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A001576
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a(n) = 1^n + 2^n + 4^n.
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100
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3, 7, 21, 73, 273, 1057, 4161, 16513, 65793, 262657, 1049601, 4196353, 16781313, 67117057, 268451841, 1073774593, 4295032833, 17180000257, 68719738881, 274878431233, 1099512676353, 4398048608257, 17592190238721, 70368752566273, 281474993487873, 1125899940397057
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OFFSET
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0,1
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COMMENTS
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Conjecture: For n > 1, if a(n) = 1^n + 2^n + 4^n is a prime number then n is of the form 3^h. For example, for h=1, n=3, a(n) = 1^3 + 2^3 + 4^3 = 73 (prime); for h=2, n=9, a(n) = 1^9 + 2^9 + 4^9 = 262657 (prime); for h=3, n=27, a(n) is not prime. - Vincenzo Librandi, Aug 03 2010
The previous conjecture was proved by Golomb in 1978. See A051154. - T. D. Noe, Aug 15 2010
Another more elementary proof can be found in Liu link. - Bernard Schott, Mar 08 2019
Fills in one quarter section of the figurate form of the Sierpinski square curve. See illustration in links and A141725. - John Elias, Mar 29 2023
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LINKS
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FORMULA
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a(n) = 6*a(n-1) - 8*a(n-2) + 3.
O.g.f.: -1/(-1+x) - 1/(-1+2*x) - 1/(-1+4*x) = ( -3+14*x-14*x^2 ) / ( (x-1)*(2*x-1)*(4*x-1) ). - R. J. Mathar, Feb 29 2008
Exp( Sum_{n >= 1} a(n)*x^n/n ) = 1 + 7*x + 35*x^2 + 155*x^3 + ... is the o.g.f. for the 2nd subdiagonal of triangle A022166, essentially A006095. - Peter Bala, Apr 07 2015
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MATHEMATICA
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Table[1^n + 2^n + 4^n, {n, 0, 24}]
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PROG
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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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