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A213859
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a(n) = 2^n mod (n+2).
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5
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1, 2, 0, 3, 4, 4, 0, 2, 6, 6, 4, 7, 8, 2, 0, 9, 16, 10, 4, 2, 12, 12, 16, 8, 14, 20, 4, 15, 16, 16, 0, 2, 18, 22, 16, 19, 20, 2, 24, 21, 16, 22, 4, 38, 24, 24, 16, 32, 6, 2, 4, 27, 34, 52, 8, 2, 30, 30, 4, 31, 32, 2, 0, 8, 16, 34, 4, 2, 46, 36, 16, 37, 38, 17
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OFFSET
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0,2
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COMMENTS
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Conjectures:
1. Indices of zeros: 2^(x+2)-2, x >= 0.
2. There are infinitely many n's such that a(n)=n.
3. Every integer k >= 0 appears in a(n) at least once.
4. Every k >= 0 appears in a(n) infinitely many times.
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LINKS
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FORMULA
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a(n) = 2^n mod (n+2).
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MATHEMATICA
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Table[PowerMod[2, n, n+2], {n, 0, 100}] (* T. D. Noe, Jun 26 2012 *)
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PROG
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(Python)
print([2**n % (n+2) for n in range(222)])
(Magma) [Modexp(2, n, n+2): n in [0..120]]; // G. C. Greubel, Jan 11 2023
(SageMath) [power_mod(2, n, n+2) for n in range(121)] # G. C. Greubel, Jan 11 2023
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CROSSREFS
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KEYWORD
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nonn,easy
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AUTHOR
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STATUS
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approved
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