A004166 -id:A004166 - cp's OEIS frontend

A118733 Numbers k such that 3^k has even digit sum.

6, 7, 8, 12, 15, 19, 23, 24, 28, 29, 33, 37, 38, 40, 42, 43, 44, 46, 47, 49, 50, 54, 55, 56, 57, 58, 64, 67, 70, 71, 72, 75, 77, 82, 83, 84, 85, 88, 90, 93, 94, 95, 96, 97, 102, 104, 106, 107, 109, 110, 111, 112, 116, 120, 122, 125, 126, 129, 132, 135, 136, 138, 139

Offset: 1

Zak Seidov, May 22 2006

Clinton Walker, Table of n, a(n) for n = 1..10000

Cf. A004166, A118730, A118734.

Select[Range[140], Mod[ Plus @@ IntegerDigits[3^# ], 2] == 0 &] (* Ray Chandler, Jun 10 2006 *)
Select[Range[150],EvenQ[Total[IntegerDigits[3^#]]]&] (* Harvey P. Dale, Mar 12 2013 *)

from gmpy2 import digits
def ok(n): return sum(map(int, digits(3**n)))&1 == 0
print([k for k in range(140) if ok(k)]) # Michael S. Branicky, May 11 2025

A239935 Numbers k such that DigitSum(3^k) > DigitSum(3^(k+1)).

Original entry on oeis.org

11, 14, 15, 18, 20, 27, 29, 31, 34, 38, 41, 43, 47, 48, 50, 53, 54, 58, 59, 63, 64, 65, 67, 69, 71, 72, 75, 77, 79, 83, 88, 90, 94, 98, 99, 102, 103, 107, 109, 112, 114, 118, 119, 123, 125, 131, 132, 134, 136, 139, 141, 142, 146, 150, 154, 159, 161, 164, 167

Offset: 1

Oliver Bel, Mar 29 2014

			For k=11, we have DigitSum(3^11) = 27 > 18 = DigitSum(3^12).

Robert Israel, Table of n, a(n) for n = 1..8009

Cf. A004166, A007953.

N:= 1000: # to get the first N terms
threen:= 3:
digsum:= 3:
count:= 0:
for n from 1 while count < N do
  threen:= 3*threen;
  oldsum:= digsum;
  digsum:= convert(convert(threen,base,10),`+`);
  if oldsum > digsum then
     count:= count+1;
     A239935[count]:= n;
  fi
od: # Robert Israel, Apr 18 2014

lis = Table[Total[IntegerDigits[3^n, 10]], {n, 1, 100}];
Flatten[Position[Greater @@@ Partition[lis, 2, 1], True]]

isok(k) = sumdigits(3^k) > sumdigits(3^(k+1)); \\ Michel Marcus, Jul 03 2021

More terms from Jon E. Schoenfield, Mar 29 2014

A358509 Sum of decimal digits of (3^n - 1)/2 (A003462).

Original entry on oeis.org

0, 1, 4, 4, 4, 4, 13, 13, 13, 22, 22, 31, 22, 31, 31, 31, 22, 31, 31, 31, 31, 31, 49, 49, 40, 40, 49, 67, 58, 58, 58, 76, 58, 76, 85, 85, 85, 85, 94, 85, 85, 94, 103, 103, 85, 94, 103, 112, 103, 112, 130, 130, 94, 121, 112, 112, 121, 103, 103, 121, 112, 121, 121, 139, 121, 148, 121, 157, 157, 157, 157, 175, 157, 157

Offset: 0

Paul Curtz, Nov 20 2022

a(n) == 4 (mod 9) for n >= 2. - Robert Israel, Nov 21 2022

			For n=5, (3^n - 1)/2 = 121 so that a(5) = 1+2+1 = 4.

Michel Marcus, Table of n, a(n) for n = 0..5000

Cf. A003462, A007953, A017209.

Cf. A004166 (of 3^n).

seq(convert(convert((3^n-1)/2,base,10),`+`),n=0..100); # Robert Israel, Nov 21 2022

a[n_] := Total[IntegerDigits[(3^n - 1)/2]]; Array[a, 100, 0] (* Amiram Eldar, Nov 20 2022 *)

a(n) = sumdigits((3^n - 1)/2); \\ Michel Marcus, Nov 20 2022

def A358509(n): return sum(map(int,str((3**n-1)>>1))) # Chai Wah Wu, Nov 21 2022

a(n) = A007953(A003462(n)).

