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failure functions

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Please note that in the case of polynomials k belongs to Z.

In the case of exponential functions k belongs to N.

I think my definition needs to be elaborated. Let me begin with polynomials. Let f(x) be a polynomial in which x and the co-efficients belong to Z. Then f(f+f(x). f(x)) is congruent to zero mod(f(x)).This is easily proved by application of Taylor's series.
A numerical example will make this clearer. Let the polynomial be f(x) = x^2 + 1. When x = 1, the relevant f.f. is x = 1 + 2k, where k belongs to Z. It can easily be verified that when the values of x generated by this f.f. are substituted in f(x) we get only failures (in this case only even numbers).
Incidentally this particular example is relevant to the topic,
" Draft proof" currently a part of the forum (Research (682)).

Sketch proof of infinitude of primes with form x^2+1: The proof depends on the fact that any value of x, in any given interval, not covered by the relevant failure functions is such that x^2 + 1 is prime.These need not be tested for primality. Example: Let us take the interval (4,21). The relevant failure functions are 1+2k, 2 + 5k, 3+5k, 4+ 17k and 5+ 13k. These do not cover 4,6,10,14,16 and 20. These values of x are such that f(x) is prime which need not be tested for primality. (To be cotinued).

Let alpha be the fraction of values of x in any given interval not covered by the relevant failure functions. Let P_0 be the largest known prime of form x^2+1. Let X_0 be the relevant value of x i.e. X_0^2+1 = P_0. Consider the interval ( X_0, x_0+P_0). (This interval is considered because f(X_0+P_0) is congruent to 0 mod(P_0)). Since alpha decreases in an asymptotic manner to a rational numberPlanetmathPlanetmathPlanetmath significantly above zero there are some values of x in the above interval not covered by the failure functions. Let the largest of such uncovered values of x be X_1 and the relevant prime i.e. f(x_1) be P_1. The iteration is repeated for the interval (X_1,X_1 +_P_1) and the largest value of x in this interval not covered by the relevant failure functions is designated X_2 rendering f(X_2) prime. Note P_i ¿ P_(i-1)¿……….P_0.. Hence the intervals considered become increasingly larger. The iteration is perpetual since alpha is asymptotic to a rational number significantly above zero. This proves the infinitude of primes of form x^2+1.

a) To go from iteration i to i + 1 it will suffice if atleast one value of x is skiped by the relevant failure functions.Since alpha is asymptotic to a rational numberPlanetmathPlanetmathPlanetmath significantly greater than zero procedure from iteration i to iteration i + 1 is certain. b) This ensures the perpetual continuation of iterations.

As in the case of the polynomial x^2 + 1, failure functions pertaining to exponential functionsPlanetmathPlanetmath not covering values of the variable exponent are such that f(x) is prime. These primes need not be tested for primality. Let me illustrate with f(x) = 2^x + 7. When x =1, the f.f. is 1 +2k. When x = 2, the f.f. is 2 + 10k. When x = 3, the non-reduntant f.f. is 3 + 4k. When x = 4, the f.f. is 4 + 22k and so on. As before k belongs to N. If you were to list x =1,2,3,4………upto x= 100 (say) and eliminate all values of x covered by the f.f.s the remaining values of x are such that each f(x) is prime and these need not be tested for primality.

An offshoot: (2^1020x7 + 1) is congruent to 0 mod(1031).

Other bigger computations can be done ( to be demonstrated).

2^(38 + 274877906950*k) + 7 is congruent to 0 (mod(274877906951); here k belongs to N.

Theses numbers run to billions of digits - one can say with confidence that the congruencePlanetmathPlanetmathPlanetmath is true.

Further examples: 1) Let our definition of a failure be a non-Shantha prime. Recall that the definition of a Shantha

prime is (3^M_p - 2 ) whenever this expression generates a prime (M_p = Mangammal prime- see OEIS A123239 ). Now let the mother function be

3^n-2. Then n = 2 + 6*k is a failure function as values of n generated by this expression, when substituted in the mother function will generate only failures. (Here k belongs to N ).

Let our definition Let our definition of a failure be a non-Carmichael number. Let the mother function be 2^n + 81. Then n = 18 + 20*k is a failure function.

Here k belongs to N.When n= 10, f(n) = 1105, a Carmichael number; however when n is generated by the failure function 18 + 20*k we get values of f(n) which are not square free and hence incapable of being Carmichael numbers i.e. failures.

failure functions play an important role in proving a conjecture indirectly. See sketch proof.

Let our definition of a failure be a non - Devarajnumber which is not a Carmichael number ( see A104017 on OEIS ).Let the mother function be 2^n + 3113. Then n = 16 + 42*k is a failure function ).

