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A344339
a(n) is the minimal number of terms of A332520 that need to be combined with the bitwise OR operator in order to give n.
1
0, 1, 1, 1, 1, 1, 2, 2, 1, 2, 1, 2, 1, 2, 2, 1, 1, 1, 2, 2, 2, 2, 3, 3, 2, 2, 2, 2, 2, 2, 3, 2, 1, 2, 1, 2, 2, 2, 2, 2, 2, 3, 2, 3, 2, 3, 2, 2, 1, 2, 2, 1, 2, 2, 3, 2, 2, 3, 2, 2, 2, 3, 3, 2, 1, 2, 2, 2, 1, 2, 2, 2, 2, 3, 2, 3, 2, 3, 2, 2, 1, 2, 2, 2, 2, 1, 3
OFFSET
0,7
COMMENTS
This sequence is related to Karnaugh maps:
- for any number n with up to 2^k binary digits (possibly with leading zeros),
- we can interpret the binary expansion of n as a truth table for a k-ary Boolean function f,
- a(n) gives the optimal number of products in a disjunctive normal form for f.
FORMULA
a(n) = 1 iff n > 0 and n belongs to A332520.
a(n) <= A000120(n).
a(A001196(n)) = a(n).
EXAMPLE
For n = 32576:
- the binary representation of 13170 is "111111101000000",
- it has 15 bits, so we can take k = 4 (15 <= 2^4),
- the corresponding 4-ary Boolean function f has the following truth table:
CD\AB| 00 01 11 10
-----+----------------
00| 0 0 1 1
01| 0 0 1 1
11| 0 0 0 1
10| 0 1 1 1
- we can express f as AC' + AB' + BCD' in optimal form,
- so a(32576) = 3.
PROG
(PARI) See Links section.
CROSSREFS
KEYWORD
nonn,base
AUTHOR
Rémy Sigrist, May 15 2021
STATUS
approved