|
%I #31 May 18 2021 18:06:19
%S 14,88,65548,33554459,387420510,1099511627800,35184372088855,
%T 3656158440063002,459986536544739960976836
%N Numbers representable as both b^c + b + c and x^y + x - y, where b, c, x, y are integers greater than 1.
%C From _Michael S. Branicky_, May 15 2021: (Start)
%C The following are terms:
%C 459986536544739960976836 = 7^28 + 7 + 28 = 49^14 + 49 - 14,
%C 1237940039285380274899124273 = 4^45 + 4 + 45 = 64^15 + 64 - 15,
%C 6362685...0216378 (46 digits) = 9^48 + 9 + 48 = 81^24 + 81 - 24, and
%C 1000000...0000070 (61 digits) = 10^60 + 10 + 60 = 100^30 + 100 - 30. (End)
%C From _Chai Wah Wu_, May 17 2021: (Start)
%C Sequence is infinite.
%C If a, b > 1 and b^a-b == 0 mod a+1 then b^c+b+c is a term for c = ab(b^(a-1)-1)/(a+1), y = c/a, x = b^a.
%C If b > 1 and b <> 2 mod 3, then b^(2b(b-1)/3)+b(2b+1)/3 is a term.
%C If b > 2, then b^((b-1)(b^(b-2)-1)) + b + (b-1)(b^(b-2)-1) is a term. (End)
%C From _Chai Wah Wu_, May 18 2021: (Start)
%C Either c>=3 or y>=3. If c=y=2, we get b^2+b+2=x^2+x-2, i.e. (x-b)(x+b+1) = 4. Since x>1 and b>1, x+b+1>4, a contradiction.
%C This allows for a faster search algorithm by assuming c>=3 and y>=3. The cases c=2 and y>=3 can be dealt with by picking y>=3 and solving for b in the quadratic equation b^2+b+2=x^y+x-y. Similarly for c>=3 and y=2. This approach was used to confirm a(9). (End)
%e a(1) = 14 = 3^2 + 3 + 2 = 2^4 + 2 - 4.
%e a(2) = 88 = 3^4 + 3 + 4 = 9^2 + 9 - 2.
%e a(3) = 65548 = 4^8 + 4 + 8 = 16^4 + 16 - 4.
%e a(4) = 33554459 = 2^25 + 2 + 25 = 32^5 + 32 - 5.
%e a(5) = 387420510 = 3^18 + 3 + 18 = 27^6 + 27 - 6.
%e a(6) = 1099511627800 = 4^20 + 4 + 20 = 32^8 + 32 - 8.
%e a(7) = 35184372088855 = 8^15 + 8 + 15 = 32^9 + 32 - 9.
%e a(8) = 3656158440063002 = 6^20 + 6 + 20 = 36^10 + 36 - 10.
%o (Python)
%o TOP = 100000000
%o a = [0]*TOP
%o for y in range(2,TOP//2):
%o if 2**y+2+y>=TOP: break
%o for x in range(2,TOP//2):
%o k = x**y+x+y
%o if k>=TOP: break
%o a[k]=1
%o for y in range(2,TOP//2):
%o if 2**y+2-y>=TOP: break
%o for x in range(2,TOP//2):
%o k = x**y+x-y
%o if k>=TOP: break
%o if k>=0: a[k]|=2
%o print([n for n in range(TOP) if a[n]==3])
%Y Cf. A254034.
%K nonn,more
%O 1,1
%A _Alex Ratushnyak_, Feb 24 2015
%E a(5)-a(8) from _Lars Blomberg_, May 19 2015
%E a(9) from _Chai Wah Wu_, May 18 2021
| 