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a(n-2) = a(n-6) + 5*(1+2*n) with a(0)=0, a(1)=2, a(2)=7, a(3)=15 for n>=4.
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%I #17 Mar 18 2020 13:01:37

%S 0,2,7,15,25,37,52,70,90,112,137,165,195,227,262,300,340,382,427,475,

%T 525,577,632,690,750,812,877,945,1015,1087,1162,1240,1320,1402,1487,

%U 1575,1665,1757,1852,1950,2050,2152,2257

%N a(n-2) = a(n-6) + 5*(1+2*n) with a(0)=0, a(1)=2, a(2)=7, a(3)=15 for n>=4.

%C a(-2)=2, a(-1)=0. 4 evens followed by 4 odds.

%C Last digit is only 0, 2, 5, 7.

%C The vertical spoke S-N of the pentagonal spiral for A004526.

%C 37

%C 37 25 25

%C 36 24 15 15 26

%C 36 24 14 7 8 16 26

%C 35 23 14 7 2 3 8 16 27

%C 35 23 13 6 2 0 0 3 9 17 27

%C 34 22 13 6 1 1 4 9 17 28

%C 34 22 12 5 5 4 10 18 28

%C 33 21 12 11 11 10 18 29

%C 33 21 20 20 19 19 29

%C 32 32 31 31 30 30

%C Rank of multiples of 10: 0, 7, 8, 15, 16, ... = A047521. Compare to A154260 in the formula.

%H Colin Barker, <a href="/A332495/b332495.txt">Table of n, a(n) for n = 0..1000</a>

%H <a href="/index/Rec#order_05">Index entries for linear recurrences with constant coefficients</a>, signature (3,-4,4,-3,1).

%F a(-1-n) = a(n).

%F a(2*n) + a(1+2*n) = 2, 22, 62, ... = A273366(n).

%F Second differences give the sequence of period 4: repeat [3, 3, 2, 2].

%F From _Colin Barker_, Feb 14 2020: (Start)

%F G.f.: x*(2 + x + 2*x^2) / ((1 - x)^3*(1 + x^2)).

%F a(n) = 3*a(n-1) - 4*a(n-2) + 4*a(n-3) - 3*a(n-4) + a(n-5) for n>4.

%F (End)

%F Multiples of 10: 10*(0, 7, 9, 30, 34, ... = A154260).

%F 4*a(n) = A087960(n) +5*n -1 +5*n^2. - _R. J. Mathar_, Feb 28 2020

%t CoefficientList[Series[x (2 + x + 2 x^2)/((1 - x)^3*(1 + x^2)), {x, 0, 42}], x] (* _Michael De Vlieger_, Feb 14 2020 *)

%o (PARI) concat(0, Vec(x*(2 + x + 2*x^2) / ((1 - x)^3*(1 + x^2)) + O(x^40))) \\ _Colin Barker_, Feb 14 2020

%Y Cf. A004526, A033429, A062786, A168668, A135706, A147874, 2*A147875 (all in the spiral).

%Y Cf. A005408, A047521, A154260

%K nonn,easy

%O 0,2

%A _Paul Curtz_, Feb 14 2020