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%I #48 May 03 2023 09:13:43
%S 1,6,26,156,626,3756,16276,97656,390626,2343756,10156276,60937656,
%T 244531876,1467191256,6357828776,38146972656,152587890626,
%U 915527343756,3967285156276,23803710937656,95520019531876,573120117191256
%N Modulo 2 binomial transform of 5^n.
%C 5^n may be retrieved through 5^n = Sum_{k=0..n} (-1)^A010060(n-k) * (binomial(n,k) mod 2)*a(k).
%H Robert Israel, <a href="/A100308/b100308.txt">Table of n, a(n) for n = 0..1429</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)*5^k.
%F From _Vladimir Shevelev_, Dec 26-27 2013: (Start)
%F Sum_{n>=0} 1/a(n)^r = Product_{k>=0} (1 + 1/(5^(2^k)+1)^r),
%F Sum_{n>=0} (-1)^A000120(n)/a(n)^r = Product_{k>=0} (1 - 1/(5^(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/(5^(2^k)+1)) = 1.2134769...;
%F Sum_{n>=0} (-1)^A000120(n)/a(n) = 0.8.
%F a(2^n) = 5^(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)) = 24*(5^(2^(t-1)+1))/(5^(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) = 26 * a(4*n),
%F a(8*n+4) = (313/13) * a(8*n+2),
%F a(16*n+8) = (195313/8138) * a(16*n+6), etc. (End)
%p f:= proc(n) local L,M;
%p L:= convert(n,base,2);
%p mul(1+5^(2^(k-1)), k = select(t -> L[t]=1, [$1..nops(L)]));
%p end proc:
%p map(f, [$0..30]); # _Robert Israel_, Aug 26 2018
%t a[n_]:= Sum[Mod[Binomial[n, k], 2] 5^k, {k, 0, n}];
%t Table[a[n], {n, 0, 30}] (* _Jean-François Alcover_, Sep 19 2018 *)
%o (Magma) [(&+[5^k*(Binomial(n, k) mod 2): k in [0..n]]): n in [0..40]]; // _G. C. Greubel_, Feb 03 2023
%o (SageMath)
%o def A100308(n): return sum(5^k*(binomial(n, k)%2) for k in range(n+1)
%o [A100308(n) for n in range(41)] # _G. C. Greubel_, Feb 03 2023
%o (Python)
%o def A100308(n): return sum((bool(~n&n-k)^1)*5**k for k in range(n+1)) # _Chai Wah Wu_, May 02 2023
%Y Cf. A001316, A001317, A038183, A100307, A100309, A100310, A100311.
%K easy,nonn
%O 0,2
%A _Paul Barry_, Dec 06 2004
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