

A335402


Numbers m such that the only normal integer partition of m whose runlengths are a palindrome is (1)^m.


2



0, 1, 2, 4, 5, 7, 11, 13, 17, 19, 23, 29, 31, 37, 41, 43, 47, 53, 59, 61, 67, 71, 73, 79, 83, 89, 97, 101, 103, 107, 109, 113, 127, 131, 137, 139, 149, 151, 157, 163, 167, 173, 179, 181, 191, 193, 197, 199, 211, 223, 227, 229, 233, 239, 241, 251, 257, 263, 269
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OFFSET

1,3


COMMENTS

An integer partition is normal if it covers an initial interval of positive integers.
Conjecture: The sequence consists of 0, 1, 4, and all primes except 3.
From Chai Wah Wu, Jun 22 2020: (Start)
The above conjecture is true.
Proof: The cases of 0, 1, 4 can be checked by inspection. Next we show that if n is prime and not equal to 3, then n is a term. Let n be prime and consider a palindromic normal partition of n covering the integers 1,...,k with k > 1. Then the multiplicity of 1 and k are the same and the multiplicities of 2 and k1 are the same, etc.
If k is even, then n is of the form (k+1)r. Since n is prime, this implies that n = k+1. Since n >= k(k+1)/2. this means that k = 2 and n = 3.
If k is odd, then n is of the form (k+1)r + w(k+1)/2. Let m = (k+1)/2, then n = m(2r+w). Since n is prime and r,w > 0, this means that m = 1, k = 1, a contradiction.
Next we show that if n is composite and not equal to 4, then n is not a term.
Suppose n = pq for 1 < p <= q. If p is odd, let k = p1 > 1.
Consider the partition covering 1,..,k where the multiplicity is 1 except for 1 and k where the multiplicity is qk/2 + 1 > 0. This is a normal palindromic partition summing up to pq = n.
If p is even, without loss of generality we can choose p = 2. Since n != 4, q >= 3. In this case, choosing k = 3 with 1 and 3 having multiplicity 1 and 2 having multiplicity q2 > 0 results in a normal palindromic partition of 2q = n. QED
It is clear that if n is not a term, then any multiple of n is also not a term.
(End)


LINKS

Table of n, a(n) for n=1..59.


FORMULA

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


EXAMPLE

There are 4 normal integer partitions of 10 whose sequence of multiplicities is a palindrome, namely (4321), (33211), (32221), (1111111111), so 10 does not belong to the sequence. The normal integer partitions of 7 are (3211), (2221), (22111), (211111), (1111111), none of which has palindromic multiplicities except the last, so 7 belongs to the sequence.


MATHEMATICA

Select[Range[0, 30], Length[Select[IntegerPartitions[#], And[Or[#=={}, Union[#]==Range[First[#]]], Length/@Split[#]==Reverse[Length/@Split[#]]]&]]==1&]


PROG

(Python)
# from definition
from sympy.utilities.iterables import partitions
from sympy import integer_nthroot
A335402_list = []
for m in range(0, 101):
for d in partitions(m, k=integer_nthroot(2*m, 2)[0]):
l = len(d)
if l > 0 and not(l == 1 and 1 in d):
k = max(d)
if l == k:
for i in range(k//2):
if d[i+1] != d[ki]:
break
else:
break
else:
A335402_list.append(m) # Chai Wah Wu, Jun 22 2020
(Python)
# from formula
from sympy import prime
A335402_list = [0, 1, 2, 4] + [prime(i) for i in range(3, 100)] # Chai Wah Wu, Jun 22 2020


CROSSREFS

Positions of 1's in A317086.
Palindromicmultiplicity partitions are counted by A317085.
Normal integer partitions are counted by A000009.
Heinz numbers of normal palindromicmultiplicity partitions are A317087.
Cf. A000041, A000837, A025065, A046022, A124010, A242414.
Sequence in context: A033160 A350147 A110924 * A192590 A028289 A307872
Adjacent sequences: A335399 A335400 A335401 * A335403 A335404 A335405


KEYWORD

nonn,easy


AUTHOR

Gus Wiseman, Jun 06 2020


EXTENSIONS

a(22)a(59) from Chai Wah Wu, Jun 22 2020


STATUS

approved



