A175435 -id:A175435 - cp's OEIS frontend

A004166 Sum of digits of 3^n.

1, 3, 9, 9, 9, 9, 18, 18, 18, 27, 27, 27, 18, 27, 45, 36, 27, 27, 45, 36, 45, 27, 45, 54, 54, 63, 63, 81, 72, 72, 63, 81, 63, 72, 99, 81, 81, 90, 90, 81, 90, 99, 90, 108, 90, 99, 108, 126, 117, 108, 144, 117, 117, 135, 108, 90, 90, 108, 126, 117, 99

Offset: 0

N. J. A. Sloane

All terms a(n), n > 1, are divisible by 9. - M. F. Hasler, Sep 27 2017

Michel Marcus, Table of n, a(n) for n = 0..10000 (terms 0..1000 from Vincenzo Librandi)
Albert Frank, Solutions of International Contest Of Logical Sequences, 2002 - 2003. (The original Contest page without solutions was removed but remains available on web.archive.org.)

Cf. A067500, A118872, A175435.

Cf. sum of digits of k^n: A001370 (k=2), this sequence (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), A175527 (k=13).

Total[IntegerDigits[#]]&/@(3^Range[0,60]) (* Harvey P. Dale, Mar 03 2013 *)
Table[Total[IntegerDigits[3^n]], {n, 0, 60}] (* Vincenzo Librandi, Oct 08 2013 *)

a(n)=sumdigits(3^n); \\ Michel Marcus, Nov 01 2013

def a(n): return sum(map(int, str(3**n)))
print([a(n) for n in range(61)]) # Michael S. Branicky, Apr 25 2022

a(n) = A007953(A000244(n)). - Michel Marcus, Nov 01 2013

Edited by M. F. Hasler, May 18 2017

A175169 Numbers k that divide the sum of digits of 2^k.

Original entry on oeis.org

1, 2, 5, 70

Offset: 1

N. J. A. Sloane, Dec 03 2010

No other terms <= 200000. - Harvey P. Dale, Dec 16 2010
No other terms <= 1320000. - Robert G. Wilson v, Dec 18 2010
There are almost certainly no further terms.

Sum of digits of k^n mod n: (k=2) A000079, A001370, A175434, A175169; (k=3) A000244, A004166, A175435, A067862; (k=5) A000351, A066001, A175456; (k=6) A000400, A066002, A175457, A067864; (k=7) A000420, A066003, A175512, A067863; (k=8) A062933; (k=13) A001022, A175527, A175528, A175525; (k=21) A175589; (k=167) A175558, A175559, A175560, A175552.

Select[Range[1000],Divisible[Total[IntegerDigits[2^#]],#]&] (* Harvey P. Dale, Dec 16 2010 *)

is(n)=sumdigits(2^n)%n==0 \\ Charles R Greathouse IV, Sep 06 2016

A175525 Numbers k that divide the sum of digits of 13^k.

Original entry on oeis.org

1, 2, 5, 140, 158, 428, 788, 887, 914, 1814, 1895, 1976, 2579, 2732, 3074, 3299, 3641, 4658, 4874, 5378, 5423, 5504, 6170, 6440, 6944, 8060, 8249, 8915, 9041, 9158, 9725, 9824, 10661, 11291, 13820, 15305, 17051, 17393, 18716, 19589, 20876, 21641, 23756, 24188, 25961, 28409, 30632, 31307, 32387, 33215, 34970, 35240, 36653, 36977, 41558, 43970, 44951, 47444, 51764, 52655, 53375, 53852, 54104, 56831, 57506, 59153, 66479, 68063, 73562, 78485, 79286, 87908, 92093, 102029, 106934, 114854, 116321, 134051, 139397, 184037, 192353, 256469, 281381, 301118, 469004

Offset: 1

T. D. Noe, Dec 03 2010

Almost certainly there are no further terms.
Comments from Donovan Johnson on the computation of this sequence, Dec 05 2010 (Start):
The number of digits of 13^k is approximately 1.114*k, so I defined an array d() that is a little bigger than 1.114 times the maximum k value to be checked. The elements of d() each are the value of a single digit of the decimal expansion of 13^k with d(1) being the least significant digit.
It's easier to see how the program works if I start with k = 2.
For k = 1, d(2) would have been set to 1 and d(1) would have been set to 3.
k = 2:
x = 13*d(1) = 13*3 = 39
y = 39\10 = 3 (integer division)
x-y*10 = 39-30 = 9, d(1) is set to 9
x = 13*d(2)+y = 13*1+3 = 16, y is the carry from previous digit
y = 16\10 = 1
x-y*10 = 16-10 = 6, d(2) is set to 6
x = 13*d(3)+y = 13*0+1 = 1, y is the carry from previous digit
y = 1\10 = 0
x-y*10 = 1-0 = 1, d(3) is set to 1
These steps would of course be inside a loop and that loop would be inside a k loop. A pointer to the most significant digit increases usually by one and sometimes by two for each successive k value checked. The number of steps of the inner loop is the size of the pointer. A scan is done from the first element to the pointer element to get the digit sum.
(End)
No other terms < 3*10^6. - Donovan Johnson, Dec 07 2010

Sum of digits of k^n mod n: (k=2) A000079, A001370, A175434, A175169; (k=3) A000244, A004166, A175435, A067862; (k=5) A000351, A066001, A175456; (k=6) A000400, A066002, A175457, A067864; (k=7) A000420, A066003, A175512, A067863; (k=8) A062933; (k=13) A001022, A175527, A175528, A175525; (k=21) A175589; (k=167) A175558, A175559, A175560, A175552.

