Deficient & Abundant Number Fun

[Turtle]

i finished the deficient-by 1's, 2's, & 3's to my arbitrary interval <=2521745 and found no further prey. then i went afield for some deficient-by 4's, 5's, 6's, & 7's, which i pursued to 40000 leagues (ok; integers :naughty:) and then i tired for the night & went looking for anomalies in the abundant-by-2 camp by searching only every 5^th on the interval between my last "regular" abundant-by-2 and modest's next of the same blame. nothing of an anomaly to 4,800,000+ yet, and no guarantee of one at that.

 

here's what i found in those new deficient leagues. :coffee_n_pc:

 

deficient by 4 <= 40000

5 (5^1) 1, 5

14 (2^1) (7^1) 1, 2, 7, 14

44 (2^2) (11^1) 1, 2, 4, 11, 22, 44

110 (2^1) (5^1) (11^1) 1, 2, 5, 10, 11, 22, 55, 110

152 (2^8) (19^1) 1, 2, 4, 8, 19, 38, 76, 152

884 (2^2) (13^1) (17^1) 1, 2, 4, 13, 17, 26, 34, 52, 68, 221, 442, 884

2144 (2^5) (67^1) 1, 2, 4, 8, 16, 32, 67, 134, 268, 536, 1072, 2144

8384 (2^6) (131^1) 1, 2, 4, 8, 16, 32, 64, 131, 262, 524, 1048, 2096, 4192, 8384

18632 (2^3) (17^1) (137^1) 1, 2, 4, 8, 17, 34, 68, 136, 137, 274, 548, 1096, 2329, 4658, 9316, 18632

 

deficient-by-5 <=40000

9 (3^2) 1, 3, 9

 

deficient-by-6 <=40000

7 (7^1) 1, 7

15 (3^1) (5^1) 1, 3, 5, 15

52 (2^2) (13^1) 1, 2, 4, 13, 26, 52

315 (3^2) (5^1) (7^1) 1, 3, 5, 7, 9, 15, 21, 35, 45, 63, 105, 315

592 (2^4) (37^1) 1, 2, 4, 8, 16, 37, 74, 148, 296, 592

1155 (3^1) (5^1) (7^1) (11^1) 1, 3, 5, 7, 11, 15, 21, 33, 35, 55, 77, 105, 165, 231, 385, 1155

 

deficient-by-7 <=40000

50 (2^1) (5^2) 1, 2, 5, 10, 25, 50

[modest]

Here's wolfram's scatter plot of the first few:

 

 

-source

 

I've also attached a list of the first few values. Strange how many are abundant by 12 :shrug:

 

~modest

[Turtle]

...

I've also attached a list of the first few values. Strange how many are abundant by 12 ;)

 

~modest

 

muchas gracias. :hihi: how hard would it be to now program a sort for that list to put the elements into the sets i'm poking 'round the brush for? :xparty: ;) i could factor them by-hand so we can try & finagle expressions for individual sets. >> All Factors of a Number

 

i was just thinking too that maybe i'd do a run on A12 to my <=2521745 interval & only keep a count of how many; then i'd find a natural density there & compare it with the ~.24 density of all abundants that i earlier referenced. :shrug:

 

right now i'm still looking for an A2 divisible-by-5 anomaly in the range up to your 8 million or so that you found with that equation. erhm...i guess that's the full monty...carry on. :ud:

[modest]

 

The easiest way was to just let excel sort it. I ran it up to 40,000.

 

The first column is the amount it is abundant (negative if it's deficient), the second is the number, and the third column are the factors. So, the format is like this:

 

-2	3	1
-2	10	 1 2 5 
-2	136	 1 2 4 8 17 34 68 
-2	32896	 1 2 4 8 16 32 64 128 257 514 1028 2056 4112 8224 16448 
-1	2	1
-1	4	 1 2 
-1	8	 1 2 4 
-1	16	 1 2 4 8 
-1	32	 1 2 4 8 16 
-1	64	 1 2 4 8 16 32 
-1	128	 1 2 4 8 16 32 64 
-1	256	 1 2 4 8 16 32 64 128 
-1	512	 1 2 4 8 16 32 64 128 256 
-1	1024	 1 2 4 8 16 32 64 128 256 512 
-1	2048	 1 2 4 8 16 32 64 128 256 512 1024 
-1	4096	 1 2 4 8 16 32 64 128 256 512 1024 2048 
-1	8192	 1 2 4 8 16 32 64 128 256 512 1024 2048 4096 
-1	16384	 1 2 4 8 16 32 64 128 256 512 1024 2048 4096 8192 
-1	32768	 1 2 4 8 16 32 64 128 256 512 1024 2048 4096 8192 16384 
0	1	1
0	6	 1 2 3 
0	28	 1 2 4 7 14 
0	496	 1 2 4 8 16 31 62 124 248 
0	8128	 1 2 4 8 16 32 64 127 254 508 1016 2032 4064 
2	20	 1 2 4 5 10 
2	104	 1 2 4 8 13 26 52 
2	464	 1 2 4 8 16 29 58 116 232 
2	650	 1 2 5 10 13 25 26 50 65 130 325 
2	1952	 1 2 4 8 16 32 61 122 244 488 976 
3	18	 1 2 3 6 9 

So... I guess there are 5 perfect numbers under 40,000, and none abundant by 1 as expected.

