A296073 Filter combining A296071(n) and A296072(n), related to the deficiencies of proper divisors of n.

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, 33

Offset: 1

Author

Antti Karttunen, Dec 04 2017

Keywords

• nonn

Comments

Construction: Pack the values of A296071(n) and A296072(n) to a single value with any injective N x N -> N packing function, like for example as f(n) = (1/2)*(2 + ((A296071(n)+A296072(n))^2) - A296071(n) - 3*A296072(n)) (the packing function here is the two-argument form of A000027). Then apply the restricted growth sequence transform to the sequence f(1), f(2), f(3), ... The transform assigns a unique increasing number for each newly encountered term of the sequence, and for any subsequent occurrences of the same term it gives the same number that term obtained for the first time.

For all i, j: a(i) = a(j) => A296074(i) = A296074(j).

Note that this is NOT restricted growth transform of A239968, which is A305800. Apart from 2's that occur at every prime, there are other duplicates also, first at a(125) = a(46) = 33.

Examples

			To see that a(46) and a(125) have the same value (33), consider the proper divisors of 46 = 1, 2, 23 and of 125 = 1, 5, 25. Their deficiencies are 1, 1, 22 and 1, 4, 19 respectively. When we look at their balanced ternary representations [as here all elements are positive, it can be obtained as A007089(A117967(n)) with 2's standing for -1's]:
   1 =    1
   1 =    1
  22 = 1211 (as 22 = 1*(3^3) + -1*(3^2) + 1*(3^1) + 1*(3^0))
and
   1 =    1
   4 =   11
  19 = 1201 (as 19 = 1*(3^3) + -1*(3^2) + 0*(3^1) + 1*(3^0)).
we see that in each column there is an equal number of 1's and an equal number of 2's. Moreover, this then implies also that the sums of those two sequences of deficiencies {1, 1, 22} and {1, 4, 19} are equal, as A296074(n) is a function of (can be computed from) a(n).
		

Links

• Antti Karttunen, Table of n, a(n) for n = 1..65536

Crossrefs

Cf. A019565, A033879, A117967, A117968, A295882, A296071, A296072, A296074.

Cf. also A293226.

Differs from A305800 for the first time at n=125.

Programs

PARI

up_to = 65536;
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; };
write_to_bfile(start_offset,vec,bfilename) = { for(n=1, length(vec), write(bfilename, (n+start_offset)-1, " ", vec[n])); }
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
A117967(n) = if(n<=1,n,if(!(n%3),3*A117967(n/3),if(1==(n%3),1+3*A117967((n-1)/3),2+3*A117967((n+1)/3))));
A117968(n) = if(1==n,2,if(!(n%3),3*A117968(n/3),if(1==(n%3),2+3*A117968((n-1)/3),1+3*A117968((n+1)/3))));
A289813(n) = { my (d=digits(n, 3)); from digits(vector(#d, i, if (d[i]==1, 1, 0)), 2); } \\ From Rémy Sigrist
A289814(n) = { my (d=digits(n, 3)); from digits(vector(#d, i, if (d[i]==2, 1, 0)), 2); } \\ From Rémy Sigrist
A295882(n) = { my(x = (2*n)-sigma(n)); if(x >= 0,A117967(x),A117968(-x)); };
A296071(n) = { my(m=1); fordiv(n,d,if(d < n,m *= A019565(A289813(A295882(d))))); m; };
A296072(n) = { my(m=1); fordiv(n,d,if(d < n,m *= A019565(A289814(A295882(d))))); m; };
Anotsubmitted3(n) = (1/2)*(2 + ((A296071(n)+A296072(n))^2) - A296071(n) - 3*A296072(n));
write_to_bfile(1,rgs_transform(vector(up_to,n,Anotsubmitted3(n))),"b296073.txt");

Extensions

Data section extended up to a(125) by Antti Karttunen, Jun 14 2018