cp's OEIS Frontend

A100309 Modulo 2 binomial transform of 6^n.

%I A100309 #31 May 03 2023 09:14:01
%S A100309 1,7,37,259,1297,9079,47989,335923,1679617,11757319,62145829,
%T A100309 435020803,2178463249,15249242743,80603140213,564221981491,
%U A100309 2821109907457,19747769352199,104381066575909,730667466031363,3658979549971729
%N A100309 Modulo 2 binomial transform of 6^n.
%C A100309 6^n may be retrieved through 6^n = Sum_{k=0..n} (-1)^A010060(n-k) * mod(binomial(n,k), 2) * a(k).
%H A100309 G. C. Greubel, <a href="/A100309/b100309.txt">Table of n, a(n) for n = 0..1000</a>
%H A100309 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 A100309 a(n) = Sum_{k=0..n} mod(binomial(n, k), 2) * 6^k.
%F A100309 From _Vladimir Shevelev_, Dec 26-27 2013: (Start)
%F A100309 Sum_{n>=0} 1/a(n)^r = Product_{k>=0} (1 + 1/(6^(2^k)+1)^r),
%F A100309 Sum_{n>=0} (-1)^A000120(n)/a(n)^r = Product_{k>=0} (1 - 1/(6^(2^k)+1)^r), where r>0 is a real number.
%F A100309 In particular,
%F A100309 Sum_{n>=0} 1/a(n) = Product_{k>=0} (1 + 1/(6^(2^k)+1)) = 1.1746508...;
%F A100309 Sum_{n>=0} (-1)^A000120(n)/a(n) = 5/6.
%F A100309 a(2^n) = 6^(2^n) + 1, n>=0.
%F A100309 Note that analogs of Stephan's limit formulas (see Shevelev link) reduce to the relations a(2^t*n+2^(t-1)) = 35*(6^(2^(t-1)+1))/(6^(2^(t-1))-1) * a(2^t*n+2^(t-1)-2), t>=2. In particular, for t=2,3,4, we have the following formulas:
%F A100309 a(4*n+2) = 37 * a(4*n);
%F A100309 a(8*n+4) = 1297/37 * a(8*n+2);
%F A100309 a(16*n+8)= 1679617/47989 * a(16*n+6), etc. (End)
%t A100309 a[n_]:= a[n]= Sum[6^k*Mod[Binomial[n,k], 2], {k,0,n}];
%t A100309 Table[a[n], {n,0,40}] (* _G. C. Greubel_, Feb 02 2023 *)
%o A100309 (Magma) [(&+[6^k*(Binomial(n,k) mod 2): k in [0..n]]): n in [0..40]]; // _G. C. Greubel_, Feb 02 2023
%o A100309 (SageMath)
%o A100309 def A100309(n): return sum(6^k*(binomial(n, k)%2) for k in range(n+1))
%o A100309 [A100309(n) for n in range(41)] # _G. C. Greubel_, Feb 02 2023
%o A100309 (Python)
%o A100309 def A100309(n): return sum((bool(~n&n-k)^1)*6**k for k in range(n+1)) # _Chai Wah Wu_, May 03 2023
%Y A100309 Cf. A000120, A000400, A001316, A001317, A010060, A038183.
%Y A100309 Cf. A100307, A100308, A100310, A100311.
%K A100309 easy,nonn
%O A100309 0,2
%A A100309 _Paul Barry_, Dec 06 2004