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> xD kaads saprot?
Blackout
post Feb 10 2007, 12:25
Post #1


Gimme More
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Posts: 294
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From: Riga
atradu sekojoshu matemaatisku formulu internetaa:
"to prove i know , the next line is

1113213211
31131211131221

Btw , interesting facts :

If you think that the sequence is "non-mathematical", I derived this mathematical expression that gives the sequence... have fun! (D is a recursive function and t is the term number.) It's a lot easier if you think verbally, isn't it?

By the way, % here is a certain non-integer remainder function. 2.1%0.1 would be 0, 2.1%0.2 would be 0.1, 2.1%0.3 would be 0 since 0.3 fits evenly into 2.1, etc.) If you really want conventional operators, you could define % with limits and modular arithmetic...

D(t+1) = (sigma(K=1,LOG(D(t)*10)-LOG(D(t)*10)%1,((D(t)-D(t)%10^(LOG(D(t))-
LOG(D(t))%1)+sigma(S=1,LOG(D(t))-LOG(D(t))%1,(((sigma(R=1,LOG(D(t)*10)
-LOG(D(t)*10)%1-1,ABS((D(t)*10-D(t)*10%10^(R+1))%10^(R+2)/10-(D(t)*10-D(t)*
10%10^R)%10^(R+1)))/10)-(sigma(R=1,LOG(D(t)*10)-LOG(D(t)*10)%1-1,ABS((D(t)*
10-D(t)*10%10^(R+1))%10^(R+2)/10-(D(t)*10-D(t)*10%10^R)%10^(R+1)))/10)
%10^(S-1))%10^S+1)%(((sigma(R=1,LOG(D(t)*10)-LOG(D(t)*10)%1-1,ABS((D(t)
*10-D(t)*10%10^(R+1))%10^(R+2)/10-(D(t)*10-D(t)*10%10^R)%10^(R+1)))/10)-(
sigma(R=1,LOG(D(t)*10)-LOG(D(t)*10)%1-1,ABS((D(t)*10-D(t)*10%10^(R+1))%10
^(R+2)/10-(D(t)*10-D(t)*10%10^R)%10^(R+1)))/10)%10^(S-1))%10^S+.5)*2*(D(t)
-D(t)%10^(S-1))%10^S))-(D(t)-D(t)%10^(LOG(D(t))-LOG(D(t))%1)+sigma(S=1,
LOG(D(t))-LOG(D(t))%1,(((sigma(R=1,LOG(D(t)*10)-LOG(D(t)*10)%1-1,ABS((D(t)
*10-D(t)*10%10^(R+1))%10^(R+2)/10-(D(t)*10-D(t)*10%10^R)%10^(R+1)))/10)
-(sigma(R=1,LOG(D(t)*10)-LOG(D(t)*10)%1-1,ABS((D(t)*10-D(t)*10%10^(R+1))%
