A337584 - OEIS

%I #28 Aug 17 2021 06:17:14

%S 1,1,1,1,1,1,1,2,1,1,1,1,1,1,1,1,2,3,2,1,1,1,1,2,2,2,1,1,1,2,2,4,3,2,

%T 1,1,1,1,3,2,4,3,2,1,1,1,2,2,4,4,4,3,2,1,1,1,1,2,3,5,3,5,3,2,1,1,1,2,

%U 3,5,5,8,5,5,3,2,1,1,1,1,2,3,5,5,8,5,5,3,2,1,1,1,2,2,4,5,7,8,8,5,5,3

%N Triangle read by rows: T(n, k) is the number of integer multisets of size k (partitions of k) that match the multiplicity multiset of some partition of n (n >= 1, 1 <= k <= n).

%C The relevant partitions of n have exactly k parts.

%C Column k is k-periodic from n = k*(k+1)/2.

%H Alois P. Heinz, <a href="/A337584/b337584.txt">Rows n = 1..125</a> (first 71 rows from Álvar Ibeas)

%H Álvar Ibeas, <a href="/A337584/a337584.txt">First 30 rows</a>

%F If k > (2*n+1)/3, T(n, k) = A088887(n - k).

%F If n >= k*(k+1)/2, T(n, k) = Sum_{d | gcd(n, k)} A000837(k/d).

%F T(n, k) = A000041(k) iff k|n and n >= k*(k+1)/2.

%e There is no partition of 5 with multiplicity multiset (3) or (1, 1, 1).

%e Indeed, both (2 = A008284(5, 3)) partitions of 5 into 3 parts (namely, (3, 1, 1) and (2, 2, 1)) have multiplicities (2, 1). Therefore, T(5, 3) = 1.

%e Triangle begins:

%e   k:  1 2 3 4 5 6 7 8 9 10

%e       --------------------

%e n=1:  1

%e n=2:  1 1

%e n=3:  1 1 1

%e n=4:  1 2 1 1

%e n=5:  1 1 1 1 1

%e n=6:  1 2 3 2 1 1

%e n=7:  1 1 2 2 2 1 1

%e n=8:  1 2 2 4 3 2 1 1

%e n=9:  1 1 3 2 4 3 2 1 1

%e n=10: 1 2 2 4 4 4 3 2 1 1

%p b:= proc(n,i) option remember; `if`(n=0 or i=1, `if`(n=0, {[]}, {[n]}),

%p      {b(n, i-1)[], seq(map(x-> sort([x[], j]), b(n-i*j, i-1))[], j=1..n/i)})

%p     end:

%p T:= n-> (p-> seq(coeff(p, x, i), i=1..n))(add(x^add(i, i=t), t=b(n$2))):

%p seq(T(n), n=1..20);  # _Alois P. Heinz_, Aug 17 2021

%Y Cf. A000041, A008284, A088887 (row sums).

%Y T(2n,n) gives A344680.

%K nonn,tabl

%O 1,8

%A _Álvar Ibeas_, Sep 02 2020