Revision History for A225922
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Showing entries 1-10
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| A225922
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a(n) is the least k such that f(a(n-1)+1) + ... + f(k) > f(a(n-2)+1) + ... + f(a(n-1)) for n > 1, where f(n) = 1/(n+7) and a(1) = 1.
(history;
published version)
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#21 by Max Alekseyev at Thu Jan 27 20:45:49 EST 2022
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#20 by Max Alekseyev at Thu Jan 27 20:45:47 EST 2022
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| DATA
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1, 16, 58, 176, 507, 1436, 4043, 11359, 31890, 89506, 251193, 704933, 1978258, 5551574, 15579326, 43720081, 122691130, 344306598, 966223316, 2711500429, 7609249779, 21353742568, 59924740904, 168166051476, 471922288540, 1324349620283, 3716505787761, 10429583743512
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| COMMENTS
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The sequence does not satisfy any linear recurrence of order below 50, which suggests it's unlikely to exist. - Max Alekseyev, Jan 27 2022
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| FORMULA
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For n>=3, a(n) = ceiling( (a(n-1)+7.5)^2 / (a(n-2)+7.5) - 7.5 ) unless the fractional part of the number inside ceiling() is very small (~ 1/a(n-2)). - Max Alekseyev, Jan 27 2022
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| KEYWORD
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nonn,more
nonn
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| EXTENSIONS
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Terms a(20) on from Max Alekseyev, Jan 27 2022
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approved
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#19 by Susanna Cuyler at Sun Jun 14 09:43:43 EDT 2020
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#18 by Joerg Arndt at Sun Jun 14 03:49:26 EDT 2020
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#17 by Michel Marcus at Sun Jun 14 02:38:47 EDT 2020
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#16 by Michel Marcus at Sun Jun 14 02:38:45 EDT 2020
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| EXAMPLE
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Successive values of a(n) yield a chain:: 1 < 1/(1+8) + ... + 1/(15+7) < 1/(15+8) + ... + 1/(58+7) < 1/(58+8) + ... + 1/(176+7) < ...
1 < 1/(1+8) + ... + 1/(15+7) < 1/(15+8) + ... + 1/(58+7) < 1/(58+8) + ... + 1/(176+7) < ...
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proposed
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#15 by Jinyuan Wang at Sun Jun 14 02:07:54 EDT 2020
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#14 by Jinyuan Wang at Sun Jun 14 02:07:10 EDT 2020
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| NAME
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a(n) = ) is the least k such that f(a(n-1)+1) + ... + f(k) > f(a(n-2)+1) + ... + f(a(n)--1) )) for n > 1, where f(n) = 1/(n+7) and a(1) = 1.
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| DATA
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1, 16, 58, 176, 507, 1436, 4043, 11359, 31890, 89506, 251193, 704933, 1978258, 5551574, 15579326, 43720081, 122691130, 344306598, 966223316
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| EXAMPLE
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a(1) = 1 by decree; a(2) = 15 because 1/9 + ... + 1/21 < 1 < 1/9 + ... + 1/(15+7), so that a(3) = 58 because 1/23 + ... + 1/57 < 1/9 + ... + 1/22 < 1/23 + ... + 1/(58+7).
1/8 + ... + 1/14 < 1 < 1/8 + ... + 1/15, so that a(3) = 58 because
1/16 + ... + 1/57 < 1/8 + ... + 1/15 < 1/16 + ... + 1/58.
1 < 1/(1/+8 + ... + ) + ... + 1//(15 < +7) < 1/16 + ... + /(15+8) + ... + 1//(58 < +7) < 1/59 + ... + /(58+8) + ... + 1//(176 < ...+7) < ...
Abbreviating this chain as b(1) = 1 < b(2) < b(3) < b(4) < ... < R = 2.806380628..., it appears that lim _{n->infinity} b(n) = log(R) = 1.03186...... .
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| EXTENSIONS
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a(19) from Jinyuan Wang, Jun 14 2020
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approved
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#13 by T. D. Noe at Fri Jun 14 13:25:54 EDT 2013
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#12 by T. D. Noe at Fri Jun 14 13:25:51 EDT 2013
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