A286512 Numbers N for which there is k > 0 such that sum of digits(N^k) = N, but the least such k is larger than the least k for which sum of digits(N^k) > N*11/10.

Original entry on oeis.org

17, 31, 63, 86, 91, 103, 118, 133, 155, 157, 211, 270, 290, 301, 338, 352, 421, 432, 440, 441, 450, 478, 513, 533, 693, 853, 1051, 1237, 1363, 1459, 1526, 1665, 2781

Offset: 1

M. F. Hasler, May 18 2017

The set of these numbers appears to be finite, and probably 2781 is its largest element.
The motivation for this sequence is the study of the behavior of the sum of digits of powers of a given number. Statistically, sumdigits(n^k) ~ 4.5*log_10(n')*k (where n' = n without trailing 0's), but typically fluctuations of some percent persist up to large values of k. (Cf. the graph of sequences n^k cited in the cross-references.)
The ratio of 11/10 is somewhat arbitrary, but larger ratios of the simple form (1 + 1/m) yield quite small subsets of this sequence (for m=2 the only element is 118, for m=3 the set is {31, 86, 118}, for m=1 it is empty), and smaller ratios yield much larger (possibly infinite?) sets. Also, the condition can be written sumdigits(N^k)-N > N/10, and 10 is the base we are using.
To compute the sequence A247889 we would like to have a rule telling us when we can stop the search for an exponent. It appears that sumdigits(N^k) >= 2*N is a limit that works for all N; the present sequence gives counterexamples to the (r.h.s.) limit of 1.1*N. The above comment mentioned the counterexamples {118} resp. {31, 86, 118}) for limits N*3/2 and N*4/3.

Cf. A247889, A007953.

Cf. sum of digits of k^n: A001370 (k=2), A004166 (k=3), A065713 (k=4), A066001 (k=5), A066002 (k=6), A066003 (k=7), A066004 (k=8), A065999 (k=9), A066005 (k=11), A066006 (k=12). (In these sequences, k is fixed and n is the index/exponent; in the present sequence it's the opposite and therefore the names k <-> n are exchanged.)

for(n=1,5000,A247889(n)&&!A247889(n,n*11\10)&&print1(n",")) \\ Here, A247889() is a variant of the function computing that sequence which accepts as second optional argument a limit m, stopping the search for the exponent as soon as the digital sum of n^k exceeds m.

A287058 Sum of decimal digits of 118^n.

Original entry on oeis.org

1, 10, 19, 19, 55, 64, 55, 64, 82, 91, 109, 100, 109, 181, 118, 145, 127, 163, 154, 172, 154, 190, 226, 190, 208, 217, 271, 289, 253, 280, 298, 307, 334, 289, 334, 280, 361, 343, 334, 379, 406, 406, 379, 424, 379, 424, 415, 406, 523, 433, 478

Offset: 0

M. F. Hasler, May 18 2017

118 is exceptional in the sense that it appears to be the only number m for which the smallest k such that sumdigits(m^k) = m occurs after the smallest k such that sumdigits(m^k) > m*3/2. If this last limit is decreased to m*4/3, then 31 and 86 also have this property. It appears that no number has this property if the limit is increased to 2m, see also A247889.

It is also remarkable that many values in the sequence are repeated (19, 55, 64, 109, 190, 154, 280, 289, 334 (3 times), 379, 406, 424, ...), while most other numbers never appear.

Cf. A007953, A247889.

Cf. sum of digits of k^n: A001370 (k=2), A004166 (k=3), A065713 (k=4), A066001(k=5), A066002 (k=6), A066003 (k=7), A066004 (k=8), A065999 (k=9), A066005 (k=11), A066006 (k=12).

Total[IntegerDigits[#]]&/@NestList[118#&,1,50] (* Harvey P. Dale, Feb 24 2022 *)

PARI
```
a(n)=sumdigits(118^n)
```

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A118733 Numbers k such that 3^k has even digit sum.

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A239935 Numbers k such that DigitSum(3^k) > DigitSum(3^(k+1)).

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A358509 Sum of decimal digits of (3^n - 1)/2 (A003462).

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A286512 Numbers N for which there is k > 0 such that sum of digits(N^k) = N, but the least such k is larger than the least k for which sum of digits(N^k) > N*11/10.

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A287058 Sum of decimal digits of 118^n.

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