The On-Line Encyclopedia of Integer Sequences (OEIS)

Revision History for A000045 (Bold, blue-underlined text is an addition; faded, red-underlined text is a ~~deletion~~.)
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A000045

Fibonacci numbers: F(n) = F(n-1) + F(n-2) with F(0) = 0 and F(1) = 1.
(history; published version)

#2225 by Alois P. Heinz at Mon Sep 09 21:04:46 EDT 2024

FORMULA

F(n) = floor(10^(floor((n+3)/4)*(n+1))/(10^(floor((n+3)/4)-1)*(100*(10^(floor((n+3)/4)-1)-1)+90)-1))-floor(floor(10^(floor((n+3)/4)*(n+1))/(10^(floor((n+3)/4)-1)*(100 *(10^(floor((n+3)/4)-1)-1)+90)-1))/10^floor((n+3)/4))*10^floor((n+3)/4). - Andrew Lenker, Sep 09 2024

STATUS

editing

approved

#2224 by Alois P. Heinz at Mon Sep 09 20:58:39 EDT 2024

STATUS

proposed

editing

Discussion

Mon Sep 09

20:59

Alois P. Heinz: this was rejected before ...

21:00

Alois P. Heinz: we do not need formulas like this ... you are free to publish this elsewhere ...

21:04

Alois P. Heinz: btw: the editor's name is Joerg Arndt ...

#2223 by Andrew Lenker at Mon Sep 09 20:42:54 EDT 2024

STATUS

editing

proposed

Discussion

Mon Sep 09

20:58

Alois P. Heinz: ugly, being terribly inelegant, does not extend to negative indices ...

#2222 by Andrew Lenker at Mon Sep 09 20:39:33 EDT 2024

FORMULA

STATUS

approved

editing

Discussion

Mon Sep 09

20:42

Andrew Lenker: Fixed it. Joerg Arnt, I'm sorry for being stubborn and for my formula that looks like a train wreck hit by a tornado, but it works on your software now. I know it is ugly and uses Rounding, as I use different methods than most. But please give my train wreck a chance.

#2221 by Joerg Arndt at Mon Sep 09 05:59:06 EDT 2024

FORMULA

F(n) = floor(10^(floor((n+2)/4)*n)/(10^(floor((n+2)/4)-1)*(100*(10^(floor((n+2)/4)-1)-1)+90)-1))-floor(floor(10^(floor((n+2)/4)*n)/(10^(floor((n+2)/4)-1)*(100 *(10^(floor((n+2)/4)-1)-1)+90)-1))/10^floor((n+2)/4))*10^floor((n+2)/4)

(Decimal extraction method: Geometrically robust, but heavy computation). - Andrew Lenker, Sep 09 2024

KEYWORD

nonn,core,nice,easy,hear,changed

STATUS

proposed

approved

#2220 by Andrew Lenker at Mon Sep 09 04:18:04 EDT 2024

STATUS

editing

proposed

Discussion

Mon Sep 09

04:32

Michel Marcus: not convinced

04:53

Andrew Lenker: 1/89, 1/9899, 1/998999, etc, produced Fibonacci Sequence in decimal. The formula iterates each four times to slow growth. Then extracts using subtraction of a reduced copy

05:59

Joerg Arndt: Pari: for(n=1,11,if(F(n)!=fibonacci(n),print1(n,", "))) gives
1, 3, 4, 5, 6, 7, 8, 9, 10, 11,
So, wrong on top of being terribly inelegant.
Reverting edit.

#2219 by Andrew Lenker at Mon Sep 09 04:16:12 EDT 2024

FORMULA

(Decimal extraction method: Geometrically robust, but heavy computation). - Andrew Lenker, Sep 09 2024

STATUS

approved

editing

#2218 by R. J. Mathar at Mon Aug 05 05:40:00 EDT 2024

STATUS

editing

approved

#2217 by R. J. Mathar at Mon Aug 05 05:39:49 EDT 2024

REFERENCES

Robert Dougherty-Bliss, The Meta-C-finite Ansatz, arXiv preprint arXiv:2206.14852, 2022

LINKS

Robert Dougherty-Bliss, <a href="https://arxiv.org/abs/2206.14852">The Meta-C-finite Ansatz</a>, arXiv preprint arXiv:2206.14852, 2022

STATUS

approved

editing

#2216 by Alois P. Heinz at Mon Jul 15 11:01:08 EDT 2024

FORMULA

Sum_{k=1..n} k^3*F(k) = (n^3-6*n^2+24*n-50)*F(n+1) - ((n+1)^3-6*(n+1)^2+24*(n+1)-50)*F(n) + 50. - Prabha Sivaramannair, Jul 15 2024

STATUS

editing

approved