A072652 -id:A072652 - cp's OEIS frontend

A072651 Table by antidiagonals used in calculating integer solutions to b^c=c^d with b,c,d>0.

1, 1, 4, 1, 16, 27, 1, 0, 0, 16, 1, 65536, 7625597484987, 256, 3125, 1, 0, 0, 0, 0, 46656, 1, 0, 0, 4294967296, 0, 10314424798490535546171949056, 823543, 1, 0, 0, 0, 0

Offset: 1

Henry Bottomley, Jun 28 2002

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.

			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.

Cf. A072652, A072653.

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.

A072653 Uniqued integer solutions n to n = b^c = c^d.

Original entry on oeis.org

1, 4, 16, 27, 64, 256, 729, 1024, 3125, 4096, 16384, 19683, 46656, 65536, 262144, 531441, 823543, 1048576, 4194304, 9765625, 14348907, 16777216, 67108864, 268435456, 387420489, 1073741824, 2176782336, 4294967296, 10000000000

Offset: 1

Henry Bottomley, Jun 28 2002

nonn

Numbers n such that (n^i)^(n^(1/i)) = (n^j)^(n^(1/j)) for some i and j.

			1 is included because of solutions of the form b^0 = 0^0, 1^c = c^0 and 1^1 = 1^d; 4 since 2^2 = 2^2; 16 since 2^4 = 4^2 and 4^2 = 2^4; 27 since 3^3 = 3^3; 64 since 8^2 = 2^6; etc.
The 10th element is n = 4096 with i = 12 and j = 6 because (4096^12)^(4096^(1/12)) = (4096^6)^(4096^(1/6)).

Cf. A072651, A072652.

See A072651 for calculation method.

Edited by N. J. A. Sloane at the suggestion of Andrew S. Plewe, Oct 07 2006, Jun 05 2007

A111260 Numbers of the form (m^n)/(n^m) with m > 0 and n>1.

Original entry on oeis.org

1, 4, 16, 27, 256, 729, 65536, 2985984, 4194304, 9765625, 134217728, 387420489, 2176782336, 24794911296, 30517578125, 104857600000, 678223072849, 2641807540224, 7625597484987, 17592186044416, 281474976710656

Offset: 1

Giorgio Balzarotti and Paolo P. Lava, Nov 12 2005

nonn

			The 4th element is a=27 with m = 3 and n = 9 because 3^9/9^3 = 27.

Cf. A072651, A072652, A072653.

a:=proc(N) local a, m,n; for m from 1 to N do for n from 2 to N do a:=(m^n)/(n^m); if(floor(a)=a)then print(a) fi; od; od; end: # convert into set # sort set

Take[ Select[ Union@Flatten@Table[(m^n)/(n^m), {m, 35}, {n, 2, 35}], IntegerQ[ # ] &], 21] (* Robert G. Wilson v, Nov 17 2005 *)

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A072651 Table by antidiagonals used in calculating integer solutions to b^c=c^d with b,c,d>0.

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A072653 Uniqued integer solutions n to n = b^c = c^d.

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A111260 Numbers of the form (m^n)/(n^m) with m > 0 and n>1.

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