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A042963
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Congruent to 1 or 2 mod 4.
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18
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1, 2, 5, 6, 9, 10, 13, 14, 17, 18, 21, 22, 25, 26, 29, 30, 33, 34, 37, 38, 41, 42, 45, 46, 49, 50, 53, 54, 57, 58, 61, 62, 65, 66, 69, 70, 73, 74, 77, 78, 81, 82, 85, 86, 89, 90, 93, 94, 97, 98, 101, 102, 105, 106, 109
(list; graph; refs; listen; history; internal format)
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OFFSET
| 1,2
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COMMENTS
| A014493(n) = A000217(a(n)); complement of A014601. [Reinhard Zumkeller, Feb 14 2012, Oct 04 2004]
Equals partial sums of A153284: (1, 1, 3, 1, 3, 1, 3,...) [From Gary W. Adamson (qntmpkt(AT)yahoo.com), Jan 03 2009]
Let S(x) = (1, 2, 2, 2,...). Then A042963 = (1/2) * ((S(x)^2 + S(x^2))
= (1/2) * ((1, 4, 8, 12, 16, 20,...) + (1, 0, 2, 0, 2, 0, 2,...))
= (1, 2, 5, 6, 9, 10,...). [From Gary W. Adamson (qntmpkt(AT)yahoo.com), Jan 03 2011]
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LINKS
| Reinhard Zumkeller, Table of n, a(n) for n = 1..1000
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FORMULA
| G.f.: (1+x+2*x^2)/((1-x)*(1-x^2)). a(n)=a(n-1)+2+(-1)^n, a(0)=1 - Michael Somos, Jan 12 2000.
a(n)=sum{k=0..n, mod(A001045(k), 4) } - Paul Barry (pbarry(AT)wit.ie), Mar 12 2004
a(n)=A005843(n)+A059841(n) . [From Philippe DELEHAM (kolotoko(AT)wanadoo.fr), Mar 31 2009]
a(n)=4*n-a(n-1)-1 (with a(0)=1).
a(n)=a(n-1)+a(n-2)-a(n-3)[From Ant King, Nov 17 2010].
If we consider the offset as 1, then a(n)=1/2 (4n-3-(-1)^n)[From Ant King, Nov 17 2010].
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MATHEMATICA
| Select[Range[109], Or[Mod[#, 4] == 1, Mod[#, 4] == 2] &].[From Ant King, Nov 17 2010].
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PROG
| (PARI) a(n)=1+2*n-n%2
(MAGMA) [ n : n in [1..165] | n mod 4 eq 1 or n mod 4 eq 2 ] [From Vincenzo Librandi, Jan 25 2011]
(Haskell)
a042963 n = a042963_list !! (n-1)
a042963_list = [x | x <- [0..], mod x 4 `elem` [1, 2]]
-- Reinhard Zumkeller, Feb 14 2012
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CROSSREFS
| A042963(n)=1+A042948(n).
A153284 [From Gary W. Adamson (qntmpkt(AT)yahoo.com), Jan 03 2009]
Cf. A046712 (subsequence).
Union of A016813 and A016825.
Sequence in context: A085183 A133759 A188258 * A166097 A187583 A000277
Adjacent sequences: A042960 A042961 A042962 * A042964 A042965 A042966
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KEYWORD
| nonn,changed
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AUTHOR
| N. J. A. Sloane (njas(AT)research.att.com).
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EXTENSIONS
| Offset corrected by Reinhard Zumkeller (reinhard.zumkeller(AT)gmail.com), Feb 14 2012
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