A237828 Number of partitions of n such that 2*(least part) + 1 = greatest part.
0, 0, 0, 1, 1, 2, 4, 4, 6, 9, 10, 12, 17, 18, 22, 27, 31, 34, 42, 45, 53, 61, 66, 72, 86, 92, 103, 113, 125, 135, 154, 163, 180, 197, 213, 229, 257, 271, 294, 318, 346, 368, 404, 426, 463, 497, 532, 564, 616, 651, 700, 747, 798, 844, 912, 962, 1033, 1097, 1167, 1231, 1327, 1397, 1486, 1576, 1677
Offset: 1
Examples
a(8) = 4 counts these partitions: 3311, 3221, 32111, 311111.
Links
- Vaclav Kotesovec, Table of n, a(n) for n = 1..10000
Programs
-
Mathematica
z = 64; q[n_] := q[n] = IntegerPartitions[n]; Table[Count[q[n], p_ /; 3 Min[p] = = Max[p]], {n, z}] (* A237825*) Table[Count[q[n], p_ /; 4 Min[p] = = Max[p]], {n, z}] (* A237826 *) Table[Count[q[n], p_ /; 5 Min[p] = = Max[p]], {n, z}] (* A237827 *) Table[Count[q[n], p_ /; 2 Min[p] + 1 = = Max[p]], {n, z}] (* A237828 *) Table[Count[q[n], p_ /; 2 Min[p] - 1 = = Max[p]], {n, z}] (* A237829 *) Table[Count[IntegerPartitions[n],?(2*Min[#]+1==Max[#]&)],{n,60}] (* _Harvey P. Dale, Jun 25 2017 *) kmax = 65; Sum[x^(3k+1)/Product[1-x^j, {j, k, 2k+1}], {k, 1, kmax}]/x + O[x]^kmax // CoefficientList[#, x]& (* Jean-François Alcover, May 30 2024, after Seiichi Manyama *) nmax = 100; p = 1; s = 0; Do[p = Simplify[p*(1 - x^(2*k - 1))*(1 - x^(2*k))/(1 - x^k)]; p = Normal[p + O[x]^(nmax + 1)]; s += x^(3*k + 1)/(1 - x^k)/(1 - x^(2*k + 1))/p;, {k, 1, nmax}]; Rest[CoefficientList[Series[s, {x, 0, nmax}], x]] (* Vaclav Kotesovec, Jun 18 2025 *) -
PARI
my(N=70, x='x+O('x^N)); concat([0, 0, 0], Vec(sum(k=1, N, x^(3*k+1)/prod(j=k, 2*k+1, 1-x^j)))) \\ Seiichi Manyama, May 17 2023
Formula
G.f.: Sum_{k>=1} x^(3*k+1)/Product_{j=k..2*k+1} (1-x^j). - Seiichi Manyama, May 17 2023
a(n) ~ sqrt(phi) * exp(Pi*sqrt(2*n/15)) / (sqrt(2)* 5^(1/4) * sqrt(n)), where phi = A001622 = (1+sqrt(5))/2 is the golden ratio. - Vaclav Kotesovec, Jun 20 2025
Comments