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A190840
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a(n+1) = 4*a(n)*(a(n)+1) for a(0) = 1.
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1
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OFFSET
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0,2
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COMMENTS
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For n>0, subsequence of A132592: both a(n)/2 and a(n)+1 are squares.
All terms (n > 0) are divisible by 8, yielding all terms of A185097, which is indexed from n=1, thus having the first term A185097(1) = 1.
For n>0, subsequence of A060355: both a(n) and a(n)+1 are powerful numbers. - Bernard Schott, Apr 24 2023
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LINKS
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FORMULA
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a(n+1) = 4*a(n)*(a(n)+1) for a(0) = 1.
a(n) = (1 + sqrt(2))^(2^(n+1))/4 + (1 - sqrt(2))^(2^(n+1))/4 - 1/2. Therefore 2*a(n) + 1 = A001601(n+1). - Bruno Berselli, Feb 01 2017
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MATHEMATICA
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CROSSREFS
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KEYWORD
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nonn
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AUTHOR
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STATUS
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approved
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