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A137397
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Number of distinct palindromic subwords in the binary representation of n.
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2
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2, 2, 3, 3, 4, 4, 4, 4, 5, 5, 5, 5, 5, 5, 5, 5, 6, 6, 6, 6, 6, 6, 6, 6, 6, 6, 6, 6, 6, 6, 6, 6, 7, 7, 7, 7, 7, 7, 7, 7, 7, 7, 7, 7, 7, 7, 7, 7, 7, 7, 7, 7, 7, 7, 7, 7, 7, 7, 7, 7, 7, 7, 7, 7, 8, 8, 8, 8, 8, 8, 8, 8, 8, 8, 8, 8, 8, 8, 8, 8, 8, 8, 8, 8, 8, 8, 8, 8, 8, 8, 8, 8, 8, 8, 8, 8, 8, 8, 8, 8, 8, 8, 8, 8, 8
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OFFSET
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0,1
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COMMENTS
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Equals A070941 from a(1) to a(202) and continues a(203)=8, a(204)=a(205)=9.
Omitting "distinct" in the definition, we get 1, 2, 4, 4, 7, 7, 7, 7, 11, 11, 11, 11, 11, 11, 11, 11, 16, 16,... which apparently is built by repeating entries of A000124 in blocks of length 2,4,8,16,32..
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LINKS
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A. Glen, J. Justin, S. Widmer, and L. Q. Zamboni, Palindromic Richness, arXiv:0801.1656 [math.CO], 2008.
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EXAMPLE
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For n=10 the binary representation is A007088(10)=1010, which contains the a(10)=5 palindromic substrings {}, {0}, {1}, {101}, {010}. The empty subword is always included in the count.
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PROG
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(Python)
def ispal(s): return s == s[::-1]
def a(n):
s = bin(n)[2:]
return 1 + len(set(s[i:j] for i in range(len(s))
for j in range(i+1, len(s)+1) if ispal(s[i:j])))
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CROSSREFS
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KEYWORD
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nonn,base
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AUTHOR
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STATUS
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approved
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