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Search: id:A130532
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| A130532 |
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a(n) + a(n - 1) is alternatively a square or a cube. |
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+0
1
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| 1, 3, 5, 4, 4, 5, 3, 1, 7, 2, 6, 3, 5, 4, 4, 5, 3, 1, 7, 2, 6, 3, 5, 4, 4, 5, 3, 1, 7, 2, 6, 3, 5, 4, 4, 5, 3, 1, 7, 2, 6, 3, 5, 4, 4, 5, 3, 1, 7, 2, 6, 3, 5, 4, 4, 5, 3, 1, 7, 2, 6, 3, 5, 4, 4, 5, 3, 1, 7, 2, 6, 3, 5, 4, 4, 5, 3, 1, 7, 2, 6, 3, 5, 4, 4, 5, 3, 1, 7, 2, 6, 3, 5, 4, 4, 5, 3, 1, 7, 2, 6, 3, 5, 4, 4
(list; graph; listen)
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OFFSET
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1,2
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COMMENT
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a(1) = 1; a(2n) is a minimal positive m such that a(2n - 1) + m is a square, a(2n + 1) is a minimal positive m such that a(2n) + m is a cube. Sequence is periodic (apparently with the same period for any a(1)).
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FORMULA
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a(n)=(1/90)*{-28*(n mod 10)+53*[(n+1) mod 10]-46*[(n+2) mod 10]+26*[(n+3) mod 10]+26*[(n+4) mod 10]-[(n+5) mod 10]+8*[(n+6) mod 10]+17*[(n+7) mod 10]-10*[(n+8) mod 10]+35*[(n+9) mod 10]}-5*{1-[(n+2) mod (n+1)]}, with n>=0. - Paolo P. Lava (ppl(AT)spl.at), Aug 31 2007
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EXAMPLE
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a(1)=1, a(2)=3 because 1+3 is a square, a(3)=5 because 3+5 is a cube, a(4)=4 because 5+4 is a square, etc.
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MATHEMATICA
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b=1; s={b}; Do[Do[bi=b+i; If[IntegerQ[Sqrt[bi]], b=i; AppendTo[s, b]; Break[]], {i, 1000}]; c=b; Do[ci=c+i; If[IntegerQ[ci^(1/3)], c=i; Break[]], {i, 1000}]; AppendTo[s, c]; b=c, {100}]; s
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CROSSREFS
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Sequence in context: A081361 A086181 A000655 this_sequence A019707 A077861 A076308
Adjacent sequences: A130529 A130530 A130531 this_sequence A130533 A130534 A130535
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KEYWORD
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nonn
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AUTHOR
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Zak Seidov (zakseidov(AT)yahoo.com), Aug 08 2007
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