blocks|key|6503818|text|Mathematica+(没有内置Fibonacci的32个字符)|type|header-two|depth|inlineStyleRanges|entityRanges|data|6503819|Tr[1/⌈(.5%2B√5/2)%5ERange@#/√5-.5⌉]&|code-block|syntax|javascript|6503820|📷|atomic|offset|length|6503821|这里，我使用了Fibonacci数的舍入公式。|unstyled|6503822|6503823|其中φ是黄金比率。|style|CODE|entityMap|0|IMAGE|mutability|IMMUTABLE|imageUrl|https://i.stack.imgur.com/yWv3U.png|imageAlt|1|https://i.stack.imgur.com/lHbH6.png^0|0|0|0|1|0|0|0|0|1|1|0|2|1^^$0|@$1|2|3|4|5|6|7|13|8|@]|9|@]|A|$]]|$1|B|3|C|5|D|7|14|8|@]|9|@]|A|$E|F]]|$1|G|3|H|5|I|7|15|8|@]|9|@$J|16|K|17|1|18]]|A|$]]|$1|L|3|M|5|N|7|19|8|@]|9|@]|A|$]]|$1|O|3|H|5|I|7|1A|8|@]|9|@$J|1B|K|1C|1|1D]]|A|$]]|$1|P|3|Q|5|N|7|1E|8|@$J|1F|K|1G|R|S]]|9|@]|A|$]]]|T|$U|$5|V|W|X|A|$Y|Z|10|-4]]|11|$5|V|W|X|A|$Y|12|10|-4]]]]

<h1>Mathematica (32 characters without built-in Fibonacci)</h1>
<pre><code>Tr[1/⌈(.5+√5/2)^Range@#/√5-.5⌉]&amp;
</code></pre>
<p><img src="https://i.stack.imgur.com/yWv3U.png" alt="enter image description here" /></p>
<p>Here I used the rounding formula for Fibonacci numbers</p>
<p><img src="https://i.stack.imgur.com/lHbH6.png" alt="enter image description here" /></p>
<p>where <code>φ</code> is the golden ratio.</p>


blocks|key|5322174|text|Javascript，33|type|header-two|depth|inlineStyleRanges|entityRanges|data|5322175|输入：n+=+10|unstyled|offset|length|style|CODE|5322176|s=f=t=1;for(;n--;)t%2B=1/(s%2B=f=s-f)|code-block|syntax|javascript|5322177|根据我在Perl中的第一篇文章，这是直接来自Google控制台的结果。|5322178|📷|atomic|entityMap|0|IMAGE|mutability|IMMUTABLE|imageUrl|https://i.stack.imgur.com/OOZIq.png|imageAlt^0|0|3|6|0|0|0|0|1|0^^$0|@$1|2|3|4|5|6|7|10|8|@]|9|@]|A|$]]|$1|B|3|C|5|D|7|11|8|@$E|12|F|13|G|H]]|9|@]|A|$]]|$1|I|3|J|5|K|7|14|8|@]|9|@]|A|$L|M]]|$1|N|3|O|5|D|7|15|8|@]|9|@]|A|$]]|$1|P|3|Q|5|R|7|16|8|@]|9|@$E|17|F|18|1|19]]|A|$]]]|S|$T|$5|U|V|W|A|$X|Y|Z|-4]]]]

<h1>Javascript, 33</h1>

<p><em>Input:</em> <code>n = 10</code></p>

<pre><code>s=f=t=1;for(;n--;)t+=1/(s+=f=s-f)
</code></pre>

<p>Based in my first post in Perl, here is the result directly from the <em>Google Chrome Console</em>.</p>

