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 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
 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 (list; graph; refs; listen; history; text; internal format)
 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

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Last modified March 1 13:16 EST 2024. Contains 370433 sequences. (Running on oeis4.)