 

It also looks like there are 860 abundant by 12. A lot :shrug:

 

Sort2.zip is a .csv file which would be good in excel or any other database program and sort.zip is a .txt file.

 

~modest

 

EDIT---->

 

Here's the code I used:

 

for ($n=1; $n <= 40000; $n++){
 $factor = 0;
 $sum = 0;
 @factors = ();
 for ($i = 1; $i < ($n/2+1); $i++){
   if (!($n % $i)){
     push(@factors, $i);
     $factor = $i;
     $sum += $factor;
   }
 }
 $abundant = $sum - $n;
 print "$abundant, $n, @factors n";
}

 

Second Edit:

 

I'm fixing the attachments... Just take a sec....

[modest]

Ok. Here are the proper attachments. I had neglected the numbers 1 and 2 for some odd reason. All's good now :shrug:

 

~modest

[modest]

Here are the first 402,192 sorted :(

 

The factors for each set are sorted too, so patters should be somewhat easy to spot. I don't think notepad will open it, but wordpad will.

 

sorted_402192.zip - File Shared from Box.net - Free Online File Storage

 

~modest

[Turtle]

Here are the first 402,192 sorted :)

 

The factors for each set are sorted too, so patters should be somewhat easy to spot. I don't think notepad will open it, but wordpad will.

 

sorted_402192.zip - File Shared from Box.net - Free Online File Storage

 

~modest

 

a veritable varietal orchard of number trees!! B) B) i'll get busy memorizing it right away. B)

[Turtle]

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.

n = 10, x = 522752

 

n = 12, x = 8382464

 

~modest

 

i searched from 522752 to 9000000 for another multiple-of-5-abundant-by-2 anomaly & found none. B) i guess at least there are a few numbers left to look at. B)

 

i thought deficient-by-4 was going to turn into something more, but your list to 400000+ added just one to the set after my search to 40000. here it is for what it's worth; i added the prime factorization for consistency.

 

deficient-by-4 <= 402192

116624 (2^4) (37^1) (197^1) 1 2 4 8 16 37 74 148 197 296 394 592 788 1576 3152 7289 14578 29156 58312

 

will get to that strange and/or unusual natural density next or such a matter. . . . . . B)

[Turtle]

okaly dokaly; got sum numbers then. for numbers abundant-by-12, my so called "strange" numbers, i found these values for the natural density.

 

2017/500031 = .0040337

3777/1000000 = .003777

5487/1500000 = .003658

7146/2000000 = .003573

8762/2521745 =. 003474

 

my take is that the strange set is the densest abundant-by that there be. :phones: 'course that smidgeon of doubt is what has me lookin' for something otherwise. ;) so, off to the races thens. . . . . . :cheer:

[modest]

my take is that the strange set is the densest abundant-by that there be.

 

Did you find in the strange number thread, or otherwheres, why that would be? It seems odd to me.

 

~modest

[Turtle]

my take is that the strange set is the densest abundant-by that there be. :phones:

Did you find in the strange number thread, or otherwheres, why that would be? It seems odd to me.

 

~modest

 

it seems odd to me as well and i did not/have not run across a why. it has taken 15 years to get to what & how. :lol: we never looked specifically at the natural density in the strange thread either, although it was a frequent general characterization. as all the unusual sets can have the majority of their elements generated by multiplying a Perfect times a Prime & Bombadil's equation gives the rest, their natural densities must be just slightly larger than that of Primes. they should also, it seems, then be of nearly equal density because of that , excepting for the variance in the number of anomalies for each set. nonetheless, the next unusal set, those abundant by 56 & my so called "bizarre" numbers, will have a lower density over the intervals i ran for stranges because 28>6. :cheer: well...no one said this was a cake walk. ;) :eek:

 

anyway, i'll leave you with a bit that q tossed into the strange salad early on and which for some strange and/or odd reason keeps nagging the back of my caranium. you know how it is with that q when he gets in the kitchen. :D ;) :cheer:

 

It is easy to show why 6p is strange for any prime p and also why it has 8 divisors. There is no reason for the opposite implications to hold, hence cases such as 304 are free to occur. As usual, it is easier to use the sum including the number itself so that perfect means S = 2n and strange means S = 2n + 12. Multiply 6 by any prime p and its divisors wil be:

 

1, 2, 3, 6, p, 2p, 3p, 6p

 

no more, no less than 8 of them. The sum is exactly 12 + 12p = 12 + 2n.