<p><img src="https://i.stack.imgur.com/OOZIq.png" alt="enter image description here"></p>


blocks|key|5322539|text|K，22|type|header-two|depth|inlineStyleRanges|entityRanges|data|5322540|..。|unstyled|5322541|{%2B/%25x#x{x,%2B/-2#x}/1+1}|code-block|syntax|javascript|entityMap^0|0|0^^$0|@$1|2|3|4|5|6|7|K|8|@]|9|@]|A|$]]|$1|B|3|C|5|D|7|L|8|@]|9|@]|A|$]]|$1|E|3|F|5|G|7|M|8|@]|9|@]|A|$H|I]]]|J|$]]

blocks|key|6503732|text|GTB，35|type|header-two|depth|inlineStyleRanges|entityRanges|offset|length|data|6503733|运行在TI-84计算器上。|unstyled|6503734|1→Y:0→X`N4;N,1,N~1/X:Y→Z:X%2BY→Y:Z→X&|code-block|syntax|javascript|6503735|将1存储在Y中，0存储在X中|unordered-list-item|style|CODE|6503736|以N为输入|6503737|初始化For循环|6503738|显示X|6503739|将Y存储在Z中，X%2BY存储在Y中|6503740|将Z存储在X中|6503741|结束For循环|entityMap|0|LINK|mutability|MUTABLE|url|http://timtechsoftware.com/?page_id=1309^0|0|3|0|0|0|0|1|1|5|1|8|1|C|1|0|1|1|0|3|3|0|2|1|0|1|1|5|1|8|3|E|1|0|1|1|5|1|0|2|3^^$0|@$1|2|3|4|5|6|7|19|8|@]|9|@$A|1A|B|1B|1|1C]]|C|$]]|$1|D|3|E|5|F|7|1D|8|@]|9|@]|C|$]]|$1|G|3|H|5|I|7|1E|8|@]|9|@]|C|$J|K]]|$1|L|3|M|5|N|7|1F|8|@$A|1G|B|1H|O|P]|$A|1I|B|1J|O|P]|$A|1K|B|1L|O|P]|$A|1M|B|1N|O|P]]|9|@]|C|$]]|$1|Q|3|R|5|N|7|1O|8|@$A|1P|B|1Q|O|P]]|9|@]|C|$]]|$1|S|3|T|5|N|7|1R|8|@$A|1S|B|1T|O|P]]|9|@]|C|$]]|$1|U|3|V|5|N|7|1U|8|@$A|1V|B|1W|O|P]]|9|@]|C|$]]|$1|W|3|X|5|N|7|1X|8|@$A|1Y|B|1Z|O|P]|$A|20|B|21|O|P]|$A|22|B|23|O|P]|$A|24|B|25|O|P]]|9|@]|C|$]]|$1|Y|3|Z|5|N|7|26|8|@$A|27|B|28|O|P]|$A|29|B|2A|O|P]]|9|@]|C|$]]|$1|10|3|11|5|N|7|2B|8|@$A|2C|B|2D|O|P]]|9|@]|C|$]]]|12|$13|$5|14|15|16|C|$17|18]]]]

<h2><a href="http://timtechsoftware.com/?page_id=1309" rel="nofollow" title="GTB">GTB</a>, 35</h2>

<p><em>Runs on a TI-84 calculator</em></p>

<pre><code>1→Y:0→X`N4;N,1,N~1/X:Y→Z:X+Y→Y:Z→X&amp;
</code></pre>

<ul>
<li>Store <code>1</code> in <code>Y</code> and <code>0</code> in <code>X</code></li>
<li>Take <code>N</code> as input</li>
<li>Initialize a <code>For</code> loop</li>
<li>Display <code>X</code></li>
<li>Store <code>Y</code> in <code>Z</code> and <code>X+Y</code> in <code>Y</code></li>
<li>Store <code>Z</code> in <code>X</code></li>
<li>End the <code>For</code> loop</li>
</ul>


blocks|key|6503845|text|BC+-+116|type|header-two|depth|inlineStyleRanges|entityRanges|data|6503846|define+f(n){if(n+<+2){return+n;}else{return+f(n-1)%2Bf(n-2);}}
define+r(n){for(i=1;i<n;%2B%2Bi){m=m%2B(1/f(i));};return+m;}|code-block|syntax|javascript|6503847|用n调用r(n)来求解。精度可以任意设定，所以这是最精确的解决方案。|unstyled|entityMap^0|0|0^^$0|@$1|2|3|4|5|6|7|K|8|@]|9|@]|A|$]]|$1|B|3|C|5|D|7|L|8|@]|9|@]|A|$E|F]]|$1|G|3|H|5|I|7|M|8|@]|9|@]|A|$]]]|J|$]]

