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A226636
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Numbers whose base-3 sum of digits is 3.
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11
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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
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1,1
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COMMENTS
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All of the entries are odd.
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
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LINKS
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FORMULA
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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
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EXAMPLE
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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.
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MAPLE
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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
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MATHEMATICA
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Select[Range@ 757, Total@ IntegerDigits[#, 3] == 3 &] (* Michael De Vlieger, Dec 23 2016 *)
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PROG
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(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 + ((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)
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CROSSREFS
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KEYWORD
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nonn,base
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AUTHOR
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STATUS
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approved
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