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 July 3rd, 2018, 01:07 AM #1 Senior Member   Joined: Nov 2011 Posts: 250 Thanks: 3 Numeric sequence I have a numeric sequence: l(n)=sum[from k=1 to n](4k-1)*l(n-k) l(0)=1 Is there a function f(x) that satisfies l(n)? And if so, what is it? Last edited by skipjack; July 3rd, 2018 at 09:01 PM.
 July 3rd, 2018, 03:13 PM #2 Math Team   Joined: Jan 2015 From: Alabama Posts: 3,264 Thanks: 902 I'm not sure what you mean by "a function f(x) that satisfies l(n). Since that formula is "well defined", of course there exist a function defined on the positive integers that satisfies that formula. l(0)= 1 so $\displaystyle l(1)= (4- 1)l(0)= 3$ $\displaystyle l(2)= (4- 1)l(1)+ (8- 1)l(0)= 3(3)+ 7(1)= 16$ $\displaystyle l(3)= (4- 1)l(2)+ (8- 1)l(1)+ (12- 1)l(0)= 4(16)+ 7(3)+ 11(1)= 64+ 21+ 11= 96$ $\displaystyle l(4)= (4- 1)l(3)+ (8- 1)l(2)+ (12- 1)l(1)+ (16- 1)l(0)= 3(96)+ 7(16)+ 11(3)+ 15(1)= 288+ 112+ 33+ 15= 448$. Continue. No simple formula jumps out at me. Last edited by skipjack; July 3rd, 2018 at 09:48 PM.
 July 3rd, 2018, 09:48 PM #3 Global Moderator   Joined: Dec 2006 Posts: 20,388 Thanks: 2015 You should have had "3(16)" instead of "4(16)", which means that l(3) = 80, l(4) = 3(80) + 7(16) + 11(3) + 15(1) = 400, etc. That is, l(1) = 3, l(2+m) = 16*5^m for m = 0, 1, 2, etc.

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