<h1>BC - 116</h1>
<pre><code>define f(n){if(n &lt; 2){return n;}else{return f(n-1)+f(n-2);}}
define r(n){for(i=1;i&lt;n;++i){m=m+(1/f(i));};return m;}
</code></pre>
<p>call r(n) with n to solve. The precision can be set arbitrarily, so this is the most accurate solution.</p>


blocks|key|5322206|text|Forth，64|type|header-two|depth|inlineStyleRanges|entityRanges|data|5322207|:+r+1+1+rot+1e+0+do+2dup+-+nip+tuck+%2B+dup+s>f+1/f+f%2B+swap+loop+;|code-block|syntax|javascript|5322208|用法：|unstyled|5322209|10+r+f.+3.34170499581934++ok
100+r+f.+3.35988566624318++ok
1000+r+f.+3.35988566624318++ok|entityMap^0|0|0|0^^$0|@$1|2|3|4|5|6|7|M|8|@]|9|@]|A|$]]|$1|B|3|C|5|D|7|N|8|@]|9|@]|A|$E|F]]|$1|G|3|H|5|I|7|O|8|@]|9|@]|A|$]]|$1|J|3|K|5|D|7|P|8|@]|9|@]|A|$E|F]]]|L|$]]

<h2>Forth, 64</h2>

<pre><code>: r 1 1 rot 1e 0 do 2dup - nip tuck + dup s&gt;f 1/f f+ swap loop ;
</code></pre>

<p>usage:</p>

<pre><code>10 r f. 3.34170499581934  ok
100 r f. 3.35988566624318  ok
1000 r f. 3.35988566624318  ok
</code></pre>


blocks|key|6504460|text|RPL+(HP+49G/G%2B/50g)，50字节|type|header-two|depth|inlineStyleRanges|entityRanges|data|6504461|«+0.+1.+DUP2+5.+ROLL
++START+SWAP+OVER+%2B+ROT+OVER+INV+%2B+UNROT
++NEXT+DROP2
»|code-block|syntax|javascript|6504462|示例：|unstyled|6504463|10个->+3.33046904076|6504464|11+->+3.34170499582|6504465|30+->+3.35988372158|6504466|55+->+3.35988566622|6504467|理论上，前55个数字之和将给出12个正确数字。最后一个数字应该是'4‘而不是'2’。这应该通过向后加法来解决，但是代码会稍微长一些。|6504468|如果使用该定义将常数计算到小数点1000位，那么和应该至少执行到第一个4790项，如果可能的话，这将在HP+50g上花费太多的时间。对于此任务，需要一种更有效的方法，例如在以下RPL程序中使用的方法(464字节，校验和#207+of+)，其参数是所需的十进制数：|6504469|%25%25HP:+T(3)A(R)F(.);
\<<+PUSH+RAD+-105+CF+-3+CF+R\->I+DUP+1+%2B+2+INV+SWAP+100+LN+*+5+LN+%2B+OVER+ASINH+/+\v/+*+CEIL+DUP+2+MOD+%2B+DUP+0+ROT
[[+1+1+]
+[+1+0+]]+SWAP+DUP2+%5E+3+GETI+UNROT+GET+4+ROLLD+4+ROLLD+DUP+%2B+%5E+1+GETI+UNROT+GET+1+%2B+0+1+8+ROLL+2+/
++START+PICK3+%2B+DUP+PICK3+*+NEG+6+PICK+SQ+%2B+/+4+PICK+SQ+*+EXPAND+ROT+PICK3+-+ROT+OVER+-+ROT+6+ROLL+6+ROLL+6+ROLL+%2B+LASTARG+*+LASTARG+5+ROLLD+5+ROLLD+/+%2B+ROT+PICK3+-+ROT+OVER+-+6+ROLL+6+ROLL+6+ROLL
++NEXT+ROT+%2B+INV+5+ROLL+%2B+EXPAND+4+ROLLD+3+DROPN+FXND+PICK3+ALOG+OVER+-+PICK3+*+SWAP+IQUOT+%2B+\->STR+DUP+HEAD+-51+FC?+{+"."+}+{+","+}+IFTE+%2B+SWAP+TAIL+%2B+1+ROT+2+%2B+SUB+POP
\>>

