Author 
Topic: 2017 (Read 4092 times) 

towr
wu::riddles Moderator Uberpuzzler
Some people are average, some are just mean.
Gender:
Posts: 13730

Does anyone have any interesting facts about the number 2017? It's a prime and part of a sexy prime pair with 2011. It's a zero of the Mertens function (2017) = (20171) + (20172)

« Last Edit: Dec 23^{rd}, 2016, 6:28am by towr » 
IP Logged 
Wikipedia, Google, Mathworld, Integer sequence DB



rloginunix
Uberpuzzler
Posts: 1026


Re: 2017
« Reply #1 on: Dec 23^{rd}, 2016, 5:35pm » 
Quote Modify

Theme: 2+0+1=3 and 7: 333 + 3 + 333 + 3 + 333 + 7 + 333 + 3 + 333 + 3 + 333


IP Logged 



rloginunix
Uberpuzzler
Posts: 1026


Re: 2017
« Reply #2 on: Dec 23^{rd}, 2016, 6:24pm » 
Quote Modify

Also, 2017 is in the center of the immediate prime neighbours: 2011 on the left and 2027 on the right. Assembled together they all form yet another prime: 201120172027


IP Logged 



dudiobugtron
Uberpuzzler
Posts: 735


Re: 2017
« Reply #3 on: Dec 27^{th}, 2016, 11:14am » 
Quote Modify

In hexadecimal it is 7e1, which in decimal scientific notation represents 7.


IP Logged 



rloginunix
Uberpuzzler
Posts: 1026


Re: 2017
« Reply #6 on: Dec 28^{th}, 2016, 4:10pm » 
Quote Modify

 for the record: 2017 = 9^{2} + 44^{2} where nine is three squared and forty four is, of course, twenty two squared,  low intensity love affair with 9: recursively subtract the sum of digits of the current number from the current number: 2017  10 = 2007 and so, downward, we subtract: occasionally 27 but mostly 9 and 18 [at some point: 181818]  interpreted as "a 24hour period" it is 84 days and one hour: 2017 = 24x84 + 1 = 2016 + 1  a busy beetle, running along a circumference, will count 5 full revolutions and then some: 2017 = 360x5 + 217 where the remainder corresponds to 7/6 past 7 on the face of an analogue clock Any interest in constructing the first 100 natural numbers from the digits in 2017 and some basic operations? 0 = 2 x 0 x 1 x 7 1 = (2 + 0  1)^{7} 2 = 2 + 0 x 1 x 7 3 = 2 + 1 + 0 x 7 4 = 7  1  0  2 5 = 7  2  0 x 1 6 = 7  1  2 x 0 7 = 7 + 2 x 0 x 1 8 = 7 + 1 + 2 x 0 9 = 7 + 2 + 0 x 1 10 = 2 + 0 + 1 + 7 11 = 2 + 0! + 1 + 7


IP Logged 



dudiobugtron
Uberpuzzler
Posts: 735


Re: 2017
« Reply #7 on: Dec 28^{th}, 2016, 5:07pm » 
Quote Modify

12 = 12 + 0 x 7 or 12 = 3 x 2 + 7  0! on Dec 28^{th}, 2016, 4:10pm, rloginunix wrote: low intensity love affair with 9: recursively subtract the sum of digits of the current number from the current number: 2017  10 = 2007 and so, downward, we subtract: occasionally 27 but mostly 9 and 18 [at some point: 181818] 
 This is a result of the fact that the digits of multiples of 9 add to a multiple of 9. (and subtracting a multiple of 9 from another yields a multiple of 9 as well.) I would be interested in whether there were any numbers (past a certain point) which didn't have a lowintensity love affair with 9.

« Last Edit: Dec 28^{th}, 2016, 5:07pm by dudiobugtron » 
IP Logged 



towr
wu::riddles Moderator Uberpuzzler
Some people are average, some are just mean.
Gender:
Posts: 13730


Re: 2017
« Reply #8 on: Dec 28^{th}, 2016, 10:24pm » 
Quote Modify

12 = 2*710! 13 = 2*710 14 = 2*71*0 15 = 2*7+1+0 16 = 2*7+1+0! 17 = 2*(7+1)+0! 18 = 2*(7+1+0!)


