login
A240872
Number of partitions p of n into distinct parts such that max(p) = 4 + min(p).
3
0, 0, 0, 0, 0, 0, 1, 0, 2, 1, 2, 2, 3, 2, 2, 3, 3, 2, 2, 2, 4, 2, 2, 2, 3, 3, 2, 2, 3, 2, 3, 2, 3, 2, 2, 3, 3, 2, 2, 2, 4, 2, 2, 2, 3, 3, 2, 2, 3, 2, 3, 2, 3, 2, 2, 3, 3, 2, 2, 2, 4, 2, 2, 2, 3, 3, 2, 2, 3, 2, 3, 2, 3, 2, 2, 3, 3, 2, 2, 2, 4, 2, 2, 2, 3, 3, 2, 2, 3, 2, 3, 2, 3, 2, 2, 3, 3, 2, 2, 2, 4
OFFSET
0,9
FORMULA
G.f.: -x^6*(x^12+2*x^11+3*x^10+5*x^9+5*x^8+6*x^7+7*x^6+6*x^5+5*x^4+4*x^3+3*x^2+x+1) / ((x-1)*(x+1)*(x^2+1)*(x^4+x^3+x^2+x+1)). - Alois P. Heinz, Jun 18 2025
EXAMPLE
a(12) counts these 3 partitions: 84, 642, 5421.
MATHEMATICA
z = 40; f[n_] := f[n] = Select[IntegerPartitions[n], Max[Length /@ Split@#] == 1 &];
Table[Count[f[n], p_ /; Max[p] == 2 + Min[p]], {n, 0, z}] (* A171182 *)
Table[Count[f[n], p_ /; Max[p] == 3 + Min[p]], {n, 0, z}] (* A240871 *)
Table[Count[f[n], p_ /; Max[p] == 4 + Min[p]], {n, 0, z}] (* A240872 *)
Table[Count[f[n], p_ /; Max[p] == 5 + Min[p]], {n, 0, z}] (* A240873 *)
CROSSREFS
KEYWORD
nonn,easy
AUTHOR
Clark Kimberling, Apr 15 2014
STATUS
approved