'RFC'+STO|6504470|1000+RFC+-->|6504471|3.3598856662431775531720113029189271796889051337319684864955538153251303189966833836154162164567900872970453429288539133041367890171008836795913517330771190785803335503325077531875998504871797778970060395645092153758927752656733540240331694417992939346109926262579646476518686594497102165589843608814726932495910794738736733785233268774997627277579468536769185419814676687429987673820969139012177220244052081510942649349513745416672789553444707777758478025963407690748474155579104200675015203410705335285129792635242062267537568055761955669720848843854407983324292851368070827522662579751188646464096737461572387236295562053612203024635409252678424224347036310363201466298040249015578724456176000319551987905969942029178866949174808096746523682654086938399069873211752166957063859411814553647364268782462926166650100098903804823359519893146150108288726392887669917149304053057745574321561167298985617729731395370735291966884327898022165047585028091806291002444277017460241040417786069190065037142835294|6504472|(22分23秒，在实际HP+50g上)|6504473|对于1000位数字，计算级数的前50项加上连续分数的另50项，每次只进行25次迭代(不过，每个迭代中有48个项就足够了)。一个等效的十进制基本程序在我的计算机上只需10毫秒(英特尔(R)+Core(TM)+Duo+E4700+@+2.6+GHz)。|entityMap^0|0|0|0|0|0|0|0|0|0|0|0|0|0^^$0|@$1|2|3|4|5|6|7|16|8|@]|9|@]|A|$]]|$1|B|3|C|5|D|7|17|8|@]|9|@]|A|$E|F]]|$1|G|3|H|5|I|7|18|8|@]|9|@]|A|$]]|$1|J|3|K|5|I|7|19|8|@]|9|@]|A|$]]|$1|L|3|M|5|I|7|1A|8|@]|9|@]|A|$]]|$1|N|3|O|5|I|7|1B|8|@]|9|@]|A|$]]|$1|P|3|Q|5|I|7|1C|8|@]|9|@]|A|$]]|$1|R|3|S|5|I|7|1D|8|@]|9|@]|A|$]]|$1|T|3|U|5|I|7|1E|8|@]|9|@]|A|$]]|$1|V|3|W|5|D|7|1F|8|@]|9|@]|A|$E|F]]|$1|X|3|Y|5|I|7|1G|8|@]|9|@]|A|$]]|$1|Z|3|10|5|I|7|1H|8|@]|9|@]|A|$]]|$1|11|3|12|5|I|7|1I|8|@]|9|@]|A|$]]|$1|13|3|14|5|I|7|1J|8|@]|9|@]|A|$]]]|15|$]]

<p><strong>RPL (HP 49G/G+/50g), 50 bytes</strong></p>

<pre><code>« 0. 1. DUP2 5. ROLL
  START SWAP OVER + ROT OVER INV + UNROT
  NEXT DROP2
»
</code></pre>

<p>Examples:</p>

<p>10 -> 3.33046904076</p>

<p>11 -> 3.34170499582</p>

<p>30 -> 3.35988372158</p>

<p>55 -> 3.35988566622</p>

<p>In theory, the sum of the first 55 digits would give 12 correct digits. The last digit should be '4' instead of '2'. This should be fixed by adding up the terms backwards, but the code would be slightly longer.</p>

<p>If the constant were to be computed to 1000 decimal places using the definition  the sum should be carried out to at least the first 4790 terms, which would take too much time on the HP 50g, if possible at all. For this task a more efficient method is required, like the one used in the following RPL program (464 bytes, checksum # 207Eh) whose argument is the desired number of decimal places: </p>

