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A252606
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Numbers j such that j + 2 divides 2^j + 2.
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2
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3, 4, 16, 196, 2836, 4551, 5956, 25936, 46775, 65536, 82503, 540736, 598816, 797476, 1151536, 3704416, 4290771, 4492203, 4976427, 8095984, 11272276, 13362420, 21235696, 21537831, 21549347, 29640832, 31084096, 42913396, 49960912, 51127259, 55137316, 56786087, 60296571, 70254724, 70836676
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OFFSET
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1,1
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COMMENTS
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Numbers j such that (2^j + 2)/(j + 2) is an integer. Numbers j such that (2^j - j)/(j + 2) is an integer.
The even members of this sequence (4, 16, 196, 2836, ...) are the numbers 2*k-2 where k>=3 is odd and 4^k == -8 (mod k).
The odd members of this sequence (3, 4551, 46775, 82503, ...) are the numbers k-2 where k>=3 is odd and 2^k == -8 (mod k). (End)
2^m is in this sequence for m = (2, 4, 16, 36, 120, 256, 456, 1296, 2556, ...), with the subsequence m = 2^k, k = (1, 2, 4, 8, 16, ...). - M. F. Hasler, Apr 09 2015
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LINKS
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EXAMPLE
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3 is in this sequence because (2^3 + 2)/(3 + 2) = 2.
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MAPLE
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select(t -> 2 &^t + 2 mod (t + 2) = 0, [$1..10^6]); # Robert Israel, Apr 09 2015
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MATHEMATICA
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Select[Range[10^6], IntegerQ[(2^#+2)/(#+2)]&] (* Ivan N. Ianakiev, Apr 17 2015 *)
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PROG
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(Magma) [n: n in [0..1200000] | Denominator((2^n+2)/(n+2)) eq 1];
(PARI) for(n=1, 10^5, if((2^n+2)%(n+2)==0, print1(n, ", "))) \\ Derek Orr, Apr 05 2015
(Python)
A252606_list = [n for n in range(10**4) if pow(2, n, n+2) == n] # Chai Wah Wu, Apr 16 2015
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CROSSREFS
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KEYWORD
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nonn,hard
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AUTHOR
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EXTENSIONS
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STATUS
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approved
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