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A100308
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Modulo 2 binomial transform of 5^n.
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7
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1, 6, 26, 156, 626, 3756, 16276, 97656, 390626, 2343756, 10156276, 60937656, 244531876, 1467191256, 6357828776, 38146972656, 152587890626, 915527343756, 3967285156276, 23803710937656, 95520019531876, 573120117191256
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OFFSET
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0,2
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COMMENTS
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5^n may be retrieved through 5^n = Sum_{k=0..n} (-1)^A010060(n-k) * (binomial(n,k) mod 2)*a(k).
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LINKS
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FORMULA
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a(n) = Sum_{k=0..n} (binomial(n, k) mod 2)*5^k.
Sum_{n>=0} 1/a(n)^r = Product_{k>=0} (1 + 1/(5^(2^k)+1)^r),
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.
In particular,
Sum_{n>=0} 1/a(n) = Product_{k>=0} (1 + 1/(5^(2^k)+1)) = 1.2134769...;
Sum_{n>=0} (-1)^A000120(n)/a(n) = 0.8.
a(2^n) = 5^(2^n) + 1, n >= 0.
Note that analogs of Stephan's limit formulas (see Shevelev link) reduce to the relations:
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.
In particular, for t=2,3,4, we have the following formulas:
a(4*n+2) = 26 * a(4*n),
a(8*n+4) = (313/13) * a(8*n+2),
a(16*n+8) = (195313/8138) * a(16*n+6), etc. (End)
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MAPLE
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f:= proc(n) local L, M;
L:= convert(n, base, 2);
mul(1+5^(2^(k-1)), k = select(t -> L[t]=1, [$1..nops(L)]));
end proc:
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MATHEMATICA
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a[n_]:= Sum[Mod[Binomial[n, k], 2] 5^k, {k, 0, n}];
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PROG
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(Magma) [(&+[5^k*(Binomial(n, k) mod 2): k in [0..n]]): n in [0..40]]; // G. C. Greubel, Feb 03 2023
(SageMath)
def A100308(n): return sum(5^k*(binomial(n, k)%2) for k in range(n+1)
(Python)
def A100308(n): return sum((bool(~n&n-k)^1)*5**k for k in range(n+1)) # Chai Wah Wu, May 02 2023
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CROSSREFS
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KEYWORD
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easy,nonn
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AUTHOR
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STATUS
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approved
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