A174061 The Lucky Tickets Problem.

1, 10, 670, 55252, 4816030, 432457640, 39581170420, 3671331273480, 343900019857310, 32458256583753952, 3081918923741896840, 294056694657804068000, 28170312778225750242100

Offset: 0

Author

Geoffrey Critzer, Mar 06 2010, Mar 13 2010

Keywords

• nonn
• base

Comments

A ticket has a 2n-digit number. (The initial digits are allowed to be zeros.) A ticket is lucky if the sum of the first n digits is equal to the sum of the last n digits. a(n) is the number of lucky tickets. a(n) is also the number of tickets in which the sum of all the digits is 9*n.

a(n) is the number of integers whose digits sum = 9*n in [0, 100^n-1]. The most common value of sums of digits of numbers in [0, 100^n-1] is 9*n. - Miquel Cerda, Jul 02 2017

Examples

			The ticket 123051 is lucky because 1 + 2 + 3 = 0 + 5 + 1.
670 is the number of integers in the [0, 100^2-1] range whose digits sum = 18 and 55252 is the number of integers in the [0, 100^3-1] range whose digits sum = 27. - _Miquel Cerda_, Jul 02 2017
		

References

• S. K. Lando, Lectures on Generating Functions, AMS, 2002, page 1.

Links

• Vincenzo Librandi, Table of n, a(n) for n = 0..200
• A. Dubey, A Simplified Analysis To A Generalized Restricted Partition Problem, Principia: The Princeton Undergraduate Mathematics Journal, Issue 2, 2016.

Programs

Mathematica

Table[Total[ CoefficientList[Series[((1 - x^10)/(1 - x))^n, {x, 0, 9*n}], x]^2], {n, 0, 15}]

PARI

a(n)=if(n==0, 1, sum(k=0, n - 1, (-1)^k*binomial(2*n, k)*binomial(11*n - 1 - 10*k, 2*n - 1))); \\ Indranil Ghosh, Jul 01 2017

Formula

a(n) = Sum_{k=0..n-1} (-1)^k * binomial(2n,k) * binomial(11n-1-10k,2n-1).

a(n) = [x^(9n)] ((1 - x^10)/(1 - x))^(2n).

a(n) = A025015(2*n). - Miquel Cerda, Jul 18 2017