A026475 a(1)=1, a(2)=3, otherwise a(n) = least positive integer > a(n-1) and not a(i) + a(j) + a(k) for 1 <= i < j < k <= n.

1, 3, 4, 5, 6, 7, 19, 20, 21, 22, 36, 37, 38, 39, 53, 54, 55, 69, 70, 71, 72, 86, 87, 88, 102, 103, 104, 105, 119, 120, 121, 135, 136, 137, 138, 152, 153, 154, 168, 169, 170, 171, 185, 186, 187, 201, 202, 203, 204, 218, 219, 220, 234, 235, 236, 237, 251, 252, 253

Offset: 1

Author

Clark Kimberling

Keywords

• nonn
• easy

Comments

Without specifying a(2)=3, a(2) would be 2 and sequence would be A026471. - Robert Israel, Aug 27 2018

Links

• Robert Israel, Table of n, a(n) for n = 1..10000
• Wieb Bosma, Rene Bruin, Robbert Fokkink, Jonathan Grube, Anniek Reuijl, and Thian Tromp, Using Walnut to solve problems from the OEIS, arXiv:2503.04122 [math.NT], 2025.
• Index entries for linear recurrences with constant coefficients, signature (1,0,0,0,0,0,1,-1).

Crossrefs

Cf. A026471.

Programs

Maple

1, 3, 4, 5, 6, 7, 19, seq(op([20, 21, 22, 36, 37, 38, 39]+k*[33$7]),k=0..10); # Robert Israel, Aug 30 2018

Mathematica

LinearRecurrence[{1,0,0,0,0,0,1,-1},{1,3,4,5,6,7,19,20,21,22,36,37,38,39,53},60] (* Harvey P. Dale, Jan 04 2022 *)

PARI

Vec(x*(1 + 2*x + x^2 + x^3 + x^4 + x^5 + 12*x^6 - x^8 + 13*x^10 - 11*x^13 + 13*x^14) / ((1 - x)^2*(1 + x + x^2 + x^3 + x^4 + x^5 + x^6)) + O(x^40)) \\ Colin Barker, Oct 10 2019

Formula

a(n+7) = a(n) + 33 for n >= 8. - Robert Israel, Aug 27 2018

From Colin Barker, Oct 10 2019: (Start)

G.f.: x*(1 + 2*x + x^2 + x^3 + x^4 + x^5 + 12*x^6 - x^8 + 13*x^10 - 11*x^13 + 13*x^14) / ((1 - x)^2*(1 + x + x^2 + x^3 + x^4 + x^5 + x^6)).

a(n) = a(n-1) + a(n-7) - a(n-8) for n>15.

(End)

Extensions

More terms from Larry Reeves (larryr(AT)acm.org), Sep 25 2000