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A206852
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Numbers N such that N/2 is a square, N/3 is a cube, and N/5 is a fifth power.
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13
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30233088000000, 32462531054272512000000, 6224724715037147546112000000, 34856377305871210027941888000000, 28156757354736328125000000000000000, 6683747269421867033919422988288000000, 681433858470444619689081338982912000000
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OFFSET
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1,1
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COMMENTS
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The terms must be of the form N = 2^a*3^b*5^c*m^(2*3*5) where gcd(m, 2*3*5) = 1 and a-1, b-1 and c-1 must be a multiple of 2, 3 and 5, respectively, and a, b, c must be a multiple of the two other prime factors, respectively. This gives (a, b, c) == (3*5, 2*5, 2*3) [mod 2*3*5], whence N = 2^15*3^10*5^6*n^30. - M. F. Hasler, Jul 22 2022
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LINKS
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Index entries for linear recurrences with constant coefficients, signature (31, -465, 4495, -31465, 169911, -736281, 2629575, -7888725, 20160075, -44352165, 84672315, -141120525, 206253075, -265182525, 300540195, -300540195, 265182525, -206253075, 141120525, -84672315, 44352165, -20160075, 7888725, -2629575, 736281, -169911, 31465, -4495, 465, -31, 1).
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FORMULA
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MATHEMATICA
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Table[30233088000000 * n^30, {n, 1, 1000}] (* Georg Fischer, Feb 07 2021 *)
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PROG
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(PARI) {is_A206852(n)=(n=divrem(n, 3^10*5^6<<15))[2]==0 && ispower(n[1], 30)} \\ replacing obsolete PARI code from 2012. - M. F. Hasler, Jul 22 2022
(Python)
for p in (2, 3, 5):
for e in range(n):
if n % p: break
n //= p
if e % 30 != 30//p: return False
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CROSSREFS
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KEYWORD
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nonn,easy
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AUTHOR
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STATUS
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approved
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