On Sat, 08 Dec 2012 19:13:02 +0000, Dr J R Stockton wrote: > ...Thu, 6 Dec 2012 23:38:29, Dr J R Stockton ... posted: > >>In that calculation, longcalc uses only elementary arithmetic, as used >>to be taught in schools in my day. Your proof, however, is 50% >>incomprehensible to me. Though it may well be right. Longcalc found an >>error in *a* printed representation of (3^349-1)/2, which you might have >>difficulty with. Fx : checks : I think it has found another one. > > Can anyone (or more) please provide here the last ten decimal digits (in > order) of ((3^349)-1)/2, freshly and independently calculated and not > copied from any other medium, and not using my LongCalc or VastCalc?
Input ((3^349)-1)/2 to bc reports 16379019558053662392174130154670449583923965684832704024983781709239\ 69468635132120415650964922608054197182470755579714456896907387777297\ 3038883717449030628887379284041
of which the last 10 digits are 7379284041.
python, with input (pow(3,349,1000000000000000)-1)/2 reports 128887379284041, giving 7379284041 also.