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A072651
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Table by antidiagonals used in calculating integer solutions to b^c=c^d with b,c,d>0.
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3
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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
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OFFSET
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1,3
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COMMENTS
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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.
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LINKS
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FORMULA
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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.
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EXAMPLE
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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.
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CROSSREFS
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KEYWORD
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AUTHOR
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STATUS
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approved
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