A226636 Numbers whose base-3 sum of digits is 3.

5, 7, 11, 13, 15, 19, 21, 29, 31, 33, 37, 39, 45, 55, 57, 63, 83, 85, 87, 91, 93, 99, 109, 111, 117, 135, 163, 165, 171, 189, 245, 247, 249, 253, 255, 261, 271, 273, 279, 297, 325, 327, 333, 351, 405, 487, 489, 495, 513, 567, 731, 733, 735, 739, 741, 747, 757

Offset: 1

Author

Tom Edgar, Aug 31 2013

Keywords

• nonn
• base

Comments

All of the entries are odd.

Subsequence of A005408. - Michel Marcus, Sep 03 2013

In general, the set of numbers with sum of base-b digits equal to b is a subset of { (b-1)*k + 1; k = 2, 3, 4, ... }. - M. F. Hasler, Dec 23 2016

Examples

			The ternary expansion of 5 is (1,2), which has sum of digits 3.
The ternary expansion of 31 is (1,0,0,2), which has sum of digits 3.
10 is not on the list since the ternary expansion of 10 is (1,0,1), which has sum of digits 2 not 3.
		

Links

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

Crossrefs

Cf. A018900, A023694, A055235, A187813.

Cf. A226969 (b = 4), A227062 (b = 5), A227080 (b = 6), A227092 (b = 7), A227095 (b = 8), A227238 (b = 9), A052224 (b = 10).

Programs

Maple

N:= 10: # for all terms < 3^(N+1)
[seq(seq(seq(3^a+3^b+3^c, c=0..`if`(b=a, b-1,b)),b = 0..a),a=0..N)]; # Robert Israel, Jun 05 2018

Mathematica

Select[Range@ 757, Total@ IntegerDigits[#, 3] == 3 &] (* Michael De Vlieger, Dec 23 2016 *)

PARI

select( is(n)=sumdigits(n,3)==3, [1..999]) \\ M. F. Hasler, Dec 23 2016

Python

from itertools import islice
def nextsod(n, base):
    c, b, w = 0, base, 0
    while True:
        d = n%b
        if d+1 < b and c:
            return (n+1)*b**w + ((c-1)%(b-1)+1)*b**((c-1)//(b-1))-1
        c += d; n //= b; w += 1
def A226636gen(sod=3, base=3): # generator of terms for any sod, base
    an = (sod%(base-1)+1)*base**(sod//(base-1))-1
    while True: yield an; an = nextsod(an, base)
print(list(islice(A226636gen(), 57))) # Michael S. Branicky, Jul 10 2022, generalizing the code by M. F. Hasler in A052224

Sage

[i for i in [0..1000] if sum(Integer(i).digits(base=3))==3]

Formula

a(k^3/6 + k^2 + 5*k/6 + j) = 3^(k+1) + A055235(j-1) for 1 <= j <= k^2/2+5*k/2+2. - Robert Israel, Jun 05 2018