20170825, 09:04  #1 
May 2004
2^{2}×7×11 Posts 
Devaraj numbers necessary and sufficient condition
Let N = p_1*p_21*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 ( N1) be a multiple of ( p_r1), r ranging from 1 to r.
For definition of Devaraj numbers see A104016 and A 104017 on OEIS. 
20170828, 05:18  #2 
May 2004
464_{8} 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:
(2I)*(3+2*I)*(I) = 1  8*I In my next post (in the software section ) will give the algorithm for search. 
20170829, 05:17  #3 
May 2004
100110100_{2} Posts 
parialgorithm 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)=((2I)*(3+2*I)*(nI)1))/((2+2*I)*(nI1))} for(n=1,10,print(p(n))) You will get three possible solutions, one of which is invalid as it is (2I) and we need squarefree composites.Both the remaining solutions result in Devaraj numbers: a) (2I)*(3+2*I)*(I) and b) (2I)*(3+2*I)*(4I). Note that in each case the test " is (NI)/(1I) a Gaussian integer? is (N1)/(2+2*I) a Gaussian integer? is (N1)/(I1) a Gaussian integer? or is(N1)/(3I) a Gaussian integer? is positive. Last fiddled with by ewmayer on 20170829 at 06:43 Reason: Merged this newthread post into existing thread  1 onanism thread is enough 
20170911, 05:27  #4 
May 2004
2^{2}·7·11 Posts 
Devaraj numbersAn improved algorithm
The Devaraj number found by new algorithm: N= (4+I)*(154*I)*(2+3*I)
In the next post I will five the algorithm, 
20170911, 09:12  #5 
May 2004
2^{2}·7·11 Posts 
Devaraj numbers  necessary and sufficient condition
This works fine when r = 3.
for r > 3 the n & s condition is that (p_r1) divides (N1)^(r2). 
20170912, 04:27  #6 
May 2004
464_{8} 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'+1I) 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 (63I) is divisible by (3+I) and that (64I) is divisible by (4+I). 4). I then run the program: {p(n) = ((4+I)*(154*I)*(n + 3*I)1)/(n+3*I1)}. I thus found the Devaraj number: N= (4+I)*(154*I)*(2+3*I)= 131 + 190*I. The above is only a brief description of search for 3factor Devaraj numbers. 
20170914, 02:44  #7 
May 2004
2^{2}×7×11 Posts 
Application of Pomerance index
Refer A 162290 on OEIS for definition of Pomerance index.
The Devaraj number (2I)*(3+2*I)*(4I) = 33  4*I has Pomerance Index equal to (15*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 (2I) and second equal to (3+2*I). Since the descent does not pass through any Gaussian integer (334*I) is the lone Devaraj number with first and second primefactor equal to (2I) and (3+2*I) respectively. 
20170923, 02:58  #8 
May 2004
2^{2}×7×11 Posts 
Devaraj numbers  necessary and sufficient condition
Sorry there are no shortcuts; we will have to test whether ((p_11)*(n1)^(r2))/(p_21)..(p_r1) is an integer or not; similarly for (p_21)*(n1)^(r2)/(p_11)*(p_31)...(p_r1)..

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