A336535 a(n) = (m(n)^2 + 3)*(m(n)^2 + 7)/32, where m(n) = 2*n - 1.
1, 6, 28, 91, 231, 496, 946, 1653, 2701, 4186, 6216, 8911, 12403, 16836, 22366, 29161, 37401, 47278, 58996, 72771, 88831, 107416, 128778, 153181, 180901, 212226, 247456, 286903, 330891, 379756, 433846, 493521, 559153, 631126, 709836, 795691, 889111, 990528, 1100386, 1219141, 1347261, 1485226, 1633528, 1792671
Offset: 1
Examples
m(2) = 2*2-1 = 3 and (3^2+3)*(3^2+7)/32 = 6, so 6 is in the sequence.
References
- David M. Burton, Elementary Number Theory, McGraw-Hill (2011), 25.
Links
- Index entries for linear recurrences with constant coefficients, signature (5,-10,10,-5,1).
Programs
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Mathematica
Table[((n^2+3)*(n^2+7))/32, {n,1,100,2}] a=Table[(n(n+1))/2,{n,0,40}];Apply[Join,Map[{(#+1)(2#+1)}&,a]] (* Miguel-Ángel Pérez García-Ortega, May 16 2025 *)
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PARI
a(n)=(1-n+n^2)*(2-n+n^2)/2 \\ Charles R Greathouse IV, Oct 21 2022
Formula
From Stefano Spezia, Jul 25 2020: (Start)
O.g.f.: x*(1 + x + 8*x^2 + x^3 + x^4)/(1 - x)^5.
a(n) = (1 - n + n^2)*(2 - n + n^2)/2.
a(n) = 5*a(n-1) - 10*a(n-2) + 10*a(n-3) - 5*a(n-4) + a(n-5) for n > 5. (End)
Comments