<pre><code>%%HP: T(3)A(R)F(.);
\&lt;&lt; PUSH RAD -105 CF -3 CF R\-&gt;I DUP 1 + 2 INV SWAP 100 LN * 5 LN + OVER ASINH / \v/ * CEIL DUP 2 MOD + DUP 0 ROT
[[ 1 1 ]
 [ 1 0 ]] SWAP DUP2 ^ 3 GETI UNROT GET 4 ROLLD 4 ROLLD DUP + ^ 1 GETI UNROT GET 1 + 0 1 8 ROLL 2 /
  START PICK3 + DUP PICK3 * NEG 6 PICK SQ + / 4 PICK SQ * EXPAND ROT PICK3 - ROT OVER - ROT 6 ROLL 6 ROLL 6 ROLL + LASTARG * LASTARG 5 ROLLD 5 ROLLD / + ROT PICK3 - ROT OVER - 6 ROLL 6 ROLL 6 ROLL
  NEXT ROT + INV 5 ROLL + EXPAND 4 ROLLD 3 DROPN FXND PICK3 ALOG OVER - PICK3 * SWAP IQUOT + \-&gt;STR DUP HEAD -51 FC? { "." } { "," } IFTE + SWAP TAIL + 1 ROT 2 + SUB POP
\&gt;&gt;

'RFC' STO
</code></pre>

<p>1000 RFC --></p>

<p>3.3598856662431775531720113029189271796889051337319684864955538153251303189966833836154162164567900872970453429288539133041367890171008836795913517330771190785803335503325077531875998504871797778970060395645092153758927752656733540240331694417992939346109926262579646476518686594497102165589843608814726932495910794738736733785233268774997627277579468536769185419814676687429987673820969139012177220244052081510942649349513745416672789553444707777758478025963407690748474155579104200675015203410705335285129792635242062267537568055761955669720848843854407983324292851368070827522662579751188646464096737461572387236295562053612203024635409252678424224347036310363201466298040249015578724456176000319551987905969942029178866949174808096746523682654086938399069873211752166957063859411814553647364268782462926166650100098903804823359519893146150108288726392887669917149304053057745574321561167298985617729731395370735291966884327898022165047585028091806291002444277017460241040417786069190065037142835294</p>

<p>(22 min 23 sec, on the real HP 50g)</p>

<hr>

<p>For one thousand digits the first 50 terms of the series plus another 50 terms of a continued fraction are evaluated, two at a time in only 25 iterations (48 terms of each would have sufficed, though). An equivalent DECIMAL BASIC program takes just 10 milliseconds on my computer (Intel(R) Core(TM) Duo CPU E4700 @ 2.6 GHz).</p>


blocks|key|5322617|text|Japt+-mx，6字节|type|header-two|depth|inlineStyleRanges|offset|length|style|CODE|entityRanges|data|5322618|1/MgUÄ|code-block|syntax|javascript|5322619|试试看|unstyled|entityMap|0|LINK|mutability|MUTABLE|url|https://ethproductions.github.io/japt/?v=1.4.6&code=MS9NZ1XE&input=MTAwMAotbXg=^0|5|3|0|0|0|3|0^^$0|@$1|2|3|4|5|6|7|U|8|@$9|V|A|W|B|C]]|D|@]|E|$]]|$1|F|3|G|5|H|7|X|8|@]|D|@]|E|$I|J]]|$1|K|3|L|5|M|7|Y|8|@]|D|@$9|Z|A|10|1|11]]|E|$]]]|N|$O|$5|P|Q|R|E|$S|T]]]]

<h1>Japt <code>-mx</code>, 6 bytes</h1>
<pre><code>1/MgUÄ
</code></pre>
<p><a href="https://ethproductions.github.io/japt/?v=1.4.6&amp;code=MS9NZ1XE&amp;input=MTAwMAotbXg=" rel="nofollow noreferrer">Try it</a></p>


