A100307 - OEIS

%I #57 May 03 2023 09:13:48

%S 1,4,10,40,82,328,820,3280,6562,26248,65620,262480,538084,2152336,

%T 5380840,21523360,43046722,172186888,430467220,1721868880,3529831204,

%U 14119324816,35298312040,141193248160,282472589764,1129890359056

%N Modulo 2 binomial transform of 3^n.

%C 3^n may be retrieved through 3^n = Sum_{k=0..n} (-1)^A010060(n-k)*(binomial(n,k) mod 2)*a(k).

%H Gheorghe Coserea, <a href="/A100307/b100307.txt">Table of n, a(n) for n = 0..200</a>

%H Vladimir Shevelev, <a href="http://arxiv.org/abs/1011.6083">On Stephan's conjectures concerning Pascal triangle modulo 2 and their polynomial generalization</a>, arXiv:1011.6083 [math.NT], 2010-2012; J. of Algebra Number Theory: Advances and Appl., 7 (2012), no.1, 11-29.

%F a(n) = Sum_{k=0..n} (binomial(n, k) mod 2)*3^k.

%F From _Vladimir Shevelev_, Dec 26-27 2013: (Start)

%F Sum_{n>=0} 1/a(n)^r = Product_{k>=0} (1 + 1/(3^(2^k)+1)^r),

%F Sum_{n>=0} (-1)^A000120(n)/a(n)^r = Product_{k>=0} (1 - 1/(3^(2^k)+1)^r), where r > 0 is a real number.

%F In particular,

%F Sum_{n>=0} 1/a(n) = Product_{k>=0} (1 + 1/(3^(2^k)+1)) = 1.391980...;

%F Sum_{n>=0} (-1)^A000120(n)/a(n) = 2/3.

%F a(2^n) = 3^(2^n)+1, n >= 0.

%F Note that analogs of Stephan's limit formulas (see Shevelev link) reduce to the relations:

%F a(2^t*n+2^(t-1)) = 8*(3^(2^(t-1)+1))/(3^(2^(t-1))-1) * a(2^t*n+2^(t-1)-2), t >= 2.

%F In particular, for t=2,3,4, we have the following formulas:

%F   a(4*n+2)  =         10 * a(4*n),

%F   a(8*n+4)  =     (41/5) * a(8*n+2),

%F   a(16*n+8) = (3281/410) * a(16*n+6), etc. (End)

%F From _Tom Edgar_, Oct 11 2015: (Start)

%F a(n) = Product_{b_j != 0} a(2^j) where n = Sum_{j>=0} b_j*2^j is the binary representation of n.

%F a(2*k+1) = 4*a(2*k). (End)

%t Table[Sum[Mod[Binomial[n,k],2]3^k,{k,0,n}],{n,0,40}] (* _Harvey P. Dale_, Aug 28 2013 *)

%o (Sage) [sum((binomial(n,k)%2)*3^k for k in [0..n]) for n in [0..50]] # _Tom Edgar_, Oct 11 2015

%o (PARI) a(n) = subst(lift((Mod(1,2)+'x)^n), 'x, 3); \\ _Gheorghe Coserea_, Jun 11 2016

%o (Magma) [(&+[3^k*(Binomial(n, k) mod 2): k in [0..n]]): n in [0..40]]; // _G. C. Greubel_, Feb 03 2023

%o (Python)

%o def A100307(n): return sum((bool(~n&n-k)^1)*3**k for k in range(n+1)) # _Chai Wah Wu_, May 02 2023

%Y Cf. A001316, A001317, A038183, A100308, A100309, A100310, A100311.

%K easy,nonn

%O 0,2

%A _Paul Barry_, Dec 06 2004