A332495 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.

0, 2, 7, 15, 25, 37, 52, 70, 90, 112, 137, 165, 195, 227, 262, 300, 340, 382, 427, 475, 525, 577, 632, 690, 750, 812, 877, 945, 1015, 1087, 1162, 1240, 1320, 1402, 1487, 1575, 1665, 1757, 1852, 1950, 2050, 2152, 2257

Offset: 0

Author

Paul Curtz, Feb 14 2020

Keywords

• nonn
• easy

Comments

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

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

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

37

37 25 25

36 24 15 15 26

36 24 14 7 8 16 26

35 23 14 7 2 3 8 16 27

35 23 13 6 2 0 0 3 9 17 27

34 22 13 6 1 1 4 9 17 28

34 22 12 5 5 4 10 18 28

33 21 12 11 11 10 18 29

33 21 20 20 19 19 29

32 32 31 31 30 30

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

Links

• Colin Barker, Table of n, a(n) for n = 0..1000
• Index entries for linear recurrences with constant coefficients, signature (3,-4,4,-3,1).

Crossrefs

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

Cf. A005408, A047521, A154260

Programs

Mathematica

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

PARI

concat(0, Vec(x*(2 + x + 2*x^2) / ((1 - x)^3*(1 + x^2)) + O(x^40))) \\ Colin Barker, Feb 14 2020

Formula

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

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

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

From Colin Barker, Feb 14 2020: (Start)

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

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

(End)

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

4*a(n) = A087960(n) +5*n -1 +5*n^2. - R. J. Mathar, Feb 28 2020