10^(R+2)/10-(D(t)*10-D(t)*10%10^R)%10^(R+1)))/10)%10^(S-1))%10^S+1)%(((
sigma(R=1,LOG(D(t)*10)-LOG(D(t)*10)%1-1,ABS((D(t)*10-D(t)*10%10^(R+1))%
10^(R+2)/10-(D(t)*10-D(t)*10%10^R)%10^(R+1)))/10)-(sigma(R=1,LOG(D(t)*
10)-LOG(D(t)*10)%1-1,ABS((D(t)*10-D(t)*10%10^(R+1))%10^(R+2)/10-(D(t)*
10-D(t)*10%10^R)%10^(R+1)))/10)%10^(S-1))%10^S+.5)*2*(D(t)-D(t)%10^(S
-1))%10^S))%10^(K-1))%10^K/10^(K-1)*100^(2*sigma(N=1,K,(((D(t)-D(t)%
10^(LOG(D(t))-LOG(D(t))%1)+sigma(S=1,LOG(D(t))-LOG(D(t))%1,(((sigma(R=1,
LOG(D(t)*10)-LOG(D(t)*10)%1-1,ABS((D(t)*10-D(t)*10%10^(R+1))%10^(R+2)/10-
(D(t)*10-D(t)*10%10^R)%10^(R+1)))/10)-(sigma(R=1,LOG(D(t)*10)-LOG(D(t)*
10)%1-1,ABS((D(t)*10-D(t)*10%10^(R+1))%10^(R+2)/10-(D(t)*10-D(t)*10%10^R)%
10^(R+1)))/10)%10^(S-1))%10^S+1)%(((sigma(R=1,LOG(D(t)*10)-LOG(D(t)*10)%
1-1,ABS((D(t)*10-D(t)*10%10^(R+1))%10^(R+2)/10-(D(t)*10-D(t)*10%10^R)%
10^(R+1)))/10)-(sigma(R=1,LOG(D(t)*10)-LOG(D(t)*10)%1-1,ABS((D(t)*10-D(t)
*10%10^(R+1))%10^(R+2)/10-(D(t)*10-D(t)*10%10^R)%10^(R+1)))/10)%10^(S-1))
%10^S+.5)*2*(D(t)-D(t)%10^(S-1))%10^S))-(D(t)-D(t)%10^(LOG(D(t))-LOG(D(t))
%1)+sigma(S=1,LOG(D(t))-LOG(D(t))%1,(((sigma(R=1,LOG(D(t)*10)-LOG(D(t)*10)
%1-1,ABS((D(t)*10-D(t)*10%10^(R+1))%10^(R+2)/10-(D(t)*10-D(t)*10%10^R)%10^
(R+1)))/10)-(sigma(R=1,LOG(D(t)*10)-LOG(D(t)*10)%1-1,ABS((D(t)*10-D(t)*10%
10^(R+1))%10^(R+2)/10-(D(t)*10-D(t)*10%10^R)%10^(R+1)))/10)%10^(S-1))%10^
S+1)%(((sigma(R=1,LOG(D(t)*10)-LOG(D(t)*10)%1-1,ABS((D(t)*10-D(t)*10%10^
(R+1))%10^(R+2)/10-(D(t)*10-D(t)*10%10^R)%10^(R+1)))/10)-(sigma(R=1,
LOG(D(t)*10)-LOG(D(t)*10)%1-1,ABS((D(t)*10-D(t)*10%10^(R+1))%10^(R+2)/
10-(D(t)*10-D(t)*10%10^R)%10^(R+1)))/10)%10^(S-1))%10^S+.5)*2*(D(t)-D(t)%
10^(S-1))%10^S))%10^(N-1))%10^N+1)%(((D(t)-D(t)%10^(LOG(D(t))-LOG(D(t))%
1)+sigma(S=1,LOG(D(t))-LOG(D(t))%1,(((sigma(R=1,LOG(D(t)*10)-LOG(D(t)*
