G.O.D
Region: IT
Saturday 01 August 2026 10:09:49 GMT
116116
7315
216
3273
Music
Download
Comments
Ellios Kiak 33 :
Trusth come to life all will see who it is Zalmoxis
2026-08-02 11:04:17
2
Ana Madalina :
Zamolxis zeul dacilor!
2026-08-01 12:27:27
62
Mario_Valahul :
Când acest POPOR v-a înțelege...dacă v-a înțelege ce RĂDĂCINII avem....Lucrurile se vor schimba...😳
2026-08-01 15:58:18
37
Lady M :
Totuși....in studiul etimologic al termenului Zamolxe....za( legătură/ uniune)....molxe/ moksha ( eliberarea spiritului). In cunoasterea Vedică, Moksha înseamnă Eliberarea spiritului din roata Samsara , ciclul reincarnarilor infinite. Moksha este un cult prin care omul in timpul vieții își dedica întreaga existență acestei treziri spirituale pt a se putea elibera . ❤️🙏
2026-08-01 10:22:22
39
Mary3ANN :
Avem atât de multe … semne și noi ce facem ? STĂM ȘI ÎL PRIVIM PE NICUȘOR HMMMM 😂ȘI SUNTEM CEEA CE SUNTEM
2026-08-02 02:26:41
15
Traian :
Sunt scrieri care spun ca Pitagora era invatacelul lui Zamolxes, nu invers
2026-08-02 06:00:11
11
user6056803402665 :
întreaga Biblie de astăzi este legata de acest cult...
2026-08-01 14:21:53
33
valeriudimciu :
Zamolxe a trăit înainte de Pitagora. și nu are nimic originar cu Grecia antică. a fost divinizat de geți și a devenit zeul lor suprem, el arătându le și explicându le o mulțime de aspecte ale vieții, în special spirituale....a inspirat și actuala idee spirituală creștină
2026-08-01 13:24:38
20
elenaradu363 :
Primul Popor al Lumii🙏🙏🙏💖💖💖
2026-08-01 11:14:55
17
sergiunistor933 :
creștinismul este foarte dezvoltat la români pentru că aveau această credință mai veche de 2000 de ani
2026-08-01 18:59:22
57
florianparvuica :
Zamolxe a fost Iisus
2026-08-01 18:49:28
13
Fane :
Broscarii afirmă chiar și în documentare ,că dacii erau barbari, cu toate că tot ei , demonstrează că ar cunoaște bine istoria columnei lui Traian...
2026-08-01 13:40:40
7
Dragomir Vlad :
dacii liberi 💪
2026-08-01 10:53:20
46
Mircea Suciu :
de Isus nu pomenește nici un filozof contemporan acelor vremuri !!!!
2026-08-01 16:04:59
5
Tase Robert :
Asta demonstreaza ca Cineva ne-a trimis semne si salvatori de multe ori dar noi 🤦🏻🤦🏻🤦🏻asta inseamna ignoranta
2026-08-01 11:17:28
16
Adina_kalea2 :
Enoch,Zamolxis si Isus au fost trimisi pe pamant pt a ne arada adevarata cale adica spiritualitatea sufletului urbana o cale dreapta,buna si demna
2026-08-01 20:30:27
14
Flavius Predut :
2026-08-01 19:33:12
8
Lady M :
Țin să menționez că postările tale....cum sunt atât de repede eliminate( 🤦♀️)....au devenit prioritatea mea nr 1 in această platformă!!! 🥰❤️😁
2026-08-01 10:23:43
26
Marian :
triblia este o poveste compusă din mai multe apocrife , convenite la Consiliiul de la Niceea !
2026-08-01 18:02:37
8
coco :
2026-08-01 13:50:17
8
Șuțu Constantin :
Corupția a înviat,restul ptr Sfânta biserica
2026-08-01 19:01:50
5
Eduard Grossmann :
Și eu le urmăresc.
2026-08-01 10:49:48
7
To see more videos from user @g.o.d6.9, please go to the Tikwm
homepage.
![Graham's number is an immense number that arose as an upper bound on the answer of a problem in the mathematical field of Ramsey theory. It is much larger than many other large numbers introduced as effective bounds in mathematics, such as Skewes's bound, which in turn is much larger than a googolplex. Graham's number is so large that the observable universe is far too small to contain its ordinary digital representation, assuming that each digit occupies one Planck volume. But even the number of digits in this digital representation of Graham's number would itself be a number so large that its digital representation cannot be represented in the observable universe. Nor even can the number of digits of that number—and so forth, for a number of times far exceeding the total number of Planck volumes in the observable universe. Thus, Graham's number cannot be expressed even by physical universe-scale power towers of the form a b c ⋅ ⋅ ⋅ {\displaystyle a^{b^{c^{\cdot ^{\cdot ^{\cdot }}}}}}, even though Graham's number is indeed a power of three. However, Graham's number can be explicitly given by computable recursive formulas using Knuth's up-arrow notation or equivalent, as was done by Ronald Graham, the number's namesake. As there is a recursive formula to define it, it is much smaller than typical busy beaver numbers, the sequence of which grows faster than any computable sequence. Though too large to ever be computed in full, the sequence of digits of Graham's number can be computed explicitly via simple algorithms; the last 10 digits of Graham's number are ...2464195387.[1] Using Knuth's up-arrow notation, Graham's number is g 64 {\displaystyle g_{64}},[2] where g n = { 3↑↑↑↑3, if n=1 and 3 ↑ g n − 1 3, if n≥2. {\displaystyle g_{n}={\begin{cases}3\uparrow \uparrow \uparrow \uparrow 3,&{\text{if }}n=1{\text{ and}}\\3\uparrow ^{g_{n-1}}3,&{\text{if }}n\geq 2.\end{cases}}} Graham's number was used by Graham in conversations with popular science writer Martin Gardner as a simplified explanation of the upper bounds of the problem he was working on. In 1977, Gardner described the number in Scientific American, introducing it to the general public. At the time of its introduction, it was the largest specific positive integer ever to have been used in a published mathematical proof. The number was described in the 1980 Guinness Book of World Records, adding to its popular interest. Other specific integers (such as TREE(3)) known to be far larger than Graham's number have since appeared in many serious mathematical proofs, for example in connection with Harvey Friedman's various finite forms of Kruskal's theorem. Additionally, smaller upper bounds on the Ramsey theory problem from which Graham's number was derived have since been proven to be valid. Context edit Example of a 2-colored 3-dimensional cube containing one single-coloured 4-vertex coplanar complete subgraph. The subgraph is shown below the cube. This cube would contain no such subgraph if, for example, the bottom edge in the present subgraph were replaced by a blue edge – thus proving by counterexample that N* > 3. Graham's number is connected to the following problem in Ramsey theory: Connect each pair of geometric vertices of an n-dimensional hypercube to obtain a complete graph on 2n vertices. Colour each of the edges of this graph either red or blue. What is the smallest value of n for which every such colouring contains at least one single-coloured complete subgraph on four coplanar vertices? In 1971, Graham and Rothschild proved the Graham–Rothschild theorem on the Ramsey theory of parameter words, a special case of which shows that this problem has a solution N*. They bounded the value of N* by 6 ≤ N* ≤ N, with N being a large but explicitly defined number N = F 7 ( 12 ) = F ( F ( F ( F ( F ( F ( F ( 12 ) ) ) ) ) ) ) , {\displaystyle N=F^{7}(12)=F(F(F(F(F(F(F(12))))))),} where F ( n ) = 2 ↑ n 3 #aiactor #allfaketiktok #nohate #aigenerated #fakescenarios](/video/cover/7668136370359094559.webp)
