Hello Dr. Rohit, as I told you there was a modified version of the program you made so that I found a relation for the x value, it comes about that it varies according to the numbers used in the comparison:

sq=Table[j,{j,100000}]
sq11=Table[j,{j,100,10000}]
n1=Select[sq,OddQ,(2002)]
n=17389
r=0
exp=(((n*x)+n)/n)==((((n*x)+x)/n))-1
eqn=Thread[exp ]
NSolve[#, x] & /@ eqn
sq=Table[j,{j,100000}]
sq11=Table[j,{j,100,10000}]
n2=17393
n1=17389
r=0
exp=(((n2*x)+n2)/n2)==((((n2*x)+x)/n2))-1
eqn=Thread[exp ]
NSolve[#, x] & /@ eqn
exp=((((n2*x)+n2)/n2)-((((n2*x)+x)/n2))/17389)-1==(((n*x)+n)/n)-((((n*x)+x)/n))-1
eqn=Thread[exp ]
NSolve[#, x] & /@ eqn

It gives true to the relation and gives me the value of x that should be considered when equaling the 2 equations, it just so happens that that number is the relation of p/p+1 p is a prime, which then led me to verify that they can be used in predicting the position of prime number without the need of a list, I Made a program to simulate the number interrelated between the 200th -400th prime with the 2000th to the 2200th prime , and I used the prime numbers selected only to have numbers to apply the formula and made a rule of 3 between them and its given position initially known of the first number, so the result of the following program gives the right position for the prime numbers n2 without having to use a list of primes, remember that I only used them to have numbers generated to try out the formula which at least for these conditions give a true result:

sq=Table[j,{j,100000}]
sq11=Table[j,{j,17389,100000}]
sq12=Table[j,{j,1223,100000}]
n1=Select[sq12,PrimeQ,(200)]
n=List[n1]
n2=Select[sq11,PrimeQ,(200)]
a=n1/(n1+1)
b=Select[sq,IntegerQ,(200)]
b1=a*b
c=b1/((n2/(n2+1)))
d=c*10
e=IntegerPart[d]
f=(e/10)+2000
IntegerPart[f]

result: {2000,2001,2002,2003,2004,2005,2006,2007,2008,2009,2010,2011,2012,2013,2014,2015,2016,2017,2018,2019,2020,2021,2022,2023,2024,2025,2026,2027,2028,2029,2030,2031,2032,2033,2034,2035,2036,2037,2038,2039,2040,2041,2042,2043,2044,2045,2046,2047,2048,2049,2050,2051,2052,2053,2054,2055,2056,2057,2058,2059,2060,2061,2062,2063,2064,2065,2066,2067,2068,2069,2070,2071,2072,2073,2074,2075,2076,2077,2078,2079,2080,2081,2082,2083,2084,2085,2086,2087,2088,2089,2090,2091,2092,2093,2094,2095,2096,2097,2098,2099,2100,2101,2102,2103,2104,2105,2106,2107,2108,2109,2110,2111,2112,2113,2114,2115,2116,2117,2118,2119,2120,2121,2122,2123,2124,2125,2126,2127,2128,2129,2130,2131,2132,2133,2134,2135,2136,2137,2138,2139,2140,2141,2142,2143,2144,2145,2146,2147,2148,2149,2150,2151,2152,2153,2154,2155,2156,2157,2158,2159,2160,2161,2162,2163,2164,2165,2166,2167,2168,2169,2170,2171,2172,2173,2174,2175,2176,2177,2178,2179,2180,2181,2182,2183,2184,2185,2186,2187,2188,2189,2190,2191,2192,2193,2194,2195,2196,2197,2198,2199} ”

For instance the program below returns `

sq=Table[j,{j,100000}]
sq11=Table[j,{j,100,10000}]
n=Select[sq,PrimeQ,(2002)]
n1=Select[sq,PrimeQ,(2002)]
r=0
exp=(((((n+r)*x)+n+r))/(n+r))/(((((n)*(x+r)+(x+r))))/(n1))
eqn=Thread[exp == 1]
NSolve[#, x] & /@ eqn



{NSolve[False,{1,3,5,7,9,11,13,15,17,19,21,23,25,27,29,31,33,35,37,39,41,43,45,47,49,51,53,55,57,59,61,63,65,67,69,71,73,75, \[CenterEllipsis]1926\[CenterEllipsis] ,3929,3931,3933,3935,3937,3939,3941,3943,3945,3947,3949,3951,3953,3955,3957,3959,3961,3963,3965,3967,3969,3971,3973,3975,3977,3979,3981,3983,3985,3987,3989,3991,3993,3995,3997,3999,4001,4003}], \[CenterEllipsis]2000\[CenterEllipsis] , \[CenterEllipsis]1\[CenterEllipsis] }

But what it calls false are the odd numbers that should be x

You are really awesome Dr Rohit, but your help made me see more than just codes, observing the results I have noticed a relation between prime numbers that will help to find the approximate position of a prime number without the need to list of all the previous prime numbers, it is a rule of three between decimal numbers, it appeared after i ran one of your codes with a very small change...but It is preliminary...as soon as I have worked it out I will let you know...God bless you.

It is still not clear to me what you are trying to do. In the above code the result from NSolve is a list like

{{{n1 -> 1.5}}, {{n1 -> 3.}}, {{n1 -> 5.}}, {{n1 -> 7.}}, {{n1 -> 10.8}}, {{n1 -> 12.8333}}, ...}

If this is closer to what you want, then what are you trying to extract from this list? Using your original problem 2 code is this what you are trying to do?

ClearAll[x, sq, sq11, n, n1, r, exp, eqn]

sq = Table[j, {j, 100000}];
sq11 = Table[j, {j, 100, 10000}];
n = Select[sq, PrimeQ, (2002)];
n1 = Select[sq, PrimeQ, (2002)];
r = 0;

exp = (((((n + r)*x) + n + r))/(n + r))/(((((n)*(x + r) + (x + r))))/(n1));

eqn = Thread[exp == 1];
sol = NSolve[#, x] & /@ eqn;

sol[[2000]]
(* {{x -> 17389.}} *)

Sorry Dr. Rohit my computer seems not to be working properly...for the equation above it gave me the results in one try then it changes the answer if I run the program again and again.

Hi Luis,

This is a list of expressions involving x

exp = (((((n+r)*x)+n+r))/(n+r))/(((((n)*(x+r)+(x+r))))/(n1))

Are you trying to find a value of x in each of those expressions that give the value 1? Then

eqn = Thread[exp == 1]
NSolve[#, x] & /@ eqn

The result is pretty obvious given how exp is constructed.

So I must be missing something. What exactly are you trying to solve and what is the relationship between your two programs?