%I #28 May 03 2023 09:14:15
%S 1,8,50,400,2402,19216,120100,960800,5764802,46118416,288240100,
%T 2305920800,13847054404,110776435232,692352720200,5538821761600,
%U 33232930569602,265863444556816,1661646528480100,13293172227840800
%N Modulo 2 binomial transform of 7^n.
%C 7^n may be retrieved through 7^n = Sum_{k=0..n} (-1)^A010060(n-k) * mod(binomial(n,k), 2) * a(k).
%H G. C. Greubel, <a href="/A100310/b100310.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} mod(binomial(n, k), 2)*7^k.
%F From _Vladimir Shevelev_, Dec 26-27 2013: (Start)
%F Sum_{n>=0} 1/a(n)^r = Product_{k>=0} (1 + 1/(7^(2^k)+1)^r),
%F Sum_{n>=0} (-1)^A000120(n)/a(n)^r = Product_{k>=0} (1 - 1/(7^(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/(7^(2^k)+1)) = 1.1479779...;
%F Sum_{n>=0} (-1)^A000120(n)/a(n) = 6/7.
%F a(2^n) = 7^(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)) = 48*(7^(2^(t-1)+1))/(7^(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) = 50 * a(4*n);
%F a(8*n+4) = 1201/25 * a(8*n+2);
%F a(16*n+8)= 2882401/60050 * a(16*n+6), etc. (End)
%t a[n_]:= a[n]= Sum[7^k*Mod[Binomial[n,k], 2], {k,0,n}];
%t Table[a[n], {n,0,40}] (* _G. C. Greubel_, Feb 01 2023 *)
%o (Magma) [(&+[7^k*(Binomial(n,k) mod 2): k in [0..n]]): n in [0..40]]; // _G. C. Greubel_, Feb 01 2023
%o (SageMath)
%o def A100310(n): return sum(7^k*(binomial(n, k)%2) for k in range(n+1))
%o [A100310(n) for n in range(41)] # _G. C. Greubel_, Feb 01 2023
%o (Python)
%o def A100310(n): return sum((bool(~n&n-k)^1)*7**k for k in range(n+1)) # _Chai Wah Wu_, May 03 2023
%Y Cf. A000120, A001316, A001317, A010060, A038183.
%Y Cf. A100307, A100308, A100309, A100311.
%K easy,nonn
%O 0,2
%A _Paul Barry_, Dec 06 2004