A187093 a(0)=0, a(1)=a(2)=1; thereafter, a(n+1) = n^2 - a(n-1).

0, 1, 1, 3, 8, 13, 17, 23, 32, 41, 49, 59, 72, 85, 97, 111, 128, 145, 161, 179, 200, 221, 241, 263, 288, 313, 337, 363, 392, 421, 449, 479, 512, 545, 577, 611, 648, 685, 721, 759, 800, 841, 881, 923, 968, 1013, 1057, 1103, 1152, 1201, 1249, 1299, 1352, 1405, 1457

Offset: 0

Author

Benjamin Coinsin, Mar 04 2011

Keywords

• nonn

Comments

The original definition was equivalent to: Let S(n) = sum_{i=0..n} a(i), then n^2+a(n)-S(n+1) = S(n-2). This in turn simplifies to the present definition.

Links

• Harvey P. Dale, 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. A194274, A081654.

Programs

Maple

A000034 := proc(n) op(1+(n mod 2),[1,2]) ; end proc:
A187093 := proc(n) (n^2-1+(-1)^floor(n/2)*A000034(n))/2 ;end proc: # R. J. Mathar

Mathematica

LinearRecurrence[{3, -4, 4, -3, 1}, {0, 1, 1, 3, 8}, 60] (* Jean-François Alcover, Mar 30 2020 *)
Join[{0},RecurrenceTable[{a[1]==a[2]==1,a[n+1]==n^2-a[n-1]},a,{n,60}]] (* Harvey P. Dale, Jan 05 2023 *)

PARI

a(n) = (n^2-1+(-1)^(n\2)*(1 + (n % 2)))/2; \\ Michel Marcus, Sep 11 2016

Python

print(0, end=',')       # a(-1)=0
prpr = prev = 1         # a(0)=a(1)=1
for n in range(2, 77):
    print(prpr, end=',')
    curr = n*n - prpr   # a(n) = n^2 - a(n-2)
    prpr = prev
    prev = curr
# from Alex Ratushnyak, Aug 05 2012

Formula

a(n) = (n^2 - 1 + (-1)^floor(n/2) * A000034(n))/2.

G.f.: x*(-1+2*x+x^3-4*x^2) / ( (x^2+1)*(x-1)^3 ).

a(2^(n+1)) = A081654(n). - Anton Zakharov, Sep 13 2016

Extensions

Edited by N. J. A. Sloane, Mar 09 2011

More terms from Alex Ratushnyak, Aug 05 2012