A333337
Indices of rows of n consecutive smallest primes in A333238, or -1 if n consecutive smallest primes do not appear in A333238.
Original entry on oeis.org
0, 1, 2, 4, 6, 8, 9, 10, 12, 15, 16, 18, 20, 21, 25, 27, 24, 28, 33, 35, 30, 39, 44, 45, 49, 51, 55, 63, 57, 65, 69, 75, 77, 81, 85, 60, 76, 87, 91, 95, 99, 105, 111, 115, 117, 119, 121, 123, 125, 135, 143, 145, 147, 153, 155, 161, 169, 159, 165, 171, 175, 177
Offset: 0
Table begins:
0: 0 1
1: 2 4
2: 6 8 9
3: 10 12 15 16
4: 18 20 21 25 27
5: 24 28 33 35
6: 30 39 44 45 49
7: 51 55 63
8: 57 65
9: 60 76 87 91 95
10: 69 75 77 81 85
11: 99 105
12: 111 115 117 119 121
13: 123 125 135
14: 143 145
15: 147 153 155 161 169
16: 159 165 171 175
17: 177 185 187
Consider the table plotting prime p in row m of A333238 at pi(p) place; intervening primes missing from row m are shown by "." as a place holder:
m Primes in row m of A333238
---------------------------------
2: 2
3: . 3
4: 2
5: 2 . 5
6: 2 3
7: 2 . . 7
8: 2 3
9: 2 3
10: 2 3 5
11: 2 3 . . 11
12: 2 3 5
13: 2 3 . . . 13
14: 2 3 . 7
15: 2 3 5
16: 2 3 5
17: 2 3 5 . . . 17
...
There are no primes in rows 0 or 1 of A333238, thus row 0 of this sequence contains {0, 1}.
The smallest prime, 2, appears alone in rows 2 and 4 of A333238, thus row 1 of this sequence contains {2, 4}.
We have the primes {2, 3} and no other primes in rows {6, 8, 9} in A333238, thus row 2 of this sequence contains {6, 8, 9}.
We have the primes {2, 3, 5} and no other primes in rows {10, 12, 15, 16} in A333238, thus row 3 of this sequence contains {10, 12, 15, 16}, etc.
-
Block[{m = 120, s, a}, a = ConstantArray[{}, m]; s = {Prime@ PrimePi@ m}; Do[If[# <= m, If[FreeQ[a[[#]], Last@ s], a = ReplacePart[a, # -> Union@ 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}]; s = {0}~Join~Map[Which[Length@ # == 0, 0, And[Length@ # == 1, First@ # == 2], 1, True, If[Union@ # == {1}, Length@ # + 1, -1] &[Differences@ PrimePi@ #, {} -> {2}]] &, a]; Array[-1 + Position[s, #][[All, 1]] /. k_ /; MissingQ@ k -> {-1} &, Max@ s + 1, 0]]
A330507
a(n) is the smallest number k having for every prime p <= prime(n) at least one prime partition with least part p, and no such partition having least part > prime(n). If no such k exists then a(n) = 0 (see comments).
Original entry on oeis.org
2, 6, 10, 18, 24, 30, 51, 57, 69, 60, 99, 111, 123, 143, 147, 159, 177, 189, 201, 213, 225, 245, 255, 267, 291, 303, 309, 321, 345, 357, 381, 393, 411, 427, 447, 465, 471, 493, 507, 519, 537, 553, 573, 583, 623, 621, 633, 669, 681, 695, 707, 729, 749, 753, 783
Offset: 1
a(1) = 2 because [2] is the only prime partition of prime(1) = 2.
a(2) = 6 because [2,2,2] and [3,3] are the only possible prime partitions of 6, namely with prime(1) and prime(2) the only least parts.
-
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; local f, k, p; p:= ithprime(n);
for k to 4*p do f:= b(k, 2, x); if degree(f)<= p and andmap(
h->0Alois P. Heinz, Mar 12 2020
-
With[{s = Array[Union@ Select[IntegerPartitions[#], AllTrue[#, PrimeQ] &][[All, -1]] &, 70]}, TakeWhile[Map[FirstPosition[s, #][[1]] &, Rest@ NestList[Append[#, Prime[Length@ # + 1]] &, {}, 12]], IntegerQ]] (* Michael De Vlieger, Mar 06 2020 *)
(* Second program: *)
Block[{a, m = 125, s}, a = ConstantArray[{}, m]; s = {Prime@ PrimePi@ m}; Do[If[# <= m, If[FreeQ[a[[#]], Last@ s], a = ReplacePart[a, # -> Union@ 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}]; TakeWhile[Map[FirstPosition[a, #][[1]] &, Rest@ NestList[Append[#, Prime[Length@ # + 1]] &, {}, Max[Length /@ a]]], IntegerQ]] (* Michael De Vlieger, Mar 11 2020 *)
A333129
Product of all distinct least part primes from all partitions of n into prime parts.
