A179894 Given the series (1, 2, 1, 2, 1, 2, ...), let (1 + 2x + x^2 + 2x^3 + ...) * (1 + 2x^2 + x^3 + 2x^4 + ...) = (1 + 2x + 3x^2 + 7x^3 + ...)

1, 2, 3, 7, 7, 12, 11, 17, 15, 22, 19, 27, 23, 32, 27, 37, 31, 42, 35, 47, 39, 52, 43, 57, 47, 62, 51, 67, 55, 72, 59, 77, 63, 82, 67, 87, 71, 92, 75, 97, 79, 102, 83, 107, 87, 112, 91, 117, 95, 122, 99, 127, 103, 132, 107, 137, 111, 142, 115, 147, 119, 152, 123, 157, 127, 162

Offset

1,2

Comments

The offset has been selected as "1" to accommodate the conjectured property of the sequence: 3 divides a(n) iff n == 0 mod 3. Example: 3 divides (3, 12, 15, 27, 27, 42, ...) but not other terms through n = 18.

Links

Table of n, a(n) for n=1..66.

Index entries for linear recurrences with constant coefficients, signature (0,2,0,-1).

Formula

(1 + 2x + 3x^2 + 7x^3 + ...) = (1 + 2x + x^2 + 2x^3 + ...) * (1 + 2x^2 + x^3 + 2x^4 + ...).

Let M = a triangle with (1, 2, 1, 2, 1, 2, ...) in every column with the leftmost column shifted upwards one row. Then A179894 = leftmost column of M^2.

a(1)=1; for odd n > 1, a(n) = 2*n - 3; for even n, a(n) = 5*n/2 - 3. So it is true that 3 divides a(n) iff 3 divides n. - Jon E. Schoenfield, Jul 31 2010

From Colin Barker, Oct 28 2012: (Start)

a(n) = ((9 + (-1)^n)*n - 12)/4 for n > 1.

a(n) = 2*a(n-2) - a(n-4) for n > 5.

G.f.: x*(2*x+1)*(x^3+x^2+1)/((x-1)^2*(x+1)^2). (End)

Maple

t1:=add(x^(2*n), n=0..50)+2*add(x^(2*n+1), n=0..50);
t2:=2*add(x^(2*n), n=0..50)-1+add(x^(2*n+1), n=0..50)-x;
t3:=t1*t2;
series(t3, x, 100);
seriestolist(%);

Crossrefs

Sequence in context: A305420 A171464 A276730 * A060215 A309666 A085420

Adjacent sequences: A179891 A179892 A179893 * A179895 A179896 A179897

Keyword

nonn,easy

Author

Gary W. Adamson, Jul 31 2010

Extensions

Edited, corrected and extended by N. J. A. Sloane and Jon E. Schoenfield, Sep 06 2010

Status

approved