{"id":75498,"date":"2025-01-29T00:13:22","date_gmt":"2025-01-29T00:13:22","guid":{"rendered":"http:\/\/e-ciftlik.com\/haberler\/egitim\/matematikte-cozulememis-problemler-goldbach-ve-riemann-ikiz-asal-sayilar-riemann-hipotezi-ve-sonsuzluk-matematikteki-en-buyuk-sirlar-eulerin-mukemmel-kuboidi-euler-mascheroni-sabiti-rasyonel-m\/"},"modified":"2025-01-29T00:13:22","modified_gmt":"2025-01-29T00:13:22","slug":"matematikte-cozulememis-problemler-goldbach-ve-riemann-ikiz-asal-sayilar-riemann-hipotezi-ve-sonsuzluk-matematikteki-en-buyuk-sirlar-eulerin-mukemmel-kuboidi-euler-mascheroni-sabiti-rasyonel-m","status":"publish","type":"post","link":"https:\/\/e-ciftlik.com\/haberler\/egitim\/matematikte-cozulememis-problemler-goldbach-ve-riemann-ikiz-asal-sayilar-riemann-hipotezi-ve-sonsuzluk-matematikteki-en-buyuk-sirlar-eulerin-mukemmel-kuboidi-euler-mascheroni-sabiti-rasyonel-m\/","title":{"rendered":"Matematikte \u00c7\u00f6z\u00fclememi\u015f Problemler: Goldbach ve Riemann \u0130kiz Asal Say\u0131lar: Riemann Hipotezi ve Sonsuzluk “Matematikteki En B\u00fcy\u00fck S\u0131rlar: Euler’in M\u00fckemmel K\u00fcboidi” Euler – Mascheroni Sabiti Rasyonel mi? Erd\u00f6s-Strauss Varsay\u0131m\u0131 \u00c7\u00f6z\u00fclemiyor! “Hodge ve Birch Swinnerton-Dyer Kesinlikle Dikkat \u00c7ekici” “Matematikteki B\u00fcy\u00fck Ke\u015fif: Sonsuz Rasyonel Noktalar!” “G\u00fcndemdeki En \u00d6nemli Haberleri \u0130\u015fte Sizler \u0130\u00e7in Derledik!” “Yeni Ara\u015ft\u0131rmaya G\u00f6re: \u0130nsanlar Neden Mutlu Oluyor?” “Viral Haber: Son Dakika Geli\u015fmeler!” “Son dakika: Olay yerindeki muhabirin g\u00f6z\u00fcnden canl\u0131 yay\u0131n!”"},"content":{"rendered":"<\/p>\n

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\u0130kiz Asal Varsay\u0131m\u0131<\/p>\n

Asal say\u0131lar, kendisinden ve 1’den ba\u015fka b\u00f6leni olmayan tam say\u0131lard\u0131r. \u00d6rne\u011fin; 2, 3, 5, 17, 29 ve 53 asal say\u0131lard\u0131r. Aralar\u0131ndaki fark\u0131 2 olan asal say\u0131lara ise “ikiz asal say\u0131lar” denir. \u00d6rne\u011fin 3 ve 5, 5 ve 7, 11 ve 13 gibi ikiz asal say\u0131lard\u0131r. Bu kavram\u0131 ilk olarak 1846 y\u0131l\u0131nda Frans\u0131z matematik\u00e7i Alphonse de Polignac ortaya atm\u0131\u015ft\u0131r. Asal say\u0131lar\u0131n da\u011f\u0131l\u0131m\u0131 d\u00fczensiz oldu\u011fundan, say\u0131lar b\u00fcy\u00fcd\u00fck\u00e7e asal say\u0131lar seyrekle\u015fir. \u00d6klid, asal say\u0131lar\u0131n sonsuz oldu\u011funu kan\u0131tlam\u0131\u015ft\u0131. Ancak ikiz asal say\u0131lar hakk\u0131nda hala bir yan\u0131t aranmaktad\u0131r.<\/p>\n