IP Logged 
Wikipedia, Google, Mathworld, Integer sequence DB



rloginunix
Uberpuzzler
Posts: 1026


Re: 2017
« Reply #9 on: Dec 29^{th}, 2016, 11:03am » 
Quote Modify

on Dec 28^{th}, 2016, 5:07pm, dudiobugtron wrote: (sorry, can't use 3 explicitly (take towr's version or 12 = 7 + (2 + 1)!  0!)) 19 = C(7, 2)  0!  1 = 2 + 0 + 17 = 20  1^{7} 20 = 7x(2 + 1)  0! 21 = 7x(2 + 1) + 0 Separately, as a sum of consecutive primes (A000040) I only managed to assemble 2011: 2011 = 157 + ... + 211 Doesn't seem to work for 2017.


IP Logged 



rloginunix
Uberpuzzler
Posts: 1026


Re: 2017
« Reply #10 on: Dec 29^{th}, 2016, 11:43am » 
Quote Modify

With the help of a (C) program: 2011 = 661 + 673 + 677 2015 = 389 + 397 + 401 + 409 + 419 2016 = 71 + ... + 157 No cigar for 2017.


IP Logged 



dudiobugtron
Uberpuzzler
Posts: 735


Re: 2017
« Reply #11 on: Dec 29^{th}, 2016, 12:13pm » 
Quote Modify

on Dec 29^{th}, 2016, 11:03am, rloginunix wrote: (sorry, can't use 3 explicitly (take towr's version or 12 = 7 + (2 + 1)!  0!)) 
 Oh gosh, I am obviously not on form in this thread.


IP Logged 



towr
wu::riddles Moderator Uberpuzzler
Some people are average, some are just mean.
Gender:
Posts: 13730


Re: 2017
« Reply #12 on: Dec 29^{th}, 2016, 1:01pm » 
Quote Modify

22 = 7x(2 + 1) + 0! 23 = (721)!  0! 24 = (721)! + 0 25 = (721)! + 0! 26 = (72) + 0! + 1 27 = _{1}^{7}x dx + 2 + 0! 28 = 7*2*(1+0!) 29 = _{1}^{7}x+(2) dx  0! 30 = _{1}^{7}x+(2) dx + 0 31 = _{1}^{7}x+(2) dx + 0! 32 = 2^{(710!)} 33 = e^{7/2} + 0*1 34 = e^{7/2} + 0 + 1 35 = e^{7/2} + 0! + 1 36 = (71)^{2} + 0 37 = (71)^{2} + 0!

« Last Edit: Dec 29^{th}, 2016, 1:27pm by towr » 
IP Logged 
Wikipedia, Google, Mathworld, Integer sequence DB



rmsgrey
Uberpuzzler
Gender:
Posts: 2861


Re: 2017
« Reply #13 on: Dec 29^{th}, 2016, 4:50pm » 
Quote Modify

on Dec 28^{th}, 2016, 5:07pm, dudiobugtron wrote:I would be interested in whether there were any numbers (past a certain point) which didn't have a lowintensity love affair with 9. 
 It's not terribly interesting. The sum of the digits of a number is the same value mod 9 as the number itself (because 10^{n}*a_{n} = 1^{n}*a_{n} = a_{n} (mod 9) ) so subtracting the sum of digits from the original number gives a multiple of 9 for any [e]positive[/e] integer

