1. Check whether the given ratios are equivalent.
d. 100:11, 1000 : 110
Answers
Answer:
100000, 1001, 110, 4, 1000000, 21, 10000000, 102, 1010, 10001, 100000000, 13, 200, 100001, 30, 1002, 1000000000, 111, 10000000000, 5, 10010, 1000001, 1100, 22, 100000000000, 10000001, 100010 (list; graph; refs; listen; history; text; internal format)
OFFSET
1,3
COMMENTS
Are there any other numbers besides n=12 for which n=a(n) ? - Ctibor O. Zizka, Oct 08 2008
The sequence is a morphism from (N*,*) into (N,+), cf. formula. Up to n=1023, the digit sum A007953(a(n)) equals Omega(n) = A001222(n). This holds whenever A051903(n)<10. Restricted to these n, the sequence is also injective. However, when n is a multiple of 2^10, 3^10, 5^10 etc, then a(n) is equal to some a(m) with m<n. - M. F. Hasler, Nov 16 2008
This has been called the "Exponential Prime Power Representation" of n by W. Nissen in a post to the sci.math newsgroup (where probably some more sophisticated convention for representing digits > 10 would be used). - M. F. Hasler, Jul 03 2016
LINKS
Robert Israel, Table of n, a(n) for n = 1..10000 (n=1..1023 from Michael De Vlieger)
Evans A Criswell, A Sequence Puzzle (Posted to rec.puzzles Jan 01 1997)
Walter Nissen, Exponential Prime Power Representation, sci.math newsgroup, May 23 1995.
FORMULA
a(m*n) = a(m) + a(n) for all m,n > 0. A007953(a(n))=A001222(n) for all n such that A051903(n) < 10. - M. F. Hasler, Nov 16 2008
a(n) = sum(10^(A049084(A027748(k))-1) * A124010(k): k = 1..A001221(n)). - Reinhard Zumkeller, Aug 03 2015
a(A054842(n)) = n for all n >= 0. - Antti Karttunen, Aug 29 2016
a(n) = Sum_{i>0} e_i*10^(i-1) when n = Product_{i>0} prime(i)^e_i. - M. F. Hasler, Mar 14 2018
EXAMPLE
a(25) = 200 because 25 = 5^2 * 3^0 * 2^0.
a(1024) = 10 = a(3), because 1024 = 2^10; but this two-digit multiplicity overflows into the 10^1 position, which encodes for powers of three.
Answer:
yes
Step-by-step explanation:
100/11
1000/110=>100/11
on reducing both the ratios, they become equal.
Therefore they are equivalent.