A093916 a(2*k-1) = (2*k-1)^2 + 2 - k, a(2*k) = 6*k^2 + 2 - k: First column of the triangle A093915.
2, 7, 9, 24, 24, 53, 47, 94, 78, 147, 117, 212, 164, 289, 219, 378, 282, 479, 353, 592, 432, 717, 519, 854, 614, 1003, 717, 1164, 828, 1337, 947, 1522, 1074, 1719, 1209, 1928, 1352, 2149, 1503, 2382, 1662, 2627, 1829, 2884, 2004, 3153, 2187, 3434, 2378, 3727, 2577, 4032, 2784, 4349, 2999, 4678, 3222, 5019, 3453, 5372
Offset: 1
Links
- Harvey P. Dale, Table of n, a(n) for n = 1..1000
- Index entries for linear recurrences with constant coefficients, signature (0,3,0,-3,0,1).
Programs
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Magma
[(n*(5*n-2) + (-1)^n*(n^2+1) + 7)/4: n in [1..70]]; // G. C. Greubel, Dec 30 2021
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Mathematica
LinearRecurrence[{0,3,0,-3,0,1},{2,7,9,24,24,53},80] (* Harvey P. Dale, Nov 24 2017 *)
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PARI
A093916(n)=((n^2+1)*(3-n%2)-n+1)/2 /* or the "experimental" version, trying out all allowed values */ A093916(n)={ local( s=(n^3+n)/2, d=(n^2-n)/2, k=ceil((2*s-d)/n)); while( (n*k+d)%s, k++ ); k } \\ M. F. Hasler, Apr 04 2009
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SageMath
[(5*n^2 -2*n +7 +(-1)^n*(n^2 +1))/4 for n in (1..70)] # G. C. Greubel, Dec 30 2021
Formula
a(n) = ((n^2+1)*b(n) - n + 1)/2 where b(n) = 3 - (n mod 2) = 2 if n odd, = 3 if n even. - M. F. Hasler, Apr 04 2009
From Colin Barker, May 01 2012: (Start)
a(n) = (n*(5*n-2) + (n^2+1)*(-1)^n + 7)/4.
a(n) = 3*a(n-2) - 3*a(n-4) + a(n-6).
G.f.: x*(2+7*x+3*x^2+3*x^3+3*x^4+2*x^5)/((1-x)^3*(1+x)^3). (End)
E.g.f.: (1/4)*( (7 +3*x +5*x^2)*exp(x) - 8 + (1 -x +x^2)*exp(-x) ). - G. C. Greubel, Dec 30 2021
Extensions
Edited and extended by M. F. Hasler, Apr 04 2009
Comments