A319693 Filter sequence combining sopfr(d) from all proper divisors d of n, where sopfr(d) is A001414(d) = sum of primes dividing d with repetition.

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, 73, 2, 81, 82, 83, 2, 84, 85

Offset: 1

Author

Antti Karttunen, Oct 02 2018

Keywords

• nonn

Comments

Restricted growth sequence transform of A319692.

For all i, j: a(i) = a(j) => A305611(i) = A305611(j).

Examples

			The proper divisors of  96 are 1, 2, 3, 4, 6, 8, 12, 16, 24, 32, 48, while
the proper divisors of 108 are 1, 2, 3, 4, 6, 9, 12, 18, 27, 36, 54.
It happens that sopfr(8) = sopfr(9), sopfr(16) = sopfr(18), sopfr(24) = sopfr(27), sopfr(32) = sopfr(36) and sopfr(48) = sopfr(54), and the rest of proper divisors (1, 2, 3, 4, 6, 12) are shared by both numbers, from which follows that by taking product of sopfr over proper divisors gives an identical result for both, thus a(96) = a(108). Here sopfr = A001414.
		

Links

• Antti Karttunen, Table of n, a(n) for n = 1..65537

Crossrefs

Cf. A001414, A319692.

Cf. also A319353.

Differs from A305800, A296073 and A317943 for the first time at n=108, as here a(108) = 73.

Programs

PARI

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; };
A001414(n) = ((n=factor(n))[, 1]~*n[, 2]); \\ From A001414.
A319692(n) = { my(m=1); fordiv(n, d, if(dA001414(d)))); (m); };
v319693 = rgs_transform(vector(up_to,n,A319692(n)));
A319693(n) = v319693[n];