cp's OEIS Frontend

A275702 Numbers n whose deficiency is 26: 2n - sigma(n) = 26.

%I A275702 #85 Nov 19 2024 03:15:23
%S A275702 58,75,328,850,1210,2848,35968,537088,549768921088,8796145451008
%N A275702 Numbers n whose deficiency is 26: 2n - sigma(n) = 26.
%C A275702 Any term x = a(m) can be combined with any term y = A275701(n) to satisfy the property (sigma(x)+sigma(y))/(x+y) = 2.  Although this property is a necessary condition for two numbers to be amicable, it is not a sufficient one.  So far, these two sequences have produced only one amicable pair: (1210,1184) = (a(5),A275701(2)) = (A063990(4),A063990(3)).  If more are ever found, then they will also exhibit x-y = 26.
%C A275702 Notice that:
%C A275702 a(1) =     58 =   29*  2 = (4^1+25)*(4^1)/2
%C A275702 a(3) =    328 =   41*  8 = (4^2+25)*(4^2)/2
%C A275702 a(6) =   2848 =   89* 32 = (4^3+25)*(4^3)/2
%C A275702 a(7) =  35968 =  281*128 = (4^4+25)*(4^4)/2
%C A275702 a(8) = 537088 = 1049*512 = (4^5+25)*(4^5)/2.
%C A275702 If p = 4^k+25 is prime and n = p*(p-25)/2, then it is not hard to show that 2*n - sigma(n) = 26. The values of k in A204388 will guarantee that p is prime (A104072). Similarly, if q = 2*4^k+25 is prime and n = q*(q-25)/2, then 2*n - sigma(n) = 26. However, q will never be prime since it will always be divisible by 3: 2*4^k+25 == (2*1^k+25) mod 3 == 27 mod 3 == 0 mod 3. So, the following values will be in this sequence and provide upper bounds for the next seven terms:
%C A275702 (4^10+25)*(4^10)/2 = 549768921088 >= a(9)
%C A275702 (4^11+25)*(4^11)/2 = 8796145451008 >= a(10)
%C A275702 (4^17+25)*(4^17)/2 = 147573952804424777728 >= a(11)
%C A275702 (4^35+25)*(4^35)/2 = 696898287454081973187748591279228938354688 >= a(12)
%C A275702 (4^46+25)*(4^46)/2 = 12259964326927110866866776279099475433218926722425028608 >= a(13)
%C A275702 (4^56+25)*(4^56)/2 = 13479973333575319897333507543509880240529303896615642871755920375808 >= a(14)
%C A275702 (4^59+25)*(4^59)/2 = 55213970774324510299478046898216207773446358605225195265697257166471168 >= a(15).
%C A275702 The rightmost digit of n = p*(p-25)/2 will always be 8. [Proof: If k is odd, then 4^k+25 == 9 mod 10 and (4^k)/2 == 2 mod 10, which implies that p*(p-25)/2 == 8 mod 10. If k is even, then 4^k+25 == 1 mod 10 and (4^k)/2 == 8 mod 10, which implies that p*(p-25)/2 == 8 mod 10.]
%C A275702 a(10) > 2.3*10^12. - _Giovanni Resta_, Aug 07 2016
%C A275702 a(11) > 10^18. - _Hiroaki Yamanouchi_, Aug 21 2018
%H A275702 D. Alpern, <a href="https://www.alpertron.com.ar/ECM.HTM">Factorization using the Elliptic Curve Method</a>.
%e A275702 a(1) = 58, since 2*58-sigma(58) = 116-90 = 26.
%e A275702 a(2) = 75, since 2*75-sigma(75) = 150-124 = 26.
%e A275702 a(3) = 328, since 2*328-sigma(328) = 656-630 = 26.
%t A275702 Select[Range[10^6], 2 # - (DivisorSigma[1, #]) == 26 &] (* _Vincenzo Librandi_, Aug 06 2016 *)
%o A275702 (Magma) [n: n in [1..2*10^6] | (2*n-SumOfDivisors(n)) eq 26]; // _Vincenzo Librandi_, Aug 06 2016
%o A275702 (PARI) is(n) = 2*n-sigma(n)==26 \\ _Felix Fröhlich_, Aug 06 2016
%Y A275702 Cf. A033879, A063990, A104072, A204388, A275701 (abundance 26).
%K A275702 nonn,more
%O A275702 1,1
%A A275702 _Timothy L. Tiffin_, Aug 05 2016
%E A275702 a(9) from _Giovanni Resta_, Aug 07 2016
%E A275702 a(10) from _Hiroaki Yamanouchi_, Aug 21 2018