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Hi,
Wants some suggestion on
If mod(Prime B ,Pn*)>1 then result will be Small Prime A in proper Range.
where Pn* is multiplication of all prime number less the nth prime number Ex 6,30,210,2310,30030,510510,9699690,….Pn*. Prime B>Pn*
Range of Prime B
3*=6 gives prime 3>=p<9
5*=30 gives prime 5>=p<=25
7*=210 gives prime 7>=p<=49
11*=2310 gives prime 11>=p<=121
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Pn*=...gives prime Pn*>=p<=(Pn)^2
Ex
6 gives prime 3>=p<9
Mod (23,6)= 5 ; Mod (29,6)= 5 ;
30 gives prime 5>=p<=25
Mod(157,30)=7: Mod(163,30)=13: Mod(167,30)=17; Mod(173,30)=23
210 gives prime 7>=p<=49
Mod(223,210)=13; Mod(227,210)=17; Mod(229,210)=19; Mod(233,210)=23; Mod(239,210)=29;Mod(241,210)=31; Mod(251,210)=41
Agnishom wrote:Very good.
I suppose 49 is prime too, by the same logic?
No dear How you have calcluted can u explain...you have made mistake.:cool
take example of consecutive prime there mod with 30,210,2310 will give prime in range .Range is given in table.
consecutive prime MOD 30 (5 < prime< 25 ) MOD 210 ( 7 < prime < 49 ) Mod 2310 (11< pprime < 121 )
2,333 23 23 23
2,339 29 29 29
2,341 1 31 31
2,347 7 37 37
2,351 11 41 41
2,357 17 47 47
2,371 1 61 61
2,377 7 67 67
2,381 11 71 71
2,383 13 73 73
2,389 19 79 79
2,393 23 83 83
2,399 29 89 89
2,411 11 101 101
2,417 17 107 107
2,423 23 113 113
2,437 7 127 127
2,441 11 131 131
2,447 17 137 137
now ok, if we have big prime we can find out smaller one.
in general big primes yeilds smaller one........:d
Very good.
I suppose 49 is prime too, by the same logic?
No dear How you have calcluted can u explain...you have made mistake.:cool
We have many number theorists here...
Is 9831655609 prime?
9831655609 ( MOD 30)=19 PRIME
Odd Prime numers always yiels a small prime number ( in proper range ).
Every odd Prime numers yiels a small prime number ( in proper range ).
Prime Number MOD 30 (5< p<25 ) MOD 210 (7< p<49 ) Mod 2310 (11< p<121 )
2,347 7 37 37
2,351 11 41 41
2,357 17 47 47
4,637 17 17 17
4,639 19 19 19
4,643 23 23 23
4,657 7 37 37
4,663 13 43 43
6,947 17 17 17
6,949 19 19 19
9,277 7 37 37
9,281 11 41 41
9,283 13 43 43
This directly Means if you have big prime number with help of mod smaller can be find out.
Hi satwnz;
You say that it is not time consuming. But I think you mean just for small numbers.
Is 1 111 111 111 111 111 111 a prime? What do you subtract from this to know?
I hv deal wit max 15 digit number.But i can go for n digit numer if some supportive system is with me.
if it is well define for numer up to 15 digit We sure it will give posative result fo n digit number.
You have a list of prime numbers in the far left column,
where they are being used for analysis/calculations.It looks as though you have to have those prime numbers
in advance to get a longer list of greater prime numbers.
Ipn series--- It is well define set of posative number.
Range of continues Prime ----It contains each and every prime number in fixed range.
No of Ipn sub series...This Ipn series have subseries it is also well define.
Common difference......As subseries are in A.P so common difference is given
% of Ipn element w.r.t. + Ve integer........it %of Ipn series wrt posative integer.i.e how many Ipn are in set of all +ve integer
Ex-I3( it is Ipn series) have one AP Serie and C.d = 2
3, 5, 7, 9, 11, 13------∞
i) Contain all primes and specific odd positive integer from 3 to infinite
ii) Continious prime 3≤P<9
ii) I3 Contain only 50.00% of positive integer
Ex-I5( it is Ipn series) have two AP Serie and C.d = 6
5 11 17 23 29 35 41 ------∞ (sub series 1)
7 13 19 25 31 37 43 -----∞ (sub series 2)
i) Contain all prime and specific odd positive integer from 5 to infinite
ii) Continuous prime 5≤P<25
ii) I5 Contains only 33.33% of positive integer
like this we can go for any series and any prime number.
