partition into parts in arithmetic progression.

[murthy amarnath]

Dear Seq fans,
Has the following  sequence idea been explored? 

Number of partitions of n into parts which can be
arranged to form an Arithmetic Progression. (even with
0 common difference).

1,2,3,4,4,7,5,7,8,9,...
 example:
a(9) = 8, and the partitions are (9), (1,8), (2,7),
(3,6), (4,5), (3,3,3), (1,3,5), (1+1+1+1+1+1+1+1+1).
a(6) = 7, and the partitions are (6), (1,5), (2,4),
(3,3), (2,2,2),(1,2,3),(1+1+1+1+1+1).

Comments:
n and ( 1+1+1...n times) are trivial partitions and
are counted.

1. a(n) >  INT [(n-1)/2)] + tau(n) +  Sum { f(d)}, 
for n  > 5.
where tau(n) is the number of divisors of n.
f(d) is a function of a divisor d  of n.
e.g. 3 is a divisor of 9 and contributes two
partitions  (3,3,3) and (1,3,5). Similarly for 6 it
contributes (2,2,2) and (1,2,3).

Open Question: find an estimate of f(d) for a given n
d divides n.

2.  for a prime p,  a(p) = 2+ INT[(p-1)/2],
3. If n is a triangular number say n =  k(k+1)/2 then
1+2+3..+k is  one such partition with length k. ( with
maximum number of parts k).

thanks
regards
sequentially
amarnath murthy




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