# Euler’s (Prime to) Prime Generating Equation

The greatest number theorist in mathematical universe, Leonhard Euler had discovered some formulas and relations in number theory, which were based on practices and were correct to limited extent but still stun the mathematicians. The *prime generating equation* by Euler is a very specific binomial equation on prime numbers and yields more primes than any other relations out there in number theory. Euler told that the equation $ f(x)=x^2+x+k$ yields many prime numbers with the values of x being input from *x=0* to* $x=k-2$ *; *k* being a prime.

Let’s see how many primes we can get by using different values of $k$ and $x$:

Serial Number | Value of k (prime) | Value of x (from x=0 to x=k-2) | Value of f(x)=(x^2+x)+k | Not a Prime? |

1 | 2 | 0 | 2 | |

2 | 3 | 0 | 3 | |

3 | 1 | 5 | ||

4 | 5 | 0 | 5 | |

5 | 1 | 7 | ||

6 | 2 | 11 | ||

7 | 3 | 17 | ||

8 | 7 | 0 | 7 | |

9 | 1 | 9 | No | |

10 | 2 | 13 | ||

11 | 3 | 19 | ||

12 | 4 | 27 | No | |

13 | 5 | 37 | ||

14 | 11 | 0 | 11 | |

15 | 1 | 13 | ||

16 | 2 | 17 | ||

17 | 3 | 23 | ||

18 | 4 | 31 | ||

19 | 5 | 41 | ||

20 | 6 | 53 | ||

21 | 7 | 67 | ||

22 | 8 | 83 | ||

23 | 9 | 101 | ||

24 | 13 | 0 | 13 | |

25 | 1 | 15 | No | |

26 | 2 | 19 | ||

27 | 3 | 25 | No | |

28 | 4 | 33 | No | |

29 | 5 | 43 | ||

30 | 6 | 55 | No | |

31 | 7 | 69 | No | |

32 | 8 | 85 | NO | |

33 | 9 | 103 | ||

34 | 10 | 123 | No | |

35 | 11 | 145 | No | |

36 | 17 | 0 | 17 | |

37 | 1 | 19 | ||

38 | 2 | 23 | ||

39 | 3 | 29 | ||

40 | 4 | 37 | ||

41 | 5 | 47 | ||

42 | 6 | 59 | ||

43 | 7 | 73 | ||

44 | 8 | 89 | ||

45 | 9 | 107 | ||

46 | 10 | 127 | ||

47 | 11 | 149 | ||

48 | 12 | 173 | ||

49 | 13 | 199 | ||

50 | 14 | 227 | ||

51 | 15 | 257 | ||

52 | 19 | 0 | 19 | |

53 | 1 | 21 | ||

54 | 2 | 25 | No | |

55 | 3 | 31 | ||

56 | 4 | 39 | No | |

57 | 5 | 49 | ||

58 | 6 | 61 | ||

59 | 7 | 75 | No | |

60 | 8 | 91 | No | |

61 | 9 | 109 | ||

62 | 10 | 129 | No | |

63 | 11 | 151 | ||

64 | 12 | 175 | No | |

65 | 13 | 201 | No | |

66 | 14 | 229 | ||

67 | 15 | 259 | ||

68 | 16 | 291 | ||

69 | 17 | 325 | No | |

70 | 23 | 0 | 23 | |

71 | 1 | 25 | No | |

72 | 2 | 29 | ||

73 | 3 | 35 | No | |

74 | 4 | 43 | ||

75 | 5 | 53 | ||

76 | 6 | 65 | No | |

77 | 7 | 79 | ||

78 | 8 | 95 | No | |

79 | 9 | 113 | ||

80 | 10. | 133 | ||

81 | 11 | 155 | No | |

82 | 12 | 179 | ||

83 | 13 | 205 | No | |

84 | 14 | 233 | ||

85 | 15 | 263 | ||

86 | 16 | 295 | No | |

87 | 17 | 329 | ||

88 | 18 | 365 | No | |

89 | 19 | 403 | ||

90 | 20 | 443 | ||

91 | 21 | 485 | No |

The above table yields many prime numbers, which again can be put at the place of k and so on the table can be progressed.

According to Euler, 41 was the most appropriate value of k yielding more prime numbers than any other k. In the list below, each value of f(x) is a prime for k=41:

k= 41 |
x= 0 |
f(x)= 41 |

1 | 43 | |

2 | 47 | |

3 | 53 | |

4 | 61 | |

5 | 71 | |

6 | 83 | |

7 | 97 | |

8 | 113 | |

9 | 131 | |

10 | 151 | |

11 | 173 | |

12 | 197 | |

13 | 223 | |

14 | 251 | |

15 | 281 | |

16 | 313 | |

17 | 347 | |

18 | 383 | |

19 | 421 | |

20 | 461 | |

21 | 503 | |

22 | 547 | |

23 | 593 | |

24 | 641 | |

25 | 691 | |

26 | 743 | |

27 | 797 | |

28 | 853 | |

29 | 911 | |

30 | 971 | |

31 | 1033 | |

32 | 1097 | |

33 | 1163 | |

34 | 1231 | |

35 | 1301 | |

36 | 1373 | |

37 | 1447 | |

38 | 1523 | |

39 | 1601 |

So, the Euler’s Prime Generating Equation can be written as

$ f(x) = x^2+x+41$ ; where $x$ is an integer ranging from 0 to 39.

What if we increase the value of x beyond the limit of 39? What will we get?

The next values of f(x) in this series would be 1681, 1763, 1847, 1933, 2021, 2111, 2203, 2297, 2393, … .

Are all these prime numbers too? The answer is no. 1681 is not a prime number, neither are 1763 and 2021. Though all others (in above sequence at-least) are prime numbers.

I believe I have a formula far better than Euler’s formulae. It will generate consecutive primes from 2 to 7993. Here is the

formula: factor(product(2*i,i,1,8000)); written in Maxima software. Try it. What do you think?

Here is a shortened version tor pagination problem:

Numerical example:

`product(i,i,1,20);`

2432902008176640000

factor(product(i,i,1,20));

2^18

3^85^47^2111317*19cadd(m)::=[ppp:factor((m)^2),zzz:(makelist(part(ppp,k),k,1,length(ppp))),xxx:factor(zzz),yyy:makelist(first(part(xxx,j)),j,1,length(xxx))]$

cadd(2432902008176640000);

Output:

`...........................,[2,3,5,7,11,13,17,19]]`