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COMMENT
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a(n) is the size of largest conjugacy class in D_2n, the dihedral group with 2n elements. - Sharon Sela (sharonsela(AT)hotmail.com), May 14 2002
a(n+1) is the composition length of the n-th symmetric power of the natural representation of a finite subgroup of SL(2,C) of type D_4 (quaternion group). - Paul Boddington (psb(AT)maths.warwick.ac.uk), Oct 23 2003
For n>1 a(n) = greatest common divisor of all permutations of {0,1,...,n} treated as base n+1 integers. - David J. Scambler (dscambler(AT)bmm.com), Nov 08 2006
a(n) is the second principal diagonal of array with rows 1) A005563, 2) (first bisection of A061037)=A142705, 3) (first trisection of A061039)=A144454, 4) first quadrisection of A061041, 5) first quintisection of A061043, 6) first hexasection A061045, 7)first heptasection of A061047, 8) first octosection of A061049, 9) , 10) , 11) , . From Rydberg-Ritz denominators of spectra of hydrogen atom: 1) 0, 3, 8, 15, 24, 35, 48, 63, 80, 99, 120, 143, 2) 0, 3, 2, 15, 6, 35, 12, 63, 20, 99, 39, 143, 3) 0, 1, 8, 5, 8, 35, 16, 7, 80, 11, 40, 143, 4) 0, 3, 1, 15, 3, 35, 3, 63, 5, 99, 15, 143, 5) 0, 3, 8, 3, 24, 7, 48, 63, 16, 99, 6) 0, 1, 2, 5, 2, 35, 4, 7, 20, 7) 0, 3, 8, 15, 24, 5, 48, 9, 8) 0, 3, 1, 15, 3, 35, 3, 63, 5, 99, 15, 143, 9) 0, 1, 8, 5, 8, 35, 16, 7, 80, 11, 40, 143, 10) 0, 3, 2, 3, 6, 7, 12, 63, 4, 99, 6, 11) 0, 3, 8, 15, 24, 35, 48, 63, 80, 9, 120, 13, Thanks to Richard Mathar, April 29, for last four rows. Note first upper principal diagonal 3, 2, 5, 3, 7, 4, 9, =A026741(n+3). [From Paul Curtzz (bpcrtz(AT)free.fr), Sep 13 2009]
a(n) = A167192(n+2,2). [From Reinhard Zumkeller (reinhard.zumkeller(AT)gmail.com), Oct 30 2009]
Contribution from Paul Curtz (bpcrtz(AT)free.fr), Nov 19 2009: (Start) The array of a(n) and its higher order differences shows essentially 0,0 followed by -3*A001792(.) on the main diagonal:
0, 1, 1, 3, 2, 5, 3, 7, 4, 9, 5, 11, 6, 13, 7, 15, 8, 17, 9, 19, 10, 21,...
1, 0, 2, -1, 3, -2, 4, -3, 5, -4, 6, -5, 7, -6, 8, -7, 9, -8, 10, -9, 11,...
-1, 2, -3, 4, -5, 6, -7, 8, -9, 10, -11, 12, -13, 14, -15, 16, -17, 18,...
3, -5, 7, -9, 11, -13, 15, -17, 19, -21, 23, -25, 27, -29, 31, -33, 35,...
-8, 12, -16, 20, -24, 28, -32, 36, -40, 44, -48, 52, -56, 60, -64, 68,...
20, -28, 36, -44, 52, -60, 68, -76, 84, -92, 100, -108, 116, -124, 132,... (END)
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FORMULA
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G.f.: (x^3+x^2+x)/(1-x^2)^2 - Len Smiley (smiley(AT)math.uaa.alaska.edu), Apr 30 2001
a(n) = n * 2^((n mod 2) - 1) - Reinhard Zumkeller (reinhard.zumkeller(AT)gmail.com), Oct 16 2001
a(n) = 2*n/(3+(-1)^n) - Benoit Cloitre (benoit7848c(AT)orange.fr), Mar 24 2002
Multiplicative with a(2^e) = 2^(e-1) and a(p^e) = p^e, p>2. - Vladeta Jovovic (vladeta(AT)eunet.rs), Apr 05 2002
a(n) = n / gcd(n, 2). a(n)/A04589(n) = n/((n+1)(n+2)).
For n>1, a(n) = denominator of sum{2/(i*(i+1))|1<=i<=n-1}, numerator=A026741. - Reinhard Zumkeller (reinhard.zumkeller(AT)gmail.com), Jul 25 2002
For n > 1, a(n) = GCD of the n-th and (n-1)th triangular numbers (A000217). - Ross La Haye (rlahaye(AT)new.rr.com), Sep 13 2003
Euler transform of finite sequence [1, 2, -1]. - Michael Somos Jun 15 2005
G.f.: x(1-x^3)/((1-x)(1-x^2)^2) = Sum_{k>0} k(x^k-x^(2k)). - Michael Somos Jun 15 2005
a(n)a(n+3) = - 1 + a(n+1)a(n+2). a(-n)=-a(n).
a(n) = Abs[ Numerator[ Det[ DiagonalMatrix[ Table[ 1/i^2 -1, {i, 1, n-1} ] ] + 1 ] ] for n>1. - Alexander Adamchuk (alex(AT)kolmogorov.com), Jun 02 2006
For n > 1, a(n) is the numerator of the average of 1,2,...,n-1; i.e., numerator of A000217(n-1)/(n-1), with corresponding denominators [1,2,1,2,...] (A000034). - Rick L. Shepherd (rshepherd2(AT)hotmail.com), Jun 05 2006
Equals A126988 * (1, -1, 0, 0, 0,...). - Gary W. Adamson (qntmpkt(AT)yahoo.com), Apr 17 2007
For n >= 1, a(n) = GCD(n,A000217(n)). - Rick L. Shepherd (rshepherd2(AT)hotmail.com), Sep 12 2007
Contribution from Johannes W. Meijer (meijgia(AT)hotmail.com), Jun 18 2009: (Start)
a(n) = numer(n/(2*n-2)) for n =>2; A022998(n-1) = denom(n/(2*n-2)) for n =>2.
(End)
a(n+1)-a(n) = (-1)^n*A028242(n) (first differences). a(n+2)-2*a(n+1)+a(n) = (-1)^(n+1)*(n+1). (2nd differences). a(n+3)-3*a(n+2)+3*a(n+1)-a(n) = (-1)^n*A144396(n+1). (3rd differences). [Paul Curtz (bpcrtz(AT)free.fr), Nov 19 2009]
a(A131577(n)) = A166444(n). [Paul Curtz (bpcrtz(AT)free.fr), Nov 19 2009]
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