A084229 Let b(1)=1, b(2)=2, b(n) = sum of digits of b(1)+b(2)+b(3)+...+b(n-1), sequence gives values of n such that b(n)=3.

3, 5, 7, 9, 17, 19, 27, 29, 87, 95, 97, 159, 591, 599, 601, 663, 1143, 4609, 4617, 4619, 4681, 5161, 8993, 13165, 38277, 38279, 38341, 38821, 42653, 46825, 75043, 79223, 327015, 327023, 327025, 327087, 327567, 331399, 335571, 363789, 367969, 642981, 647153, 2847029, 2847031, 2847093, 2847573

Offset: 1

Author

Benoit Cloitre, Jun 21 2003

Keywords

• base
• nonn

Comments

The {b(n)} sequence is A084228. - N. J. A. Sloane, Jun 26 2014

Note that b(k)==0 (mod 3) for n>2.

Links

• Robert G. Wilson v, Table of n, a(n) for n = 1..107

Crossrefs

Cf. A065075, A084228.

Programs

Mathematica

k = 3; lst = {}; a = 3; While[k < 100000001, b = a + Total@ IntegerDigits@ a; If[b == a + 3, AppendTo[lst, k]; Print@ k]; a = b; k++]; lst (* Robert G. Wilson v, Jun 27 2014 *)

PARI

upto(n)={my(L=List(), s=3, k=3); while(k<=n, my(t=sumdigits(s)); if(t==3, listput(L,k)); s+=t; k++); Vec(L)} \\ Andrew Howroyd, Oct 16 2024

Formula

Conjecture : a(n)/n^3 is bounded.

Extensions

a(23) onward from Robert G. Wilson v, Jun 27 2014