A366525 Irregular triangular array read by rows: T(n,k) = number of partitions p of n such that f(p) = k >= 0, where f is defined in Comments.
1, 1, 1, 2, 1, 2, 3, 4, 3, 3, 6, 2, 6, 6, 3, 5, 9, 5, 3, 7, 9, 9, 4, 1, 7, 13, 12, 6, 4, 10, 12, 15, 12, 5, 2, 7, 16, 19, 16, 12, 5, 2, 12, 16, 24, 22, 18, 6, 3, 11, 20, 28, 29, 24, 14, 6, 3, 12, 19, 31, 34, 36, 24, 13, 4, 3, 12, 23, 36, 42, 50, 30, 25, 8, 4
Offset: 1
Examples
First 18 rows: 1 1 1 2 1 2 3 4 3 3 6 2 6 6 3 5 9 5 3 7 9 9 4 1 7 13 12 6 4 10 12 15 12 5 2 7 16 19 16 12 5 2 12 16 24 22 18 6 3 11 20 28 29 24 14 6 3 12 19 31 34 36 24 13 4 3 12 23 36 42 50 30 25 8 4 1 16 23 42 54 59 45 34 15 5 4 13 28 47 57 74 61 52 28 16 5 4 Row 6 represents 3 partitions p that are self-repeating (i.e., k = 0), 6 such that f(p) = 1, and 2 such that f(p) = 2. Specifically, f(p) = 0 for these partitions: [6], [2,2,1,1], [2,1,1,1]. f(p) = 1 for these: [4,2], [3,3], [3,2,1], [3,1,1,1], [2,2,2], [1,1,1,1,1,1]. f(p) = 2 for these: [5,1], [4,1,1].
Links
- John Tyler Rascoe, Rows n = 1..60, flattened
Programs
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Mathematica
r[list_] := If[Length[list] == 1, list, Reverse[Sort[# + Join[{-1}, ConstantArray[0, Length[#] - 2], {1}]] &[Reverse[Sort[list]]]]]; f[list_] := NestWhileList[r, Reverse[Sort[list]], Unequal, All]; t = Table[BinCounts[#, {0, Max[#] + 1, 1}] &[Map[-1 + Length[Union[#]] &[f[#]] &, IntegerPartitions[n]]], {n, 1,20}] Map[Length, t]; t1 = Take[t, 18]; TableForm[t1] Flatten[t1] (* Peter J. C. Moses, Oct 10 2023 *)
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Python
from sympy .utilities.iterables import ordered_partitions from collections import Counter def A366525_rowlist(row_n): A = [] for i in range(1,row_n+1): A.append([]); p,C = list(ordered_partitions(i)),Counter() for j in range(0,len(p)): x,a1,a,b = 0,[],list(p[j]),list(p[j]) while i: b[-1] -= 1; b[0] += 1 if b[-1] == 0: b.pop(-1) b = sorted(b); x += 1 if a == b or a1 == b: C.update({x}); break else: a1 = a.copy(); a = b.copy() for z in range(1,len(C)+1): A[i-1].append(C[z]) return(A) # John Tyler Rascoe, Nov 09 2023
Comments