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? >___>

tu gribi lai tavai pamatskolas galvinjai paskaidro kas ir sigmas un logaritmiskaas funkcujas, vai arii veelies padaliities ar citiem pokemoniem shajaa KrUtAa FoRmUlAa?

absoluuti vienalga.

augstaakaa matemaatika. Man gan vislielaako neizpratni izraisiija taa anglju valoda.

atradu vienaa anglju forumaa, so yeah >_>

ko tu domaa ar "anglju forumu" ? Vietu kur cilveeki maacaas anglju valodu?

Interesanta formulinja 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)

nee, forums ir taads, kur ir vienkaarshi cilveeki kas jau runaa angliski -.- vispaar apspriezhas. par jebko. no rules.

Varbuut iesaakumaa tur kaut kas ir bijis...
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))+

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

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 .

Nee, uzrakstiiju fikso progu XDD

Paarbaudiiju : excelis nemaz nenejm to briinumu pretii : 'formula to long'

Arpraats, taksh tas nebija jaareekjina



Paskaties pirmo formulu vairaak un papeeti taas uzbuuvi, tas ir tas pats copy paste
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"...

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.

Kur tu izraavi taadu rezultaatu?

veicot elemtaaro matemaatiku. Aciimredzot tavai programai kautkas iisti nepatiik, vai arii vinja straadaa tikai ar intedzheriem.

Es RTU visaadu diferenciaalo uzdevumu un interaalju rekjinaashanai izmantoju Matcad...Citaadi taadi cilveeki kaa es ar iQ<60 buutu aatri vien pie matenes atbirushi...

Droshi velc Progu - bet neaizmirsti 500lpp manuaali, jo formulas un uzdevuma ieadiishana ir piedziivojums pats par sevi

Nee es atklaaju kas par geelu.. vinja nestraadaa tikai ar integeriem bet gan ar kaartiigu double probleema vareetu buut faktaa ka sin un cos parametrs tiek nejemts nevis kaa graadi bet radiaani XD

EDIT : un proga sigmas reekinsasanai ir visparastakais cikls ar skaitlju sakaitiishanu ^^

lol, kautkaada muljkjiiga tev taa programma.

Shumins, es no augstaakaas matematiikas sajeedzu tikpat cik no baletdejoshanas, man no taadas programmas nebuutu jeegas. Manas vajadziibas aprobezhojas ar vektoriem un sin-cos formulaam. Tobish, man neko vairaak par gjeometriju nevajag. Es gan neatteiktos iemaaciities, kas taa par sigmu, zinu tikai to, ka mana programmatuura taadas nezin.

Sigma a.k.a. summas zīme - respektīvi saskaita visu kopā, piemēram, lai nav jāraksta F1 (gomen, slinkums salikt apakšējos indeksus) + F2 + F3 + ... + Fvvz lieto ∑Fi.

jaa, to ka ar griekju sigmu apziimee fun. summu es zinaaju, bet ja jau vinja nav lietota pamatformulaa, vai tad tas nebuus kas cits? paskatiijos wikipedijaa, itkaa nekaa dizhi cita nav. Mosh arii ir vnk parastaa summa.

Spamerus naktiis apciemo lietuveens!!!

Topic closed.