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 A100308 Modulo 2 binomial transform of 5^n. 7

%I

%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 be retrieved through 5^n=sum{k=0..n, (-1)^A010060(n-k)*mod(binomial(n,k),2)A100308(k)}.

%H Robert Israel, <a href="/A100308/b100308.txt">Table of n, a(n) for n = 0..1429</a>

%H V. Shevelev, <a href="http://arxiv.org/abs/1011.6083">On Stephan's conjectures concerning Pascal triangle modulo 2 and their polynomial generalization</a>, 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)5^k}.

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

%F sum{n>=0}1/a(n)^r = prod{k>=0}(1 + 1/(5^(2^k)+1)^r),

%F sum{n>=0}(-1)^A000120(n)/a(n)^r = prod{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) = prod{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.

%F (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 *)

%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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Last modified August 6 15:44 EDT 2020. Contains 336254 sequences. (Running on oeis4.)