A263982 Number of partitions of n with a palindromicity of 3.

1, 1, 3, 4, 8, 10, 18, 22, 36, 44, 67, 81, 119, 142

Offset: 6

Author

Gregory L. Simay, Nov 01 2015

Keywords

• nonn
• more

Comments

A non-strict partition of n can be arranged into two consecutive sequences that are mirror images of each other, separated by a central sequence of k distinct summands, k>=0. Strict partitions (ref. A000009) only have the central sequence. In both cases, the palindromicity of the partition is k.

Palindromic partitions (ref. A025065) have palindromicity 0 (no central summand) or 1 (central summand). Non-palindromic partitions have palindromicities >=2.

Examples

			If considered unordered rather than a nonincreasing sequence, the partition 9,7,7,5,4,4,2,2,1,1,1 can be arranged as 7,4,2,1 [central sequence of 9,5,1] 1,2,4,7. Therefore the palindromicity of this particular partition is 3.
w(14,3) = w(11,2) + w(11,3) = A014153(4) + 10 = 26 + 10 = 36.
w(15,3) = w(2*8-1,3) = A014153(4) + A014153(3) + A014153(1) + A014153(0) = 26 + 14 + 3 + 1 = 44.
		

Crossrefs

Cf. A000070, A014153, A025065.

Formula

p(n,k) = number of partitions of n with palindromicity k.

If k*(k+1)/2 <= p(n) < (k+1)*(k+2)/2, then p(n) = p(n,0) + .. + p(n,k)

Let q(n,k)= number of strict partitions of n (ref. A000009) with exactly k parts. Then p(n,k) = Sum_{j>=0} q(n-2j,k)*p(j), which affords another way to demonstrate that the convolution of q(2n-j) with p(j) equals p(2n).

p(2n,0) = p(n) and p(2n+1,0) = 0 (ref. A025065).

p(2n,1) = p(2n+1,1) = A000070(n-1), the first partial sum of A000041 (ref. A025065).

p(2n,2) = p(2n-1,2) = A014153(n-2), the second partial sum of A000041.

p(2n,3) = A014153(n-3) + A014153(n-5) + A014153(n-6) + A014153(n-8) + A014153(n-9) + A014153(n-11) + A014153(n-12) + ...

G.f. for p(2n,3): p(x)* x^3*(1+x+x^2+x^3)/(1-x)*(1-x^2)*(1-x^3) where p(x) is the g.f. for A000041.

p(2n-1,3) = A014153(n-4) + A014153(n-5) + A014153(n-7) + A014153(n-8) + A014153(n-10) + A014153(n-11) + A014153(n-13) + A014153(n-14) + ...

G.f. for p(2n-1,3): p(x)* x^3*(1+2x+x^2)/(1-x)*(1-x^2)*(1-x^3) where p(x) is the g.f. for A000041.

More generally, p(n,k>=3) = p(n-k,k-1) + p(n-2k, k-1) + p(n-3k,n-1) + ... for k>=3 = p(n-k, k-1) + p(n-k,k).