

A226636


Numbers whose base3 sum of digits is 3.


11



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
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OFFSET

1,1


COMMENTS

All of the entries are odd.
In general, the set of numbers with sum of baseb digits equal to b is a subset of { (b1)*k + 1; k = 2, 3, 4, ... }.  M. F. Hasler, Dec 23 2016


LINKS



FORMULA

a(k^3/6 + k^2 + 5*k/6 + j) = 3^(k+1) + A055235(j1) for 1 <= j <= k^2/2+5*k/2+2.  Robert Israel, Jun 05 2018


EXAMPLE

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.


MAPLE

N:= 10: # for all terms < 3^(N+1)
[seq(seq(seq(3^a+3^b+3^c, c=0..`if`(b=a, b1, 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 *)


PROG

(Sage) [i for i in [0..1000] if sum(Integer(i).digits(base=3))==3]
(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 + ((c1)%(b1)+1)*b**((c1)//(b1))1
c += d; n //= b; w += 1
def A226636gen(sod=3, base=3): # generator of terms for any sod, base
an = (sod%(base1)+1)*base**(sod//(base1))1
while True: yield an; an = nextsod(an, base)


CROSSREFS



KEYWORD

nonn,base


AUTHOR



STATUS

approved



