A177186 a(n+1) = a(n) + p, where p is the largest prime dividing a(n) but not a(n-1), or 1 if there is no such prime.

1, 2, 4, 5, 10, 12, 15, 20, 22, 33, 36, 38, 57, 60, 65, 78, 81, 82, 123, 126, 133, 152, 154, 165, 170, 187, 198, 201, 268, 270, 275, 286, 299, 322, 329, 376, 378, 385, 396, 399, 418, 429, 442, 459, 462, 473, 516, 519, 692, 694, 1041, 1044, 1073, 1110, 1115, 1338

Offset

1,2

Comments

Setting a(1) = 1 is arbitrary; since 1 has no prime divisors, we must then have p = 1 => a(2) = 2, without reference to a(0).

There are three cases where p = 1: n=1 (a(n)=1); n=3 (a(n)=4); n=17 (a(n)=81); and no others through n = 10000. Except that there cannot be such cases consecutively, p=1 iff a(n) is a prime power.

Links

Harvey P. Dale, Table of n, a(n) for n = 1..1000

Example

After a(9)=22, a(10)=33, the prime divisors of a(10) are 3 and 11; 11 divides 22, so p=3, and a(11)=36.

Mathematica

p[n1_, n2_] := If[pp = Complement[Transpose[FactorInteger[n2]][[1]],
Transpose[FactorInteger[n1]][[1]]]; pp == {}, 1, Last[pp]]; a[1] = 1; a[2] = 2; a[n_] := a[n] = a[n-1] + p[a[n-2], a[n-1]]; Table[a[n], {n, 56}] (* Jean-François Alcover, Sep 02 2011 *)
nxt[{a_, b_}]:={b, b+Max[1, Complement[FactorInteger[b][[All, 1]], FactorInteger[ a] [[All, 1]]]]}; NestList[nxt, {1, 2}, 60][[All, 1]] (* Harvey P. Dale, Dec 17 2022 *)

Prog

(PARI) invec(v, x)=for(i=1, #v, if(v[i]==x, return(1))); 0
lastnotin(vi, vx, dft)=forstep(i=#vi, 1, -1, if(!invec(vx, vi[i]), return(vi[i]))); dft
al(n)=local(r); r=vector(n); r[1]=1; r[2]=2; for(k=3, n, r[k]=r[k-1]+lastnotin(factor(r[k-1])[, 1]~, factor(r[k-2])[, 1]~, 1)); r

Crossrefs

Cf. A076271, A060735, A002620.

Sequence in context: A047611 A325095 A120491 * A260385 A355149 A022944

Adjacent sequences: A177183 A177184 A177185 * A177187 A177188 A177189

Keyword

nonn

Author

Franklin T. Adams-Watters, May 04 2010

Status

approved