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%I #24 Apr 26 2021 20:53:40
%S 0,0,525,3105,18939,114381,693129,4195557,25405586,153820395,
%T 931359050,5639156409,34143908573,206733865761,1251728824798,
%U 7578945799704,45888871327435,277847147039527,1682304127857000,10185986079451152
%N Number of n-digit bouncy numbers in which every pair of adjacent digits are distinct.
%C We might call such numbers "strictly bouncy numbers"; they exclude most n-digit "bouncy numbers" (cf. A152054) for n >= 4.
%C As n increases, a(n) approaches c/(2*cos(Pi*9/19))^n,
%C where c is 2.32290643963522604128193759601...
%C Is c the result of some simple expression?
%C From _Jon E. Schoenfield_, Dec 16 2008: (Start)
%C We could define the recursive formula
%C f(n) = 5*f(n-1) + 10*f(n-2) - 20*f(n-3) - 15*f(n-4) + 21*f(n-5) + 7*f(n-6) - 8*f(n-7) - f(n-8) + f(n-9)
%C and use a(n)=f(n) for n > 2 (a(n)=0 otherwise). Working backwards, given the terms f(11)=a(11) down through f(3)=a(3), the recursive formula would yield f(2)=81, f(1)=17 and f(0)=1, followed by the values 2, -1, 2, -2, 4, -5, 10, -14, 28, -42, 84, -132, etc., for negative values of n; these values are negative Catalan numbers for even n and twice (positive) Catalan numbers for odd n, down to f(-16).
%C The above results apply for numbers in base 10. In general, for base m+1 (so that the largest possible value for a digit is m), we can write
%C a(n) = f(n) for n > 2, 0 otherwise, where
%C f(n) = Sum_{j=1..m} (-1)^floor((j-1)/2)*binomial(floor((m+j)/2),j)*f(n-j) for n > 2,
%C f(2) = m^2, f(1) = 2*m - 1, f(0)=1,
%C f(n) = 2*Catalan((-1-n)/2) for odd n, 2 - 2m < n < 0 and
%C f(n) = -Catalan(-n/2) for even n, 2 - 2m <= n < 0.
%C (The expressions for n < 0 work more than far enough down to give enough terms to begin generating f(3), f(4), etc.) (End)
%H <a href="/index/Rec#order_09">Index entries for linear recurrences with constant coefficients</a>, signature (5,10,-20,-15,21,7,-8,-1,1).
%F a(n) = Sum_{i=1..9} (u(n,i) + d(n,i)) for n > 2 (0 otherwise), where
%F u(n,i) = Sum_{j=i+1..9} d(n-1,j) for n > 1,
%F d(n,i) = Sum_{j=0..i-1} u(n-1,j) for n > 1,
%F u(1,i) = 1, and
%F d(1,i) = 1.
%Y Cf. A043096, A152054.
%K base,nonn
%O 1,3
%A _Jon E. Schoenfield_, Dec 05 2008
%E Correction to formula for odd negative n by _Jon E. Schoenfield_, Dec 22 2008
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