|
%I #5 Mar 30 2012 18:51:38
%S 1,1,4,1,16,27,1,0,0,16,1,65536,7625597484987,256,3125,1,0,0,0,0,
%T 46656,1,0,0,4294967296,0,10314424798490535546171949056,823543,1,0,0,
%U 0,0
%N Table by antidiagonals used in calculating integer solutions to b^c=c^d with b,c,d>0.
%C There are also trivial values involving 0 to give 1; for any b and c: b^0=0^0 and 1^c=c^0. For b=p prime, the solutions occur when k is a power of p, say p^x and so c=p^(p^x), d=p^(p^(p^x)-k) and the value is p^(p^(p^x))). For p=2 this seems to give all but the first two terms of A051285.
%F T(n, k)=n^(A052410(n)^k) if A052409(n)*A052410(n)^k/k is an integer [and with T(1, k)=1] but otherwise T(n, k)=0 if A052409(n)*A052410(n)^k/k is not an integer. b=n=r^m where r=A052410(n) is the smallest root of n and m=A052409(n) is the power of r that n is; c=r^k; and d=m*r^k/k providing this is an integer [while if n=1 then b=1, c=1 and d=k]. In effect, there is a solution if the largest divisor of k which is coprime to all powers of r is also a divisor of m.
%e Rows start: 1, 1, 1, ...; 4, 16, 0, 65536, 0, 0, 0, 115792089237316195423570985008687907853269984665640564039457584007913129639936, 0, ...; 27, 0, 7625597484987, 0, 0, ...; 16, 256, 0, 4294967296, 0, ... etc. with the nonzero values corresponding to 1^1=1^1, 1^1=1^2, 1^1=1^3; 2^2=2^2, 2^4=4^2, 2^16=16^4, 2^256=256^32; 3^3=3^3, 3^27=27^9; 4^2=2^4, 4^4=4^4, 4^16=16^8; etc. For b=10^9: r=10 and m=9 so there solutions with c=10^k and d=9*10^k/k providing k is in A070023, i.e. if 1/k has period 0 or 1 in base 10.
%Y Cf. A072652, A072653.
%K nonn,tabl
%O 1,3
%A _Henry Bottomley_, Jun 28 2002
|
