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Number of powers of 5 having exactly n digits '0' (in base 10), conjectured.
2

%I #6 Jun 23 2018 09:19:07

%S 16,16,12,11,21,12,17,14,16,17,14,13,16,18,13,14,10,10,21,7,19,13,15,

%T 13,10,15,12,15,11,11,15,10,9,15,17,16,13,12,12,11,14,9,14,15,16,14,

%U 13,14,15,24,14

%N Number of powers of 5 having exactly n digits '0' (in base 10), conjectured.

%C a(0) = 16 is the number of terms in A008839 and in A195948, which includes the power 5^0 = 1.

%C These are the row lengths of A305925. It remains an open problem to provide a proof that these rows are complete (as are all terms of A020665), but the search has been pushed to many orders of magnitude beyond the largest known term, and the probability of finding an additional term is vanishing, cf. Khovanova link.

%H M. F. Hasler, <a href="/wiki/Zeroless_Powers">Zeroless powers</a>, OEIS Wiki, March 2014, updated 2018.

%H T. Khovanova, <a href="https://blog.tanyakhovanova.com/2011/02/86-conjecture/">The 86-conjecture</a>, Tanya Khovanova's Math Blog, Feb. 2011.

%H W. Schneider, <a href="http://web.archive.org/web/20050407120908/http://www.wschnei.de:80/digit-related-numbers/nozeros.html">No Zeros</a>, 2000, updated 2003. (On web.archive.org--see A007496 for a cached copy.)

%o (PARI) A305945(n,M=99*n+199)=sum(k=0,M,#select(d->!d,digits(5^k))==n)

%o (PARI) A305945_vec(nMax,M=99*nMax+199,a=vector(nMax+=2))={for(k=0,M,a[min(1+#select(d->!d,digits(5^k)),nMax)]++);a[^-1]}

%Y Cf. A030701 (= row 0 of A305925): k such that 5^k has no 0's; A195948: these powers 4^k.

%Y Cf. A020665: largest k such that n^k has no '0's.

%Y Cf. A063585 (= column 1 of A305925): least k such that 5^k has n digits '0' in base 10.

%Y Cf. A305942 (analog for 2^k), ..., A305947, A305938, A305939 (analog for 9^k).

%K nonn,base

%O 0,1

%A _M. F. Hasler_, Jun 22 2018