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.
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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 *)
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.
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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 *)
A332861
Primes p with the property that if q
Original entry on oeis.org
2, 3, 7, 13, 23, 31, 41, 79, 101, 107, 149, 163, 173, 191, 197, 269, 271, 293, 347, 419, 443, 523, 557, 647, 761, 769, 787, 1013, 1153, 1373, 1613, 1619, 1669, 1693, 1777, 1783, 1873, 2153, 2161, 2207, 2399, 2447, 2801, 2939, 2999, 3011, 3049, 3253, 3319, 3413
Offset: 1
Prime 13 is a member, because the minimal primes in partitions of 13 into prime parts smaller than 13 occur at least twice: [2,2,2,2,2,3], [2,2,3,3,3], [2,2,2,2,5], [2,3,3,5], [2,2,2,7], [2,11], [3,3,7], [3,5,5]; 3 occurs twice, 2 occurs 6 times.
Prime 11 is not a member, because 3 occurs only once as a minimal prime in partitions of 11 into smaller primes: [2,2,2,2,3], [2,3,3,3], [2,2,2,5], [2,2,7], [3,3,5].
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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 p; p:= a(n-1); do
p:= nextprime(p); if (f-> andmap(i-> coeff(f, x, i)
<>1, [$2..p-1]))(b(p, 2, x)) then return p fi od
end: a(1):=2:
seq(a(n), n=1..33); # Alois P. Heinz, Mar 13 2020
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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]]]];
a[n_] := a[n] = Module[{p = a[n - 1]}, While[True, p = NextPrime[p]; If[AllTrue[Range[2, p-1], SeriesCoefficient[b[p, 2, x], {x, 0, #}] != 1&], Return [p]]]];
a[1] = 2;
Table[Print[n, " ", a[n]]; a[n], {n, 1, 33}] (* Jean-François Alcover, Nov 23 2020, after Alois P. Heinz *)
A335106
Irregular triangle T(n,k) is the number of times that prime(k) is the greatest part in a partition of n into prime parts; Triangle T(n,k), n>=0, 1 <= k <= max(1,A000720(A335285(n))), read by rows.
Original entry on oeis.org
0, 0, 1, 0, 1, 1, 0, 1, 1, 1, 1, 0, 1, 1, 1, 1, 1, 1, 0, 2, 1, 1, 1, 1, 2, 1, 0, 2, 2, 1, 1, 1, 2, 2, 2, 0, 2, 3, 2, 1, 1, 1, 2, 3, 3, 1, 0, 3, 4, 3, 1, 1, 1, 2, 4, 4, 2, 1, 0, 3, 5, 5, 2, 1, 1, 1, 3, 5, 5, 3, 2, 0, 3, 6, 7, 3, 2, 1, 1, 1, 3, 7, 7, 4, 3, 1, 0, 4
Offset: 0
A000607(10) = 5 and the prime partitions of 10 are: (2,2,2,2,2), (2,2,3,3), (2,3,5), (5,5) and (3,7). Thus G(10) = {2,3,5,5,7}, and consequently row 10 is [1,1,2,1]. In the table below, for n >= 2, 0 is used to indicate when prime(k) is not in G(n) and is less than the greatest member of G(n), otherwise the entry for prime(k) not in G(n) is left empty. For n >= 2 the sum of entries in the n-th row is |G(n)| = A000607(n). Triangle T(n,k) begins:
0;
0;
1;
0, 1;
1;
0, 1, 1;
1, 1;
0, 1, 1, 1;
1, 1, 1;
0, 2, 1, 1;
1, 1, 2, 1;
0, 2, 2, 1, 1;
1, 2, 2, 2;
0, 2, 3, 2, 1, 1;
1, 2, 3, 3, 1;
0, 3, 4, 3, 1, 1;
1, 2, 4, 4, 2, 1;
0, 3, 5, 5, 2, 1, 1;
...
