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Number of partitions satisfying cn(0,5) = cn(2,5) + cn(3,5).
0

%I #6 Mar 30 2012 17:20:56

%S 1,1,1,2,2,3,4,6,7,8,10,15,19,23,27,35,44,56,67,82,100,124,150,184,

%T 221,268,321,388,464,558,664,792,939,1119,1326,1573,1850,2181,2567,

%U 3027,3547,4155,4853,5681,6631

%N Number of partitions satisfying cn(0,5) = cn(2,5) + cn(3,5).

%C For a given partition cn(i,n) means the number of its parts equal to i modulo n.

%C Short: 0 = 2 + 3 (BBpEZ).

%K nonn

%O 1,4

%A _Olivier GĂ©rard_