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There is sequence $a_n = a_{n - 1} + 2 a_{n - 2} + \dots + n a_0$, and $a_0 = 1$. I found the generating function for this sequence: $(3t^2-3t+1)/(1-2t)^2$ but I do not know what to do next. How can I solve this? Should be reduced to a finite sum? Thank you in advance

Lex
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1 Answers1

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You expand the generating function as 1+x+3*x^2+8*x^3+20*x^4+... and look the coefficients up in http://oeis.org/A001792 .

R. J. Mathar
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  • Could you tell me how can I expand my generating function into this series? – Lex Oct 19 '14 at 18:26
  • @Lex: one way is to set $t=0$ and find what $(3t^2-3t+1)/(1-2t)^2$ equals, which is 1. This gives the first term of the generating function. Then think of the generating function that $t$ gives: $(3t^2-3t+1)/(1-2t)^2 = c_0 + c_1 t + c_2 t^2 + c_3 t^3 + \dots$. If we subtract $c_0$, which we now know is $c_0=1$, we get: $(3t^2-3t+1)/(1-2t)^2 - c_0 = c_0 - c_0 + c_1 t + c_2 t^2 + c_3 t^3 + \dots$ which equals $(3t^2-3t+1)/(1-2t)^2 - c_0 = c_1 t + c_2 t^2 + c_3 t^3 + \dots$. Divide by $t$ to get $c_1 + c_2 t + c_3 t^2 + \dots$, and set $t=0$ to get $c_1$. Continue subtracting and diving. – Matt Groff Oct 21 '14 at 02:26