# Determining sequence of regular numbers

I'm interested in the set of numbers that are regular in the sexagesimal (base 60) system, that is, the numbers that are expressible as $2^a3^b5^c$, where $a,b,c\in\mathbb{N}$. So the regular numbers so defined are $x=1,2,3,4,5,6,8,9,10,12,15,18,20,...$ There's no OEIS entry in particular for that sequence.

Just starting from $x=1$ and looping through a sieve-like procedure (repeated division by 2,3 and 5) generates this list of regular numbers:

n a b c x
1 0 0 0 1
2 1 0 0 2
3 0 1 0 3
4 2 0 0 4
5 0 0 1 5
6 1 1 0 6
7 3 0 0 8
9 0 2 0 9


...

My question: is there a more efficient way to generate the list of $x$ (or the coefficients $a,b,c$) other than by testing each number by repeated division?

• Well, repeated division seems unnecessary. Fixing some size $N$ it is not hard to enumerate the triples $(a,b,c)$ with $a,b,c$ each $≤ N$. Now, you just have to sort the output here. – lulu Sep 3 '17 at 17:59
• Better--still kind of involved (generating all the triples and sorting). – Joe Knapp Sep 3 '17 at 19:03
• Yes, well I fear the sorting is necessary. Given the triple $(a,b,c)$ I don't believe there is a simple way to determine which triple is "next". – lulu Sep 3 '17 at 19:14

==========================================================

Sun Sep  3 16:37:06 PDT 2017

1       2       3       4       5       6       8       9      10      12
15      16      18      20      24      25      27      30      32      36
40      45      48      50      54      60      64      72      75      80
81      90      96     100     108     120     125     128     135     144
150     160     162     180     192     200     216     225     240     243
250     256     270     288     300     320     324     360     375     384
400     405     432     450     480     486     500     512     540     576
600     625     640     648     675     720     729     750     768     800
810     864     900     960     972    1000    1024    1080    1125    1152
1200    1215    1250    1280    1296    1350    1440    1458    1500    1536
1600    1620    1728    1800    1875    1920    1944    2000    2025    2048
2160    2187    2250    2304    2400    2430    2500    2560    2592    2700
2880    2916    3000    3072    3125    3200    3240    3375    3456    3600
3645    3750    3840    3888    4000    4050    4096    4320    4374    4500
4608    4800    4860    5000    5120    5184    5400    5625    5760    5832
6000    6075    6144    6250    6400    6480    6561    6750    6912    7200
7290    7500    7680    7776    8000    8100    8192    8640    8748    9000
9216    9375    9600    9720   10000   10125   10240   10368   10800   10935
11250   11520   11664   12000   12150   12288   12500   12800   12960   13122
13500   13824   14400   14580   15000   15360   15552   15625   16000   16200
16384   16875   17280   17496   18000   18225   18432   18750   19200   19440
19683   20000   20250   20480   20736   21600   21870   22500   23040   23328
24000   24300   24576   25000   25600   25920   26244   27000   27648   28125
28800   29160   30000   30375   30720   31104   31250   32000   32400   32768
32805   33750   34560   34992   36000   36450   36864   37500   38400   38880
39366   40000   40500   40960   41472   43200   43740   45000   46080   46656
46875   48000   48600   49152   50000   50625   51200   51840   52488   54000
54675   55296   56250   57600   58320   59049   60000   60750   61440   62208
62500   64000   64800   65536   65610   67500   69120   69984   72000   72900
73728   75000   76800   77760   78125   78732   80000   81000   81920   82944
84375   86400   87480   90000   91125   92160   93312   93750   96000   97200
98304   98415  100000  101250  102400  103680  104976  108000  109350  110592
112500  115200  116640  118098  120000  121500  122880  124416  125000  128000
129600  131072  131220  135000  138240  139968  140625  144000  145800  147456
150000  151875  153600  155520  156250  157464  160000  162000  163840  164025
165888  168750  172800  174960  177147  180000  182250  184320  186624  187500
192000  194400  196608  196830  200000  202500  204800  207360  209952  216000
218700  221184  225000  230400  233280  234375  236196  240000  243000  245760
248832  250000  253125  256000  259200  262144  262440  270000  273375  276480
279936  281250  288000  291600  294912  295245  300000  303750  307200  311040
312500  314928  320000  324000  327680  328050  331776  337500  345600  349920
354294  360000  364500  368640  373248  375000  384000  388800  390625  393216
393660  400000  405000  409600  414720  419904  421875  432000  437400  442368
450000  455625  460800  466560  468750  472392  480000  486000  491520  492075
497664  500000  506250  512000  518400  524288  524880  531441  540000  546750
552960  559872  562500  576000  583200  589824  590490  600000  607500  614400
622080  625000  629856  640000  648000  655360  656100  663552  675000  691200
699840  703125  708588  720000  729000  737280  746496  750000  759375  768000
777600  781250  786432  787320  800000  810000  819200  820125  829440  839808
843750  864000  874800  884736  885735  900000  911250  921600  933120  937500
944784  960000  972000  983040  984150  995328 1000000

Sun Sep  3 16:37:06 PDT 2017


==============================================================

int main()
{

set<mpz_class> smooth;
set<mpz_class>::iterator iter;

mpz_class bound = 1000000;
system("date");
cout << endl << endl;

for(mpz_class a = 1; a <= bound; a *= 2) {
for(mpz_class b = 1; a * b <= bound; b *= 3){
for( mpz_class c = 1; a * b * c <= bound; c *= 5){
smooth.insert(a * b * c);

}}}

int count = 0;
for(iter = smooth.begin(); iter != smooth.end(); ++iter)
{
++count;
cout << setw(8) << *iter;
if( count % 10 == 0) cout << endl;
}

cout << endl << endl;
system("date");
cout << endl << endl;
return 0;
}


===============================================================

• Nice, but finding the regular numbers to 1,000,000 with a sieve takes about 0.05 seconds on my machine too. To 100,000,000 takes 4.4 seconds. What would be ideal is, given a regular number, a more direct calculation of the next regular number. Maybe that's not possible though. – Joe Knapp Sep 4 '17 at 0:17
• @JoeKnapp you gave the impression you did not know how to program this: on your revised problem, it is more like a double loop, as, given fixed $n,$ and some $2^a 3^b \leq n,$ there is just one suitable exponent of $5.$ – Will Jagy Sep 4 '17 at 0:24