Turtle Posted June 21, 2010 Report Share Posted June 21, 2010 in the thread Strange Numbers we poked, probed, prodded, & pinned into submission a class of Abundant numbers, which i termed "Unusual Sets", that were Abundant by 2 times a Perfect number. the idea now is to search for & fill out sets of numbers Abundant or Deficient by other specific amounts & see whether they exhibit any of the the regularities of the Unusuals. :) logically...to my pea brain at any rate, we should start with numbers Abundant by 1 and go from there. seems however that others have already taken a peek at this & not a single 1 is to be found so far. seems they already have a special name, and that be quasiperfect numbers. :shrug:Quasiperfect number - Wikipedia, the free encyclopediaNo quasiperfect numbers have been found so far, but if a quasiperfect number exists, it must be an odd square number greater than 1035 and have at least seven distinct prime factors. so thens...first one to list the first 30 or so numbers Abundant by 2 gets...erhm...something nice. :D ready, set, ...program! :) Updated Sets List Set Name___{Set Elements}______________________________<= Range Limit_______ A1 {?...} [#1] A2 {20, 104, 464, 650, 1952, 130304, 522752, 8382464, 134193152, 549754241024, 8796086730752, 140737463189504, 144115187270549504, 196159429230833773869868419445529014560349481041922097152, 3450873173395281893717377931138512601610429881249330192849350210617344, 14134776518227074636666380005943348118644503200295456427550007248658890752, 1018517988167243043134222844204689080525734194692096586259130403089761695707425648517578752, 286687326998758938951352611912760867599570623646035104549333658790952417373021521733808280 74567534641152, 866459279412754643618254432544713657323886586054942679740774868744608611810713282141395700 00066155755750644954940903259472134144, 567842753355942883241659224912503542463782313036967234594914218109368944838583053102436751 8372461675192150235469728157765532148301824, 952682052708737863580809701474965303268004804280081527972154833870047320663602522345191510 94886831234103907878649603962944581817093108793344...?} <= 2521745 [#2, #4, #16] A3 {18, ...?} <= 2521745 [#7, #8] A4 {12, 70, 88, 1888, 4030, 5830, 32128, 521728, 1848964 ...?} <=2521745 [#8, #10, #11] A5 {ø...?} <=2521745 [#12] A6 {8925, 32445, 442365, ...?} <=2521745 [#12] A7 {196, ...?}<=2521745 [#12] ****************************************************************D1 {2, 4, 8, 16, 32, 64, 128, 256, 512, 1024, 2048, 4096, 8192, 16384, 32768, 65536, 131072, 262144, 524288, 1048576, 2097152, ... ?} <=2521745 [#15] D2 {3, 10, 136, 32896 ...?} <=2521745 [#15, #17] D3 {ø, ...?} <=2521745 [#15] D4 {5, 14, 44, 110, 152, 884, 21544, 8384, 18632, 116624 , ...?} <=402192 [#18, #23, #24] D5 {9, ...?} <=40000 [#18] D6 {7, 15 52, 315, 592, 1155, ...?} <=40000 [#18] D7 {50, ...?} <=40000 [#18] modest 1 Quote Link to comment Share on other sites More sharing options...
Turtle Posted June 22, 2010 Author Report Share Posted June 22, 2010 alrighty thens, or such a matter. i finally dragged out the dinosaur machine & software & started the hunt for Abundants by 2. just to put the beasts in perspective, it took 1 min 11 sec to search between 7,000 & 8,000. :eek: :doh: it's rounding 20,000 just now & i have 5 in the bag. here they be: >> Set of Numbers Abundant by 2:20 (2^2) (5^1) 1, 2, 4, 5, 10, 20104 (2^3) (13^1) 1, 2, 4, 8, 13, 26, 52, 104464 (2^4) (29^1) 1, 2, 4, 8, 16, 29, 58, 116, 232, 464650 (2^1) (5^2) (13^1) 1, 2, 5, 10, 13, 25, 26, 50, 65, 130, 325, 6501952 (2^5) (61^1) 1, 2, 4, 8, 16, 32, 61, 122, 244, 488, 976, 1952... smoke 'em if ya got 'em. B) Quote Link to comment Share on other sites More sharing options...
