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About newreadtest28a.c

newreadtest28a.c computes the logarithms of some colossally abundant numbers:

[david2@localhost eratosthenes10]$ tail newreadtest28a.txt
p = 990999999973 epsilon = 3.653183885026180310242468860377e-14 sumlog = 991000328030.7521840041349258218183689
p = 991999999999 epsilon = 3.649367988977524534022013659376e-14 sumlog = 992000305969.8131669915172919502946597
p = 992999999983 epsilon = 3.645559922508440300966127655797e-14 sumlog = 993000403998.5058883802442669692510009
p = 993999999997 epsilon = 3.641759661414506257864442341368e-14 sumlog = 994000343836.5476306542283131617881152
p = 994999999999 epsilon = 3.637967182022573332237327579364e-14 sumlog = 995000415040.4985166201990123610689142
p = 995999999987 epsilon = 3.634182460603590021375694809873e-14 sumlog = 996000434206.3010629958101936344713337
p = 996999999997 epsilon = 3.630405473381214673873020363065e-14 sumlog = 997000460391.2291432726170884306987827
p = 997999999961 epsilon = 3.626636197067484207099474991832e-14 sumlog = 998000421458.2776765211077950579268171
p = 998999999993 epsilon = 3.622874607783614393365893006510e-14 sumlog = 999000434086.3161680586739901545384991
p = 999999999989 epsilon = 3.619120682566540114722976255853e-14 sumlog = 1000000442519.6083178943627706249117172

For p=999999999989, we have the smallest colossally abundant number divisible by 999999999989.  The data in the file test_data has the sums of logs of primes in intervals of length 10^9.

For p=999999999989, the least colossally abundant number divisible by 999999999989 has prime exponents:  ………. 1, 1, 1, 1, 1,  …. 1   the last 1 being the exponent of 999999999989. epsilon is a critical exponent for p, using a formula of Erdos and Alaoglu. The file primes1meg02a_tmp.txt contains the first million primes. The critical exponent epsilon is used to compute the exponents of 2, 3, 5, …. in the least colossally abundant number divisible by 999999999989.  The file lastprimes.txt has the largest prime less than k*10^9 for k=1… 1000.

__float128 datalog[1000];

fscanf(in, “%ld”, &low);
fscanf(in, “%ld”, &high);
fscanf(in, “%ld”, &count);
fscanf(in, “%s”, &buf1);
fscanf(in, “%s”, &buf2);
r = strtoflt128(buf2, NULL);

to read-in 1000007655.0495319095806686265977084 from the file test_data, first we read it as a string, via:

fscanf(in, “%s”, &buf2);

then we takes the string and convert it to a “float128” number via:

r = strtoflt128(buf2, NULL);

(libquadmath): “The function strtoflt128 converts a string into a __float128 number. ”

[david2@localhost eratosthenes10]$ tail test_data
990000000000 991000000000 36199430 1.0000365472937546499511140 999881674.7754749531487137072245348 990999999973
991000000000 992000000000 36201568 1.0000365125886500074621567 999977260.0600607772866329260618576 991999999999
992000000000 993000000000 36204598 1.0000364788519648209976829 1000097453.2757808979303762626266836 992999999983
993000000000 994000000000 36197548 1.0000364350387814193174497 999939158.9920624920826767515634271 993999999997
994000000000 995000000000 36200984 1.0000364018563934663325811 1000070496.5828500338643069387810857 994999999999
995000000000 996000000000 36197781 1.0000363620712353678219426 1000018392.2570007504864186850774648 995999999987
996000000000 997000000000 36196723 1.0000363245187239850619526 1000025505.8069259366166040624561963 996999999997
997000000000 998000000000 36193050 1.0000362844207603347530094 999960331.3078604391568357486432332 997999999961
998000000000 999000000000 36193600 1.0000362486319180248149473 1000011793.2258025649434780740817093 998999999993
999000000000 1000000000000 36192139 1.0000362109025271018407632 1000007655.0495319095806686265977084 999999999989

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Written by meditationatae

May 31, 2015 at 6:32 pm

Posted in History

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