A192107 Sum of all the n-digit numbers whose digits are all odd.

25, 1375, 69375, 3471875, 173609375, 8680546875, 434027734375, 21701388671875, 1085069443359375, 54253472216796875, 2712673611083984375, 135633680555419921875, 6781684027777099609375, 339084201388885498046875, 16954210069444427490234375

Offset: 1

Author

Bernard Schott, Dec 30 2012

Keywords

• nonn
• base
• easy

Comments

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.

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

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

Examples

			a(1) = 1 + 3 + 5 + 7 + 9 = 25.
a(2) = 11 + 13 + ... + 19 + 31 + ... + 79 + 91 + ... + 99 = 1375.
		

References

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

Links

• The IMO compendium, Problem 4, Finnish High School Mathematics Contest 1997.
• Index entries for linear recurrences with constant coefficients, signature (55,-250).
• Index to sequences related to Olympiads.

Crossrefs

Cf. A220094, A192370, A014261.

Programs

Maple

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

Mathematica

Table[((10^n - 1)*5^(n + 1))/9, {n, 20}] (* T. D. Noe, Dec 31 2012 *)
LinearRecurrence[{55,-250},{25,1375},20] (* Harvey P. Dale, Oct 11 2018 *)

PARI

a(n) = (10^n-1) * 5^(n+1)/9 \\ Charles R Greathouse IV, Jul 06 2017

Formula

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.

From Colin Barker, Jan 04 2013: (Start)

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

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