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A372286
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a(n) = number of terms of A086893 that are in the interval [n, A371094(n)].
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6
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1, 5, 2, 4, 2, 7, 1, 2, 2, 4, 2, 3, 2, 6, 1, 3, 1, 4, 2, 3, 2, 9, 1, 2, 1, 3, 1, 2, 2, 4, 2, 2, 2, 4, 2, 3, 2, 6, 2, 3, 2, 4, 2, 3, 2, 5, 2, 3, 2, 4, 2, 3, 2, 8, 1, 2, 1, 3, 1, 3, 1, 5, 1, 3, 1, 3, 1, 3, 1, 6, 1, 3, 2, 4, 2, 3, 2, 5, 2, 3, 2, 4, 2, 3, 2, 11, 1, 2, 1, 3, 1, 2, 1, 4, 1, 2, 1, 3, 1, 2, 1, 5, 1, 2, 1, 3
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OFFSET
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0,2
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COMMENTS
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a(n) is the number of nonzero terms of A096773 that are found in the interval [n, A371094(n)], inclusive.
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LINKS
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EXAMPLE
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A086893 = 1, 3, 5, 13, 21, 53, 85, 213, 341, 853, 1365, 3413, 5461, ...
For n=0, A371094(0) = 1, and only the initial term of A086893 is in the interval [0, 1], therefore a(0) = 1.
For n=1, A371094(1) = 21, so the first five terms of A086893 are in the interval [1, 21], thus a(1) = 5.
For n=6, A371094(6) = 19, and in the interval [6, 19] there is only term 13 of A086893, thus a(6) = 1.
For n=35, A371094(35) = 213, so we count terms 53, 85, 213 of A086893, therefore a(35) = 3.
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PROG
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(PARI)
A371094(n) = { my(m=1+3*n, e=valuation(m, 2)); ((m*(2^e)) + (((4^e)-1)/3)); };
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CROSSREFS
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Cf. A002450 (seems to give the positions of records).
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KEYWORD
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nonn
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AUTHOR
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STATUS
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approved
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