« Last Edit: Dec 30^{th}, 2016, 5:40am by rmsgrey » 
IP Logged 



rloginunix
Uberpuzzler
Posts: 1026


Re: 2017
« Reply #14 on: Dec 29^{th}, 2016, 7:57pm » 
Quote Modify

on Dec 29^{th}, 2016, 4:50pm, rmsgrey wrote:It's not terribly interesting. 
 I concur, we made it too easy for ourselves. Will the following "conservation of ordinal position" constraint make it more interesting:  the digits in the construction must keep their yearly position The construct with the smallest number of operat[ions]/[ors] shall win: 0 = 2 x 0 x 1 x 7 1 = (2  0  1)^{7} 2 = 2  0 x 1 x 7 3 = 2  0!  1 + 7 4 = 2  0  1 + 7 5 = 2  0 x 1 + 7 6 = 2  0 + 1 + 7 7 = 2 x 0 x 1 + 7 8 = 2 x 0 + 1 + 7 9 = 2  0 x 1 + 7 10 = 2  0 + 1 + 7 11 = 2 + 0! + 1 + 7 12 = 20  1  7 13 = (2  0 + 1)! + 7 14 = 20 + 1  7 15 = 2 + 0 + 17 16 = 2 + 0! + 17 17 = 2 x 0 + 17 18 = 2  0! + 17 19 = 2 + 0 + 17


IP Logged 



towr
wu::riddles Moderator Uberpuzzler
Some people are average, some are just mean.
Gender:
Posts: 13730


Re: 2017
« Reply #15 on: Dec 29^{th}, 2016, 11:12pm » 
Quote Modify

on Dec 29^{th}, 2016, 7:57pm, rloginunix wrote: I concur, we made it too easy for ourselves. 
 He was talking about something else, though... 20 = 20 * 1^{7} 21 = (20+1)*7 20 + 1^{7} 22 = log(20!) * 1 + log(7!) 23 = log(20!) * 1 + log(7!) 24 = log(20!) + 1 + log(7!) 25 = exp( (2+0) * ln(1+ (7)) ) 26 = 20  1 + 7 27 = 20 + 1 * 7 28 = 20 + 1 + 7 29 = 20 + 1 * ln(7!) 30 = 20 + 1 + ln(7!) 31 = 20 + ln((1+7)!) 32 = 2^{(0!  1 + 7)} 33 = exp(2^{01}*7) 34 = (2 + 0) * 17 35 = exp(2)* (0!1+7) 36 = exp(2)  0!* (1+7) 37 = 20 + 17 38 = ((201))  ln(7) 39 = ((201))  ln(7) 40 = ((201)) * ln(7) 41 = ((201)) + ln(7) 42 = ((201)) + ln(7) 43 = ((201)) + log(exp(7)) 44 = ((201)) + log(exp(7)) 45 = exp(2)* ln((0*1+7)!) I'll leave the winning to someone else, it's hard enough getting this far.

« Last Edit: Jan 1^{st}, 2017, 11:01am by towr » 
IP Logged 
Wikipedia, Google, Mathworld, Integer sequence DB



rloginunix
Uberpuzzler
Posts: 1026


Re: 2017
« Reply #16 on: Dec 30^{th}, 2016, 3:10pm » 
Quote Modify

Nice. I am sure it's just a typo  for 21 you meant 20 + 1^{7}. on Dec 29^{th}, 2016, 11:12pm, towr wrote:He was talking about something else, though... 
 Yeah, I know  programmer's habit to reuse things. Speaking of reuse  let's reuse! 46 = 2  0 + (1 + 7)!! 47 = 2 + 0! + (1 + 7)!! = ((2 + 0!)!)!!  1^{7} 48 = (2 + 0!)! x (1 + 7) 49 = ((2 + 0!)! + 1) x 7 50 = 2 + 0 + (1 + 7)!! 51 = (2 + 0!) x 17 = 20 + (1 + 7!) = !((2 + 0!)!  1) + 7 52 = 2^{0! + 1}! + 7!!! 2 x (0! + 1) x F_{7} 53 = 2^{h(0! + 1)} + 7!!!! (e^{2} 0!)!!  e^{1}+ 7 54 = (2 + 0!)! + (1 + 7)!! 55 = ((2 + 0 + 1)!)!! + 7 56 = ((2 + 0! + 1)!! x 7 Reference: h(n)  hexagonal numbers A000384 n!! is A006882 n!!! is A007661 n!!!! is A007662 !n is A000166 [e] see towr's comment below [/e]

