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A335402
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Numbers m such that the only normal integer partition of m whose run-lengths are a palindrome is (1)^m.
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2
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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
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1,3
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COMMENTS
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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.
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)
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LINKS
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FORMULA
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n is a term if and only if n = 0, 1, 2, 4 or a prime > 3. - Chai Wah Wu, Jun 22 2020
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EXAMPLE
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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.
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MATHEMATICA
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Select[Range[0, 30], Length[Select[IntegerPartitions[#], And[Or[#=={}, Union[#]==Range[First[#]]], Length/@Split[#]==Reverse[Length/@Split[#]]]&]]==1&]
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PROG
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(Python)
# from definition
from sympy.utilities.iterables import partitions
from sympy import integer_nthroot
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:
(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
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CROSSREFS
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Palindromic-multiplicity partitions are counted by A317085.
Normal integer partitions are counted by A000009.
Heinz numbers of normal palindromic-multiplicity partitions are A317087.
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KEYWORD
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nonn,easy
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AUTHOR
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EXTENSIONS
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STATUS
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approved
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