Failure functions predict the values of the variable, which when substituted in the mother function, yield failures, in accordance with our definition of a failure. Normally we cannot say anything about the the other values of the variable - they may yield failures or successes. However in the case of the quadratic x^2 + 1 we can definitely say that the values of of x not covered by the failure functions yield only successes - (see sketch proof ).

Although it has been proved ( by Carl Pomerance and others) that there are infinitely many Carmichael numbers, there can only be finitely many Carmichael numbers a) with a given number of primefactors and b) with a given primenumber as one of its factors. This is evident when we apply Pomerance index (see A 162290 on OEIS ). How to apply the Pomerance index in arriving at upper bound for the number of Carmichael numbers as qualified above? Steps will be furnished in subsequent messages.

Background: Carl Pomerance had proved a conjecture of mine pertaining to Carmichael numbers (via snail mail). The conjecture: if N = p_1p_2p_3 be a Carmichael number then

(p_1-1)(N-1)/(p_2-1)(p_3-1) is an integer (called Pomerance index). Subsequently I generalised this conjecture as follows:

Let N = p_1p_2….p_r be a Carmichael number with r factors.Then

(p_1-1)(n-1)^(r - 2)/(p_2-1)(p_3-1)…..(p_r-1) is an integer. This was proved by Max Alleksyev.(His proof was on the lines of Carl Pomerance’s proof. (To be continued).

Asymptotic property of Pomerance index: Pomerance index of 561, the smallest Carmichael number, is 7. If we keep 3, the smallest factor of 561 fixed and increase the other two indefinitely the relevant Pomerance index decreases asymptotically to 6 i.e. it never reaches 6. This proves that 561 is the only 3 - factor Carmichael number having 3 as one of its factors. This is because Pomerance’s proof of my conjecture ( mentioned earlier) requires the index to be an integer ( to be continued).

Asymptotic property of Pomerance index is utilised in computing upper bound for the number of Carmichael numbers with a given number of prime factorsPlanetmathPlanetmath and and a given prime as a factor. Procedure: Let us consider 3-factor Carmichael numbers with 11 as one of the factors. We begin with the composite 11*13*17. The relevant Pomerance index is 126.56. Keeping 11 fixed we increase the other two prime factors indefinitely resulting in Pomerance index becoming asymptotic to 110. Hence the maximum number of possible 3-factor Carmichael numbers with 11 as p_1 = 126- 110 = 16. After adding the solitary case of 561 (where 11 is not the minimum factor) we get upper bound upper bound = 17.

If p is a quadratic non- residue then (p-1)! + 1 is not congruent to 0 mod (p^2. However there is a value of x, say x’ in the range p+1 to p^2 such that (p-1)! + x’ is congruent to 0 (mod(p^2). Examples: a) let p = 7. Then 6! + 15 is congruent to 0 (mod 49 );b) let p = 17. 16! + 205 is congruent to 0 (mod 289).

Formerly we could see papers on this site. Nowadays there seems to be no section entitled ”papers”. Perhaps R.S.Puzio could clarify.

Formerly there was a section devoed to papers. Nowadays it is missing. Could eith R, PUZIO or C.WOO kindly enlighten me? This is the II message on this topic.

The papers, books, and exposition sections went away when we switched over to the new software. Unlike the old Noosphere, Planetary does not have support for these. We are also making some experiments on seeing how the collections feature could support papers and books.

Thank you!

Let our definition of a failure be a composite number. Let the mother function be 2^n + 7 (n belongs to N ). Then n = 2 + 10*k is a failure function i.e. values of n generated by this function are such that f(n) is composite. The f(n)s so formed constitute a group isomorphicPlanetmathPlanetmathPlanetmath with Z_11. Here k belongs to W.

The quotients f(n)/11 constitute a group isomorphicPlanetmathPlanetmathPlanetmathPlanetmath with Z_11.

Perhaps members may not agree with me; however would like to submit what I feel are unhealthy aspects of the site: a) PSUEDONYMS: Why cant members contribute under their own name? When I contribute something it is always under my own name: A.K. Devaraj. Who is Bc1? who is Mathsprof b) GRABING CREDIT: Better to take credit for one’s original contribution rather than grabing credit for someone else’s contribution c) ADMINISTRATORS DO NOT READ MESSSAGES: I had asked something about papers - no reply received.

The above is not exhaustive.

Let the mother function be the quadratic x^2+1. Let our definition of a failure be a composite number. Then when x = 4, f(x) = 17; x = 4 + 17k is a failure function i.e. values of x generated by this failue function will be such that f(x) is composite ( in fact a multiplePlanetmathPlanetmath of 17). Such f(x) when divided by 17 constitute a group isomorphicPlanetmathPlanetmathPlanetmathPlanetmath with Z_17.

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