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.

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

Offset: 0

Author

Michael De Vlieger, Mar 16 2020

Keywords

• nonn
• look

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.

Examples

			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:
    n      Row n of A333238    binary   a(n)
    ---------------------------------------
    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
    ...
		

Links

• Alois P. Heinz, Table of n, a(n) for n = 0..10000

Crossrefs

Cf. A000607, A000720, A330507, A333129, A333238.

Programs

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:
seq(a(n), n=0..63);  # Alois P. Heinz, Mar 16 2020

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]]]