blocks|key|5322816|text|C#+(.NET+Core)，66字节|type|header-two|depth|inlineStyleRanges|entityRanges|offset|length|data|5322817|a=>{double+r=1,x=1,y=1,i=0;for(;%2B%2Bi<a;y%2B=x,x=y-x)r%2B=1/y;return+r;}|code-block|syntax|javascript|5322818|在网上试试！|unstyled|5322819|不能使用浮标，因为不精确。|5322820|输入=8的示例：|5322821|双人:+3.28287545787546+(回合到3.282875轮)|unordered-list-item|5322822|浮动:+3.282876+(减去0.000001)|5322823|未高尔夫球：|5322824|a+=>+{
++++double+r+=+1,+++++++++++++++++++++++//+initialize+r+(sum+of+reciprocals)+-+start+at+1+to+save+one+byte+in+for+loop
++++++++++++x+=+1,+y+=+1,+++++++++++++++++++//+initialize+x+(2nd+largest+Fibonacci)+and+y+(largest+Fibonacci)
++++++++++++i+=+0;++++++++++++++++++++++++++//+initialize+i+(loop+counter)
++++for(;+%2B%2Bi+<+a;+y+%2B=+x,+x+=+y+-+x)+++//+from+1+to+a,+while+incrementing+the+Fibonacci+numbers
++++++++r+%2B=+1+/+y;+++++++++++++++++++++++++//+increment+the+reciprocal+Fibonacci
++++return+r;+++++++++++++++++++++++++++//+return+the+reciprocal+Fibonacci
}|entityMap|0|LINK|mutability|MUTABLE|url|https://www.microsoft.com/net/core/platform|1|https://tio.run/##NY3LCsIwFETX9ivuMqX1Ud0IaQQRBEE3KriO6a0EagI3qTSUfnsNlA4chsMsRrmlsoRj67T5wCM4j18OqpHOwRH6xHnptYJza1RZ2fbdYA5TH6AGMUpx6CcHEkXeRUJEiw2vLTGeZbqUPGSii1tYdillolgHTuhbMkB8GPl88rO6gpvUhqV9sjhZ42yDqxdpj6xm0X9IfvW0F@N3Wzbvd5TVVRtkaQxPhmHc/wE^0|0|E|0|0|0|0|6|1|0|0|0|0|0|0^^$0|@$1|2|3|4|5|6|7|17|8|@]|9|@$A|18|B|19|1|1A]]|C|$]]|$1|D|3|E|5|F|7|1B|8|@]|9|@]|C|$G|H]]|$1|I|3|J|5|K|7|1C|8|@]|9|@$A|1D|B|1E|1|1F]]|C|$]]|$1|L|3|M|5|K|7|1G|8|@]|9|@]|C|$]]|$1|N|3|O|5|K|7|1H|8|@]|9|@]|C|$]]|$1|P|3|Q|5|R|7|1I|8|@]|9|@]|C|$]]|$1|S|3|T|5|R|7|1J|8|@]|9|@]|C|$]]|$1|U|3|V|5|K|7|1K|8|@]|9|@]|C|$]]|$1|W|3|X|5|F|7|1L|8|@]|9|@]|C|$G|H]]]|Y|$Z|$5|10|11|12|C|$13|14]]|15|$5|10|11|12|C|$13|16]]]]

<h1><a href="https://www.microsoft.com/net/core/platform" rel="nofollow noreferrer">C# (.NET Core)</a>, 66 bytes</h1>



<pre class="lang-cs prettyprint-override"><code>a=&gt;{double r=1,x=1,y=1,i=0;for(;++i&lt;a;y+=x,x=y-x)r+=1/y;return r;}
</code></pre>

<p><a href="https://tio.run/##NY3LCsIwFETX9ivuMqX1Ud0IaQQRBEE3KriO6a0EagI3qTSUfnsNlA4chsMsRrmlsoRj67T5wCM4j18OqpHOwRH6xHnptYJza1RZ2fbdYA5TH6AGMUpx6CcHEkXeRUJEiw2vLTGeZbqUPGSii1tYdillolgHTuhbMkB8GPl88rO6gpvUhqV9sjhZ42yDqxdpj6xm0X9IfvW0F@N3Wzbvd5TVVRtkaQxPhmHc/wE" rel="nofollow noreferrer" title="C# (.NET Core) – Try It Online">Try it online!</a></p>

<p>Can't use floats because of imprecision.</p>

<p>Example where input = 8:</p>

<ul>
<li>double: 3.28287545787546 (rounds to 3.282875)</li>
<li>float:  3.282876 (off by 0.000001)</li>
</ul>

