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a(n) = 5n + 26*floor(n/5).
1

%I #7 Apr 02 2017 20:59:06

%S 0,5,10,15,20,51,56,61,66,71,102,107,112,117,122,153,158,163,168,173,

%T 204,209,214,219,224,255,260,265,270,275,306,311,316,321,326,357,362,

%U 367,372,377,408,413,418,423,428,459,464,469,474,479,510,515,520,525,530,561,566

%N a(n) = 5n + 26*floor(n/5).

%H Robert Israel, <a href="/A137935/b137935.txt">Table of n, a(n) for n = 0..10000</a>

%H <a href="/index/Rec#order_06">Index entries for linear recurrences with constant coefficients</a>, signature (1,0,0,0,1,-1).

%F a(n) = 5n + 26*floor(n/5) = 5n + 26*A002266(n)

%F G.f.: (5*x+5*x^2+5*x^3+5*x^4+31*x^5)/(1-x-x^5+x^6). - _Robert Israel_, Apr 02 2017

%e a(0) = 5(0) + 26*floor(0/5) = 0

%e a(3) = 5(3) + 26*floor(3/5) = 15

%p seq(5*n + 26*floor(n/5), n=0..200); # _Robert Israel_, Apr 02 2017

%o (Python) a = lambda n: 5*n + 26*floor(n/5)

%Y Cf. A002266.

%K nonn,easy

%O 0,2

%A _William A. Tedeschi_, Mar 06 2008