« Last Edit: Dec 31^{st}, 2016, 7:02pm by rloginunix » 
IP Logged 



towr
wu::riddles Moderator Uberpuzzler
Some people are average, some are just mean.
Gender:
Posts: 13730


Re: 2017
« Reply #17 on: Dec 30^{th}, 2016, 11:56pm » 
Quote Modify

Hmm, editing that mistake screwed up the whole post. The symbol script doesn't seem to work on this computer. I also think that when you start needing references, it may be going a bit over the top. You can probably find some function f(x) that does exactly what you want somewhere on the web, or otherwise put it somewhere. Maybe it's an idea to limit it to what wolframalpha will accept? That should be broad enough. So n!! is ok, but not n!!!


IP Logged 
Wikipedia, Google, Mathworld, Integer sequence DB



rloginunix
Uberpuzzler
Posts: 1026


Re: 2017
« Reply #18 on: Dec 31^{st}, 2016, 7:47pm » 
Quote Modify

Deal! I've checked  it computes subfactorials (!4 = 9, !5 = 44) and Fibonacci numbers as 'fibonacci 7' returning 13. Before the New Year rolls through the East Coast (4!! = 8, 6!! = 48, 7!! = 105, (1 + 7!) = (5041) = 71): 57 = (((2 + 0 + 1)!)!! + 7!! 58 = 59 = 60 = 61 = (!((2 + 0!)!  1)) + 7!! 62 = 63 = !(2 + 0! + 1) x 7 64 = 2^{0  1 + 7} 65 = (2 + 0!)! + (1 + 7!) 66 = 67 = 68 = 2  0! + (1 + 7!) 69 = 2  0 + (1 + 7!) 70 = 2 + 0! + (1 + 7!) 71 = 2 x 0 + (1 + 7!) 72 = 2  0! + (1 + 7!) 73 = 2 + 0 + (1 + 7!) 74 = 2 + 0! + (1 + 7!) 75 = 76 = 77 = (2 + 0!)! + (1 + 7!) 78 = 79 = 80 = 81 = 82 = 83 = 84 = 20  1 + 7!! 85 = 20/1 + 7!! 86 = 20 + 1 + 7!! 87 = 88 = 2 x !(0!  1 + 7) 89 = 90 = 91 = 92 = 93 = 94 = 2 x (0! + (1 + 7)!!) 95 = 96 = (2 + 0) x (1 + 7)!! 97 = ((2^{0! + 1})!!) + 7!! 98 = ((2 + 0!)! + 1) + 7!! 99 = ((2 + 0!)! x 1) + 7!! 100 = ((2 + 0!)!  1) + 7!!


IP Logged 



towr
wu::riddles Moderator Uberpuzzler
Some people are average, some are just mean.
Gender:
Posts: 13730


Re: 2017
« Reply #19 on: Jan 2^{nd}, 2017, 12:13pm » 
Quote Modify

57 = (2%)^(0  1) + 7 .. 60 = ldexp(2, 0!) + ldexp(1, totient(7)) .. 62 = 2 + 0 + ldexp(1, totient(7)) .. 66 = 2 + 0 + ldexp(1, totient(7)) 67 = 2 + 0! + ldexp(1, totient(7)) .. 80 = gcd(round(ldexp(2, 0!) / (1%))), 7!) .. 91 = 2 + 0! / (1%)  7 92 = 2 + 0! / (1%)  totient(7) 93 = lb(2) * 0! / (1%)  7 95 = 2 + 0! / 1%  7 .. 101 = lb(2) + 0! / ((1^7)%) 102 = 2 + 0! / ((1^7)%) 103 = ? 104 = 2 + ((0! / 1%) + totient(7)) 105 = 2 + ((0! / 1%) + 7) 106 = lb(2) * (0! / (1%) + totient(7)) 107 = lb(2) + (0! / (1%) + totient(7)) 108 = 2 + (0! / (1%) + totient(7)) 109 = 2 + (0! / (1%) + 7)


