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Modulo 2 binomial transform of 8^n.
8

%I #42 May 03 2023 09:13:52

%S 1,9,65,585,4097,36873,266305,2396745,16777217,150994953,1090519105,

%T 9814671945,68736258049,618626322441,4467856773185,40210710958665,

%U 281474976710657,2533274790395913,18295873486192705,164662861375734345

%N Modulo 2 binomial transform of 8^n.

%C 8^n may be retrieved through 8^n = Sum_{k=0..n} (-1)^A010060(n-k) * (binomial(n,k) mod 2) * A100311(k).

%H G. C. Greubel, <a href="/A100311/b100311.txt">Table of n, a(n) for n = 0..1000</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)*8^k.

%F Conjecture: a(0)=1, a(n+1) = (a(n)*8) XOR a(n), where XOR is the bitwise exclusive-or operator. - _Alex Ratushnyak_, Apr 22 2012

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

%F Sum_{n>=0} 1/a(n)^r = Product_{k>=0} (1 + 1/(8^(2^k)+1)^r),

%F Sum_{n>=0} (-1)^A000120(n)/a(n)^r = Product_{k>=0} (1 - 1/(8^(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/(8^(2^k)+1)) = 1.1284805...;

%F Sum_{n>=0} (-1)^A000120(n)/a(n) = 7/8.

%F a(2^n) = 8^(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)) = 63*(8^(2^(t-1)+1))/(8^(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) = 65 * a(4*n);

%F a(8*n+4) = 4097/65 * a(8*n+2);

%F a(16*n+8) = (16777217/266305) * a(16*n+6), etc. (End)

%t A100311[n_]:= A100311[n]= Sum[Mod[Binomial[n, k], 2]*8^k, {k, 0, n}];

%t Table[A100311[n], {n, 0, 30}] (* _G. C. Greubel_, Jan 25 2023 *)

%o (Python)

%o a=1

%o for i in range(33):

%o print(a, end=", ")

%o a = (a*8) ^ a

%o # _Alex Ratushnyak_, Apr 22 2012

%o (Python)

%o def A100311(n): return sum((bool(~n&n-k)^1)<<3*k for k in range(n+1)) # _Chai Wah Wu_, May 02 2023

%o (Magma) [(&+[(Binomial(n,k) mod 2)*8^k: k in [0..n]]): n in [0..40]]; // _G. C. Greubel_, Jan 25 2023

%o (SageMath)

%o def A100311(n): return sum( (binomial(n, k)%2)*8^k for k in range(n+1))

%o [A100311(n) for n in range(41)] # _G. C. Greubel_, Jan 25 2023

%Y Cf. A001316, A001317, A010060, A038183, A047999.

%Y Cf. A100307, A100308, A100309, A100310.

%K easy,nonn

%O 0,2

%A _Paul Barry_, Dec 06 2004