<p>Ungolfed:</p>

<pre class="lang-cs prettyprint-override"><code>a =&gt; {
    double r = 1,                       // initialize r (sum of reciprocals) - start at 1 to save one byte in for loop
            x = 1, y = 1,                   // initialize x (2nd largest Fibonacci) and y (largest Fibonacci)
            i = 0;                          // initialize i (loop counter)
    for(; ++i &lt; a; y += x, x = y - x)   // from 1 to a, while incrementing the Fibonacci numbers
        r += 1 / y;                         // increment the reciprocal Fibonacci
    return r;                           // return the reciprocal Fibonacci
}
</code></pre>


blocks|key|5322856|text|果冻，5字节|type|header-two|depth|inlineStyleRanges|entityRanges|offset|length|data|5322857|RÆḞİS|code-block|syntax|javascript|5322858|在网上试试！|unstyled|5322859|是如何工作的|5322860|RÆḞİS++++Main+link+(monad).+Input:+integer
R++++++++Range
+ÆḞ++++++nth+Fibonacci+number
+++İ+++++Reciprocal
++++S++++Sum|entityMap|0|LINK|mutability|MUTABLE|url|https://github.com/DennisMitchell/jelly|1|https://tio.run/##y0rNyan8/z/ocNvDHfOObAj@//@/KQA^0|0|2|0|0|0|0|6|1|0|0^^$0|@$1|2|3|4|5|6|7|Y|8|@]|9|@$A|Z|B|10|1|11]]|C|$]]|$1|D|3|E|5|F|7|12|8|@]|9|@]|C|$G|H]]|$1|I|3|J|5|K|7|13|8|@]|9|@$A|14|B|15|1|16]]|C|$]]|$1|L|3|M|5|6|7|17|8|@]|9|@]|C|$]]|$1|N|3|O|5|F|7|18|8|@]|9|@]|C|$G|H]]]|P|$Q|$5|R|S|T|C|$U|V]]|W|$5|R|S|T|C|$U|X]]]]

<h1><a href="https://github.com/DennisMitchell/jelly" rel="nofollow noreferrer">Jelly</a>, 5 bytes</h1>

<pre><code>RÆḞİS
</code></pre>

<p><a href="https://tio.run/##y0rNyan8/z/ocNvDHfOObAj@//@/KQA" rel="nofollow noreferrer" title="Jelly – Try It Online">Try it online!</a></p>

<h3>How it works</h3>

<pre><code>RÆḞİS    Main link (monad). Input: integer
R        Range
 ÆḞ      nth Fibonacci number
   İ     Reciprocal
    S    Sum
</code></pre>


<p>Seeing as there have been an awful lot of normal Fibonacci challenges, I decided that it might be interesting to calculate the <a href="http://en.wikipedia.org/wiki/Reciprocal_Fibonacci_constant" rel="nofollow">Reciprocal Fibonacci constant</a> - namely, the sum of the reciprocals of the Fibonacci sequence.</p>

<p>The challenge is to calculate the Reciprocal Fibonacci constant with the number of Fibonacci series to use digit given as input, i.e. an input of 10 means to calculate based on the reciprocals of the first 10 Fibonacci numbers.  In the likely case of a tie, shortest code wins.</p>

<p>The winner will be chosen in three weeks via standard code-golf rules.</p>


Reciprocal Fibonacci constant

翻译质量差，导致语言生硬或混乱。

没有提供实际的解决方法或示例。

解答不清晰，无法理解或解决问题。

页面排版不美观，阅读体验差。

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功能1上新10个字符

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功能2上新100个字符功能2上新100个字符功能2上新100个字符功能2上新100个字符功能2上新100个字符功能2上新100个字符功能2上新100个字符功能2上新100个字符功能2上新100个字符。

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考虑到有很多正常的斐波纳契挑战，我决定计算倒数Fibonacci常数 --也就是斐波纳契序列倒数之和--可能会很有趣。挑战是用Fibonacci级数的数目计算倒数Fibonacci常数，以数字作为输入，即输入10种方法，根据前10个Fibonacci数的倒数来计算。在可能出现平局的情况下，最短的代码将获胜。获胜者将在三周内通过标准规则-高尔夫规则被选中。

问倒数Fibonacci常数
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Mathematica (没有内置Fibonacci的32个字符)

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Javascript，33

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K，22

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