Original entry on oeis.org
1, 1, 2, 3, 2, 10, 6, 14, 6, 6, 30, 66, 30, 78, 42, 30, 30, 510, 210, 570, 210, 210, 330, 690, 2310, 210, 2730, 210, 2310, 6090, 30030, 6510, 2730, 2310, 39270, 2310, 46410, 85470, 3990, 30030, 39270, 94710, 570570, 1291290, 30030, 30030, 903210, 1411410, 746130
Offset: 0
a(2) = 2 because [2] is the only prime partition of 2. a(5) = 10 because the prime partitions of 5 are [2,3] and [5], so the products of all distinct least part primes is 2*5 = 10.
-
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:= n-> (p-> mul(`if`(coeff(p, x, i)>0, i, 1), i=2..n))(b(n, 2, x)):
seq(a(n), n=0..55); # Alois P. Heinz, Mar 12 2020
-
a[0] = 1; a[n_] := Times @@ Union[Min /@ IntegerPartitions[n, All, Prime[ Range[PrimePi[n]]]]];
a /@ Range[0, 55] (* Jean-François Alcover, Nov 01 2020 *)
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.
Original entry on oeis.org
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
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
...
-
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
-
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]]]
A333365
T(n,k) is the number of times that prime(k) is the least part in a partition of n into prime parts; triangle T(n,k), n >= 0, 1 <= k <= max(1,A000720(A331634(n))), read by rows.
Original entry on oeis.org
0, 0, 1, 0, 1, 1, 1, 0, 1, 1, 1, 2, 0, 0, 1, 2, 1, 3, 1, 3, 1, 1, 4, 1, 0, 0, 1, 5, 1, 1, 6, 2, 0, 0, 0, 1, 7, 2, 0, 1, 9, 2, 1, 10, 3, 1, 12, 3, 1, 0, 0, 0, 1, 14, 3, 1, 1, 17, 4, 1, 0, 0, 0, 0, 1, 19, 5, 1, 1, 23, 5, 1, 1, 26, 6, 2, 0, 1, 30, 7, 2, 0, 0, 0, 0, 0, 1
Offset: 0
In the A000607(11) = 6 partitions of 11 into prime parts, (11), 335, 227, 2225, 2333, 22223 the least parts are 11 = prime(5) (once), 3 = prime(2)(once), and 2 = prime(1) (four times), whereas 5 and 7 (prime(3) and prime(4)) do not occur. Thus row 11 is [4,1,0,0,1].
Triangle T(n,k) begins:
0 ;
0 ;
1 ;
0, 1 ;
1 ;
1, 0, 1 ;
1, 1 ;
2, 0, 0, 1 ;
2, 1 ;
3, 1 ;
3, 1, 1 ;
4, 1, 0, 0, 1 ;
5, 1, 1 ;
6, 2, 0, 0, 0, 1 ;
7, 2, 0, 1 ;
9, 2, 1 ;
10, 3, 1 ;
12, 3, 1, 0, 0, 0, 1 ;
14, 3, 1, 1 ;
17, 4, 1, 0, 0, 0, 0, 1 ;
19, 5, 1, 1 ;
...
Indices of rows without 1's:
A330433.
-
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:
T:= proc(n) option remember; (p-> seq(`if`(isprime(i),
coeff(p, x, i), [][]), i=2..max(2,degree(p))))(b(n, 2, x))
end:
seq(T(n), n=0..23);
-
b[n_, p_, t_] := b[n, p, t] = If[n == 0, 1, If[p > n, 0, With[{q = NextPrime[p]}, Sum[b[n - p*j, q, 1], {j, 1, n/p}]*t^p + b[n, q, t]]]];
T[n_] := If[n < 2, {0}, MapIndexed[If[PrimeQ[#2[[1]]], #1, Nothing]&, Rest @ CoefficientList[b[n, 2, x], x]]];
T /@ Range[0, 23] // Flatten (* Jean-François Alcover, Mar 30 2021, after Alois P. Heinz *)
A333266
a(n) is the smallest number k such that for all m >= k there is at least one prime partition of m with prime(n) as least part.