Don Blazys Posted June 22, 2010 Report Share Posted June 22, 2010 I got em... This is gonna be good ! Quote Link to comment Share on other sites More sharing options...
Turtle Posted June 22, 2010 Author Report Share Posted June 22, 2010 I got em... This is gonna be good ! thnx donster! :eek: at the very least, it has distracted a turtle. i found 2 more overnight, but due to sloppy/hasty/lazy programming, i have to go back & make sure i didn't miss some in-betweeners. here's the new bagged 'ems thens: >> 130304 (2^8) (509^1) 1, 2, 4, 8, 16, 32, 64, 128, 256, 509, 1018, 2036, 4072, 8144, 16288, 32576, 65152, 130304522752 (2^9) (1021^1) 1, 2, 4, 8, 6, 32, 64, 128, 256, 512, 1021, 2042, 4084, 8168, 16336, 32672, 65344, 130688, 261376, 522752 if i missed some, they are between these 2. after 522, 752 there are no more out to 1,000,000+, where i stopped to go back. i hope to clean things up a bit at my usual hasty pace. . . . . . . . . . B) Quote Link to comment Share on other sites More sharing options...
Turtle Posted June 23, 2010 Author Report Share Posted June 23, 2010 reporting in with nothing to report in on. :ud: well, no new abundant-by-2 numbers to report that is. B) i shut down all non-essential displays & got a little speed gain, but you can't make a silk calculator out of a sow's ear. :dog: anyway, i'm at 1,800,000 integers searched & just the seven set elements to show for it. i'm filling in that earlier mentioned short gap just now and will then let 'er run on this set overnight & then tomorrow have a go @ numbers abundant-by-3. B) Quote Link to comment Share on other sites More sharing options...
Rade Posted June 23, 2010 Report Share Posted June 23, 2010 Are you not missing numbers derived from (2^6) and (2^7) ? I mean, all other possibilities from (2^1) to (2^9) are represented in your sequence: 20 (2^2) ...104 (2^3) ...464 (2^4) ...650 (2^1) ...1952 (2^5) ...130304 (2^8) ...522752 (2^9) ... what about # (2^6) & (2^7) .... ? Turtle 1 Quote Link to comment Share on other sites More sharing options...
Turtle Posted June 24, 2010 Author Report Share Posted June 24, 2010 Are you not missing numbers derived from (2^6) and (2^7) ? I mean, all other possibilities from (2^1) to (2^9) are represented in your sequence: 20 (2^2) ...104 (2^3) ...464 (2^4) ...650 (2^1) ...1952 (2^5) ...130304 (2^8) ...522752 (2^9) ... what about # (2^6) & (2^7) .... ? well, they are missing, but that's why i left them out. :lol: good eye though! :hihi: my algorithm is exhaustive, taking each successive integer & factoring it, summing the factors, and comparing the sum to the number from whence they came. the results are what they are, as well as what they are not. mind you this is not to say 2^6 and/or 2^7 may not show up later paired with suitably large primes. :shrug: this is what we don't know; this is why we are looking. so, i stopped the search for abundant-by-2 at the integer 2,521,745. in that interval i found only the 7 numbers. (i'll figure out an index/list page for all the set elements soon.) so we have:set of numbers abundant-by-2 <= 2521745{20, 104, 464, 650, 1952, 130304, 522752, ...?} i have let the search run all day for abundant-by-3 and have found only 1 after checking the first 1,460,000 integers. it is... >>> :D .... 18: (2^1) (3^2) 1 2 3 6 9 18 :D set of numbers abundant-by-3 <= 1460000{18, ...?} that's all i got. . . . :D Quote Link to comment Share on other sites More sharing options...
