A208277 Smallest number of multiplicative persistence n in factorial base.

0, 2, 5, 633, 443153013

Offset: 0

Author

Charles R Greathouse IV, Feb 28 2012

Keywords

• nonn
• base
• more

Comments

a(n) exists for all n, unlike (conjecturally) its decimal equivalent A003001. In particular, with k = a(n-1), a(n) <= k * k! + (k-1)! + ... + 2! + 1! < (a(n-1)+1)! for n > 1. Diamond & Reidpath ask if this upper bound can be improved.

a(5) <= 255429978433810461138446192454297813.

Examples

			5 = 1*1!+2*2!, and so is 21 in factorial base; the product of its digits is 2*1 = 10_! and the product of its digits in factorial base is 0*1 = 0, so 5 has multiplicative persistence 2. Since it is the smallest, a(2) = 5.
633 = 51111_! -> 21_! -> 10_! -> 0_! is the least chain of length 3 and so a(3) = 633.
		

Links

• M. R. Diamond and D. D. Reidpath, A counterexample to conjectures by Sloane and Erdos concerning the persistence of numbers, Journal of Recreational Mathematics 29:2 (1998), pp. 89-92.

Crossrefs

Cf. A003001, A007623, A031346, A208575, A208576.

Programs

PARI

pr(n)=my(k=1,s=1);while(n,s*=n%k++;n\=k);s
persist(n)=my(t); while(n>1, t++; n=pr(n)); t
a(n)=my(k=0);while(persist(k)!=n, k++); k \\ Charles R Greathouse IV, Jan 21 2013