A030050 Numbers from the Conway-Schneeberger 15-theorem.
1, 2, 3, 5, 6, 7, 10, 14, 15
Offset: 1
Examples
a(2*1) = Sum{d|(1+1)} mu(d) t_{(1+1)/d} = 1 * t_2 + (-1) * t_1 = 3 - 1 = 2;
a(2*2) = Sum{d|(2+1)} mu(d) t_{(2+1)/d} = 1 * t_3 + (-1) * t_1 = 6 - 1 = 5;
a(2*3) = Sum{d|(3+1)} mu(d) t_{(3+1)/d} = 1 * t_4 + (-1) * t_2 + 0 * t_1 = 10 - 3 = 7;
a(2*4) = Sum{d|(4+1)} mu(d) t_{(4+1)/d} = 1 * t_5 + (-1) * t_1 = 15 - 1 = 14.
References
- Manjul Bhargava, On the Conway-Schneeberger fifteen theorem, Contemporary Mathematics 272 (1999), 27-37.
- J. H. Conway, The Sensual (Quadratic) Form, M.A.A., 1997, p. 141.
- J. H. Conway, Universal quadratic forms and the fifteen theorem, Contemporary Mathematics 272 (1999), 23-26.
- J. H. Conway and W. A. Schneeberger, personal communication.
Links
- Manjul Bhargava, The Fifteen Theorem and Generalizations
- Ivars Peterson, All Square: Science News Online (subscription required)
- Wikipedia, 15 and 290 theorems.
Programs
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Mathematica
a[n_] := If[OddQ[n], (n+1)*(n+3)/8, DivisorSum[n/2+1, MoebiusMu[#]*(n+2#+2)*(n+2)/(8#^2) &]]; Array[a, 9] (* Jean-François Alcover, Dec 03 2015 *)
Formula
From Daniel Forgues, Feb 24 & 26 2015: (Start)
a(2n-1) = t_n = n*(n+1)/2 = A000217(n), 1 <= n <= 5;
a(2n) = Sum{d|(n+1)} mu(d) t_{(n+1)/d} = A007438(n+1), 1 <= n <= 4. (End)
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