Turtle Posted June 24, 2010 Author Report Share Posted June 24, 2010 ...i have let the search run all day for abundant-by-3 and have found only 1 after checking the first 1,460,000 integers. it is... >>> :phones: .... 18: (2^1) (3^2) 1 2 3 6 9 18 :shrug: set of numbers abundant-by-3 <= 1460000{18, ...?} that's all i got. . . . ;) okaly dokaly. ran it all night again looking for abundant-by-3's & just came up on 2521745 where i stopped the abundant-by-2 search. i found no more quarry in that interval. :eek: of course there is the possibility there are no more i suppose. :doh: so thens, desisting abundant-by-3 search, commencing abundant-by-4. this is looking populous. :soccerb: set of numbers abundant-by-4 <= 403012 (2^2) (3^1) 1 2 3 4 6 1270 (2^1) (5^1) (7^1) 1 2 5 7 10 14 35 7088 (2^3) (11^1) 1 2 4 8 11 22 44 881888 (2^5) (59^1) 1, 2, 4, 8, 16, 32, 59, 118, 236, 472, 944, 18884030 (2^1) (5^1) (13^1) (31^1) 1, 2, 5, 10, 13, 26, 31, 62, 65, 130, 155, 310, 403, 806, 2015, 4030 Quote Link to comment Share on other sites More sharing options...
Qfwfq Posted June 24, 2010 Report Share Posted June 24, 2010 Wouldn't it make sense to have the loop checking for each number having less than some chosen abundance of interest? This way, the same CPU time would find all of them and report each with its abundance value. Quote Link to comment Share on other sites More sharing options...
Turtle Posted June 24, 2010 Author Report Share Posted June 24, 2010 Wouldn't it make sense to have the loop checking for each number haviong less than some chosen abundance of interest? This way, the same CPU time would find all of them and report each with its abundance value. it would if i had the manual for my software or could remember the instructions & syntax for making files. :phones: alas, neither case is the case. :shrug: (i'm using a compiled basic that runs in DOS with an integer variable limit of just over 2 billion.) :soccerb: even so, the delay time to write to such files ultimately slows down the filling out of any one set so i decided to focus on one set at a time to get some under my belt. as always i welcome discussions on different algorithms & efficiency here, as well as the actual subject of the numbers in question. speaking of which, i have a few more abundant-by-4's to add.>> 5830 (2^1) (5^1) (11^1) (53^1) 1, 2, 5, 10, 11, 22, 53, 55, 106, 110, 265, 530, 583, 1166, 2915, 5830 32128 (2^7) (251^1) 1, 2, 4, 8, 16, 32, 64, 128, 251, 502, 1004, 2008, 4016, 8032, 16064, 32128 521728 (2^9) (1019^1) 1, 2, 4, 8, 16, 32, 64, 128, 256, 512, 1019, 2038, 4076, 8152, 16304, 32608, 65216, 130432, 260864, 521728 ps i have appended an appendix listing the sets & their elements to post #1 & i will try & keep it updated as i get new data. Quote Link to comment Share on other sites More sharing options...
Turtle Posted June 25, 2010 Author Report Share Posted June 25, 2010 i guess i can make some arrays & not use files. :shrug: as soon as i finish this run i'll try 5, 6, & 7 all at once. :soccerb: anyways, found one more abundant-by-4 overnight. ;) what an odd-ball looking thing. :phones: 1848964 (2^2) (13^1) (31^2) (37^1) 1, 2, 4, 13, 26, 31, 37, 52, 62, 74, 124, 148, 403, 481, 806, 961, 962, 1147, 1612, 1922, 1924, 2294, 3844, 4588, 12493, 14911, 24986, 29822, 35557, 49972, 59644, 71114, 142228, 462241, 924482, 1848964 Quote Link to comment Share on other sites More sharing options...
