A306570 - OEIS

%I #64 Oct 21 2021 01:15:30

%S 5,25,125,3125,203125,8203125,408203125,8408203125,18408203125,

%T 618408203125,2618408203125,52618408203125,152618408203125,

%U 3152618408203125,93152618408203125,493152618408203125,7493152618408203125,17493152618408203125,117493152618408203125,7117493152618408203125,87117493152618408203125

%N Values of n such that 5^n ends in n, or expomorphic numbers relative to "base" 5.

%C Definition: For positive integers b (as base) and n, the positive integer (allowing initial zeros) k(n) is expomorphic relative to base b (here 5) if k(n) has exactly n decimal digits and if b^k(n) == k(n) (mod 10^n) or, equivalently, b^k(n) ends in k(n). [See Crux Mathematicorum link.]

%C For sequences in the OEIS, no term is allowed to begin with a digit 0 (except for the 1-digit number 0 itself). However, in the problem as defined in the Crux Mathematicorum article, leading 0 digits are allowed; under that definition a(n) = k(n) until the first k(n) which begins with digit 0. When k(n) begins with 0, then, a(n) is the next term of the sequence k(n) which doesn't begin with digit 0.

%C Under that definition, the term after a(4) = 3125 is not "03125" but a(5) = 203125. [Comments from _Jon E. Schoenfield_ in A288845 and discussion with _Rémy Sigrist_]

%C Conjecture: if k(n) is expomorphic relative to "base" b, then, the next one in the sequence, k(n+1), consists of the last n+1 digits of b^k(n).

%C a(n) is the backward concatenation of A133615(0) through A133615(n-1). So, a(1) is 5, a(2) is 25, and so on, with recognition of the comments about the OEIS and terms beginning with 0 (for example, when n = 5, A133615(n-1) = 0, so the next nonzero digit is concatenated as well, reducing the amount subtracted from n by 1). - _Davis Smith_ Mar 07 2019

%H Davis Smith, <a href="/A306570/b306570.txt">Table of n, a(n) for n = 1..894</a>

%H Charles W. Trigg, <a href="https://cms.math.ca/crux/backfile/Crux_v7n06_Jun.pdf">Problem 559</a>, Crux Mathematicorum, page 192, Vol. 7, Jun. 81.

%e 5^5 = 25 ends in 5, so 5 is a term; 5^25 = ...125 ends in 25, so 25 is another term.

%o (PARI) is(n) = my(t=#digits(n)); lift(Mod(5, 10^t)^n)==n

%o for(n=1, oo, my(x=n*5); if(lift(Mod(5, 10)^x)==x%10, if(is(x), print1(x, ", ")))) \\ _Felix Fröhlich_, Feb 24 2019

%o (PARI) tetrmod(b,n,m)=my(t=b); for(i=1, n, if(i>1, t=lift(Mod(b,m)^t), t)); t

%o for(n=1, 21,if(tetrmod(5,n,10^n)!=tetrmod(5,n-1,10^(n-1)),print1(tetrmod(5,n,10^(n-1)),", "))) \\ _Davis Smith_, Mar 09 2019

%Y Cf. A064541 (base 2), A183613 (base 3), A288845 (base 4), A290788 (base 6), A321970 (base 7), A306686 (base 9), A289138 (smallest expomorphic number in base n).

%Y Cf. A003226 (automorphic numbers), A033819 (trimorphic numbers).

%Y Cf. A133615 (leading digits).

%K nonn,base

%O 1,1

%A _Bernard Schott_, Feb 24 2019

%E a(5)-a(7) from _Felix Fröhlich_, Feb 24 2019

%E a(8) from _Michel Marcus_, Mar 02 2019

%E a(9)-a(21) from _Davis Smith_, Mar 07 2019