Patreon<\/p>\n

Patreon destek\u00e7ilerimiz, Evrim A\u011fac\u0131’na destek olduklar\u0131 s\u00fcre boyunca, destek miktar\u0131ndan ba\u011f\u0131ms\u0131z olarak reklams\u0131z deneyime eri\u015febilirler. Patreon destek\u00e7ilerimizin reklams\u0131z deneyimi, destek sa\u011flad\u0131klar\u0131 anda devreye girer ve ek bir i\u015flem yapmalar\u0131 gerekmez. Patreon destek\u00e7ilerinin Evrim A\u011fac\u0131’na ba\u011f\u0131\u015f yapt\u0131klar\u0131 e-posta hesaplar\u0131, \u00fcyelik e-postalar\u0131yla ayn\u0131 olmal\u0131d\u0131r. Reklams\u0131z deneyim, destek sa\u011fland\u0131ktan sonra 24 saat i\u00e7inde devreye girebilir.<\/p>\n

YouTube<\/p>\n

YouTube destek\u00e7ilerimiz \u015fu anda otomatik olarak reklams\u0131z deneyime eri\u015fememektedirler. Farkl\u0131 seviyelerde sunulan ayr\u0131cal\u0131klar\u0131 \u00f6\u011frenmek i\u00e7in YouTube Destek Sistemi’ndeki a\u00e7\u0131klamalar\u0131 okuyabilirsiniz. E\u011fer se\u00e7ti\u011finiz seviye reklams\u0131z deneyim sunuyorsa, destek sa\u011flad\u0131ktan sonra YouTube taraf\u0131ndan g\u00f6nderilen ba\u011flant\u0131daki formu doldurarak reklams\u0131z deneyime eri\u015febilirsiniz. YouTube destek\u00e7ilerimizin reklams\u0131z deneyimi, formu doldurduktan sonra 24-72 saat i\u00e7inde devreye girebilir.<\/p>\n

Di\u011fer Platformlar<\/p>\n

Patreon ve YouTube d\u0131\u015f\u0131ndaki platformlarda destek olan destek\u00e7ilerimize maalesef reklams\u0131z deneyim ayr\u0131cal\u0131\u011f\u0131 sunamamaktay\u0131z. Ancak destekleriniz sayesinde sistemleri geli\u015ftirmeye devam ediyor ve zamanla bu ayr\u0131cal\u0131klar\u0131 geni\u015fletebilece\u011fimizi umuyoruz.<\/p>\n

Giri\u015f Yapmay\u0131 Unutmay\u0131n!<\/p>\n

Reklams\u0131z deneyim i\u00e7in, maddi destekle ili\u015fkilendirilmi\u015f olan Evrim A\u011fac\u0131 hesab\u0131n\u0131za giri\u015f yapman\u0131z gerekmektedir. Giri\u015f yapmad\u0131\u011f\u0131n\u0131z takdirde reklamlar\u0131 g\u00f6rmeye devam edeceksinizdir.<\/p>\n

Collatz Problemi<\/p>\n

Collatz Problemi, herhangi bir say\u0131 se\u00e7ildi\u011finde, \u00e7iftse 2’ye b\u00f6l\u00fcn\u00fcp, tekse 3 ile \u00e7arp\u0131l\u0131p 1 eklenerek devam eden bir matematiksel problemdir. Bu i\u015flem sonucunda 4, 2, 1 d\u00f6ng\u00fcs\u00fcne ula\u015f\u0131ld\u0131\u011f\u0131 g\u00f6zlemlenmi\u015ftir. Bu problemin ke\u015ffi 1932 y\u0131l\u0131nda Lothar Collatz taraf\u0131ndan yap\u0131lm\u0131\u015ft\u0131r. G\u00fcn\u00fcm\u00fczde bu i\u015flemin sonucunda 4, 2, 1 d\u00f6ng\u00fcs\u00fcn\u00fc elde edemeyen bir say\u0131 hen\u00fcz bulunamam\u0131\u015ft\u0131r.<\/p>\n

