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%I #32 Dec 07 2019 12:18:28
%S 4,7,10,15,18,23,29,35,40,47,54,60,68,75,83,90,99,107,116,125,134,143,
%T 152,162,172,182,193,203,214,225,236,248,259,271,283,295,307,320,332,
%U 345,358,372,385,398,412,426,440,454,469,483,498,513,528,543,559,574,590,606,622,638,654,671,688,704
%N a(n) is the least m such that A008284(m,n+1) > A008284(m,n).
%C 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.
%e 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).
%t 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 *)
%o (Python)
%o element = 1
%o goal = 64
%o n = 1
%o p = [[]]
%o while element <= goal:
%o # fill in the n-th row of the table
%o p.append([0]*(goal+2))
%o for k in range(1, min(n,goal+1)+1):
%o if (k == 1) or (k == n):
%o p[n][k] = 1
%o else:
%o p[n][k] = p[n-1][k-1] + p[n-k][k]
%o # see if we can increment element
%o if p[n][element+1] > p[n][element]:
%o print "p[{}][{}]={} and p[{}][{}]={} so a[{}] = {}".format(
%o n,element,p[n][element],n,element+1,p[n][element+1],element,n)
%o element = element+1
%o n = n+1
%Y Cf. A008284.
%K nonn
%O 1,1
%A _Glen Whitney_, Jun 23 2016
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