A206852 Numbers N such that N/2 is a square, N/3 is a cube, and N/5 is a fifth power.
30233088000000, 32462531054272512000000, 6224724715037147546112000000, 34856377305871210027941888000000, 28156757354736328125000000000000000, 6683747269421867033919422988288000000, 681433858470444619689081338982912000000
Offset: 1
Links
- Georg Fischer, Table of n, a(n) for n = 1..1000
- Shyam Sunder Gupta, Do you know, as of Feb 15 2012.
- Michael Penn, a sunny number puzzle!, YouTube video, 2021.
- 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).
Programs
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Mathematica
Table[30233088000000 * n^30, {n,1,1000}] (* Georg Fischer, Feb 07 2021 *)
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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
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PARI
a(n)=30233088000000*n^30 \\ Charles R Greathouse IV, Apr 25 2012
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Python
def A206852(n): return 30233088000000*n**30 # M. F. Hasler, Jul 24 2022
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Python
def is_A206852(n): for p in (2, 3, 5): for e in range(n): if n % p: break n //= p if e % 30 != 30//p: return False return is_A122971(n) # M. F. Hasler, Jul 24 2022
Formula
a(n) = 30233088000000 * n^30 = 2^15 * 3^10 * 5^6 * n^30. - Charles R Greathouse IV, Apr 25 2012
Comments