196 Say\u0131s\u0131 Problemi<\/p>\n

Palindromik say\u0131lar, tersten ve d\u00fcz olarak yaz\u0131ld\u0131klar\u0131nda ayn\u0131 olan say\u0131lard\u0131r. \u00d6rne\u011fin, 1221 bir palindromik say\u0131d\u0131r. Baz\u0131 say\u0131lar kullan\u0131larak palindromik say\u0131lar elde edilebilir. \u00d6rne\u011fin, 54 ve 45’i toplad\u0131\u011f\u0131m\u0131zda 99’a ula\u015f\u0131r\u0131z. Bu palindromik olmayan say\u0131y\u0131 kullanarak i\u015fleme devam edebiliriz. Bu \u015fekilde, palindromik say\u0131lara ula\u015fabiliriz. Matematik d\u00fcnyas\u0131nda ilgin\u00e7 bir problem g\u00fcndemde: Palindromik say\u0131lar. \u0130ki say\u0131n\u0131n toplam\u0131n\u0131n, elde edilen toplam\u0131n rakamlar\u0131 ters s\u0131raland\u0131\u011f\u0131nda ayn\u0131 say\u0131y\u0131 veren palindromik say\u0131lar olduk\u00e7a ilgin\u00e7 bir konu. \u00d6rne\u011fin, 726 ve 627 say\u0131lar\u0131n\u0131n toplam\u0131 1353’t\u00fcr ancak bu say\u0131 palindromik de\u011fildir.<\/p>\n

Ancak, 1353 ve 3531 say\u0131lar\u0131 topland\u0131\u011f\u0131nda 4884, yani bir palindromik say\u0131 elde edilir. Bu durumda i\u015flem sona erer.<\/p>\n

196 say\u0131s\u0131, palindromik say\u0131 olmayan en k\u00fc\u00e7\u00fck say\u0131 olarak bilinir. Ayr\u0131ca, 196’dan ba\u015fka palindromik say\u0131 olmayan di\u011fer say\u0131lar da vard\u0131r: 295, 394, 493, 592, 689, 691, 788, 790, 879, 887…<\/p>\n

1990 y\u0131l\u0131nda John Walker isimli bir programc\u0131, 196 say\u0131s\u0131 i\u00e7in bu i\u015flemi 2.415.836 kez tekrarlam\u0131\u015f ve milyonlarca basamaktan olu\u015fan, palindromik olmayan bir say\u0131 bulmu\u015ftur. 2012 y\u0131l\u0131nda yap\u0131lan bir ara\u015ft\u0131rma ise bu i\u015flemin devam edilmesi halinde, palindromik bir say\u0131ya ula\u015f\u0131lmas\u0131 durumunda bu say\u0131n\u0131n 600 milyondan fazla basamaktan olu\u015faca\u011f\u0131n\u0131 ortaya koymu\u015ftur.<\/p>\n

Matematik d\u00fcnyas\u0131ndaki bir di\u011fer ilgin\u00e7 konu ise Mutlu Son Problemi’dir. Bu probleme bu ismin verilmesinin sebebi, problem \u00fczerinde \u00e7al\u0131\u015fan iki matematik\u00e7i Esther Klein ve George Szekeres’in evlenmeleridir. Problemi basit\u00e7e ifade etmek gerekirse; rastgele da\u011f\u0131lm\u0131\u015f 5 noktadan olu\u015fan bir d\u00fczlemde, bu noktalardan d\u00f6rd\u00fc kullan\u0131larak daima bir konveks d\u00f6rtgen elde edilebilir. Ancak, be\u015fgen olu\u015fturmak i\u00e7in en az 9 noktaya, alt\u0131gen olu\u015fturmak i\u00e7in ise en az 17 noktaya ihtiya\u00e7 vard\u0131r. Ancak, yedigen ve sonras\u0131nda ne kadar noktaya ihtiya\u00e7 duyulaca\u011f\u0131 hala bir bilinmezdir.<\/p>\n

