A198391 Numbers m such that Sum_{i=1..k} (1-1/p_i) + Product_{i=1..k} (1-1/p_i) is an integer, where p_i are the k prime factors of m (with multiplicity).

2, 15, 20, 272, 476, 19024, 47425, 65792, 125172, 216900, 539280, 1222976, 1372736, 2770496, 3494336, 5321808, 5844528, 6177168, 7032528, 8885808, 20670768, 60727876, 69081344, 82724356, 95579136, 544382208, 907440192, 1657497600, 4295032832, 5048574976

Offset: 1

Author

Paolo P. Lava, Oct 24 2011

Keywords

• nonn

Comments

The numbers of the sequence are solutions of the differential equation m’=(k-a)*m+b, which can be written as A003415(m)=(k-a)*m+A003958(m), where k is the number of prime factors of m, and a is the integer Sum_{i=1..k} (1-1/p_i) + Product_{i=1..k} (1-1/p_i).

If k = a we have m’ = b or A003415(m) = A003958(m). For instance 15 has prime factors 3, 5; its arithmetic derivative is 15’ = 8 and b = 3*5 - 3 - 5 + 1 = 8. The term 47425 has prime factors 5, 5, 7, 271. Its arithmetic derivative is 47425’ = 25920 and b = 5*5*7*271 - 5*5*7 - 5*5*271 - 5*7*271 - 5*7*271 + 5*5 + 5*7 + 5*271 + 5*7 + 5*271 + 7*271 - 5 - 5 - 7 - 271 + 1 = 25920.

The numbers of the sequence satisfy also Sum_{i=1..k} (1+1/p_i) - Product_{i=1..k} (1-1/p_i) = some integer.

Examples

			125172 has prime factors 2, 2, 3, 3, 3, 19, 61. 1 - 1/2 + 1 - 1/2 + 1 - 1/3 + 1 - 1/3 + 1 - 1/3 + 1 - 1/19 + 1 - 1/61 = 5715/1159 is the sum over the 1-1/p_i. (1-1/2) * (1-1/2) * (1-1/3) * (1-1/3) * (1-1/3) * (1-1/19) * (1-1/61) = 80/1159 is the product of the 1-1/p_i. The sum over sum and product is 5715/1159 + 80/1159 = 5, an integer.
		

Links

• Giovanni Resta, Table of n, a(n) for n = 1..48 (terms < 10^12)
• J. M. Borwein and E. Wong, A survey of results relating to Giuga’s conjecture on primality, May 8, 1995.
• Romeo Mestrovic, On a Congruence Modulo n^3 Involving Two Consecutive Sums of Powers, Journal of Integer Sequences, Vol. 17 (2014), 14.8.4.

Crossrefs

Cf. A003415, A003958, A007850, A199767.

Programs

Maple

isA198391 := proc(n)
    p := ifactors(n)[2] ;
    add(op(2,d)-op(2,d)/op(1,d),d=p) + mul((1-1/op(1,d))^op(2,d),d=p) ;
    type(%,'integer') ;
end proc:
for n from 2 to 20000000 do
    if isA198391(n) then
        printf("%d,\n",n);
    end if;
end do: # R. J. Mathar, Nov 26 2011

Mathematica

Select[Range[2, 10^5], IntegerQ[(Plus @@ # + Times @@ #) &@ (1 - 1/ Flatten[ Table[#1, {#2}] & @@@ FactorInteger@#])] &] (* Giovanni Resta, May 23 2016 *)

Extensions

Missing a(23) and a(26)-a(30) from Giovanni Resta, May 23 2016