OFFSET
1,1
COMMENTS
A008284(m,n) is the number of partitions of the integer m into n parts; p(m,n) in the following. It is numerically and intuitively clear that for any fixed n, for sufficiently large m, p(m,n+1) > p(m,n). Moreover, from examining the table of p(m,n) for small values of n, it appears that for any fixed n, once it has occurred for some m that p(m,n+1) > p(m,n), then it holds for all larger m. However, I did not see a simple proof of this, nor could I easily find one on the net. Presuming it is true, then the m at which p(m,n+1) first overtakes p(m,n) is of intrinsic interest.
EXAMPLE
a(1) = 4 since p(4,2) = 2, which is greater than p(4,1) = 1, whereas for any lesser integer, e.g. 3, p(3,2) <= p(3,1).
MATHEMATICA
t[n_, 1] = 1; t[n_, k_] := t[n, k] = If[n >= k, Sum[t[n - i, k - 1], {i, 1, n - 1}] - Sum[t[n - i, k], {i, 1, k - 1}], 0]; Table[m = 1; While[t[m, n + 1] <= t[m, n], m++]; m, {n, 0, 50}] (* Michael De Vlieger, Jun 23 2016, after Mats Granvik at A008284 *)
PROG
(Python)
element = 1
goal = 64
n = 1
p = [[]]
while element <= goal:
# fill in the n-th row of the table
p.append([0]*(goal+2))
for k in range(1, min(n, goal+1)+1):
if (k == 1) or (k == n):
p[n][k] = 1
else:
p[n][k] = p[n-1][k-1] + p[n-k][k]
# see if we can increment element
if p[n][element+1] > p[n][element]:
print("p[{}][{}]={} and p[{}][{}]={} so a[{}] = {}".format(
n, element, p[n][element], n, element+1, p[n][element+1], element, n))
element = element+1
n = n+1
CROSSREFS
KEYWORD
nonn
AUTHOR
Glen Whitney, Jun 23 2016
STATUS
approved