A248098 -id:A248098 - cp's OEIS frontend

A213967 a(n)=n for n<=3; thereafter a(n)=a(n-1)+a(n-2)+a(n-3)+1.

0, 1, 2, 3, 7, 13, 24, 45, 83, 153, 282, 519, 955, 1757, 3232, 5945, 10935, 20113, 36994, 68043, 125151, 230189, 423384, 778725, 1432299, 2634409, 4845434, 8912143, 16391987, 30149565, 55453696, 101995249, 187598511, 345047457, 634641218, 1167287187

Offset: 0

N. J. A. Sloane, Jun 30 2012

Atanassov, K. T.; Atanassova, V.; Shannon, A. G.; Turner, J. C. New visual perspectives on Fibonacci numbers. With a foreword by A. F. Horadam. World Scientific Publishing Co., Inc., River Edge, NJ, 2002. xvi+313 pp. ISBN: 981-238-134-1 MR1932564 (2003h:11015). See p. 68.

Bruno Berselli, Table of n, a(n) for n = 0..1000
Index entries for linear recurrences with constant coefficients, signature (2,0,0,-1).

Cf. A001590.

Cf. A248098.

a213967 n = a213967_list !! n
a213967_list = 0 : xs where
               xs = 1 : 2 : 3 : map (+ 1)
                    (zipWith3 (((+) .) . (+)) xs (tail xs) (drop 2 xs))
-- Reinhard Zumkeller, Dec 29 2014

[n le 3 select n else Self(n)+Self(n-1)+Self(n-2)+1: n in [0..35]]; // Bruno Berselli, Jul 02 2012

f:=proc(n) option remember; if n <= 3 then n else f(n-1)+f(n-2)+f(n-3)+1; fi; end:
seq(f(n),n=0..60);

Join[{0}, LinearRecurrence[{2, 0, 0, -1}, {1, 2, 3, 7}, 40]] (* Jean-François Alcover, Feb 13 2018 *)
nxt[{a_,b_,c_}]:={b,c,a+b+c+1}; Join[{0},NestList[nxt,{1,2,3},40][[All,1]]] (* Harvey P. Dale, Sep 07 2020 *)

G.f.: x*(1-x^2+x^3)/(1-2*x+x^4). - Bruno Berselli, Jul 02 2012

A360464 a(n) = a(n-1) + a(n-2) - a(n-3) + gcd(a(n-1), a(n-3)), with a(1) = a(2) = a(3) = 1.

Original entry on oeis.org

1, 1, 1, 2, 3, 5, 7, 10, 17, 21, 29, 34, 43, 49, 59, 66, 77, 85, 97, 106, 119, 129, 143, 154, 169, 193, 209, 234, 251, 277, 295, 322, 341, 369, 389, 418, 439, 469, 491, 522, 545, 577, 601, 634, 659, 693, 719, 754, 781, 817, 845, 882, 911, 949, 979, 1018, 1049

Offset: 1

Jack Braxton, Feb 08 2023

All terms beyond a(458) are divisible by 5. - Jack Braxton, Feb 14 2023
From Robert Israel, Feb 15 2023: (Start)
a(n) is divisible by 25 for n >= 8857.
a(n) is divisible by 125 for n >= 8861.
a(n) is divisible by 625 for n >= 8945.
a(n) is divisible by 1875 for n >= 9060.
a(n) is divisible by 5625 for n >= 9064.
Do there exist N > 9064 and m > 5625 such that a(n) is divisible by m for n >= N? If so, N >= 2*10^7. (End)
From Pontus von Brömssen, Feb 17 2023: (Start)
(Answer to the question above.) Yes:
  a(n) has an additional factor  5 for n >=  64423404 (so a(n) is divisible by 28125);
  a(n) has an additional factor  5 for n >=  64423410;
  a(n) has an additional factor  3 for n >=  64424073;
  a(n) has an additional factor 21 for n >=  64424144;
  a(n) has an additional factor  3 for n >=  64428745;
  a(n) has an additional factor  7 for n >=  64428748;
  a(n) has an additional factor  3 for n >=  64428756;
  a(n) has an additional factor  3 for n >=  64428821;
  a(n) has an additional factor  3 for n >=  64514757;
  a(n) has an additional factor  5 for n >=  64514783;
  a(n) has an additional factor  3 for n >= 797299454;
  a(n) has an additional factor  3 for n >= 797299480;
  a(n) has an additional factor  5 for n >= 797299487;
  a(n) has an additional factor  3 for n >= 797299490;
  a(n) has an additional factor  5 for n >= 797299652;
  a(n) has an additional factor  3 for n >= 797299667;
  a(n) has an additional factor  7 for n >= 797299846;
  a(n) has an additional factor  3 for n >= 797299933.
The index for which the next additional factor occurs (if it exists) is larger than 2*10^10.
(End)

			a(5) = 2 + 1 - 1 + gcd(2, 1) = 3.

Robert Israel, Table of n, a(n) for n = 1..10000

Cf. A083658, A248098.

A:= Vector(200):
A[1]:= 1: A[2]:= 1: A[3]:= 1:
for n from 4 to 200 do
  A[n]:= A[n-1] + A[n-2] - A[n-3] + igcd(A[n-1],A[n-3])
od:
convert(A,list); # Robert Israel, Feb 15 2023

a[1] = a[2] = a[3] = 1; a[n_] := a[n] = a[n-1] + a[n-2] - a[n-3] + GCD[a[n-1], a[n-3]]; Array[a, 100] (* Amiram Eldar, Feb 08 2023 *)

from math import gcd
a = [0, 1, 1, 1]
[a.append(a[n-1]+a[n-2]-a[n-3]+gcd(a[n-1], a[n-3])) for n in range(4, 58)]
print(a[1:]) # Michael S. Branicky, Feb 09 2023

a(n) = a(n-1) + a(n-2) - a(n-3) + gcd(a(n-1), a(n-3)).

cp's OEIS Frontend

A213967 a(n)=n for n<=3; thereafter a(n)=a(n-1)+a(n-2)+a(n-3)+1.

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