OFFSET
1,1
COMMENTS
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
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).
FORMULA
a(n) = 30233088000000 * n^30 = 2^15 * 3^10 * 5^6 * n^30. - Charles R Greathouse IV, Apr 25 2012
MATHEMATICA
Table[30233088000000 * n^30, {n, 1, 1000}] (* Georg Fischer, Feb 07 2021 *)
PROG
(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
(PARI) a(n)=30233088000000*n^30 \\ Charles R Greathouse IV, Apr 25 2012
(Python) def A206852(n): return 30233088000000*n**30 # M. F. Hasler, Jul 24 2022
(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
CROSSREFS
KEYWORD
nonn,easy
AUTHOR
M. F. Hasler, Feb 15 2012
STATUS
approved