Original entry on oeis.org
4, 8, 15, 24, 39, 49, 67, 83, 89, 115, 127, 143, 163, 179, 193, 223, 235, 249, 271, 281, 295, 333, 349, 363, 387, 403, 409, 427, 461, 483, 515, 535, 545, 565, 595, 625, 643, 659, 685, 703, 725, 733, 759, 805, 813, 835, 851, 895, 907, 923, 937, 965, 989, 1033
Offset: 1
For any k >= 4 there exists a prime partition of k having least part 2, hence a(1)=4.
A333417
a(n) is the greatest number k having for every prime <= prime(n) at least one prime partition with least part p, and no such partition having least part > prime(n). If no such k exists then a(n) = 0.
Original entry on oeis.org
4, 9, 16, 27, 35, 49, 63, 65, 85, 95, 105, 121, 135, 145, 169, 175, 187, 203, 207, 221, 253, 265, 273, 289, 301, 305, 319, 351, 369, 387, 403, 407, 425, 445, 473, 485, 495, 517, 529, 545, 551, 567, 611, 615, 629, 637, 671, 679, 693, 697, 725, 747, 781, 793, 799
Offset: 1
a(1) = 4 because [2,2] is the only prime partition of 4, and no greater number n has only 2 as least part in any partition of n into primes.
From _Michael De Vlieger_, Mar 20 2020: (Start)
Looking at this sequence as the first position of 2^n - 1 in A333259, which in binary is a k-bit repunit, we look for the last occasion of such in A333259, indicated by the arrows. a(k) = n for rows n that have an arrow. In the chart, we reverse the portrayal of the binary rendition of A333259(n), replacing zeros with "." for clarity:
n A333259(n) k
------------------------------
2 1 1
3 . 1
4 1 -> 1
5 1 . 1
6 1 1 2
7 1 . . 1
8 1 1 2
9 1 1 -> 2
10 1 1 1 3
11 1 1 . . 1
12 1 1 1 3
13 1 1 . . . 1
14 1 1 . 1
15 1 1 1 3
16 1 1 1 -> 3
17 1 1 1 . . . 1
18 1 1 1 1 4
19 1 1 1 . . . . 1
20 1 1 1 1 4
... (End)
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With[{s = TakeWhile[Import["https://oeis.org/A333259/b333259.txt", "Data"], Length@ # > 0 &][[All, -1]]}, Array[If[Length[#] == 0, 0, #[[-1, 1]] - 1] &@ Position[s, 2^# - 1] &, 55]] (* Michael De Vlieger, Mar 20 2020, using the b-file at A333259 *)
A333636
a(n) is the greatest least part of a partition of n into prime parts which does not divide n, or 0 if no such prime exists.
Original entry on oeis.org
0, 0, 0, 2, 0, 2, 3, 2, 3, 3, 5, 3, 3, 2, 5, 5, 7, 5, 7, 5, 5, 5, 11, 7, 7, 7, 11, 7, 13, 7, 13, 7, 11, 11, 17, 11, 7, 11, 17, 11, 19, 13, 13, 13, 17, 13, 19, 13, 19, 13, 23, 17, 23, 17, 19, 17, 17, 17, 29, 19, 19, 17, 23, 19, 29, 19, 31, 19, 29, 19, 31, 19, 31, 23, 29, 23, 37, 19, 37, 23, 29, 23, 41
Offset: 2
The only prime partition of 2 is [2], but 2|2, so a(2) = 0. Also, since [2,2,2] and [3,3] are the prime partitions of 6, with 2|6 and 3|6, a(6) = 0. The prime partitions of 5 are [2,3] and [5], but 2 does not divide 5 so a(5) = 2.
From _Michael De Vlieger_, Apr 01 2020: (Start)
Chart showing terms k in rows 5 <= n <= 24 of A333238, plotted at pi(k), with "." replacing terms k | n. In the table, we place a(n) in parenthesis:
n k
-------------------
5 (2) .
6 . .
7 (2) .
8 . (3)
9 (2) .
10 . (3) .
11 2 (3) .
12 . . (5)
13 2 (3) .
14 . (3) .
15 (2) . .
16 . 3 (5)
17 2 3 (5) .
18 . . 5 (7)
19 2 3 (5) .
20 . 3 . (7)
21 2 . (5) .
22 . 3 (5) .
23 2 3 (5) .
24 . . 5 7 (11)
... (End)
-
Block[{m = 84, s, a}, a = ConstantArray[{}, m]; s = {Prime@ PrimePi@ m}; Do[If[# <= m, If[And[FreeQ[a[[#]], Last[s]], Mod[#, Last[s]] != 0], a = ReplacePart[a, # -> Union@ 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}]; Map[If[Length[#] == 0, 0, Last@ #] &, Rest@ a]] (* Michael De Vlieger, Apr 01 2020 *)
Showing 1-8 of 8 results.
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