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A061583 a(1) = 1, a(n) is the number obtained by replacing each digit of a(n-1) with five times its value.

%I A061583 #18 Oct 27 2023 22:05:37
%S A061583 1,5,25,1025,501025,250501025,10250250501025,501025010250250501025,
%T A061583 2505010250501025010250250501025,
%U A061583 1025025050102502505010250501025010250250501025
%N A061583 a(1) = 1, a(n) is the number obtained by replacing each digit of a(n-1) with five times its value.
%C A061583 Number of digits of each term is the sequence A038718. - _Dmitry Kamenetsky_, Jan 17 2009
%H A061583 John Cerkan, <a href="/A061583/b061583.txt">Table of n, a(n) for n = 1..17</a>
%p A061583 a:= proc(n) option remember; `if`(n=1, 1, (s-> parse(cat(
%p A061583       seq(parse(s[i])*5, i=1..length(s)))))(""||(a(n-1))))
%p A061583     end:
%p A061583 seq(a(n), n=1..10);  # _Alois P. Heinz_, May 25 2018
%t A061583 NestList[FromDigits[Flatten[IntegerDigits/@(5*IntegerDigits[#])]]&,1,10] (* _Harvey P. Dale_, Dec 31 2013 *)
%o A061583 (Python)
%o A061583 def A061583_first(n):
%o A061583     an = "1"
%o A061583     a061583 = []
%o A061583     while n > 1:
%o A061583         a061583.append(int(an))
%o A061583         newan = ""
%o A061583         for i in an:
%o A061583             newan += str(5*int(i))
%o A061583         an = newan
%o A061583         n -= 1
%o A061583     a061583.append(int(an))
%o A061583     return a061583 # _John Cerkan_, May 25 2018
%Y A061583 Cf. A061581, A061582, A061584, A061585, A061586, A061587.
%K A061583 nonn,base,easy
%O A061583 1,2
%A A061583 _Amarnath Murthy_, May 13 2001
%E A061583 More terms from Larry Reeves (larryr(AT)acm.org) and _Asher Auel_, May 15 2001