A192107 - OEIS

%I #38 Jan 12 2021 02:04:44

%S 25,1375,69375,3471875,173609375,8680546875,434027734375,

%T 21701388671875,1085069443359375,54253472216796875,

%U 2712673611083984375,135633680555419921875,6781684027777099609375,339084201388885498046875,16954210069444427490234375

%N Sum of all the n-digit numbers whose digits are all odd.

%C The idea for this sequence comes from question 4 of the Final Round of the Finnish High School Mathematics Contest in 1997 (see link IMO Compendium and Crux reference) where the question was asked regarding only 4-digit numbers.

%C A192370 is the similar sequence when all the digits are even: 2, 4, 6, 8.

%C A220094 is the similar sequence with the digits belonging to {1, 2, 3, 4, 5, 6, 7, 8, 9}.

%D Finnish High School Mathematics Contest, Final Round, 1997, problem 4. [Crux Mathematicorum, v22 n3, Apr. 2002, p. 143]

%H The IMO compendium, <a href="https://imomath.com/othercomp/Fin/FinMO97.pdf">Problem 4</a>, Finnish High School Mathematics Contest 1997.

%H <a href="/index/Rec#order_02">Index entries for linear recurrences with constant coefficients</a>, signature (55,-250).

%H <a href="/index/O#Olympiads">Index to sequences related to Olympiads</a>.

%F a(n) = ((10^n-1) * 5^(n+1))/9 = 5^(n+1) * R_n with R_n is the repunit with n times the digit 1.

%F From _Colin Barker_, Jan 04 2013: (Start)

%F a(n) = 55*a(n-1) - 250*a(n-2).

%F G.f.: 25*x/((5*x-1)*(50*x-1)). (End)

%e a(1) = 1 + 3 + 5 + 7 + 9 = 25.

%e a(2) = 11 + 13 + ... + 19 + 31 + ... + 79 + 91 + ... + 99 = 1375.

%p A:=seq((10^n-1)*5^(n+1)/9,n=1..20);

%t Table[((10^n - 1)*5^(n + 1))/9, {n, 20}] (* _T. D. Noe_, Dec 31 2012 *)

%t LinearRecurrence[{55,-250},{25,1375},20] (* _Harvey P. Dale_, Oct 11 2018 *)

%o (PARI) a(n) = (10^n-1) * 5^(n+1)/9 \\ _Charles R Greathouse IV_, Jul 06 2017

%Y Cf. A220094, A192370, A014261.

%K nonn,base,easy

%O 1,1

%A _Bernard Schott_, Dec 30 2012