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A158973
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a(n) = count of numbers k <= n such that all proper divisors of k are divisors of n.
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3
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1, 2, 3, 4, 4, 6, 5, 7, 6, 7, 6, 11, 7, 9, 9, 10, 8, 12, 9, 13, 11, 11, 10, 17, 11, 12, 12, 14, 11, 18, 12, 16, 14, 14, 14, 20, 13, 15, 15, 20, 14, 20, 15, 19, 20, 17, 16, 25, 17, 20, 18, 20, 17, 23, 19, 24, 19, 19, 18, 29, 19, 21, 24, 24, 21, 25, 20, 24, 22, 27, 21, 32, 22, 24, 26
(list; graph; refs; listen; history; internal format)
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OFFSET
| 1,2
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COMMENTS
| For primes p, a(p) = A036234(p) = A000720(p) + 1.
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FORMULA
| a(n) = A000005(n) + A004788(n). [From Vladeta Jovovic (vladeta(AT)eunet.yu), Apr 07 2009]
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EXAMPLE
| For n = 8 we have the following proper divisors for k <= n: {1}, {1}, {1}, {1, 2}, {1}, {1, 2, 3}, {1}, {1, 2, 4}. Only k = 6 has a proper divisor that is not a divisor of 8, viz. 3. Hence a(8) = 7.
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PROG
| (MAGMA) [ #[ k: k in [1..n] | forall(t){ d: d in Divisors(k) | d eq k or d in Divisors(n) } ]: n in [1..75] ];
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CROSSREFS
| Cf. A000040, A036234, A000720.
Sequence in context: A008328 A091860 A181833 * A071323 A071324 A063655
Adjacent sequences: A158970 A158971 A158972 * A158974 A158975 A158976
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KEYWORD
| nonn
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AUTHOR
| Jaroslav Krizek (jaroslav.krizek(AT)atlas.cz), Apr 01 2009
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EXTENSIONS
| Edited and extended by Klaus Brockhaus (klaus-brockhaus(AT)t-online.de), Apr 06 2009
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