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 A116525 a(0)=1, a(1)=1, a(n)=11a(n/2) for n=2,4,6,..., a(n)=10a((n-1)/2)+a((n+1)/2) for n=3,5,7,.... 0

%I

%S 0,1,11,21,121,131,231,331,1331,1341,1441,1541,2541,2641,3641,4641,

%T 14641,14651,14751,14851,15851,15951,16951,17951,27951,28051,29051,

%U 30051,40051,41051,51051,61051,161051,161061,161161,161261,162261

%N a(0)=1, a(1)=1, a(n)=11a(n/2) for n=2,4,6,..., a(n)=10a((n-1)/2)+a((n+1)/2) for n=3,5,7,....

%C An 11-divide version of A084230.

%C The Harborth : f(2^k)=3^k suggests that a family of sequences of the form: f(2^k)=Prime[n]^k There does indeed seem to be an infinite family of such functions.

%D Harborth, H. Number of Odd Binomial Coefficients. Proc. Amer. Math. Soc. 62, 19-22, 1977

%H Eric Weisstein's World of Mathematics, <a href="http://mathworld.wolfram.com/Stolarsky-HarborthConstant.html">Stolarsky-Harborth Constant</a>

%p a:=proc(n) if n=0 then 0 elif n=1 then 1 elif n mod 2 = 0 then 11*a(n/2) else 10*a((n-1)/2)+a((n+1)/2) fi end: seq(a(n),n=0..42);

%t b[0] := 0 b[1] := 1 b[n_?EvenQ] := b[n] = 11*b[n/2] b[n_?OddQ] := b[n] = 10*b[(n - 1)/2] + b[(n + 1)/2] a = Table[b[n], {n, 1, 25}]

%Y Cf. A006046, A077465.

%K nonn

%O 0,3

%A _Roger Bagula_, Mar 15 2006

%E Edited by _N. J. A. Sloane_, Apr 16 2005

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