[math-fun] Chunks to be erased = summing digits
Hello Math-Fun, The idea is to read the decimalization of Pi from left to right and to delete _all_ the read digits with the hereunder rule: - Sum, one by one, the digits you encounter; - When the partial sum is visible immediately to the right of the summed digits, erase the said digits; - Start summing again, from zero. Example: Pi decimalization starts: 3.1.4.1.5.9.2.6.5.3.5.8.9.7.9.3.2.3.8.4.6.2.6.4.3.3.8.3.2.7.9.5.0.2.8.8.4.1.9.7.1.6.9.3.9.9.3.7.5.1.0.5.8.2.0.9.7.4.9.4.4.5.9.2.3.0.7.8.1.6.4.0.6.2.8.6.2.0.8.9.9.8.6.2.8.0.3.4.8.2.5.3.4.2.1.1.7.0.6.7.9... We read from left to right; as 3+1 = 4 we erase the first two digits; we are left with: 4.1.5.9.2.6.5.3.5.8.9.7.9.3.2.3.8.4.6.2.6.4.3.3.8.3.2.7.9.5.0.2.8.8.4.1.9.7.1.6.9.3.9.9.3.7.5.1.0.5.8.2.0.9.7.4.9.4.4.5.9.2.3.0.7.8.1.6.4.0.6.2.8.6.2.0.8.9.9.8.6.2.8.0.3.4.8.2.5.3.4.2.1.1.7.0.6.7.9... We start summing again, from left to right; as 4+1 = 5 we erase the first two digits; we are left with: 5.9.2.6.5.3.5.8.9.7.9.3.2.3.8.4.6.2.6.4.3.3.8.3.2.7.9.5.0.2.8.8.4.1.9.7.1.6.9.3.9.9.3.7.5.1.0.5.8.2.0.9.7.4.9.4.4.5.9.2.3.0.7.8.1.6.4.0.6.2.8.6.2.0.8.9.9.8.6.2.8.0.3.4.8.2.5.3.4.2.1.1.7.0.6.7.9... We start summing again, from left to right; as ... [and here I must confess that I don't have the tools to find the next chunk of digits; the sum is bigger than 100] This erasing technique can be summarized by the size (in digits) of each erased chunk. We would thus have(for Pi): [2,2,...] Best, É. e size of each chunk is the signature of th
When the partial sum is visible immediately to the right of the summed digits, erase the said digits;
If after dropping some initial digits we are left with say 4.4.8…, the first 4 matches the second 4 without being technically a sum. We would drop the one digit and continue with 4.8… Correct? By "visible immediately to the right of the summed digits", you mean the *digits of the sum* are immediately to the right of the summed digits? So if after dropping some initial digits we are left with say 7.8.9.0.0.2.4.1…, we find a match after 5 (but not after 3) digits. Correct? Neither of the above situations (repeating-digit start, leading zeros) occurs in my search but being explicit is always helpful in creating an algorithm. For the number-of-dropped-digits sequence, I have seven terms: 2, 2, 10654, 61938, 59585723, 2, 195. The respective sums are: 4, 5, 47894, 278677, 268108667, 8, 933.
Hans:
So if after dropping some initial digits we are left with say 7.8.9.0.0.2.4.1…, we find a match after 5 (but not after 3) digits. Correct?
... yes, correct, good example Hans! And thanks for the first terms of the sequence! Best, É. Propulsé d'un aPhone Le 21 juin 2013 à 14:29, "Hans Havermann" <gladhobo@teksavvy.com<mailto:gladhobo@teksavvy.com>> a écrit : When the partial sum is visible immediately to the right of the summed digits, erase the said digits; If after dropping some initial digits we are left with say 4.4.8…, the first 4 matches the second 4 without being technically a sum. We would drop the one digit and continue with 4.8… Correct? By "visible immediately to the right of the summed digits", you mean the *digits of the sum* are immediately to the right of the summed digits? So if after dropping some initial digits we are left with say 7.8.9.0.0.2.4.1…, we find a match after 5 (but not after 3) digits. Correct? Neither of the above situations (repeating-digit start, leading zeros) occurs in my search but being explicit is always helpful in creating an algorithm. For the number-of-dropped-digits sequence, I have seven terms: 2, 2, 10654, 61938, 59585723, 2, 195. The respective sums are: 4, 5, 47894, 278677, 268108667, 8, 933. _______________________________________________ math-fun mailing list math-fun@mailman.xmission.com<mailto:math-fun@mailman.xmission.com> http://mailman.xmission.com/cgi-bin/mailman/listinfo/math-fun
Hans again:
If after dropping some initial digits we are left with say 4.4.8…, the first 4 matches the second 4 without being technically a sum. We would drop the one digit and continue with 4.8… Correct?
... yes, correct too. Sorry, I should have provided those nice examples myself :-(( Best, É. Propulsé d'un aPhone Le 21 juin 2013 à 14:29, "Hans Havermann" <gladhobo@teksavvy.com<mailto:gladhobo@teksavvy.com>> a écrit : If after dropping some initial digits we are left with say 4.4.8…, the first 4 matches the second 4 without being technically a sum. We would drop the one digit and continue with 4.8… Correct?
participants (2)
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Eric Angelini -
Hans Havermann