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Search: id:A141829
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| A141829 |
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a(n) = the number of positive divisors of (p(n)-1) that are each <= p(n+1)-p(n), where p(n) is the nth prime. |
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+0
3
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| 1, 2, 2, 3, 2, 4, 2, 3, 2, 2, 5, 4, 2, 3, 2, 3, 2, 6, 3, 2, 5, 3, 2, 4, 4, 2, 3, 2, 4, 6, 3, 3, 2, 4, 2, 5, 5, 3, 2, 3, 2, 8, 2, 4, 2, 6, 7, 3, 2, 4, 3, 2, 8, 3, 3, 2, 2, 5, 4, 2, 4, 3, 3, 2, 4, 3, 5, 7, 2, 4, 3, 2, 4, 5, 3, 2, 3, 4, 5, 6, 2, 8, 2, 5, 3, 2, 5, 4, 2, 3, 2, 2, 3, 4, 3, 2, 3, 2, 6, 6, 5, 3, 2, 2, 5
(list; graph; listen)
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OFFSET
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1,2
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COMMENT
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a(n) also equals the number of positive integers k, k <= p(n+1)-p(k), that divide (p(n)+k-1).
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EXAMPLE
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The 16th prime is 53, and the 17th prime is 59. So the divisors of 53-1=52 that are <= 59-53=6 are 1,2,4. There are three such divisors.
Also, 53 is divisible by 1. 54 is divisible by 2. 55 is not divisible by 3. 56 is divisible by 4. 57 is not divisible by 5. And 58 is not divisible 6. So in the span of integers p(16)=53 to p(17)-1=58, there are 3 integers k where k divides (p(16)+k-1). So a(16) = 3.
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MAPLE
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A141829 := proc(n) local p, q, a, d ; p := ithprime(n) ; q := nextprime(p) ; a := 0 ; for d in numtheory[divisors](p-1) do if d <= q-p then a :=a+1 ; fi; od: RETURN(a) ; end: for n from 1 to 200 do printf("%a, ", A141829(n)) ; od: [From R. J. Mathar (mathar(AT)strw.leidenuniv.nl), Aug 08 2008]
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CROSSREFS
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Cf. A141830, A141831.
Sequence in context: A035213 A083901 A038148 this_sequence A111336 A083902 A106491
Adjacent sequences: A141826 A141827 A141828 this_sequence A141830 A141831 A141832
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KEYWORD
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nonn
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AUTHOR
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Leroy Quet (q1qq2qqq3qqqq(AT)yahoo.com), Jul 09 2008
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EXTENSIONS
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Extended beyond a(17) by R. J. Mathar (mathar(AT)strw.leidenuniv.nl), Aug 08 2008
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