Turtle Posted June 26, 2010 Author Report Share Posted June 26, 2010 i put sum pride in my stride & pimped my ride> :) now have drove to 550,000 on the hunt for abundant-by-5's, 6's, & 7's & i got 4 nabbed in my grill. :lol: Abundant-by-5 <=550,000ø, nada, zip, zilch, goose egg Abundant-by-6 <=550,0008925 (3^1) (5^2) (7^1) (17^1) 1, 3, 5, 7, 15, 17, 21, 25, 35, 51, 75, 85, 105, 119, 175, 255, 357, 425, 525, 595, 1275, 1785, 2975, 8925 32445 (3^2) (5^1) (7^1) (103^1) 1, 3, 5, 7, 9, 15, 21, 35, 45, 63, 103, 105, 309, 315, 515, 721, 927, 1545, 2163, 3605, 4635, 6489, 10815, 32445 442365 (3^1) (5^1) (7^1) (11^1) (383^1) 1, 3, 5, 7, 11, 15, 21, 33, 35, 55, 77, 105, 165, 231, 383, 385, 1149, 1155, 1915, 2681, 4213, 5745, 8043, 12639, 13405, 21065, 29491, 40215, 63195, 88473, 147455, 442365Abundant-by-7 <=550,000196 (2^2) (7^2) 1, 2, 4, 7, 14, 28, 49, 98, 196 Quote Link to comment Share on other sites More sharing options...
Don Blazys Posted June 26, 2010 Report Share Posted June 26, 2010 Quoting Turtle:Abundant-by-5 <=550,000ø, nada, zip, zilch, goose egg That is yet another "strange feather" that the number [math]5[/math] can put in its cap or hat ! [math]5[/math] really is a most remarkable integer,with properties both unique and important. For instance... most of us have [math]5[/math] senses,and I personally have [math]5[/math] fingers on my left hand,which, amazingly enough, just "happens to be" the exact same number of fingersthat I have on my right hand ! :shrug: Moreover... There are [math]5[/math] Platonic solids, and the formula for the "Golden Mean" is: [math](\sqrt{5}+1)/2=1.6180339887499...[/math] which involves the square root of [math]5[/math]. Then, there is the following crazy result,discovered by Leonard Euler (when I was still quite young) which relates the number [math]5[/math] to this entire topic in a most surprising way. Here it is, taken from one of my books... The formula for pentagonal numbers is: [math]\frac{1}{2}*n*(3*n-1)[/math], where if we let [math]n[/math] represent both positive and negative integersand arrange the results in ascending order, we get: [math]1, 2, 5, 7, 12, 15, 22, 26, 35, 40, 51, 57, 70, 77, 92, 100...[/math]. Now, if [math]Q(n)[/math] is the sum of divisors of [math]n[/math], then: [math]Q(n)=Q(n-1)+Q(n-2)-Q(n-5)-Q(n-7)+Q(n-12)+Q(n-15)-Q(n-22)-...[/math] (Note the between terms pattern: [math]++--++--++--...[/math]). The sum continues as long as the terms represent the sum of the factors of positive integers.If [math]Q(0)[/math] appears as the last term, then it must be replaced by [math]n[/math]. For example: [math]Q(12)=Q(11)+Q(10)-Q(7)-Q(5)+Q(0)=[/math] [math]12+18-8-6+12=28[/math]. This relationship can be used to calculate the value of [math]Q(n)[/math]if you know the appropriate previous values,which is itself quite curious, since to find the sum of the divisors of a number,you apparently need to know it's factors,and therefore whether or not it is prime, but none of this information is needed to use the formula! In other words, this relationship allows us to sum the factors of [math]n[/math]without ever knowing the factors of [math]n[/math]. Thus, at least in principle, The Great Turtle's results can be duplicatedby using the above relationship and subtracting [math]n[/math]instead of factoring [math]n[/math]. How wierd is that !?!? Don Quote Link to comment Share on other sites More sharing options...
