A366429 a(n) = maximum degree of vertices in the distance graph of the partitions of n.
0, 1, 2, 3, 4, 6, 7, 8, 8, 12, 13, 14, 14, 15, 20, 21, 22, 22, 23, 23, 30, 31, 32, 32, 33, 33, 34, 42, 43, 44, 44, 45, 45, 46, 46, 56, 57, 58, 58, 59, 59, 60, 60, 60, 72, 73, 74, 74, 75, 75, 76, 76, 76, 77, 90, 91, 92, 92, 93, 93, 94, 94, 94, 95, 95, 110, 111
Offset: 1
Keywords
Examples
Enumerate the 7 partitions (= vertices) of 5 as follows: 1: 5 2: 4,1 3: 3,2 4: 3,1,1 5: 2,2,1 6: 2,1,1,1 7: 1,1,1,1,1 Call q a neighbor of p if d(p,q)=2. The set of neighbors for vertex k, for k = 1..7, is given by vertex 1: {2} vertex 2: {1,3,4} vertex 3: {2,4,5} vertex 4: {2,3,5,6} vertex 5: {3,4,6} vertex 6: {4,5,7} vertex 7: {6} The maximal degree is 4, which is the degree of vertex 4, so that a(5) = 4.
Crossrefs
Programs
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Mathematica
c[n_] := PartitionsP[n]; q[n_, k_] := q[n, k] = IntegerPartitions[n][[k]]; r[n_, k_] := r[n, k] = Join[q[n, k], ConstantArray[0, n - Length[q[n, k]]]]; d[u_, v_] := d[u, v] = Total[Abs[u - v]]; s[n_, k_] := s[n, k] = Select[Range[c[n]], d[r[n, k], r[n, #]] == 2 &] t[n_] := t[n] = Table[s[n, k], {k, 1, c[n]}] a[n_] := Max[Map[Length, t[n]]] Table[a[n], {n, 1, 30}]
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Python
from math import isqrt, comb def A366429(n): return isqrt(n-comb((m:=isqrt(k:=n+1<<1))+(k>m*(m+1)),2)<<2|1)+(r:=(isqrt(k<<2)+1>>1)-1)*(r-1)-1 # Chai Wah Wu, Jun 21 2025
Formula
Extensions
More terms from Pontus von Brömssen, Oct 24 2023
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