December 2007 - mathfun@mailman.xmission.com

[math-fun] Strange answer
by Eric Angelini 03 Dec '07

03 Dec '07

http://www.cetteadressecomportecinquantesignes.com/AutoGooo.htm Best, É. (Dylan fan)

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RE: [math-fun] Add first diff of digits and repeat
by Eric Angelini 03 Dec '07

03 Dec '07

Thanks a lot, Mike. EA: > Are there integers leading to infinite sequences? Mike: > no, you can never increase the number of digits. (...) --- Edwin Clark has kindly computed this: n = 1, max num iterations jump = 1, final value = 1 n = 10, max num iterations jump = 2, final value = 11 n = 12, max num iterations jump = 7, final value = 33 n = 18, max num iterations jump = 8, final value = 44 n = 100, max num iterations jump = 20, final value = 222 n = 200, max num iterations jump = 42, final value = 444 n = 201, max num iterations jump = 43, final value = 444 n = 1000, max num iterations jump = 377, final value = 4444 n = 10000, max num iterations jump = 2926, final value = 44444 n = 100000, max num iterations jump = 23541, final value = 444444 Best, É.

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Re: [math-fun] Add first diff of digits and repeat
by Michael Reid 03 Dec '07

03 Dec '07

> Let's say a(n) = 329441. How do we build a(n+1)? > First, sum, from left to right, the absolute differences of a(n)'s digits: > a(n) = 3 2 9 4 4 1 > abs. diff = 1+7+5+0+3 = 16 > Now add this sum to a(n) so to produce a(n+1): > 329441 + 16 = 329457 = a(n+1) > And repeat: [ ... ] > The sequence stops when all digits off a(n) are the same, of course: > If we start with 18, for instance, we have: > S = 18, 25, 28, 34, 35, 37, 41, 44, 44, 44, ... > Are there integers leading to infinite sequences? no, you can never increase the number of digits. for 1-digit numbers, this is clear. for 2-digit numbers, you add <= 9 at every step, so if you reach 3-digits, the previous number is 9a for some digit a , but this would make the next term 99 . for 3-digit numbers, you add <= 18 at every step, so if you reach 4-digits, the last 3-digit number is in the range 982 - 999 . but for these numbers, you add at most 10 . thus the last 3-digit number is at least 990 . for any number of the form 99a , the next term is 999 . for n >= 4 digits, at each step, you increase by at most 9 (n - 1) , which is less than 10^(n-2) . if you reach n + 1 digits, the last n-digit number has the form 99abc...x . for such a number, you increase it by the same amount as you increase 9abc...x , which means that the sequence is the same, except for putting an extra digit of 9 in front. since n-1 digit numbers can't reach n digits, the same holds for n digit numbers by induction. mike

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Re: [math-fun] what's special about 786433 ?
by Michael Reid 02 Dec '07

02 Dec '07

> what is the least prime p, such that mod(p, k)=1 for 18 values of k, k<= floor( > sqrt(n) ) ? > Answer: n=786433 > > Why would, among the first 100,000 primes, > it be so rare for p mod k to occur just 18 times? you are counting the number of divisors, d , of p-1 that are in the range 1 < d <= sqrt(p) . if you were to include 1 in the range, you would have half the divisors of p-1 , with the convention that if p-1 is a square, you include the divisor sqrt(p-1) . this means that your count is equivalent to floor((number_of_divisors(p-1) - 1)/ 2) . in order to get an 18 , you need p - 1 to have either 37 or 38 divisors. this means p - 1 has the form a^36 or a^18 b , where a and b are primes. since a^36 + 1 is divisible by a^4 + 1 , the first form cannot occur if p is prime. mike

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[math-fun] what's special about 786433 ?
by wouter meeussen 02 Dec '07

02 Dec '07

what is the least prime p, such that mod(p, k)=1 for 18 values of k, k<= floor( sqrt(n) ) ? Answer: n=786433 Why would, among the first 100,000 primes, it be so rare for p mod k to occur just 18 times? the count {k , frequency of mod(p,k)=1 } among the first 10^5 primes is {0,2},{1,5380},{2,2897},{3,15798},{4,823},{5,10186},{6,232}, {7,17325},{8,1398},{9,2239},{10,28},{11,12585},{12,9},{13,602}, {14,550},{15,9327},{16,3},{17,2828},{18,1},{19,2043},{20,149}, {21,36},{22,22},{23,6659},{24,42},{25,10},{26,236},{27,422}, {28,0},{29,766},{30,0},{31,2190},{32,13},{33,0}, ... does it mean anything special that 786433 mod k = 1 for k= {2, 3, 4, 6, 8, 12, 16, 24, 32, 48, 64, 96, 128, 192, 256, 384, 512, 768} and each of these factors into 2's and 3's? (Pierpont primes A005109) This 786433 also occurs in A070806 (and on the 19th place). Hmmm. Wouter. ------------------------------------ for Mathematica Affectionado's: it2=Table[{Floor[Sqrt[Prime[n]]],Count[Mod[Prime[n],Range@Floor[Sqrt[Prime[n]]]],1]},{n,100000}]; {#,Count[it2,{_,#}]}&/@Range[0,Max[Last/@it2]] ------------------------------------

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[math-fun] Add first diff of digits and repeat
by Eric Angelini 01 Dec '07

01 Dec '07

Hello Math-fun, Let's say a(n) = 329441. How do we build a(n+1)? First, sum, from left to right, the absolute differences of a(n)'s digits: a(n) = 3 2 9 4 4 1 abs. diff = 1+7+5+0+3 = 16 Now add this sum to a(n) so to produce a(n+1): 329441 + 16 = 329457 = a(n+1) And repeat: 329457 + 16 = 329473 329473 + 20 = 329493 329493 + 24 = 329517 329517 + 22 = 329539 etc. The sequence stops when all digits off a(n) are the same, of course: If we start with 18, for instance, we have: S = 18, 25, 28, 34, 35, 37, 41, 44, 44, 44, ... Are there integers leading to infinite sequences? If yes, what would be the smallest one? Best, É. (hoping, as usual, that this is not old hat)

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