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A073161
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Nested floor product of n and fractions (p_k + 1)/p_k where p_k is k-th prime.
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0
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1, 4, 6, 10, 12, 20, 21, 26, 28, 36, 38, 46, 50, 56, 57, 64, 69, 80, 81, 87, 92, 99, 104, 112, 116, 122, 126, 133, 140, 146, 147, 155, 169, 172, 176, 185, 188, 198, 202, 210, 213, 222, 228, 237, 242, 253, 254, 264, 270, 279, 280, 289, 298, 303, 305, 318, 323
(list; graph; refs; listen; history; internal format)
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OFFSET
| 1,2
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COMMENTS
| Sequence consists only of composite numbers.
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FORMULA
| a(n)=[...[[[[n(1+1/2)](1+1/3)](1+1/5)]...(1+1/p_k)]..., where [x] = floor of x and p_k is the k-th prime; this infinite nested floor product will eventually level-off at a(n). Alternatively, for n>=1, let b(0, n)=n and b(k, n) = floor( b(k-1, n)*(1 + 1/p_k) ), for k=1..m, then a(n)=b(m, n) where m is smallest integer such that p_m > a(n).
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EXAMPLE
| a(4)=10 since [[[[4(1+1/2)](1+1/3)](1+1/5)](1+1/7)]= [[[6(1+1/3)](1+1/5)](1+1/7)]=[[8(1+1/5)](1+1/7)]=[9(1+1/7)]=10.
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CROSSREFS
| Sequence in context: A086298 A144761 A122171 * A177929 A086493 A190370
Adjacent sequences: A073158 A073159 A073160 * A073162 A073163 A073164
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KEYWORD
| easy,nonn
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AUTHOR
| Paul D. Hanna (pauldhanna(AT)juno.com), Jul 29 2002
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