A317086 Number of normal integer partitions of n whose sequence of multiplicities is a palindrome.

1, 1, 1, 2, 1, 1, 3, 1, 2, 2, 4, 1, 4, 1, 4, 5, 4, 1, 7, 1, 8, 6, 6, 1, 10, 5, 7, 8, 11, 1, 20, 1, 9, 12, 9, 13, 25, 1, 10, 17, 21, 1, 37, 1, 21, 36, 12, 1, 44, 16, 23, 30, 33, 1, 53, 17, 55, 38, 15, 1, 103, 1, 16, 95, 51, 28, 69, 1, 73, 57, 82

Offset: 0

Author

Gus Wiseman, Jul 21 2018

Keywords

• nonn
• nice

Comments

A partition is normal if its parts span an initial interval of positive integers.

a(n) = 1 if and only if n = 0, 1, 2, 4 or a prime > 3. - Chai Wah Wu, Jun 22 2020

From David A. Corneth, Jul 08 2020: (Start)

Let [f_1, f_2, ,..., f_i, ..., f_m] be the multiplicities of parts i in a partition of Sum_{i=1..m} (f_i * i). Then, as the sequence of multiplicities is a palindrome, we have f_1 = f_m, ..., f_i = f_(m+1-i). So the sum is f_1 * (1 + m) + f_2 * (2 + m-1) + ... + f_(floor(m/2)) * m/2 (the last term depending on the parity of m.). This way it becomes a list of Diophantine equations for which we look for the number of solutions.

For example, for m = 4 we look for solutions to the Diophantine equation 5 * (c + d) = n where c, d are positive integers >= 1. A similar technique is used in A254524. (End)

Examples

			The a(20) = 8 partitions:
  (44432111), (44332211), (43332221),
  (3333221111), (3332222111), (3322222211), (3222222221),
  (11111111111111111111).
		

Links

• David A. Corneth, Table of n, a(n) for n = 0..9999 (first 215 terms from Chai Wah Wu)
• Wikipedia, Palindrome

Crossrefs

Cf. A000009, A000041, A000837, A025065, A055932, A242414, A254524, A317085, A317087.

Programs

Mathematica

Table[Length[Select[IntegerPartitions[n],And[Union[#]==Range[First[#]],Length/@Split[#]==Reverse[Length/@Split[#]]]&]],{n,30}]

Python

from sympy.utilities.iterables import partitions
from sympy import integer_nthroot, isprime
def A317086(n):
    if n > 3 and isprime(n):
        return 1
    else:
        c = 1
        for d in partitions(n,k=integer_nthroot(2*n,2)[0],m=n*2//3):
            l = len(d)
            if l > 0:
                k = max(d)
                if l == k:
                    for i in range(k//2):
                        if d[i+1] != d[k-i]:
                            break
                    else:
                        c += 1
        return c # Chai Wah Wu, Jun 22 2020