==ns, break()); n = bitxor(n, ns); s <<= 1; ); return (n); } \\ From A006068
A284003(n) = A019565(A006068(n)); \\ (and use A019565 from above) - Antti Karttunen, Apr 16 2018

Scheme

(define (A284003 n) (A007913 (A283477 n)))

Formula

a(n) = A007913(A283477(n)).

Other identities. For all n >= 0:

A048675(a(n)) = A006068(n).

A046523(a(n)) = A284004(n).

It seems that A001222(a(n)) = A209281(n).

a(n) = A019565(A006068(n)) = A302033(A064707(n)). - Antti Karttunen, Apr 16 2018

Extensions

Name amended with a second formula by Antti Karttunen, Apr 16 2018