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A050488
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3*(2^n-1) - 2*n.
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14
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0, 1, 5, 15, 37, 83, 177, 367, 749, 1515, 3049, 6119, 12261, 24547, 49121, 98271, 196573, 393179, 786393, 1572823, 3145685, 6291411, 12582865, 25165775, 50331597, 100663243, 201326537, 402653127, 805306309, 1610612675, 3221225409
(list; graph; refs; listen; history; internal format)
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OFFSET
| 0,3
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COMMENTS
| Number of words of length n+1 where first element is from {0,1,2}, other elements are from {0,1} and sequence does not decrease (for n=2 there are 3*2^2 sequences, but 000,100,110,111,200,210,211 decrease, so a(2) = 12-7 = 5).
Number of subgroups of C_(2^n) X C_(2^n) (see A060724).
Starting with 1 = row sums of triangle A054582. - Gary W. Adamson (qntmpkt(AT)yahoo.com), Jun 23 2008
Contribution from Gary W. Adamson (qntmpkt(AT)yahoo.com), Jul 24 2010: (Start)
Starting with "1" equals the eigensequence of a triangle with integer squares
(1, 4, 9, 16,...) as the left border and the rest 1's. (End)
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FORMULA
| Row sums of A125165: (1, 5, 15, 37...). Binomial transform of [1, 4, 6, 6, 6...] = [1, 5, 15, 37,...]. 4-th diagonal from the right of A126777 = (1, 5, 15,...). - Gary W. Adamson (qntmpkt(AT)yahoo.com), Dec 23 2006
a(n) = 2*a(n-1) + (2n-1); e.g. a(4) = 37 = 2*15 + 7. - Gary W. Adamson (qntmpkt(AT)yahoo.com), Sep 30 2007
Contribution from Johannes W. Meijer (meijgia(AT)hotmail.com), Feb 20 2009: (Start)
a(n) = 4a(n-1)-5a(n-2)+2a(n-3) for n>2 with a(0) = 0, a(1) = 1, a(2) = 5.
G.f.: z*(1+z)/((1-z)^2*(1-2*z))
(End)
a(n)=2*n+2*a(n-1)-1 (with a(0)=0) [From Vincenzo Librandi (vincenzo.librandi(AT)tin.it), Aug 06 2010]
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EXAMPLE
| a(1)=2*1+2*0-1=1; a(2)=2*2+2*1-1=5; a(3)=2*3+2*5-1=15 [From Vincenzo Librandi (vincenzo.librandi(AT)tin.it), Aug 06 2010]
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MATHEMATICA
| s=0; lst={s}; Do[s+=n+=s-1; AppendTo[lst, s], {n, 2, 5!, 2}]; lst [From Vladimir Orlovsky (4vladimir(AT)gmail.com), Nov 07 2008]
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CROSSREFS
| A050487(2^m-1).
Equals (1/2) A051667.
Cf. A126277, A125165.
Cf. A054852.
Contribution from Johannes W. Meijer (meijgia(AT)hotmail.com), Feb 20 2009: (Start)
Cf. A156925
A050488(n+1) equals A156920(n+1,1)
A050488(n+1) equals A156919(n+1,1)/2^n
A050488(n+1) equals A142963(n+2,1)/2
(End)
Sequence in context: A109818 A146797 A005491 * A142964 A188282 A014316
Adjacent sequences: A050485 A050486 A050487 * A050489 A050490 A050491
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KEYWORD
| nonn
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AUTHOR
| James A. Sellers (sellersj(AT)math.psu.edu), Dec 26, 1999.
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