A270142 a(n) = product of first k composites, with the i-th composite raised to the d-th power, where k = A055642(n) and d is the i-th digit of n.

4, 16, 64, 256, 1024, 4096, 16384, 65536, 262144, 4, 24, 144, 864, 5184, 31104, 186624, 1119744, 6718464, 40310784, 16, 96, 576, 3456, 20736, 124416, 746496, 4478976, 26873856, 161243136, 64, 384, 2304, 13824, 82944, 497664, 2985984, 17915904, 107495424

Offset: 1

Author

Felix Fröhlich, Mar 12 2016

Keywords

• nonn
• base
• easy

Comments

All terms are multiples of 4, since A002808(1) = 4 and the most significant digit of n is always nonzero.

Does a term exist such that a(n) = n? Such a number would be the analog of a Meertens number when raising composites to the powers of the digits of n instead of raising primes to the powers of the digits.

From Chai Wah Wu, Dec 15 2022: (Start)

If a(n) is defined using digits of n in base b, then there are bases b and numbers n such that a(n) = n. For instance:

base b n

------------------------------------------------

2 4, 24, 36, 24192000, 85155840

3 2592

4 4, 103680

6 20736

8 16, 256, 13824

12 1327104

16 21233664

23 24

24 746496

(End)

Examples

			a(12) = 144, since A002808(1) = 4, A002808(2) = 6 and 4^1 * 6^2 = 144.
		

Links

• Felix Fröhlich, Table of n, a(n) for n = 1..10000

Crossrefs

Cf. A189398, A246532.

Programs

PARI

composite(n) = my(i=0, c=2); while(1, if(!ispseudoprime(c), i++); if(i==n, return(c)); c++)
compopowerprod(n) = my(d=digits(n)); for(k=1, #d, p=prod(i=1, #d, composite(i)^d[i])); p
a(n) = compopowerprod(n)

Python

from math import prod
from sympy import composite
def A270142(n): return prod(composite(i)**int(d) for i, d in enumerate(str(n),1)) # Chai Wah Wu, Dec 09 2022