20200213, 12:22  #1 
Jan 2020
2^{2}×43 Posts 
Twin Primes with 128 Decimal Digits
2^{191}*3^{90}*82589933*40463441610*232792560±1
Click the attached thumbnail to view all the digits in decimal base. Would like to know why when the number consists 82,589,933 as the factor will result in twin primes at certain power combinations. Last fiddled with by tuckerkao on 20200213 at 12:24 
20200215, 04:06  #2 
∂^{2}ω=0
Sep 2002
República de California
11609_{10} Posts 
@tuckertao: It would help if you explained how you chose/found the above pair, and why you think it is somehow special. You have a largeish smooth part in form of 2^191*3^90, then the largestknown Mersenne prime exponent, but I see nothing special about the remaining multiplier 40463441610*232792560 = 9419588158802421600. If I define a = 2^191*3^90*82589933, set n to your prime, then successively increment n by a, I can easily find the nextlarger pair of this form,
Code:
n = 21310867332998259383016283883353020788678719772398547918516685244168507150940923363729663389961621332915816701945558484016693248+1 = 2^191*3^90*82589933*9419588158802425428+1 = 2^193*3^91*82589933*11*29*89*27648398432609+1. 
20200216, 06:23  #3  
Jan 2020
2^{2}·43 Posts 
Quote:
My formula is 2^{n}*3^{m}*prime*large abundant number±1 I'm wondering whether it's still possible to find the next larger pair when both n and m are in the millions or larger with a megaprime possibly the largest known Mersenne Prime. I go from 2^{0}*3^{m} to 2^{n}*3^{0}, so I have a whole Arc rotation on a given orbit(fix # of numerical digits), doesn't have to be 128 digits, it can be 24 millions+ total. Last fiddled with by tuckerkao on 20200216 at 06:24 

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