A384904 Number of integer partitions of n with all equal lengths of maximal runs of consecutive parts decreasing by 1 but not by 0.
1, 1, 2, 3, 4, 5, 9, 9, 14, 17, 23, 25, 40, 41, 59, 68, 92, 99, 140, 151, 204, 229, 296, 328, 433, 476, 606, 685, 858, 955, 1203, 1336, 1654, 1858, 2266, 2537, 3102, 3453, 4169, 4680, 5611, 6262, 7495, 8358, 9927, 11105, 13096, 14613, 17227, 19179, 22459
Offset: 0
Examples
The partition (6,5,5,4,2,1) has maximal runs ((6,5),(5,4),(2,1)), with lengths (2,2,2), so is counted under a(23). The a(1) = 1 through a(8) = 14 partitions: (1) (2) (3) (4) (5) (6) (7) (8) (11) (21) (22) (32) (33) (43) (44) (111) (31) (41) (42) (52) (53) (1111) (311) (51) (61) (62) (11111) (222) (331) (71) (321) (511) (422) (411) (4111) (611) (3111) (31111) (2222) (111111) (1111111) (3221) (3311) (5111) (41111) (311111) (11111111)
Links
- John Tyler Rascoe, Table of n, a(n) for n = 0..100
Programs
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Mathematica
Table[Length[Select[IntegerPartitions[n],SameQ@@Length/@Split[#,#2==#1-1&]&]],{n,0,30}]
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PARI
A_q(N) = {Vec(1+sum(k=1,floor(-1/2+sqrt(2+2*N)), sum(i=1,(N/(k*(k+1)/2))+1, q^((k*i*(2+i*(k-1)))/2)/(1-q^(k*i))*prod(j=1,i-1, 1 + q^(2*k*j)/(1 - q^(k*j))))) + O('q^(N+1)))} \\ John Tyler Rascoe, Aug 20 2025
Formula
G.f.: 1 + Sum_{i,k>0} q^((i*k*(2 + i*(k-1)))/2) * Product_{j=1..i-1} ( 1 + q^(2*k*j)/(1 - q^(k*j)) ) / (1 - q^(i*k)). - John Tyler Rascoe, Aug 20 2025