A274383 a(n) is the least m such that A008284(m,n+1) > A008284(m,n).

4, 7, 10, 15, 18, 23, 29, 35, 40, 47, 54, 60, 68, 75, 83, 90, 99, 107, 116, 125, 134, 143, 152, 162, 172, 182, 193, 203, 214, 225, 236, 248, 259, 271, 283, 295, 307, 320, 332, 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

Offset: 1

Author

Glen Whitney, Jun 23 2016

Keywords

• nonn

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.

Examples

			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).
		

Crossrefs

Cf. A008284.

Programs

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 *)

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