@ouss.barber: Si le client gagne à sa coupe gratuite 😂😂😂

OUSSBARBER ✪
OUSSBARBER ✪
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Region: FR
Monday 03 November 2025 20:34:23 GMT
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maeldu_78
mael78 :
« Il y en a un il a combattu au 1er étage et l’autre au rez-de-chaussée »🤣🤣
2025-11-04 17:07:49
19162
samirke93
Samir :
Le p’tit y c fais mélanger miskine
2025-11-04 16:43:50
17344
babacarrr.13
BS’13 :
ca c’est du divertissement
2025-11-04 16:04:44
6152
abdell9378
abdell9378 :
le petit il était grave content
2025-11-04 15:20:34
8073
ar0.mnch
￴ ￴ ￴ ￴ ￴ ￴ ￴ ￴ ￴ ￴ ￴ ￴ ￴ ￴ ￴ :
Imagine tu va te couper tu vois ton coiffeur entrain de faire un sparing
2025-11-04 16:54:30
5965
blackgoku932
Ibk93🐺🇲🇦 :
Js mort ca a dit le bb
2025-11-04 15:11:56
3640
hks.champtierducoq
hks.champtierducoq :
Le bb il kiffe sa vie mdr 😂😂🔥
2025-11-04 15:01:19
2341
rebeumelancolik
￴ ￴ ￴ ￴￴￴ ￴ ￴ ￴ ￴ ￴ ￴ ￴ :
Le p’tit derrière
2025-11-04 17:42:15
2199
anaissarahgzi
Anaïs Sarah Gzi :
Il m’a trop fait penser à l’acteur Kevin Hart 😂😂😂😂😂
2025-11-05 04:24:44
1183
k_213_dz
❄️☃️ :
Si on met ko y’a coupe gratuite a vie 😂
2025-11-04 15:48:16
578
shynz.4z
𝐒𝐇𝐀𝐘🪭 愛 :
Le petit il vas raconter sa a l’école Sv le prendre pour un fou alors que sa c’est vraiment passée 😂😂😂
2025-11-04 19:35:12
532
zyann_xx
𖣂 Zyann 𖣂 :
Meilleur salon du monde carrement
2025-11-04 17:51:03
438
user157471953
MLK :
Pour une coupe gratuite il va mettre l'employé en ITT 45 jours 🤣🤣🤣😂😂😂😂
2025-11-04 22:47:21
305
moi_ezeee
🇬🇦ezzzz🇬🇦 :
il aime trop
2025-11-04 23:44:17
166
rayanef.06
Rayane A Rome 🇮🇹 :
La france prime du divertissement ce salon mérite plus de visite 😭😭😭😭
2025-11-04 17:04:29
381
farid13k
Farid🇩🇿 :
Wsh il l'a voulait vrm sa coupe gratuite 🤣
2025-11-04 20:25:05
159
dciss9350
Diabé 🇲🇱 🇲🇷 :
Le caméraman il veut pas s’arrêter de rigoler
2025-11-05 02:03:19
125
inconnuuee1
inconuuu1 :
ppptdr le cameraman il vit sa best life 😭🤣🤣🤣
2025-11-05 15:48:08
29
david260520
davw :
meilleur salon de France je veux rien savoir 😂
2025-11-04 21:36:25
63
jahell_92
Ja🦶🏿 :
« Il y en a un il a combattu au 1er étage et l’autre au rez-de-chaussée »🤣🤣
2025-11-04 16:18:07
413
3thvnj
່ :
Ça c’est du concept
2025-11-05 09:19:57
12
oweisssyed1
oweiss.syed :
Kevin Hart et The Rock mdrrr
2025-11-04 17:37:52
48
tm_chrn
t0〽️! :
ko il voulait trop sa coupe gratuite 😂😂😂
2025-11-04 16:23:20
175
alasko_91
alasko_91🧨 :
mdr le grand il tape il esquive après wsh 😂😂
2025-11-04 15:59:16
123
nanou.nnl
nanou.nnl :
jsuis en fou rire 😹😹
2025-11-04 16:47:41
16
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#skittles #omarmattenbased #49 #troon #transition  ‎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.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 which contains three tetrations.[4] In 2019 this was further improved to[5]The lower bound of 6 was later improved to 11 by Geoffrey Exoo in 2003,[6] and to 13 by Jerome Barkley in 2008.[7] Thus, the best known bounds for N* are 13 ≤ N* ≤ N''. ‎ ‎Graham's number, G, is much larger than N: it is  ‎f ‎64 ‎( ‎4 ‎) ‎{\displaystyle f^{64}(4)}, where  ‎f ‎( ‎n ‎) ‎= ‎3 ‎↑ ‎n ‎3
#skittles #omarmattenbased #49 #troon #transition ‎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.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 which contains three tetrations.[4] In 2019 this was further improved to[5]The lower bound of 6 was later improved to 11 by Geoffrey Exoo in 2003,[6] and to 13 by Jerome Barkley in 2008.[7] Thus, the best known bounds for N* are 13 ≤ N* ≤ N''. ‎ ‎Graham's number, G, is much larger than N: it is ‎f ‎64 ‎( ‎4 ‎) ‎{\displaystyle f^{64}(4)}, where ‎f ‎( ‎n ‎) ‎= ‎3 ‎↑ ‎n ‎3

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