10)%1-1,ABS((D(t)*10-D(t)*10%10^(R+1))%10^(R+2)/10-(D(t)*10-D(t)*10%10^
R)%10^(R+1)))/10)-(sigma(R=1,LOG(D(t)*10)-LOG(D(t)*10)%1-1,ABS((D(t)*10-
D(t)*10%10^(R+1))%10^(R+2)/10-(D(t)*10-D(t)*10%10^R)%10^(R+1)))/10)%10^
(S-1))%10^S+1)%(((sigma(R=1,LOG(D(t)*10)-LOG(D(t)*10)%1-1,ABS((D(t)*10-
D(t)*10%10^(R+1))%10^(R+2)/10-(D(t)*10-D(t)*10%10^R)%10^(R+1)))/10)-
(sigma(R=1,LOG(D(t)*10)-LOG(D(t)*10)%1-1,ABS((D(t)*10-D(t)*10%10^(R+1))%
10^(R+2)/10-(D(t)*10-D(t)*10%10^R)%10^(R+1)))/10)%10^(S-1))%10^S+.5)*2*
(D(t)-D(t)%10^(S-1))%10^S))-(D(t)-D(t)%10^(LOG(D(t))-LOG(D(t))%1)+
sigma(S=1,LOG(D(t))-LOG(D(t))%1,(((sigma(R=1,LOG(D(t)*10)-LOG(D(t)*10
)%1-1,ABS((D(t)*10-D(t)*10%10^(R+1))%10^(R+2)/10-(D(t)*10-D(t)*10%10^
R)%10^(R+1)))/10)-(sigma(R=1,LOG(D(t)*10)-LOG(D(t)*10)%1-1,ABS((D(t)*
10-D(t)*10%10^(R+1))%10^(R+2)/10-(D(t)*10-D(t)*10%10^R)%10^(R+1)))/
10)%10^(S-1))%10^S+1)%(((sigma(R=1,LOG(D(t)*10)-LOG(D(t)*10)%1-1,
ABS((D(t)*10-D(t)*10%10^(R+1))%10^(R+2)/10-(D(t)*10-D(t)*10%10^R)%
10^(R+1)))/10)-(sigma(R=1,LOG(D(t)*10)-LOG(D(t)*10)%1-1,ABS((D(t)*
10-D(t)*10%10^(R+1))%10^(R+2)/10-(D(t)*10-D(t)*10%10^R)%10^(R+1)))
/10)%10^(S-1))%10^S+.5)*2*(D(t)-D(t)%10^(S-1))%10^S))%10^(N-1))%10^
N+.5))))/100)+(sigma(K=1,LOG(D(t)*10)-LOG(D(t)*10)%1,100^(1+sigma(N=
1,K-1,2*((((sigma(R=1,LOG(D(t)*10)-LOG(D(t)*10)%1-1,ABS((D(t)*10-D(t)*
10%10^(R+1))%10^(R+2)/10-(D(t)*10-D(t)*10%10^R)%10^(R+1)))/10)-(
sigma(R=1,LOG(D(t)*10)-LOG(D(t)*10)%1-1,ABS((D(t)*10-D(t)*10%10^(R+1))%
10^(R+2)/10-(D(t)*10-D(t)*10%10^R)%10^(R+1)))/10)%10^(N-1))%10^N/10^(N-
1)+1)%(((sigma(R=1,LOG(D(t)*10)-LOG(D(t)*10)%1-1,ABS((D(t)*10-D(t)*10%
10^(R+1))%10^(R+2)/10-(D(t)*10-D(t)*10%10^R)%10^(R+1)))/10)-(sigma(R=
1,LOG(D(t)*10)-LOG(D(t)*10)%1-1,ABS((D(t)*10-D(t)*10%10^(R+1))%10^(R+
2)/10-(D(t)*10-D(t)*10%10^R)%10^(R+1)))/10)%10^(N-1))%10^N/10^(N-1)+
.5)))))/10)"