Turtle Posted June 26, 2010 Author Report Share Posted June 26, 2010 Abundant-by-5 <=550,000ø, nada, zip, zilch, goose egg That is yet another "strange feather" that the number [math]5[/math] can put in its cap or hat ! [math]5[/math] really is a most remarkable integer,with properties both unique and important. For instance... most of us have [math]5[/math] senses,and I personally have [math]5[/math] fingers on my left hand,which, amazingly enough, just "happens to be" the exact same number of fingersthat I have on my right hand ! :shrug: :hyper: while i can't say if, or if not, any numbers abundant by 5 exist, i did notice that amazingly i also have 5 toes on each foot!! :D Thus, at least in principle, The Great Turtle's results can be duplicatedby using the above relationship and subtracting [math]n[/math]instead of factoring [math]n[/math]. How wierd is that !?!? Don gotta love leonhard! i'm going to have to cogitate on this one. by all means if you can make the list before i find it, please do so & post. thought this would add some fuel to our fire as well. Abundant number - Wikipedia, the free encyclopediaThe smallest abundant number not divisible by two, i.e. odd, is 945, and the smallest not divisible by 2 or by 3 is 5391411025 whose prime factors are 52, 7, 11, 13, 17, 19, 23, and 29. An algorithm given by Iannucci in 2005 shows how to find the smallest abundant number not divisible by the first k primes.[1] If A(k) represents the smallest abundant number not divisible by the first k primes then for all ε > 0 we have:(1 − ε)(klnk)2 − ε < lnA(k) < (1 + ε)(klnk)2 + ε for k sufficiently large. Infinitely many even and odd abundant numbers exist. Marc Deléglise showed in 1998 that the natural density of abundant numbers is between 0.2474 and 0.2480.[2] Every proper multiple of a perfect number, and every multiple of an abundant number, is abundant. Also, every integer greater than 20161 can be written as the sum of two abundant numbers.[3] after running to 1,500,000 overnight, i have found still no abundants-by-5 and no further abundants-by-6 or by-7. when i finish this run @ around 2,255,000 i think i'll have a look at deficients by 1, 2, 3, 4, 5, 6, & 7 for balance. . . . Quote Link to comment Share on other sites More sharing options...
Turtle Posted June 27, 2010 Author Report Share Posted June 27, 2010 no more abundants-by 5, 6, or 7 in the interval <=2521745, so on to deficients. this is cool. :) all powers of 2 are deficient by 1. also, up to 32768 no other numbers but powers of 2 are deficient by 1. * :lol: deficient-by-1 <= 327682 (2^1) 1, 24 (2^2) 1, 2, 48 (2^3) 1, 2, 4, 816 (2^4) 1, 2, 4, 8, 1632 (2^5) 1, 2, 4, 8, 16, 3264 (2^6) 1, 2, 4, 8, 16, 32, 64128 (2^7) 1, 2, 4, 8, 16, 32, 64, 128256 (2^8) 1, 2, 4, 8, 16, 32, 64, 128, 256512 (2^9) 1, 2, 4, 8, 16, 32, 64, 128, 256, 5121024 (2^10) 1, 2, 4, 8, 16, 32, 64, 128, 256, 512, 10242048 (2^11) 1, 2, 4, 8, 16, 32, 64, 128, 256, 512, 1024, 20484096 (2^12) 1, 2, 4, 8, 16, 32, 64, 128, 256, 512, 1024, 2048, 40968192 (2^13) 1, 2, 4, 8, 16, 32, 64, 128, 256, 512, 1024, 2048, 4096, 819216384 (2^14) 1, 2, 4, 8, 16, 32, 64, 128, 256, 512, 1024, 2048, 4096, 8192, 1638432 768 (2^15) 1, 2, 4, 8, 16, 32, 64, 128, 256, 512, 1024, 2048, 4096, 8192, 16384, 32768 deficient-by-2 <= 327683 (3^1) 1, 310 (2^1) (5^1) 1, 2, 5, 10136 (2^3) (17^1) 1, 2, 4, 8, 17, 34, 68, 136 deficient-by-3 <= 32768ø *note: id:A000079 - OEIS Search Results Least deficient or near-perfect numbers (i.e. n such that sigma(n)=A000203(n)=2n-1). - Lekraj Beedassy (blekraj(AT)yahoo.com), Jun 03 2004. Comment from Max Alekseyev (maxale(AT)gmail.com), Jan 26 2005: All the powers of 2 are least deficient numbers but it is not known if there exists a least deficient number not a power of 2. ... Quote Link to comment Share on other sites More sharing options...
