June 20th, 2019, 03:01 AM  #11  
Senior Member Joined: Aug 2008 From: Blacksburg VA USA Posts: 351 Thanks: 7 Math Focus: primes of course  Quote:
Specifically, from an empty list, +1, +3, +5,...etc. You can use Wolfram Alpha input "digitsum(xxxx)" if that helps. I did it in a spreadsheet. Unless I'm wrong, the pattern is there. Now, for any # of 1's, it would be just slight variations on what I showed, allowing for the last set not being the full nine, but having only the first 4 of the 9 for example.  
June 20th, 2019, 03:19 AM  #12 
Newbie Joined: Mar 2017 From: Norway Posts: 26 Thanks: 0 
I made something like this: 760cf057ab2743f28e017b3a1926f278_A.jpg 
June 20th, 2019, 03:32 AM  #13 
Senior Member Joined: Mar 2019 From: iran Posts: 316 Thanks: 14 
i found a formula s(n) = sum of the digits of (111...1)^2 with n 1s s(n) = (integer part of n/10)*100 + m^2 where m is the last digit of n s(1989) = 198*100 + 9^2 = 19881 
June 20th, 2019, 03:47 AM  #14 
Newbie Joined: Mar 2017 From: Norway Posts: 26 Thanks: 0 
I am sorry but this is not correct. The correct answer is 17901, but how? For example with your formula when there are 12 "1"s: s(12)=1*100+2^2=104 but the correct answer is: (111111111111)^2=12345679012320987654321 and the sum is 90 Last edited by Farzin; June 20th, 2019 at 04:06 AM. 
June 20th, 2019, 03:58 AM  #15 
Senior Member Joined: Mar 2019 From: iran Posts: 316 Thanks: 14 
this one is true: s(n) = (integer part of n/9)*81 + m^2 where m is the last digit of n in base 9 
June 20th, 2019, 04:08 AM  #16 
Newbie Joined: Mar 2017 From: Norway Posts: 26 Thanks: 0 
Can you explain where that formula came from?

June 20th, 2019, 04:22 AM  #17 
Senior Member Joined: Mar 2019 From: iran Posts: 316 Thanks: 14 
i checked for smaller bases and i saw this pattern: s(n) = sum of the digits of (111..1)^2 with n 1s in base k s(n) = (integer part of n/(k1))*(k1)^2 + m^2 where m is the last digit of n in base (k  1) 
June 20th, 2019, 04:28 AM  #18 
Senior Member Joined: Aug 2008 From: Blacksburg VA USA Posts: 351 Thanks: 7 Math Focus: primes of course  
June 20th, 2019, 07:47 AM  #19 
Newbie Joined: Mar 2017 From: Norway Posts: 26 Thanks: 0 
I do see that 17901 is equal to that product, but is it the general formula? We need to find a general formula which works for any # of "1"s. 
June 20th, 2019, 07:50 AM  #20 
Senior Member Joined: Aug 2008 From: Blacksburg VA USA Posts: 351 Thanks: 7 Math Focus: primes of course 
Yes, I explained it, I led you to the water. I didn't see you post the 20 or so values I suggested...


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