A324863 Binary length of A324866(n), where A324866(n) = A156552(n) OR (A323243(n) - A156552(n)).

0, 1, 2, 2, 3, 3, 4, 3, 3, 4, 5, 4, 6, 5, 4, 4, 7, 4, 8, 5, 5, 6, 9, 5, 5, 7, 4, 6, 10, 5, 11, 5, 6, 8, 5, 5, 12, 9, 7, 6, 13, 6, 14, 7, 5, 10, 15, 6, 6, 5, 8, 8, 16, 5, 6, 7, 9, 11, 17, 6, 18, 12, 6, 6, 7, 7, 19, 9, 10, 6, 20, 6, 21, 13, 5, 10, 6, 8, 22, 7, 6, 14, 23, 7, 8, 15, 11, 8, 24, 6, 7, 11, 12, 16, 9, 7, 25, 6, 7, 6, 26, 9, 27, 9, 6

Offset

1,3

Comments

Differs from A324861 [binary length of A324876(n)] for the first time at n=50.

Provided that the maximal value that A324861(d) attains among divisors d of n is attained an odd number of times, then a(n) gives that maximal value. It is conjectured that this always holds. Among n = 1..10000, there are only two such cases, where the maximal value occurs more than once among the divisors: 3675 and 7623, where it occurs three times in both (see the examples).

Links

Antti Karttunen, Table of n, a(n) for n = 1..10000 (based on Hans Havermann's factorization of A156552)

Index entries for sequences related to binary expansion of n

Index entries for sequences computed from indices in prime factorization

Index entries for sequences related to sigma(n)

Formula

a(1) = 0; for n > 1, a(n) = A070939(A324866(n)) = 1 + A000523(A324866(n)).

a(A000040(n)) = n.

a(n) = Max_{d|n} A324861(d) [conjectured].

Example

For n = 50, we have A156552(50) = 25 and A323243(50) = 31. Taking bitwise-OR (A003986) of 25 and 31-25 = 6, we get 31, in binary "11111", with length 5, thus a(50) = 5.
The rest of examples pertain to the conjectured interpretation of this sequence:
Divisors of 8 are [1, 2, 4, 8]. A324861 applied to these gives values [0, 1, 2, 3], of which the largest is 3, thus a(8) = 3.
Divisors of 25 are [1, 5, 25]. A324861 applied to these gives values [0, 3, 5], of which the largest is 5, thus a(25) = 5.
Divisors of 50 are [1, 2, 5, 10, 25, 50]. A324861 applied to these gives values [0, 1, 3, 4, 5, 4], of which the largest is 5, thus a(50) = 5.
Divisors of 88 are [1, 2, 4, 8, 11, 22, 44, 88]. A324861 applied to these gives values [0, 1, 2, 3, 5, 6, 7, 8], of which the largest is 8, thus a(88) = 8.
Divisors of 3675 are [1, 3, 5, 7, 15, 21, 25, 35, 49, 75, 105, 147, 175, 245, 525, 735, 1225, 3675]. A324861 applied to these gives values [0, 2, 3, 4, 4, 5, 5, 5, 6, 4, 6, 5, 6, 5, 8, 7, 8, 8], of which the largest is 8 (occurs three times), thus a(3675) = 8.
Divisors of 7623 are [1, 3, 7, 9, 11, 21, 33, 63, 77, 99, 121, 231, 363, 693, 847, 1089, 2541, 7623]. A324861 applied to these gives values [0, 2, 4, 3, 5, 5, 6, 6, 6, 7, 7, 7, 6, 8, 6, 9, 9, 9], of which the largest is 9 (occurs three times), thus a(7623) = 9.

Prog

(PARI)
A156552(n) = {my(f = factor(n), p2 = 1, res = 0); for(i = 1, #f~, p = 1 << (primepi(f[i, 1]) - 1); res += (p * p2 * (2^(f[i, 2]) - 1)); p2 <<= f[i, 2]); res}; \\ From A156552 by David A. Corneth
A324866(n) = { my(k=A156552(n)); bitor(k, (A323243(n)-k)); }; \\ Needs also code from A323243.
A324863(n) = #binary(A324866(n));
(PARI)
A324876(n) = { my(v=0); fordiv(n, d, if(issquarefree(n/d), v=bitxor(v, A324866(d)))); (v); };
A324861(n) = #binary(A324876(n));
A324863(n) = { my(m=0, w, c=0); fordiv(n, d, w=A324861(d); if(w>=m, if(w==m, c++, c=1; m=w))); (m); };

Crossrefs

Cf. A000523, A003986, A070939, A156552, A252464, A323243, A324861, A324864, A324866, A324874, A324879.

Differs from A252464 for the first time at n=25, A324870 gives the differences.

Sequence in context: A253555 A252464 A324861 * A332894 A269989 A057935

Adjacent sequences: A324860 A324861 A324862 * A324864 A324865 A324866

Keyword

nonn

Author

Antti Karttunen, Mar 21 2019

Status

approved