IP Logged 
Wikipedia, Google, Mathworld, Integer sequence DB



dudiobugtron
Uberpuzzler
Posts: 735


Re: 2017
« Reply #20 on: Jan 2^{nd}, 2017, 12:16pm » 
Quote Modify

... 2017 = 2017

« Last Edit: Jan 2^{nd}, 2017, 12:16pm by dudiobugtron » 
IP Logged 



rloginunix
Uberpuzzler
Posts: 1026


Re: 2017
« Reply #21 on: Jan 2^{nd}, 2017, 7:07pm » 
Quote Modify

103 = 2  0 x 1 + 7!! I don't think I have the patience to slug it out to 2017! (love % idea)


IP Logged 



towr
wu::riddles Moderator Uberpuzzler
Some people are average, some are just mean.
Gender:
Posts: 13730


Re: 2017
« Reply #22 on: Jan 3^{rd}, 2017, 11:37am » 
Quote Modify

58 = floor((20 + 1) * lb(7)) 59 = ceil((20 + 1) * lb(7)) 75 = floor(201 / sqrt(7)) 76 = floor(sqrt(20) * 17) 78 = ceil(ldexp(20, 1) * ln(7)) 79 = ceil((sqrt(20) + 1) / (7%)) 81 = floor(20 * lb(17)) 82 = floor(20 * sqrt(17)) 83 = ceil(20 * sqrt(17)) 87 = floor(log(20)^17) 89 = ceil(201^log(7)) 90 = (totient(20)^1) * gamma(7) 94 = ((2^0) / 1%)  totient(7) 103 = floor(201 / ln(7)) 110 = ceil(lb(201) / (7%)) 111 = floor(totient(201) * log(7)) 112 = ldexp(totient(20), 1) * 7 113 = ceil(ldexp(20, 1) * lb(7)) 114 = (20  1) * totient(7) 115 = 20! mod (1 + gamma(7)) 116 = floor(ln(20) * exp(1) / (7%)) 117 = floor(20 / (17%)) 118 = ceil(20 / (17%)) 119 = floor((20  (1%)) * totient(7)) 120 = 20 * (1 * totient(7)) 121 = ceil((20  exp(1)) * 7) 122 = ceil((2^((0!  1%) * 7))) 123 = floor((2  (0 + 1%))^7) 124 = ldexp(2, 0!) + ldexp(1, 7) 125 = totient(201)  7 126 = (20 + 1) * totient(7) 127 = !20 mod (1 + gamma(7)) 128 = ldexp(gcd(20, 1), 7) 129 = (2^0) + ldexp(1, 7) 130 = 2 + 0 + ldexp(1, 7) 131 = 2 + 0! + ldexp(1, 7) 132 = totient(201) mod floor(exp(7)) 133 = (20  1) * 7 134 = floor((totient(201) + lb(7))) 135 = ceil((totient(201) + lb(7))) 136 = totient(20) * 17 137 = ceil(((20 + exp(1)) * totient(7))) 138 = totient(201) + totient(7) 139 = totient(201) + 7 140 = 20 * 1 * 7 141 = ceil((20 * (1% + 7))) 142 = ceil((log(2)^(0  sqrt(17)))) 143 = ceil((ldexp(20, 1) / 7%)) 144 = 20% / 1 / gamma(7) 145 = ceil(20% * (1 + gamma(7))) 146 = 20 / (1%)  !7 147 = (20 + 1) * 7 148 = (20 + ldexp(1, 7)) 149 = floor(log(201)^totient(7)) 150 = (ldexp(2%, 0!)^1) * totient(7) 151 = ceil(ln(20) * (1%) * 7!) 