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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.
4
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
1,1
COMMENTS
The sequence was initially defined as the first column of the triangle A093915, constructed by trial and error. It is however easy to prove that the sum of the r-th row of A093915, A093917(r), equals twice A006003(r) when r is odd, and three times A006003(r) when r is even. Given the expression of the row sum A093917(r) in terms of the first element a(r), one obtains the explicit formula for a(r). - M. F. Hasler, Apr 04 2009
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
MATHEMATICA
LinearRecurrence[{0, 3, 0, -3, 0, 1}, {2, 7, 9, 24, 24, 53}, 80] (* Harvey P. Dale, Nov 24 2017 *)
PROG
(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
(Magma) [(n*(5*n-2) + (-1)^n*(n^2+1) + 7)/4: n in [1..70]]; // G. C. Greubel, Dec 30 2021
(SageMath) [(5*n^2 -2*n +7 +(-1)^n*(n^2 +1))/4 for n in (1..70)] # G. C. Greubel, Dec 30 2021
CROSSREFS
KEYWORD
nonn,easy
AUTHOR
Amarnath Murthy, Apr 25 2004
EXTENSIONS
Edited and extended by M. F. Hasler, Apr 04 2009
STATUS
approved