A323178 - OEIS

%I #21 Feb 18 2019 16:43:32

%S 1,101,401,901,1601,2501,3601,4901,6401,8101,10001,12101,14401,16901,

%T 19601,22501,25601,28901,32401,36101,40001,44101,48401,52901,57601,

%U 62501,67601,72901,78401,84101,90001,96101,102401,108901

%N a(n) = 1 + 100*n^2 for n >= 0.

%C Terms of A261327 ending in 1 (01 for n > 0.)

%C a(n) mod 9 = period 9: repeat [1, 2, 5, 1, 8, 8, 1, 5, 2] = A275704(n+3).

%C (Analogous sequence: b(n) = 29 + 100*n*(n+1) = A261327(A017329) = 29, 229, 629, ... .)

%F a(n) = A261327(A008602(n)).

%F Recurrence: a(n) = 3*a(n-1) - 3*a(n-2) + a(n-3) for n > 2 with initial values a(0) = 1, a(1) = 101 and a(2) = 401.

%F From _Stefano Spezia_, Jan 06 2019: (Start)

%F O.g.f.: (-1 - 98*x - 101*x^2)/(-1 + x)^3.

%F E.g.f.: exp(x)*(1 + 100*x + 100*x^2).

%F (End)

%t a[n_] := 1 + 100*n^2 ; Array[a, 50, 0] (* or *)

%t CoefficientList[Series[(-1 - 98 x - 101 x^2)/(-1 + x)^3, {x, 0, 50}], x] (* or *)

%t CoefficientList[Series[E^x (1 + 100 x + 100 x^2), {x, 0, 50}], x]*Table[n!, {n, 0, 50}] (* _Stefano Spezia_, Jan 06 2019 *)

%Y Cf. A000290, A008602 (20*n), A261327, A275704.

%Y Subsequence of A017281.

%K nonn

%O 0,2

%A _Paul Curtz_, Jan 06 2019

%E Corrected and extended (recurrence formula) by _Werner Schulte_, Feb 18 2019