modest Posted June 27, 2010 Report Share Posted June 27, 2010 so thens...first one to list the first 30 or so numbers Abundant by 2 gets...erhm...something nice. :D ready, set, ...program! :) Updated Sets List Set Name___{Set Elements}______________________________<= Range Limit_______ A1 {?...} [#1] A2 {20, 104, 464, 650, 1952, 130304, 522752, ...?} <= 2521745 [#2, #4] Interesting topic! As far as abundant by 2, according to OEIS: if 2^n - 3 is prime then 2^(n-1)*(2^n-3) is abundant by 2 :lol: I don't think the method will give all those abundant by 2, but it looks like it gives most. The first few numbers n where 2^n - 3 is prime are: 3, 4, 5, 6, 9, 10, 12, 14, 20, 22, 24, 29, 94, 116, 122, 150, 174, 213, 221, 233 which, assuming I've programmed the algorithm correctly, give the following numbers abundant by 2: n = 3, x = 20 n = 4, x = 104 n = 5, x = 464 n = 6, x = 1952 n = 9, x = 130304 n = 10, x = 522752 n = 12, x = 8382464 n = 14, x = 134193152 n = 20, x = 549754241024 n = 22, x = 8796086730752 n = 24, x = 140737463189504 n = 29, x = 144115187270549504 n = 94, x = 196159429230833773869868419445529014560349481041922097152 n = 116, x = 3450873173395281893717377931138512601610429881249330192849350210617344 n = 122, x = 14134776518227074636666380005943348118644503200295456427550007248658890752 n = 150, x = 1018517988167243043134222844204689080525734194692096586259130403089761695707425648517578752 n = 174, x = 28668732699875893895135261191276086759957062364603510454933365879095241737302152173380828074567534641152 n = 213, x = 86645927941275464361825443254471365732388658605494267974077486874460861181071328214139570000066155755750644954940903259472134144 n = 221, x = 5678427533559428832416592249125035424637823130369672345949142181093689448385830531024367518372461675192150235469728157765532148301824 n = 233, x = 95268205270873786358080970147496530326800480428008152797215483387004732066360252234519151094886831234103907878649603962944581817093108793344 ~modest Turtle 1 Quote Link to comment Share on other sites More sharing options...
Turtle Posted June 27, 2010 Author Report Share Posted June 27, 2010 Interesting topic! As far as abundant by 2, according to OEIS: if 2^n - 3 is prime then 2^(n-1)*(2^n-3) is abundant by 2 :) I don't think the method will give all those abundant by 2, but it looks like it gives most. The first few numbers n where 2^n - 3 is prime are: 3, 4, 5, 6, 9, 10, 12, 14, 20, 22, 24, 29, 94, 116, 122, 150, 174, 213, 221, 233 which, assuming I've programmed the algorithm correctly, give the following numbers abundant by 2: n = 3, x = 20 n = 4, x = 104 n = 5, x = 464 n = 6, x = 1952 ......n = 233, x = 95268205270873786358080970147496530326800480428008152797215483387004732066360252234519151094886831234103907878649603962944581817093108793344 ~modest oh dear! i promised something nice, but what to get the guy who has everything! :eek: :hyper: i'll start with deep genuflection, , in your direction while i make some time to shop. i will add your additional numbers to the master list in the mean time to go with my thnx. while the expression does generate 20, i think it's because the prime is 5 which will happen justy that oncey and i suggest that like 650, any further anomalies the expression misses will be of the factor of five flava. on my deficient end nothing but powers-of-two in my bag, except i did find another deficient-by-2 that i overlooked listing yesterday. here's that today then: deficient-by-2 <=104857632896 (2^7) (257^1) , 2, 4, 8, 16, 32, 64, 128, 257, 514, 1028, 2056, 4112, 8224, 16448, 32896 on the interval <=1048576 that i have searched, we have no numbers deficient-by-3. :naughty: :cry: :hyper: i think that's it for now; i better get sum coffee in me. :cup: :cup: :coffee_n_pc: Quote Link to comment Share on other sites More sharing options...
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