br//
satish
PREVIOUS POST WAS APPLICATION OF PATTERN ONLY
Ipn series Range of continues Prime No of Ipn sub series Common difference % of Ipn element w.r.t. + Ve integer
I2 2 ≤ P < 4 1 2 100%
I3 3 ≤ P < 9 1 2 50%
I5 5 ≤ P < 25 2 6 33.33%
I7 7 ≤ P < 49 8 30 22%
I11 11 ≤ P < 121 48 210 20%
I13 13 ≤ P < 169 480 2310 19%
I17 17 ≤ P < 289 5760 30030 18%
I19 19 ≤ P < 361 92160 510510 17%
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IpN................
Ex-I3 have one AP Serie and C.d = 2
3, 5, 7, 9, 11, 13------∞
i) Contain all rpimes and specific odd positive integer from 3 to infinite
ii) Continious prime 3≤P<9
ii) I3 Contain only 50.00% of positive integer
Ex-I5 have two AP Serie and C.d = 6
5 11 17 23 29 35 41 ------∞
7 13 19 25 31 37 43 -----∞
i) Contain all prime and specific odd positive integer from 5 to infinite
ii) Continuous prime 5≤P<25
ii) I5 Contains only 33.33% of positive integer
Hi satwnz;
Welcome to the forum.
2310-2179=131
2310-2173=137
2310-2171=139
2310-2161=149
2310-2159=151
2310-2153=157
2310-2147=163
2310-2143=167The numbers you are using to subtract, how are you creating them?
how are you creating them...........There two step algo to fin it.If u r intrested in it plz contact me on [email removed by moderator]
Ah I see, that's an interesting conjecture.
satwnz you can try publishing it.
really...........will u help me.
Hi GeorgeY;
I know about that result and there are others.
2310-2173=137
That still does not explain where the 2173 ( along with the rest ) comes from. That is what I hoping to get him to reveal.
In fact i am researching in field of prime number pattern..........and number u hd underlied is element of welldfine set.
Hi satwnz;
Welcome to the forum.
2310-2179=131
2310-2173=137
2310-2171=139
2310-2161=149
2310-2159=151
2310-2153=157
2310-2147=163
2310-2143=167The numbers you are using to subtract, how are you creating them?
Hi it is well define sets of number .........and can go for larger number also..........but my laptop support to 15 digits only.
My question is that, is it any pseudo pattern gives prime number in proper range? If some set of number can give you each and every prime number in fixed range? Range is like this
2 to 4
3 to 9
5 to 25
11 to 121
13 to 169
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n to n*n
I have way to find out prime number. It is simple. Most important things is that its not a time taking. Just i will have to subtract a unique set of number from fixed number.
example no 1 ( This give us prime no grater then equal to 11 and less then equal to 121).
210-199=11
210-197=13
210-193=17
210-191=19
210-187=23
210-181=29
210-179=31
210-173=37
210-169=41
210-167=43
210-163=47
210-157=53
210-151=59
210-149=61
210-143=67
210-139=71
210-137=73
210-131=79
210-127=83
210-121=89
210-113=97
210-109=101
210-107=103
210-103=107
210-101=109
210- 97=113
I can find out any prime number.
I have used unique set of number and subtract it from fixed number as in example no 1.
example no 2.( This give us prime no grater then equal to 13 and less then equal to 169)
2310-2297=13
2310-2293=17
2310-2291=19
2310-2287=23
2310-2281=29
2310-2279=31
2310-2273=37
2310-2269=41
2310-2267=43
2310-2263=47
2310-2257=53
2310-2251=59
2310-2249=61
2310-2243=67
2310-2239=71
2310-2237=73
2310-2231=79
2310-2227=83
2310-2221=89
2310-2213=97
2310-2209=101
2310-2207=103
2310-2203=107
2310-2201=109
2310-2197=113
2310-2183=127
2310-2179=131
2310-2173=137
2310-2171=139
2310-2161=149
2310-2159=151
2310-2153=157
2310-2147=163
2310-2143=167.
Sir I can find any prime number of multi digits. But calculating limit of computer do not allow me to work further. So please help me in this regards.
Sir i had already solved so many things in prime number representation is required.
So my request is if possible plz suggest me the platform where i can represent it.
SATISH KUMAR SINGH
GTL NPD PUNE
09011056492
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