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A335402 Numbers m such that the only normal integer partition of m whose run-lengths 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 (list; graph; refs; listen; history; text; internal format)
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 k-1 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 = p-1 > 1.

Consider the partition covering 1,..,k where the multiplicity is 1 except for 1 and k where the multiplicity is q-k/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 q-2 > 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[k-i]:

                        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.

Palindromic-multiplicity partitions are counted by A317085.

Normal integer partitions are counted by A000009.

Heinz numbers of normal palindromic-multiplicity 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

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Last modified September 28 14:44 EDT 2022. Contains 357073 sequences. (Running on oeis4.)