A100881 Number of partitions of n in which the sequence of frequencies of the summands is decreasing.

1, 1, 2, 2, 3, 3, 4, 4, 6, 5, 7, 8, 8, 9, 13, 10, 13, 15, 16, 18, 21, 17, 24, 28, 26, 26, 36, 32, 38, 42, 40, 46, 52, 48, 63, 63, 59, 63, 85, 77, 81, 92, 89, 102, 116, 98, 122, 134, 130, 140, 157, 145, 165, 182, 190, 191, 207, 195, 235, 259, 232, 252, 293, 279

Offset: 0

Author

David S. Newman, Nov 21 2004

Keywords

• nonn

Examples

			a(7) = 4 because in each of the four partitions [7], [3,3,1], [2,2,2,1], [1,1,1,1,1,1,1] the frequency with which a summand is used decreases as the summand decreases.
		

Links

• Alois P. Heinz, Table of n, a(n) for n = 0..1000

Crossrefs

Cf. A100882, A100883, A100884.

Cf. A100471, A098859.

Programs

Haskell

a100881 = p 0 0 1 where
   p m m' k x | x == 0    = if m > m' || m == 0 then 1 else 0
              | x < k     = 0
              | m == 0    = p 1 m' k (x - k) + p 0 m' (k + 1) x
              | otherwise = p (m + 1) m' k (x - k) +
                            if m > m' then p 0 m (k + 1) x else 0
-- Reinhard Zumkeller, Dec 27 2012

Maple

b:= proc(n, i, t) option remember;
      if n<0 then 0
    elif n=0 then 1
    elif i=0 then 0
    else      b(n, i-1, t)
         +add(b(n-i*j, i-1, j), j=1..min(t-1, floor(n/i)))
      fi
    end:
a:= n-> b(n, n, n+1):
seq(a(n), n=0..60);  # Alois P. Heinz, Feb 21 2011

Mathematica

b[n_, i_, t_] := b[n, i, t] = Which[n<0, 0, n==0, 1, i==0, 0, True, b[n, i-1, t] + Sum[b[n-i*j, i-1, j], {j, 1, Min[t-1, Floor[n/i]]}]]; a[n_] := b[n, n, n+1]; Table[a[n], {n, 0, 60}] (* Jean-François Alcover, Jul 15 2015, after Alois P. Heinz *)

Extensions

More terms from Alois P. Heinz, Feb 21 2011