Kaads vispaar saprot par ko iet runa, un kaa viss notiek? >___>


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ETM
post Feb 10 2007, 13:46
Post #2


Very Old Fart
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Group: 40K
Posts: 1177
Joined: 20-January 05
From: Man pajāt
tu gribi lai tavai pamatskolas galvinjai paskaidro kas ir sigmas un logaritmiskaas funkcujas, vai arii veelies padaliities ar citiem pokemoniem shajaa KrUtAa FoRmUlAa?


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Blackout
post Feb 10 2007, 14:04
Post #3


Gimme More
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Group: weirdo
Posts: 294
Joined: 4-November 05
From: Riga
absoluuti vienalga.


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Sigfa
post Feb 10 2007, 14:24
Post #4


Kami-sama
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Group: ALK
Posts: 2738
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From: Starp pudelēm un četrām sienām
augstaakaa matemaatika. Man gan vislielaako neizpratni izraisiija taa anglju valoda.
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Blackout
post Feb 10 2007, 14:25
Post #5


Gimme More
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Group: weirdo
Posts: 294
Joined: 4-November 05
From: Riga
atradu vienaa anglju forumaa, so yeah >_>


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Sigfa
post Feb 10 2007, 14:28
Post #6


Kami-sama
**********

Group: ALK
Posts: 2738
Joined: 17-May 04
From: Starp pudelēm un četrām sienām
ko tu domaa ar "anglju forumu" ? Vietu kur cilveeki maacaas anglju valodu?
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anti
post Feb 10 2007, 14:35
Post #7


Draņķītis
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From: Latvija, Rīga
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ALWAYS, I WANNA BE WITH YOU
AND MAKE BELIEVE WITH YOU
AND LIVE IN HARMONY, HARMONY, OH LOVE ♫
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Storm
post Feb 10 2007, 14:38
Post #8


Danshaku
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Group: Chuunin
Posts: 290
Joined: 12-April 06
From: nya
Interesanta formulinja biggrin.gif jam man buutu piecas minuutes laika sadaam mulkiibaam uztaisiitu vienu prodzinju kas shito jautriibu izreekina. Ideja ir dabuut naakoso skaitli rindaa ja zinaams iepriekseejais tb. D(t+1)


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Blackout
post Feb 10 2007, 15:29
Post #9


Gimme More
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Group: weirdo
Posts: 294
Joined: 4-November 05
From: Riga
nee, forums ir taads, kur ir vienkaarshi cilveeki kas jau runaa angliski -.- vispaar apspriezhas. par jebko. no rules.


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Shumins
post Feb 10 2007, 16:03
Post #10


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Varbuut iesaakumaa tur kaut kas ir bijis... dry.gif
Bet iespeejams katrs idiots, kas to meegjinaajis paarkopeet un paarakstiit, nesaprotot rakstiito, to sabojaajis tiktaal, ka nekaa nesaprotu.

Ar garaam formulaam ir taapat kaa gariem postiem, neviens nesaka, ka tur a vidu kaut kas ir. Taapat arii te, nekas nav atvasinaats, integreets, nekaadu rundu un lauku - domaaju katram pokemonam jau beidzot videni buutu jaazin.

Buutu labaak kaadu augstaaku skolu amekleejushi, nevis ticeetu uber formulaam un spokiem + citplaneetieshiem + un burvestiibaam

Es jau arii varu uzrakstiit uber apreekjinu:

(2/3)+(4*8)-(3/9)/((22)+(2/2)+(3*7)+sin(120)+cos(30))+((2/3)+(4*8)-(3/9)/((22)+(2/2)+(3*7)+sin(120)+cos(30)))+(2/3)+(4*8)-(3/9)/((22)+(2/2)+(3*7)+sin(120)+cos(30))+(2/3)+(4*8)-(3/9)/((22)+(2/2)+(3*7)+sin(120)+cos(30))+(2/3)+(4*8)-(3/9)/((22)+(2/2)+(3*7)+sin(120)+cos(30))+((2/3)+(4*8)-(3/9)/((22)+(2/2)+(3*7)+sin(120)+cos(30)))+(2/3)+(4*8)-(3/9)/((22)+(2/2)+(3*7)+sin(120)+cos(30))+(2/3)+(4*8)-(3/9)/((22)+(2/2)+(3*7)+sin(120)+cos(30))+(2/3)+(4*8)-(3/9)/((22)+(2/2)+(3*7)+sin(120)+cos(30))+((2/3)+(4*8)-(3/9)/((22)+(2/2)+(3*7)+sin(120)+cos(30)))+(2/3)+(4*8)-(3/9)/((22)+(2/2)+(3*7)+sin(120)+cos(30))+(2/3)+(4*8)-(3/9)/((22)+(2/2)+(3*7)+sin(120)+cos(30))+(2/3)+(4*8)-(3/9)/((22)+(2/2)+(3*7)+sin(120)+cos(30))+((2/3)+(4*8)-(3/9)/((22)+(2/2)+(3*7)+sin(120)+cos(30)))+(2/3)+(4*8)-(3/9)/((22)+(2/2)+(3*7)+sin(120)+cos(30))+(2/3)+(4*8)-(3/9)/((22)+(2/2)+(3*7)+sin(120)+cos(30))+(2/3)+(4*8)-(3/9)/((22)+(2/2)+(3*7)+sin(120)+cos(30))+((2/3)+(4*8)-(3/9)/((22)+(2/2)+(3*7)+sin(120)+cos(30)))+(2/3)+(4*8)-(3/9)/((22)+(2/2)+(3*7)+sin(120)+cos(30))+(2/3)+(4*8)-(3/9)/((22)+(2/2)+(3*7)+sin(120)+cos(30))+(2/3)+(4*8)-(3/9)/((22)+(2/2)+(3*7)+sin(120)+cos(30))+((2/3)+(4*8)-(3/9)/((22)+(2/2)+(3*7)+sin(120)+cos(30)))+(2/3)+(4*8)-(3/9)/((22)+(2/2)+(3*7)+sin(120)+cos(30))+(2/3)+(4*8)-(3/9)/((22)+(2/2)+(3*7)+sin(120)+cos(30))+(2/3)+(4*8)-(3/9)/((22)+(2/2)+(3*7)+sin(120)+cos(30))+((2/3)+(4*8)-(3/9)/((22)+(2/2)+(3*7)+sin(120)+cos(30)))+(2/3)+(4*8)-(3/9)/((22)+(2/2)+(3*7)+sin(120)+cos(30))+(2/3)+(4*8)-(3/9)/((22)+(2/2)+(3*7)+sin(120)+cos(30))+
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Storm
post Feb 10 2007, 16:39
Post #11


