A338482 Least number of centered triangular numbers that sum to n.

1, 2, 3, 1, 2, 3, 4, 2, 3, 1, 2, 3, 4, 2, 3, 4, 5, 3, 1, 2, 3, 4, 2, 3, 4, 5, 3, 4, 2, 3, 1, 2, 3, 4, 2, 3, 4, 2, 3, 4, 2, 3, 4, 5, 3, 1, 2, 3, 4, 2, 3, 4, 5, 3, 4, 2, 3, 4, 5, 3, 4, 2, 3, 1, 2, 3, 4, 2, 3, 4, 5, 3, 4, 2, 3, 4, 2, 3, 4, 5, 3, 4, 2, 3, 1, 2, 3, 4, 2, 3, 4, 2, 3, 4, 2, 3, 4, 5, 3, 4, 5

Offset: 1

Author

Ilya Gutkovskiy, Oct 29 2020

Keywords

• nonn

Comments

It appears that a(n) = 3 for n == 0 (mod 3), 1 <= a(n) <= 4 for n == 1 (mod 3), and 2 <= a(n) <= 5 for n == 2 (mod 3). - Robert Israel, Nov 13 2020

Links

• Robert Israel, Table of n, a(n) for n = 1..10000
• Eric Weisstein's World of Mathematics, Centered Triangular Number
• Index entries for sequences related to centered polygonal numbers

Crossrefs

Cf. A005448, A061336, A101412, A104246, A234533, A338484.

Programs

Maple

f:= proc(n) option remember; local r,i;
    r:= sqrt(24*n-15)/6+1/2;
    if r::integer then return 1 fi;
    1+min(seq(procname(n-(3*i*(i-1)/2+1)),i=1..floor(r)))
end proc:
map(f, [$1..200]); # Robert Israel, Nov 13 2020

Mathematica

f[n_] := f[n] = Module[{r}, r = Sqrt[24n-15]/6+1/2; If[IntegerQ[r], Return[1]]; 1+Min[Table[f[n-(3i*(i-1)/2+1)], {i, 1, Floor[r]}]]];
Map[f, Range[200]] (* Jean-François Alcover, Sep 16 2022, after Robert Israel *)