Euler’in M\u00fckemmel K\u00fcboidi ise matematik d\u00fcnyas\u0131ndaki di\u011fer bir ilgin\u00e7 konudur. Bu konuda aranan \u00f6zellik; a, b, c ve g olmak \u00fczere k\u00fcboidin kenar uzunluklar\u0131 ve hacim k\u00f6\u015fegeni ile y\u00fczey k\u00f6\u015fegenlerinin tam say\u0131 olmas\u0131d\u0131r. Bu konu hen\u00fcz \u00e7\u00f6z\u00fclememi\u015f olsa da, matematik\u00e7iler bu konuda \u00e7al\u0131\u015fmalar\u0131na devam etmektedirler. Charles Dickens’in “American Notes for General Circulation” adl\u0131 eseri, 1842 y\u0131l\u0131nda Ocak ay\u0131ndan Haziran ay\u0131na kadar Kuzey Amerika’ya yapt\u0131\u011f\u0131 seyahati detayl\u0131 bir \u015fekilde anlatmaktad\u0131r. Burada, Kuzey Amerika toplumunu ele\u015ftirel bir g\u00f6zlemci olarak g\u00f6rev yapm\u0131\u015f ve neredeyse ilerleme durumlar\u0131 hakk\u0131nda bir durum raporu sunmu\u015f gibi davranm\u0131\u015ft\u0131r. Bu, d\u00f6rt y\u0131l sonra yazd\u0131\u011f\u0131 Italya Resimleri tarz\u0131yla kar\u015f\u0131la\u015ft\u0131r\u0131labilir, burada daha \u00e7ok bir turist gibi yazm\u0131\u015ft\u0131r. Amerika seyahati ayr\u0131ca Martin Chuzzlewit adl\u0131 roman\u0131n\u0131n da ilham kayna\u011f\u0131 olmu\u015ftur. Boston’a var\u0131\u015f yapt\u0131ktan sonra Lowell, New York ve Philadelphia’y\u0131 ziyaret etmi\u015f ve Richmond’e, St. Louis’e kadar ve Quebec’e kadar g\u00fcneye do\u011fru seyahat etmi\u015ftir. Dickens’in en \u00e7ok be\u011fendi\u011fi Amerikan \u015fehri Boston olmu\u015ftur – “hava o kadar temizdi, evler o kadar parlakt\u0131 ve ne\u015feliydi. […] \u015eehir g\u00fczel bir \u015fehir ve t\u00fcm yabanc\u0131lar\u0131 olduk\u00e7a olumlu etkilemekte ba\u015far\u0131s\u0131z olamazd\u0131.” Ayr\u0131ca, Perkins Kurumu ve Massachusetts K\u00f6rl\u00fcler Asylum’unun yak\u0131n\u0131nda bulunan Laura Bridgman ile tan\u0131\u015ft\u0131\u011f\u0131 yer olan Boston, onu b\u00fcy\u00fck \u00f6l\u00e7\u00fcde etkilemi\u015ftir.<\/p>\n

Leonhard Euler taraf\u0131ndan 1734 y\u0131l\u0131nda ke\u015ffedilen ve \u03b3 sembol\u00fc ile temsil edilen Euler – Mascheroni sabiti, Euler’in 16 ondal\u0131k basama\u011fa kadar hesaplad\u0131\u011f\u0131 bir sabittir. 1790 y\u0131l\u0131nda Lorenzo Mascheroni taraf\u0131ndan 32 ondal\u0131k basama\u011fa kadar hesaplanarak say\u0131 geni\u015fletilmi\u015ftir. Euler – Mascheroni sabiti, harmonik serilerle do\u011fal logaritman\u0131n fark\u0131na veya s\u0131n\u0131ra e\u015fittir ve a\u015fa\u011f\u0131daki gibi ifade edilir: \u03b3 = lim (n \u2192 \u221e) (\u2211 k = 1 ^ n 1 \/ k – ln (n)). Bu sabit, matemati\u011fin bir\u00e7ok alan\u0131nda kullan\u0131lsa da, say\u0131n\u0131n rasyonel mi yoksa irrasyonel mi oldu\u011fu hala belirsizli\u011fini korumaktad\u0131r.<\/p>\n

