

A174239


a(n) = (3*n + 1 + (1)^n*(n+3))/4.


4



1, 0, 3, 1, 5, 2, 7, 3, 9, 4, 11, 5, 13, 6, 15, 7, 17, 8, 19, 9, 21, 10, 23, 11, 25, 12, 27, 13, 29, 14, 31, 15, 33, 16, 35, 17, 37, 18, 39, 19, 41, 20, 43, 21, 45, 22, 47, 23, 49, 24, 51, 25, 53, 26, 55, 27, 57, 28, 59, 29, 61, 30, 63, 31, 65, 32, 67, 33, 69, 34, 71, 35, 73, 36, 75, 37, 77, 38, 79, 39, 81
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OFFSET

0,3


COMMENTS

Obtained from A026741 by swapping pairs of consecutive entries.
The main diagonal of an array with this sequence in the top row and further rows defined by the first differences of their previous row is essentially 1 followed by 3*A045623(.):
1, 0, 3, 1, 5, 2, 7, 3, 9, 4, 11, 5, 13, 6, 15, 7, 17, 8, ...
1, 3, 2, 4, 3, 5, 4, 6, 5, 7, 6, 8, 7, 9, 8, 10, 9, ...
4, 5, 6, 7, 8, 9, 10, 11, 12, 13, 14, 15, 16, 17, ...
9, 11, 13, 15, 17, 19, 21, 23, 25, 27, 29, 31, ...
20, 24, 28, 32, 36, 40, 44, 48, 52, 56, 60, 64, ...
44, 52, 60, 68, 76, 84, 92, 100, 108, 116, 124, 132, ...
96, 112, 128, 144, 160, 176, 192, 208, 224, 240, ...
Also, numerator of (Nimsum n+1)/2 = A004442(n)/2.  Wesley Ivan Hurt, Mar 21 2015


LINKS

Table of n, a(n) for n=0..80.
Index entries for linear recurrences with constant coefficients, signature (0,2,0,1).


FORMULA

a(2n) = 2n+1; a(2n+1) = n.
a(n) = 2*a(n2)  a(n4).
a(2n+1)  2*a(2n) = A016789(n+1).
a(2n+2)  2*a(2n+1) = 3.
G.f.: ( 1+x^2+x^3 ) / ( (x1)^2*(1+x)^2 ).  R. J. Mathar, Feb 07 2011


MAPLE

A174239:=n>(3*n+1+(1)^n*(n+3))/4: seq(A174239(n), n=0..100); # Wesley Ivan Hurt, Mar 21 2015


MATHEMATICA

Table[(3 n + 1 + (1)^n*(n + 3))/4, {n, 0, 100}] (* Wesley Ivan Hurt, Mar 21 2015 *)
LinearRecurrence[{0, 2, 0, 1}, {1, 0, 3, 1}, 90] (* Harvey P. Dale, Jul 16 2018 *)


PROG

(MAGMA) [(3*n+1 +(1)^n*(n+3))/4: n in [0..80]]; // Vincenzo Librandi, Feb 08 2011


CROSSREFS

Cf. A004442.
Sequence in context: A225080 A276505 A201901 * A066249 A065168 A318581
Adjacent sequences: A174236 A174237 A174238 * A174240 A174241 A174242


KEYWORD

nonn,easy


AUTHOR

Paul Curtz, Mar 13 2010


STATUS

approved



