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A240871
Number of partitions p of n into distinct parts such that max(p) = 3 + min(p).
3
0, 0, 0, 0, 0, 1, 0, 2, 1, 1, 2, 2, 0, 2, 2, 1, 1, 2, 1, 2, 1, 1, 2, 2, 0, 2, 2, 1, 1, 2, 1, 2, 1, 1, 2, 2, 0, 2, 2, 1, 1, 2, 1, 2, 1, 1, 2, 2, 0, 2, 2, 1, 1, 2, 1, 2, 1, 1, 2, 2, 0, 2, 2, 1, 1, 2, 1, 2, 1, 1, 2, 2, 0, 2, 2, 1, 1, 2, 1, 2, 1, 1, 2, 2, 0, 2, 2, 1, 1, 2, 1, 2, 1, 1, 2, 2, 0, 2, 2, 1, 1, 2, 1, 2, 1, 1, 2, 2, 0
OFFSET
0,8
FORMULA
From Alois P. Heinz, Jan 13 2025: (Start)
G.f.: -x^5*(x^7+2*x^6+3*x^5+3*x^4+3*x^3+3*x^2+x+1)/((x-1)*(x+1)*(x^2+1)*(x^2+x+1)).
a(n) = a(n-12) for n>=19. (End)
EXAMPLE
a(7) counts these 2 partitions: 5+2, 4+2+1.
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 *)
PROG
(PARI)
A240871aux(n, minp=0, maxp=0) = if(0==n, (minp>0 && (maxp==3+minp)), sum(i=1+maxp, min(n, 3+minp), A240871aux(n-i, if(!minp, i, minp), i)));
A240871(n) = sum(i=1, floor(n/2), A240871aux(n-i, i, i)); \\ Antti Karttunen, Jan 13 2025
CROSSREFS
KEYWORD
nonn,easy
AUTHOR
Clark Kimberling, Apr 15 2014
EXTENSIONS
More terms from Antti Karttunen, Jan 13 2025
STATUS
approved