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OEIS A212558: Proof of Unproven Conjecture? Proven! +- HP Forums (http://www.hpmuseum.org/forum) +-- Forum: HP Calculators (and very old HP Computers) (/forum-3.html) +--- Forum: General Forum (/forum-4.html) +--- Thread: OEIS A212558: Proof of Unproven Conjecture? Proven! (/thread-9719.html) |
OEIS A212558: Proof of Unproven Conjecture? Proven! - Gerald H - 12-18-2017 02:23 AM The sequence https://oeis.org/A212558 defined as a(n) = ((n - s)^2 mod (n + s)) - ((n + s)^2 mod (n - s)), where s is the sum of the decimal digits of n has the value 0 for 2,999 < n < 20,000,000 & for random input 20,000,000 =< n < 10^94. Can anyone suggest a proof of 0 value for all integer input > 2,999 ? RE: OEIS A212558: Proof of Unproven Conjecture? - stored - 12-18-2017 04:21 AM (n - s)^2 = (n + s)^2 - 4 n s = (n + s)^2 - 4 s (n + s) + 4 s^2 (n + s)^2 = (n - s)^2 + 4 n s = (n - s)^2 + 4 s (n - s) + 4 s^2 Then source problem is equivalent 4 s^2 mod (n+s) = 4 s^2 mod (n-s) This is true for all s, that 4 s^2 < n-s. RE: OEIS A212558: Proof of Unproven Conjecture? Proven! - stored - 12-19-2017 12:21 AM Gerald, thanks you for the interesting quiz! Small addendum about sum of the decimal digits of n. Let s(n) is the sum of the decimal digits of n. Consider condition 4*s(n)^2 < n-s(n), (*) 4*s(n)^2 + s(n) < n Function in left is monotonically increasing function, then in respect that s(n) <= 9*log10(n+1) we get estimation 4*s(n)^2 + s(n) <= 4 * (9 * log10(n+1))^2 + 9*log10(n+1) = 324 * log10(n+1)^2 + 9*log10(n+1) Maximal solution of the equation 324 * log10(n+1)^2 + 9*log10(n+1) = n is n0 ~ 4313.68. (I use Wolfram Alpha for getting this value.) Hence, condition (*) is true for all n > n0. For n from 2999 to 4313 source statement may be checked by direct computations. RE: OEIS A212558: Proof of Unproven Conjecture? Proven! - Gerald H - 12-19-2017 02:10 AM & a programme for the 49G: Code: :: |