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A317086
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Number of normal integer partitions of n whose sequence of multiplicities is a palindrome.
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15
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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
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OFFSET
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0,4
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COMMENTS
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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
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)
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LINKS
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EXAMPLE
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The a(20) = 8 partitions:
(44432111), (44332211), (43332221),
(3333221111), (3332222111), (3322222211), (3222222221),
(11111111111111111111).
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MATHEMATICA
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Table[Length[Select[IntegerPartitions[n], And[Union[#]==Range[First[#]], Length/@Split[#]==Reverse[Length/@Split[#]]]&]], {n, 30}]
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PROG
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(Python)
from sympy.utilities.iterables import partitions
from sympy import integer_nthroot, isprime
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
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CROSSREFS
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KEYWORD
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nonn,nice
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AUTHOR
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STATUS
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approved
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