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A113497
Ascending descending base exponent transform of sequence A000034(n) = 1 + n mod 2.
2
1, 3, 6, 6, 11, 9, 16, 12, 21, 15, 26, 18, 31, 21, 36, 24, 41, 27, 46, 30, 51, 33, 56, 36, 61, 39, 66, 42, 71, 45, 76, 48, 81, 51, 86, 54, 91, 57, 96, 60, 101, 63, 106, 66, 111, 69, 116, 72, 121, 75, 126, 78, 131, 81, 136, 84, 141, 87, 146, 90, 151, 93, 156, 96, 161, 99, 166, 102, 171
OFFSET
1,2
COMMENTS
A000034 = 1, 2, 1, 2, 1, 2, 1, 2, 1, 2, 1, 2, 1, 2, 1, 2, 1, 2, 1, 2, 1, ... = continued fraction for (sqrt(3)+1)/2 (cf. A040001) = base 3 digital root of n+1. In general, the ascending descending base exponent transform of any simple periodic sequence can be written as a periodic set of interleaved sequences.
FORMULA
a(n) = Sum_{i=1..n} A000034(i)^A000034(n-i+1).
a(2*n) = 3*n; a(2*n+1) = 5*n+1.
From Colin Barker, Jun 16 2012: (Start)
a(n) = (-3+3*(-1)^n+8*n-2*(-1)^n*n)/4.
a(n) = 2*a(n-2)-a(n-4).
G.f.: x*(1+3*x+4*x^2)/((1-x)^2*(1+x)^2). (End)
E.g.f.: (1/2)*(3*(x-1)*sinh(x) + 5*x*cosh(x)). - G. C. Greubel, Mar 12 2017
EXAMPLE
a(1) = 1^1 = 1.
a(2) = 1^2 + 2^1 = 3.
a(3) = 1^1 + 2^2 + 1^1 = 6.
a(4) = 1^2 + 2^1 + 1^2 + 2^1 = 6.
a(5) = 1^1 + 2^2 + 1^1 + 2^2 + 1^1 = 11.
a(6) = 1^2 + 2^1 + 1^2 + 2^1 + 1^2 + 2^1 = 9.
MATHEMATICA
Table[(-3 + 3*(-1)^n + 8*n - 2*(-1)^n*n)/4, {n, 1, 50}] (* G. C. Greubel, Mar 12 2017 *)
PROG
(PARI) x='x +O('x^50); Vec(x*(1+3*x+4*x^2)/((1-x)^2*(1+x)^2)) \\ G. C. Greubel, Mar 12 2017
KEYWORD
nonn,easy
AUTHOR
Jonathan Vos Post, Jan 10 2006
EXTENSIONS
Definition improved by M. F. Hasler, Jan 13 2012
STATUS
approved