[Turtle]

okaly dokaly; got sum more numbers then. for numbers abundant-by-56, my so called "bizarre" numbers, i found these values for the natural density over the same intervals i earlier took the stranges. ;)

 

520/500000 = .00104

952/1000000 = .000952

1370/1500000 = .000913

1711/2000000 = .000855

2124/2521745 =. 000842

 

just for fun i looked to see if the densities were proportional between strange & bizarre to the ratio of their respective perfect generators, i.e. , 28/6 = 4.66666

 

i found:

2017/520 = 3.87884

3777/952 = 3.96743

5487/1370 = 4.00510

7146/1711 = 4.17650

8762/2124 = 4.12523

 

mmmm... i expected it to keep increasing toward the 4.66666, then that last one dropped a squeek. :eek: must be the anomalies or sumpin'. :cheer: if i recall, the strange set has fewer anomalies than its cousins. :lol:

 

well, i digress me thinks. i will get to reviewing modest's list for some prospective probeees outside of the unusuals. :cheer: . . . . . :phones:

[Turtle]

abundant-by-8 looks promisingly populous per pmodes't plist sorted_402192.txt on the interval <=402192. :cheer: i'll have a run at confirming it & extending it to <=2521745. here's modesterno's numbers; prime factorizations my addition. :cheer: :phones:

 

abundant-by-8 <=402192

56 (2^3) (7^1) 1 2 4 7 8 14 28

 

368 (2^4) (23^1) 1 2 4 8 16 23 46 92 184

 

836 (2^2) (11^1) (19^1) 1 2 4 11 19 22 38 44 76 209 418

 

11096 (2^3) (19^1) (73^1) 1 2 4 8 19 38 73 76 146 152 292 584 1387 2774 5548

 

17816 (2^3) (17^1) (131^1) 1 2 4 8 17 34 68 131 136 262 524 1048 2227 4454 8908

 

45356 (2^2) (17^1) (23^1) (29^1) 1 2 4 17 23 29 34 46 58 68 92 116 391 493 667 782 986 1334 1564 1972 2668 11339 22678

 

77744 (2^4) (43^1) (113^1) 1 2 4 8 16 43 86 113 172 226 344 452 688 904 1808 4859 9718 19436 38872

 

91388 (2^2) (11^1) (31^1) (67^1) 1 2 4 11 22 31 44 62 67 124 134 268 341 682 737 1364 1474 2077 2948 4154 8308 22847 45694

 

128768 (2^8) (503^1) 1 2 4 8 16 32 64 128 256 503 1006 2012 4024 8048 16096 32192 64384

 

254012 (2^2) (11^1) (23^1) (251^1) 1 2 4 11 22 23 44 46 92 251 253 502 506 1004 1012 2761 5522 5773 11044 11546 23092 63503 127006

 

388076 (2^2) (13^1) (17^1) (439^1) 1 2 4 13 17 26 34 52 68 221 439 442 878 884 1756 5707 7463 11414 14926 22828 29852 97019 194038

[Turtle]

i would have more to report but the last 24 has been a living hell. :evil: seems some too-clever-for-their-own-good douche-bag left the "increment by 5" command in place on the main loop after a certain anomaly search & so resulting in a severe mismatch between the programs abundant-by-8 output & modestino's & so further resulting in a desperate search for the problemo. :doh: good grief! :rant: what's it take to get decent help around here!!! well, i think i'm back on track & have the first 7 elements matching & in the bag. i need some beer & an m-80 or 2. :friday: :ebomb: :ebomb:

[Turtle]

so nows... have got to 1,850,000+ overnight & no new A8's to show for it. if i hadn't lost a day in confusion i could well have twice that many to show. :friday: anywhos, i've listed the new finds below & should hit the arbitrary interval <=2521745 by even time & will report again then on the apparent dead end.

[Turtle]

found 2 more A8's. please sir; may i have sum more? :friday:

 

abundant-by-8 <=2521745

2087936 (2^10) (2039^1) 1, 2, 4, 8, 16, 32, 64, 128, 256, 512, 1024, 2039, 4078, 8156, 16312, 32624, 65248, 130496, 260992, 521984, 1043968

 

2291936 (2^5) (67^1) (1069^1) 1, 2, 4, 8, 16, 32, 67, 134, 268, 536, 1069, 1072, 2138, 2144, 4276, 8552, 17104, 34208, 71623, 143246, 286492, 572984, 1145968

[modest]

found 2 more A8's.

 

Brilliant.... as they say across the pond.

 

With some time in hand tomorrow, I'll see if I can get that list a bit longer. Hopefully I'll even be able to look at it :friday:

 

~modest