A033114 Base-4 digits are, in order, the first n terms of the periodic sequence with initial period 1,0.
1, 4, 17, 68, 273, 1092, 4369, 17476, 69905, 279620, 1118481, 4473924, 17895697, 71582788, 286331153, 1145324612, 4581298449, 18325193796, 73300775185, 293203100740, 1172812402961, 4691249611844, 18764998447377, 75059993789508
Offset: 1
Links
- Vincenzo Librandi, Table of n, a(n) for n = 1..1000
- Thomas Baruchel, Properties of the cumulated deficient binary digit sum, arXiv:1908.02250 [math.NT], 2019.
- Madeleine Goertz and Aaron Williams, The Quaternary Gray Code and How It Can Be Used to Solve Ziggurat and Other Ziggu Puzzles, arXiv:2411.19291 [math.CO], 2024. See p. 17.
- Index entries for linear recurrences with constant coefficients, signature (4,1,-4).
Programs
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Magma
[Round((8*4^n-5)/30): n in [1..30]]; // Vincenzo Librandi, Jun 25 2011
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Maple
seq(floor((4^(n+1)-1)/15),n=1..25) # Mircea Merca, Dec 28 2010
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Mathematica
Join[{a=1,b=4},Table[c=3*b+4*a+1;a=b;b=c,{n,100}]] (* Vladimir Joseph Stephan Orlovsky, Jan 17 2011 *)
Formula
a(n) = floor(4^(n+1)/15) = 4^(n+1)/15 - 1/6 - (-1)^n/10. - Benoit Cloitre, Apr 18 2003
G.f.: 1/((1-x)*(1+x)*(1-4*x)); a(n) = 3*a(n-1) + 4*a(n-2)+1. Partial sum of A015521. - Paul Barry, Nov 12 2003
a(n) = Sum_{k=0..floor(n/2)} 4^(n-2*k); a(n) = Sum_{k=0..n} Sum_{j=0..k} (-1)^(j+k)*4^j. - Paul Barry, Nov 12 2003
Convolution of A000302 and A059841 (4^n and periodic{1, 0}). a(n) = Sum_{k=0..n} (1 + (-1)^(n-k))*4^k/2. - Paul Barry, Jul 19 2004
a(n) = Sum_{k=0..n} (-1)^(n-k)*(J(2*k+1)-1)/2, J(n)=A001045(n). - Paul Barry, Mar 06 2008
a(n) = round((8*4^n-5)/30) = ceiling((4*4^n-4)/15) = round((4*4^n-4)/15); a(n) = a(n-2) + 4^(n-1), n > 1. - Mircea Merca, Dec 28 2010
a(n) = A117616(n)/2. - J. M. Bergot, Apr 22 2015
a(n)+a(n+1) = A002450(n+1). - R. J. Mathar, Feb 27 2019