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A309776
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Form a triangle: first row is n in base 2, next row is sums of pairs of adjacent digits of previous row, repeat until get a single number which is a(n).
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1
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0, 1, 1, 2, 1, 2, 3, 4, 1, 2, 4, 5, 4, 5, 7, 8, 1, 2, 5, 6, 7, 8, 11, 12, 5, 6, 9, 10, 11, 12, 15, 16, 1, 2, 6, 7, 11, 12, 16, 17, 11, 12, 16, 17, 21, 22, 26, 27, 6, 7, 11, 12, 16, 17, 21, 22, 16, 17, 21, 22, 26, 27, 31, 32, 1, 2, 7, 8, 16, 17, 22, 23, 21, 22
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OFFSET
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0,4
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COMMENTS
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a(n) = 1 occurs at n = 2^k for nonnegative integers k.
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LINKS
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FORMULA
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a(2^k + 1) = 2 for k >= 1 where 2^k+1 = 1000..0001_2.
a(2^k - 1) = 2^(k-1) for k >= 2 where 2^k-1 = 111..111_2.
a((4^k-1)/3) = 2^(2*k-3) for k >= 2 where (4^k-1)/3 = 10101..0101_2.
(End)
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EXAMPLE
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For n=5 the triangle is
1 0 1
1 1
2
so a(5)=2.
For n=14 we get
1 1 1 0
2 2 1
4 3
7
so a(14)=7.
For n=26=11010_2; (n1+n2, n2+n3, n3+n4, n4+n5) = 2111; (n1'+n2', n2'+n3', n3'+n4') = 322; (n1''+n2'', n2''+n3'') = 54; (n1'''+n2''') = 9; a(26)= 9.
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PROG
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(PARI) a(n) = my (b=binary(n)); sum(k=1, #b, b[k]*binomial(#b-1, k-1)) \\ Rémy Sigrist, Aug 20 2019
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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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EXTENSIONS
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STATUS
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approved
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