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Flatten@ Block[{nn = 22, t}, t = Block[{s = {Prime@ PrimePi@ nn}}, KeySort@ Merge[#, Identity] &@ Join[{0 -> {}, 1 -> {}}, Reap[Do[If[# <= nn, Sow[# -> s]; 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}]][[-1, -1]] ] ]; Array[Function[p, If[! IntegerQ@ First@ p, {0}, Array[Count[p, Prime@ #] &, PrimePi@ Max@ p]]]@ Map[Max, t[[#]]] &, Max@ Keys@ t]] (* Michael De Vlieger, May 23 2020 *)
row[0]={0}; row[k_] := Join[If[OddQ@k, {0}, {}], Last /@ Tally@ Sort[ First /@ IntegerPartitions[k, All, Prime@ Range@ PrimePi@ k]]]; Join @@ Array[row, 20, 0] (* Giovanni Resta, May 31 2020 *)
A335285
a(n) is the greatest possible greatest part of any partition of n into prime parts.
Original entry on oeis.org
2, 3, 2, 5, 3, 7, 5, 7, 7, 11, 7, 13, 11, 13, 13, 17, 13, 19, 17, 19, 19, 23, 19, 23, 23, 23, 23, 29, 23, 31, 29, 31, 31, 31, 31, 37, 31, 37, 37, 41, 37, 43, 41, 43, 43, 47, 43, 47, 47, 47, 47, 53, 47, 53, 53, 53, 53, 59, 53, 61, 59, 61, 61, 61, 61, 67, 61, 67
Offset: 2
a(9) = 7, the greatest prime < 9 (prime partitions of 9 are: [2,7], [2,2,5], [2,2,2,3] and [3,3,3], in which 7 is greatest of greatest parts).
a(12) = 7 (greatest prime < 11).
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Array[Max@ Select[IntegerPartitions[#], AllTrue[#, PrimeQ] &][[All, 1]] &, 68, 2] (* Michael De Vlieger, May 30 2020 *)
Array[Prime[PrimePi@ # - Boole[And[PrimeQ[# - 1], # != 3]]] &, 68, 2] (* Michael De Vlieger, May 31 2020 *)
A330433
Numbers k such that if there is a prime partition of k with least part p, then there exists at least one other prime partition of k with least part p.
Original entry on oeis.org
63, 161, 195, 235, 253, 425, 513, 581, 611, 615, 635, 667, 767, 779, 791, 803, 959, 1001, 1015, 1079, 1095, 1121, 1127, 1251, 1253, 1265, 1267, 1547, 1557, 1595, 1617, 1625, 1647, 1649, 1681, 1683, 1687, 1771, 1817, 1829, 1915, 1921, 2071, 2125, 2159, 2185
Offset: 1
9 is not a term because [3,3,3] is the only prime partition of 9 having 3 as least part.
63 is a term because every possible prime partition is accounted for as follows, where (m,p) means m partitions of 63 with least part p: (2198,2), (323,3), (60,5), (15,7), (5,11), (2,13), (2,17), (sum of m values = 2605 = A000607(63)). 63 must be in the sequence because (1,p) does not appear in this list, and is the smallest such number because every odd composite < 63 has at least one prime partition with unique least part (as for 9 above).
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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 k; for k from a(n-1)+1
while 1 in {coeffs(b(k, 2, x))} do od; k
end: a(0):=1:
seq(a(n), n=1..40); # Alois P. Heinz, Mar 21 2020
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b[n_, p_, t_] := b[n, p, t] = If[n == 0, 1, If[p > n, 0, Function[q, Sum[b[n - p j, q, 1], {j, 1, n/p}] t^p + b[n, q, t]][NextPrime[p]]]];
a[0] = 1;
a[n_] := a[n] = Module[{k}, For[k = a[n-1]+1, MemberQ[CoefficientList[b[k, 2, x], x], 1], k++]; k];
Table[Print[n, " ", a[n]]; a[n], {n, 1, 40}] (* Jean-François Alcover, Nov 26 2020, after Alois P. Heinz *)
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)
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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.
Comments