Claude Louis Navier ve George Gabriel Stokes’un ad\u0131n\u0131 ta\u015f\u0131yan Navier – Stokes denklemleri, ak\u0131\u015fkanlar\u0131n hareketini a\u00e7\u0131klayan diferansiyel denklemlerdir. Bu denklemler, muslu\u011funuzdan akan suyun hareketini veya u\u00e7an bir u\u00e7a\u011f\u0131n kanad\u0131n\u0131n etraf\u0131ndaki hava ak\u0131\u015f\u0131n\u0131 tan\u0131mlamak i\u00e7in kullan\u0131labilir. Ancak, bu denklemler her zaman do\u011fru sonu\u00e7lar vermeyebilir. Navier – Stokes denklemleri, yaln\u0131zca belirli bir sistemin temsili fiziksel uzunluk \u00f6l\u00e7e\u011fi, ak\u0131\u015fkan\u0131 olu\u015fturan molek\u00fcllerin ortalama serbest yolundan \u00e7ok daha b\u00fcy\u00fck oldu\u011funda ge\u00e7erlidir. Matematik d\u00fcnyas\u0131nda 1948 y\u0131l\u0131nda Paul Erd\u00f6s ve Ernst Strauss taraf\u0131ndan ortaya at\u0131lan bir varsay\u0131m, hala \u00e7\u00f6z\u00fclememi\u015f durumda. Erd\u00f6s – Strauss Varsay\u0131m\u0131 olarak bilinen bu problemde, en az 2 say\u0131s\u0131ndan olu\u015fan n de\u011feri i\u00e7in, 4\/n = 1\/a + 1\/b + 1\/c e\u015fitli\u011fini sa\u011flayan a, b ve c pozitif tam say\u0131lar\u0131n\u0131n varl\u0131\u011f\u0131 ara\u015ft\u0131r\u0131l\u0131yor. Matematik\u00e7iler bu problemi \u00e7\u00f6zmek i\u00e7in \u00e7aba g\u00f6steriyor ancak hen\u00fcz ba\u015far\u0131l\u0131 olabilmi\u015f de\u011filler. Erd\u00f6s – Strauss Varsay\u0131m\u0131’n\u0131n \u00e7\u00f6z\u00fcm\u00fc, matematik d\u00fcnyas\u0131nda b\u00fcy\u00fck bir ilgiyle bekleniyor. Matematik\u00e7iler, 4\/n4\/n<\/span>4\/<\/span>n<\/span><\/span><\/span><\/span><\/span><\/span> kesirli ifadeyi \u00fc\u00e7 pozitif birim kesrin toplam\u0131 olarak yazabilir miydi? \u00d6rne\u011fin n=5n=5<\/span>n<\/span>=<\/span><\/span>5<\/span><\/span><\/span><\/span><\/span><\/span> \u015feklinde i\u015flem yapt\u0131\u011f\u0131m\u0131zda, iki farkl\u0131 \u00e7\u00f6z\u00fcm yolu elde ederiz:<\/p>\n

\u0130lk olarak, 45=12+14+120\\frac{4}{5} = \\frac{1}{2} +\\frac{1}{4} + \\frac{1}{20}<\/span>5<\/span><\/span><\/span><\/span><\/span>4<\/span><\/span><\/span><\/span><\/span>\u200b<\/span><\/span><\/span><\/span><\/span><\/span><\/span>=<\/span><\/span>2<\/span><\/span><\/span><\/span><\/span>1<\/span><\/span><\/span><\/span><\/span>\u200b<\/span><\/span><\/span><\/span><\/span><\/span><\/span>+<\/span><\/span>4<\/span><\/span><\/span><\/span><\/span>1<\/span><\/span><\/span><\/span><\/span>\u200b<\/span><\/span><\/span><\/span><\/span><\/span><\/span>+<\/span><\/span>20<\/span><\/span><\/span><\/span><\/span>1<\/span><\/span><\/span><\/span><\/span>\u200b<\/span><\/span><\/span><\/span><\/span><\/span><\/span><\/span><\/span><\/span><\/span><\/span><\/p>\n

\u0130kinci olarak, 45=12+15+110\\frac{4}{5} = \\frac{1}{2} +\\frac{1}{5} + \\frac{1}{10}<\/span>5<\/span><\/span><\/span><\/span><\/span>4<\/span><\/span><\/span><\/span><\/span>\u200b<\/span><\/span><\/span><\/span><\/span><\/span><\/span>=<\/span><\/span>2<\/span><\/span><\/span><\/span><\/span>1<\/span><\/span><\/span><\/span><\/span>\u200b<\/span><\/span><\/span><\/span><\/span><\/span><\/span>+<\/span><\/span>5<\/span><\/span><\/span><\/span><\/span>1<\/span><\/span><\/span><\/span><\/span>\u200b<\/span><\/span><\/span><\/span><\/span><\/span><\/span>+<\/span><\/span>10<\/span><\/span><\/span><\/span><\/span>1<\/span><\/span><\/span><\/span><\/span>\u200b<\/span><\/span><\/span><\/span><\/span><\/span><\/span><\/span><\/span><\/span><\/span><\/span><\/p>\n

Bu durum t\u00fcm pozitif tam say\u0131lar i\u00e7in ge\u00e7erli midir? Matematik\u00e7ilerin \u00e7o\u011fu buna “evet” diyor, ancak bu varsay\u0131m hala \u00e7\u00f6z\u00fclememi\u015f en \u00f6nemli problemlerden biri olarak kalmaya devam ediyor.<\/p>\n

