|
| |
|
|
A333259
|
|
a(n) = Sum_{p in L(n)} 2^(pi(p) - 1) where L(n) is the set of all least primes in partitions of n into prime parts.
|
|
6
|
|
|
|
0, 0, 1, 2, 1, 5, 3, 9, 3, 3, 7, 19, 7, 35, 11, 7, 7, 71, 15, 135, 15, 15, 23, 263, 31, 15, 47, 15, 31, 527, 63, 1039, 47, 31, 95, 31, 111, 2079, 143, 63, 95, 4127, 191, 8255, 63, 63, 351, 16447, 223, 63, 191, 127, 319, 32895, 383, 127, 191, 255, 639, 65663
(list;
graph;
refs;
listen;
history;
text;
internal format)
|
|
|
|
OFFSET
|
0,4
|
|
|
COMMENTS
|
In other words, convert the indices of primes p_i in row n of A333238 to 1s in the (i - 1)-th place to create a binary number m; convert m to decimal.
The number of prime partitions of n is shown by A000607(n), which in terms of this sequence equates to the number of 1s in a(n), written in binary.
For prime p, row p of A333238 includes p itself as the largest term, since p is the sum of (p); here we find a(p) >= 2^(p - 1). More specifically, a(2) = 1, a(p) > 2^(p - 1) for p odd.
For n = A330507(m), a(n) = 2^m - 1, the smallest n with this value in this sequence.
|
|
|
LINKS
|
|
|
|
EXAMPLE
|
The least primes among the prime partitions of 5 are 2 and 5, cf. the 2 prime partitions of 5: (5) and (3, 2), thus row 5 of A333238 lists {2, 5}. Convert these to their indices gives us {1, 3}, take the sum of 2^(1 - 1) and 2^(3 - 1) = 2^0 + 2^2 = 1 + 4 = 5, thus a(5) = 5.
The least primes among the prime partitions of 6 are 2 and 3, cf. the two prime partitions of 6, (3, 3), and (2, 2, 2), thus row 6 of A333238 lists {2, 3}. Convert these to their indices: {1, 2}, take the sum of 2^(1 - 1) and 2^(2 - 1) = 2^0 + 2^1 = 1 + 2 = 3, thus a(6) = 3.
Row 7 of A333238 contains {2, 7} because there are 3 prime partitions of 7: (7), (5, 2), (3, 2, 2). Note that 2 is the smallest part of the latter two partitions, thus only 2 and 7 are distinct. Convert to indices: {1, 4}, sum 2^(1 - 1) and 2^(4 - 1) = 2^0 + 2^3 = 1 + 8 = 9, therefore a(7) = 9.
Table plotting prime p in row n of A333238 at pi(p) place, intervening primes missing from row n are shown by "." as a place holder. We convert the indices of these primes into a binary number to obtain the terms of this sequence:
---------------------------------------
2: 2 => 1 => 1
3: . 3 => 10 => 2
4: 2 => 1 => 1
5: 2 . 5 => 101 => 5
6: 2 3 => 11 => 3
7: 2 . . 7 => 1001 => 9
8: 2 3 => 11 => 3
9: 2 3 => 11 => 3
10: 2 3 5 => 111 => 7
11: 2 3 . . 11 => 10011 => 19
12: 2 3 5 => 111 => 7
...
|
|
|
MAPLE
|
b:= proc(n, p, t) option remember; `if`(n=0, 1, `if`(p>n, 0, (q->
add(b(n-p*j, q, 1), j=1..n/p)*t^p+b(n, q, t))(nextprime(p))))
end:
a:= proc(n) option remember; (p-> add(`if`(isprime(i) and coeff(p, x,
i)>0, 2^(numtheory[pi](i)-1), 0), i=2..degree(p)))(b(n, 2, x))
end:
|
|
|
MATHEMATICA
|
Block[{a, m = 59, s}, a = ConstantArray[{}, m]; s = {Prime@ PrimePi@ m}; Do[If[# <= m, If[FreeQ[a[[#]], Last@ s], a = ReplacePart[a, # -> Append[a[[#]], Last@ s]], Nothing]; AppendTo[s, Last@ s], If[Last@ s == 2, s = DeleteCases[s, 2]; If[Length@ s == 0, Break[], s = MapAt[Prime[PrimePi[#] - 1] &, s, -1]], s = MapAt[Prime[PrimePi[#] - 1] &, s, -1]]] &@ Total[s], {i, Infinity}]; {0, 0}~Join~Map[Total[2^(-1 + PrimePi@ #)] &, Rest[Union /@ a]]]
|
|
|
CROSSREFS
|
|
|
|
KEYWORD
|
|
|
|
AUTHOR
|
|
|
|
STATUS
|
approved
|
| |
|
|
