A281151 a(n) = floor(4*n*(n+1)/5).
0, 1, 4, 9, 16, 24, 33, 44, 57, 72, 88, 105, 124, 145, 168, 192, 217, 244, 273, 304, 336, 369, 404, 441, 480, 520, 561, 604, 649, 696, 744, 793, 844, 897, 952, 1008, 1065, 1124, 1185, 1248, 1312, 1377, 1444, 1513, 1584, 1656, 1729, 1804, 1881, 1960, 2040, 2121, 2204, 2289
Offset: 0
Links
- Bruno Berselli, Table of n, a(n) for n = 0..1000
- Index entries for linear recurrences with constant coefficients, signature (2,-1,0,0,1,-2,1).
Crossrefs
Programs
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Magma
[4*n*(n+1) div 5: n in [0..60]];
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Mathematica
Table[Floor[4 n (n + 1)/5], {n, 0, 60}] -
Maxima
makelist(floor(4*n*(n+1)/5), n, 0, 60);
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PARI
vector(60, n, n--; floor(4*n*(n+1)/5))
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Python
[int(4*n*(n+1)/5) for n in range(60)]
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Sage
[floor(4*n*(n+1)/5) for n in range(60)]
Formula
O.g.f.: x*(1 + x^2)*(1 + x)^2/((1 - x)^3*(1 + x + x^2 + x^3 + x^4)).
a(n) = a(-n-1) = 2*a(n-1) - a(n-2) + a(n-5) - 2*a(n-6) + a(n-7) = a(n-5) + 8*(n-2).
a(5*k+r) = 20*k^2 + 4*(2*r+1)*k + r^2, where 0 <= r <= 4. Example: for r=3, a(5*k+3) = (2*k+1)*(10*k+9), which gives: 9, 57, 145, 273, 441, 649 etc. Also, a(n) belongs to A047462, in fact: for r = 0 or 4, a(n) == 0 (mod 8); for r = 1 or 3, a(n) == 1 (mod 8); for r = 2, a(n) == 4 (mod 8).
a(n) = a(-n) + A047462(n).
a(n) = n^2 - floor((n-2)^2/5).
Comments