### Hodge Varsay\u0131m\u0131<\/p>\n

Hodge varsay\u0131m\u0131, yedi milenyum probleminden biri olarak bilinir ve 1950 y\u0131l\u0131nda William Hodge taraf\u0131ndan ortaya at\u0131lm\u0131\u015ft\u0131r. Bu teori, harmonik formlar\u0131 homoloji elemanlar\u0131yla ili\u015fkilendirerek matematiksel analiz ve topoloji aras\u0131nda bir ba\u011flant\u0131 kurmaya \u00e7al\u0131\u015f\u0131r. Karma\u015f\u0131k alt manifoldlar\u0131n karma\u015f\u0131k manifoldlar i\u00e7indeki varolu\u015funu a\u00e7\u0131klamak i\u00e7in do\u011fal bir durum \u00f6nerisi sunar. G\u00fcn\u00fcm\u00fczde bu teori, geometri, analiz ve matematiksel fizi\u011fin geli\u015fimine \u00f6nemli katk\u0131lar sa\u011flamaktad\u0131r.<\/p>\n

Hodge varsay\u0131m\u0131, baz\u0131 geometrik yap\u0131 t\u00fcrlerini daha iyi incelemek ve s\u0131n\u0131fland\u0131rmak i\u00e7in cebirsel kar\u015f\u0131l\u0131klar oldu\u011funu \u00f6ne s\u00fcrer. Bu yakla\u015f\u0131m matematik\u00e7iler aras\u0131nda tart\u0131\u015fmal\u0131 olsa da, genel olarak matematikte ilerlemeye b\u00fcy\u00fck katk\u0131 sa\u011flamaktad\u0131r.<\/p>\n

### Birch ve Swinnerton-Dyer Kestirimi<\/p>\n

Eliptik e\u011frileri incelemek i\u00e7in y2=x3+ax+by^2=x^3+ax+b<\/span>\n","protected":false},"excerpt":{"rendered":"

\u0130kiz Asal Varsay\u0131m\u0131 Asal say\u0131lar, kendisinden ve 1’den ba\u015fka b\u00f6leni olmayan tam say\u0131lard\u0131r. \u00d6rne\u011fin; 2,…<\/p>\n","protected":false},"author":1,"featured_media":75499,"comment_status":"open","ping_status":"open","sticky":false,"template":"","format":"standard","meta":{"footnotes":""},"categories":[12],"tags":[],"class_list":["post-75498","post","type-post","status-publish","format-standard","has-post-thumbnail","hentry","category-egitim"],"yoast_head":"\nMatematikte \u00c7\u00f6z\u00fclememi\u015f Problemler: Goldbach ve Riemann \u0130kiz Asal Say\u0131lar: Riemann Hipotezi ve Sonsuzluk "Matematikteki En B\u00fcy\u00fck S\u0131rlar: Euler'in M\u00fckemmel K\u00fcboidi" Euler - Mascheroni Sabiti Rasyonel mi? Erd\u00f6s-Strauss Varsay\u0131m\u0131 \u00c7\u00f6z\u00fclemiyor! "Hodge ve Birch Swinnerton-Dyer Kesinlikle Dikkat \u00c7ekici" "Matematikteki B\u00fcy\u00fck Ke\u015fif: Sonsuz Rasyonel Noktalar!" "G\u00fcndemdeki En \u00d6nemli Haberleri \u0130\u015fte Sizler \u0130\u00e7in Derledik!" "Yeni Ara\u015ft\u0131rmaya G\u00f6re: \u0130nsanlar Neden Mutlu Oluyor?" "Viral Haber: Son Dakika Geli\u015fmeler!" "Son dakika: Olay yerindeki muhabirin g\u00f6z\u00fcnden canl\u0131 yay\u0131n!" - E-Ç\u0130FTL\u0130K HABER<\/title>\n<meta name=\"description\" content=\"\u0130kiz Asal Varsay\u0131m\u0131 Asal say\u0131lar, kendisinden ve 1'den ba\u015fka b\u00f6leni olmayan tam say\u0131lard\u0131r. \u00d6rne\u011fin; 2, 3, 5, 17, 29 ve 53 asal say\u0131lard\u0131r. 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