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 2017-08-25, 09:04 #1 devarajkandadai     May 2004 22×79 Posts Devaraj numbers- necessary and sufficient condition Let N = p_1*p_2-1*p_3....p_r be a r factor (square free ) composite number. Then the necessary and sufficient condition for it to be a Devaraj number is that ( N-1) be a multiple of ( p_r-1), r ranging from 1 to r. For definition of Devaraj numbers see A104016 and A 104017 on OEIS.
 2017-08-28, 05:18 #2 devarajkandadai     May 2004 22·79 Posts Devaraj numbers - necessary and sufficient condition Applied the above necessary and sufficient condition to search for Devaraj numbers in the ring of Gaussian integers. The second Devaraj ( the first one was discovered accidentally) number by applying the above: (2-I)*(3+2*I)*(-I) = 1 - 8*I In my next post (in the software section ) will give the algorithm for search.
 2017-08-29, 05:17 #3 devarajkandadai     May 2004 31610 Posts pari-algorithm for finding 3 factor Devaraj numbers I will illustrate this with an example. Select two primes in the ring of Gaussian integers, say, (2 - I) and (3 + 2*I). In pari run the programme : {p(n)=((2-I)*(3+2*I)*(n-I)-1))/((2+2*I)*(n-I-1))} for(n=1,10,print(p(n))) You will get three possible solutions, one of which is invalid as it is (2-I) and we need square-free composites.Both the remaining solutions result in Devaraj numbers: a) (2-I)*(3+2*I)*(-I) and b) (2-I)*(3+2*I)*(4-I). Note that in each case the test " is (N-I)/(1-I) a Gaussian integer? is (N-1)/(2+2*I) a Gaussian integer? is (N-1)/(-I-1) a Gaussian integer? or is(N-1)/(3-I) a Gaussian integer? is positive. Last fiddled with by ewmayer on 2017-08-29 at 06:43 Reason: Merged this new-thread post into existing thread - 1 onanism thread is enough
 2017-09-11, 05:27 #4 devarajkandadai     May 2004 4748 Posts Devaraj numbers-An improved algorithm The Devaraj number found by new algorithm: N= (4+I)*(15-4*I)*(2+3*I) In the next post I will five the algorithm,
 2017-09-11, 09:12 #5 devarajkandadai     May 2004 22·79 Posts Devaraj numbers - necessary and sufficient condition This works fine when r = 3. for r > 3 the n & s condition is that (p_r-1) divides (N-1)^(r-2).
 2017-09-12, 04:27 #6 devarajkandadai     May 2004 1001111002 Posts pari - Devaraj numbers- improved algorithm In the previous algorithm, two prime numbers in the ring of Gaussian integers were selected at random.The problem with this approach is that the second prime number may not be compatible with the first to form a Devaraj number. Now we choose on;y one prime number. I will illustrate the algorithm with an example. 1) I choose (4 + I). 2) I run the program {p(n) = (n - I)/3+I)}; for n = 1,25,print(p(n))) 3)Out all possible solutions I choose valid ones by eliminating invalid ones as follows: a) Let n' represent one of the possible solutions. Is (n'+1-I) composite? If not go to the next possible solution. If the next is composite test whether (n'+1 -I) is divisible by (4+I). In the example I found that (63-I) is divisible by (3+I) and that (64-I) is divisible by (4+I). 4). I then run the program: {p(n) = ((4+I)*(15-4*I)*(n + 3*I)-1)/(n+3*I-1)}. I thus found the Devaraj number: N= (4+I)*(15-4*I)*(2+3*I)= 131 + 190*I. The above is only a brief description of search for 3-factor Devaraj numbers.
 2017-09-14, 02:44 #7 devarajkandadai     May 2004 22×79 Posts Application of Pomerance index Refer A 162290 on OEIS for definition of Pomerance index. The Devaraj number (2-I)*(3+2*I)*(4-I) = 33 - 4*I has Pomerance Index equal to (1-5*I) Its norm is 26. If we were to keep the first two prime factors fixed and increase the third indefinitely the norm of the P.i. decreases asymptotically to 16.This proves that there can be only a finite number of Devaraj numbers with the first prime factor equal to (2-I) and second equal to (3+2*I). Since the descent does not pass through any Gaussian integer (33-4*I) is the lone Devaraj number with first and second primefactor equal to (2-I) and (3+2*I) respectively.
 2017-09-23, 02:58 #8 devarajkandadai     May 2004 22·79 Posts Devaraj numbers - necessary and sufficient condition Sorry there are no shortcuts; we will have to test whether ((p_1-1)*(n-1)^(r-2))/(p_2-1)..(p_r-1) is an integer or not; similarly for (p_2-1)*(n-1)^(r-2)/(p_1-1)*(p_3-1)...(p_r-1)..

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