Danshaku
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Group: Chuunin
Posts: 290
Joined: 12-April 06
From: nya
Rezultaats : 896 XDD

Bet par teemu runaajot kaads sakars vipaar postot kko taadu? Bet varbuut butu jauka iedeja aizsaakt vienu interesantu topiku, kur kaads uzdod aakiigo uzdevumu un kurs tad pirmais izdomaa biggrin.gif


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Sigfa
post Feb 10 2007, 16:42
Post #12


Kami-sama
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From: Starp pudelēm un četrām sienām
tev to visu noteikti vajadzeeja rakstiit vienaa rindaa shumin... ar copy+paste tu savu domu nepieraadiisi.

Storm, ieliki excelii, lai izreekjina? EDIT: un tu kljuudijies, rezultaats ir 914.458029984138 .

This post has been edited by Sigfa: Feb 10 2007, 16:55
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Storm
post Feb 10 2007, 16:44
Post #13


Danshaku
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Group: Chuunin
Posts: 290
Joined: 12-April 06
From: nya
Nee, uzrakstiiju fikso progu XDD

Paarbaudiiju : excelis nemaz nenejm to briinumu pretii : 'formula to long' biggrin.gif

This post has been edited by Storm: Feb 10 2007, 16:48


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Shumins
post Feb 10 2007, 16:52
Post #14


Unregistered
Arpraats, taksh tas nebija jaareekjina mad.gif

QUOTE(Sigfa @ Feb 10 2007, 17:42) *

ar copy+paste tu savu domu nepieraadiisi.


Paskaties pirmo formulu vairaak un papeeti taas uzbuuvi, tas ir tas pats copy paste laugh.gif
Garas formulas ne vienmeer briesmiigas, vienkaarshi daudz mazas saliktas kopaa, un taapat nekaada doma nav pirmaa formulaa, tas ir sviets ar kaadu mainiigo or constanti "t"... mad.gif
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Sigfa
post Feb 10 2007, 17:03
Post #15


Kami-sama
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Group: ALK
Posts: 2738
Joined: 17-May 04
From: Starp pudelēm un četrām sienām
konstante nevar buut mainiiga.

storm, es gan gribeetu redzeet programmu, kura reekjina sigmas. EDIT: excelim ir arii taada forsha fiicha kaa macros, kurus var jauki izmantot garaam formulaam.

This post has been edited by Sigfa: Feb 10 2007, 17:08
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