152 = floor(20 * exp(1) * lb(7)) 153 = floor(ldexp(20% + 1, 7)) 154 = ceil(ldexp(20% + 1, 7)) 155 = floor((sqrt(20)^exp(1)) * sqrt(7)) 156 = floor(totient(201) / log(7)) 157 = ceil(totient(201) / log(7)) 158 = floor((ldexp(2, 0!) / 1%)^log(7)) 159 = floor((20 + exp(1)) * 7) 160 = 20 * (1 + 7) 161 = ceil(sqrt(20)^1 * gamma(7)) 162 = (gamma(20) * (1%)) mod !7 163 = floor(lb(20) / (1% * sqrt(7))) 164 = ceil(lb(20) / (1% * sqrt(7))) 165 = floor(ldexp((log(20)  1%), 7)) 166 = floor(ldexp(log(20), 1 * 7)) 167 = ceil(ldexp(log(20), 1 * 7)) 168 = 20! mod (1 + gamma(7)) 169 = floor(201 * log(7)) 170 = gamma(20) mod (1 + !7) 171 = ceil(log(2)^(0 + exp(1)  7)) 172 = floor((totient(20)  1)^sqrt(7)) 173 = ceil((totient(20)  1)^sqrt(7)) 174 = floor(sqrt(201)^ln(7)) 175 = (ldexp(2%, 0!)^1) * 7 176 = ceil(2 + 0 + ldexp(exp(1), totient(7))) 177 = floor(ldexp(ln(2), (0 + 1 + 7))) 178 = ceil(ldexp(ln(2), (0 + 1 + 7))) 179 = floor(ldexp(sqrt(2)  0  (1%), 7)) 180 = (ldexp(2, 0!)^1) * gamma(7) 181 = floor(ldexp(sqrt(2), 0 + 1 * 7)) 182 = ceil(ldexp(sqrt(2), 0 + 1 * 7)) 183 = ceil(ldexp(sqrt(2) + 0 + (1%), 7)) 184 = floor(ldexp(ln(2)^(0  1), 7)) 185 = ceil(ldexp(ln(2)^(0  1), 7)) 186 = 2 * (0! / (1%)  7) 187 = ceil((ldexp(ln(2), 0!)^totient(17))) 188 = 2 * (0! / (1%)  totient(7)) 189 = floor(ldexp(gamma(20)^(1%), 7)) 190 = ceil(ldexp(gamma(20)^(1%), 7)) 191 = floor(ldexp(ln(20), 1 * totient(7))) 192 = ldexp(2 + 0 + 1, totient(7)) 193 = 2 / (0 + (1%))  7 194 = 201  7 195 = 201  totient(7) 196 = floor((20 / exp(1))^sqrt(7)) 197 = ceil((20 / exp(1))^sqrt(7)) 198 = floor(201  lb(7)) 199 = ceil((201  lb(7))) 200 = 2 / (0 + (1^7)%) (I haven't thrown them all through wolframalpha yet. it sometimes reacts a bit different than python)

« Last Edit: Jan 3^{rd}, 2017, 11:58am by towr » 
IP Logged 
Wikipedia, Google, Mathworld, Integer sequence DB



towr
wu::riddles Moderator Uberpuzzler
Some people are average, some are just mean.
Gender:
Posts: 13730


Re: 2017
« Reply #23 on: Jan 3^{rd}, 2017, 12:02pm » 
Quote Modify

in general 2k+0 = lb(2)+log(0!/(1%....%))+7 2k+1 = 2+log(0!/(1%....%))+7 (Which is definitely not the least number of operators for most)


IP Logged 
Wikipedia, Google, Mathworld, Integer sequence DB



rloginunix
Uberpuzzler
Posts: 1026


Re: 2017
« Reply #24 on: Jan 3^{rd}, 2017, 1:30pm » 
Quote Modify

I've checked till 120  wolframalpha has issues with 87, 89, 94 and 111. With 94 it's just a typo  extra parens are needed around 1%: "((2^0) / (1%))  totient(7)"


IP Logged 