Select[Range[1000], Mod[Total[IntegerDigits[13^#]], #] == 0 &]

a(47)-a(79) from N. J. A. Sloane, Dec 04 2010

a(80)-a(85) from Donovan Johnson, Dec 05 2010

A067862 Numbers k that divide the sum of digits of 3^k.

Original entry on oeis.org

1, 3, 6, 9, 27, 54, 180, 216, 225, 486

Offset: 1

Shyam Sunder Gupta and Amarnath Murthy, Feb 16 2002

No further terms below 200000. - Sascha Kurz, Mar 19 2002
No further terms below 1000000 - Harvey P. Dale, Dec 01 2010
Almost certainly there are no further terms.
All terms greater than 1 are multiples of 3. - Alonso del Arte, Oct 08 2013
Numbers k such that A175435(k) = 0. - Michel Marcus, Oct 09 2013

			6 divides the sum of digits of 3^6 (i.e., 7 + 2 + 9 = 18), so it belongs to the sequence.

Cf. A004166, A175169, A175435.

Select[Range[1000000], Divisible[Total[IntegerDigits[3^#]], #] &] (* Harvey P. Dale, Dec 01 2010 *)

Checked and found no further terms up to n = 1000000. - Harvey P. Dale, Dec 01 2010

A175552 Numbers k such that the digit sum of 167^k is divisible by k.

Original entry on oeis.org

1, 2, 5, 7, 22, 490, 724, 778, 868, 994, 1109, 1390, 1415, 1462, 1642, 1739, 1829, 2146, 2362, 3136, 4954, 6437, 6628, 7103, 11200, 12424, 12863, 14242, 14249, 15059, 15203, 16222, 17140, 18353, 19192, 21233, 22853, 24106, 24574, 24833, 26896, 27652, 28253, 30323, 31306, 31594, 32386, 33790, 34985, 36184, 36310, 40673, 42196, 43931, 45911, 45983

Offset: 1

N. J. A. Sloane, Dec 03 2010

From Donovan Johnson, Dec 03 2010: (Start)
To generate the additional terms I used PFGW.exe to get the decimal expansion for each number of the form 167^n (n <= 50000). Then I wrote a program in powerbasic to read the pfgw.out file and get the digit sums.
The digit sum is 10 times the n value for terms a(5) to a(56). (End)
I believe that this sequence is finite. - N. J. A. Sloane, Dec 05 2010

Donovan Johnson, Table of n, a(n) for n = 1..129

Sum of digits of k^n mod n: (k=2) A000079, A001370, A175434, A175169; (k=3) A000244, A004166, A175435, A067862; (k=5) A000351, A066001, A175456; (k=6) A000400, A066002, A175457, A067864; (k=7) A000420, A066003, A175512, A067863; (k=8) A062933; (k=13) A001022, A175527, A175528, A175525; (k=21) A175589; (k=167) A175558, A175559, A175560, A175552.

Select[Range[10000], Mod[Total[IntegerDigits[167^#]], #] == 0 &]

a(25)-a(56) from Donovan Johnson, Dec 03 2010

A175434 (Digit sum of 2^n) mod n.

Original entry on oeis.org

0, 0, 2, 3, 0, 4, 4, 5, 8, 7, 3, 7, 7, 8, 11, 9, 14, 1, 10, 11, 5, 3, 18, 13, 4, 14, 8, 15, 12, 7, 16, 26, 29, 27, 24, 28, 19, 29, 32, 21, 9, 4, 13, 14, 17, 24, 21, 25, 16, 26, 29, 27, 24, 28, 37, 29, 23, 12, 18, 22, 13, 23, 26, 24, 21, 43, 43, 35, 20, 0, 15, 37, 37, 56, 50, 30, 27, 22, 31, 32, 26, 42, 39, 34, 43, 26, 20, 27, 24, 28, 55, 47, 32, 57, 45, 31, 40, 14, 8, 15

Offset: 1

N. J. A. Sloane, Dec 03 2010

			For n = 1,2,3,4,5,6, the digit-sum of 2^n is 2,4,8,7,5,10, so
a(1) through a(6) are 0,0,2,3,0,4. - _N. J. A. Sloane_, Aug 12 2014

Sum of digits of k^n mod n: (k=2) A000079, A001370, A175434, A175169; (k=3) A000244, A004166, A175435, A067862; (k=5) A000351, A066001, A175456; (k=6) A000400, A066002, A175457, A067864; (k=7) A000420, A066003, A175512, A067863; (k=8) A062933; (k=13) A001022, A175527, A175528, A175525; (k=21) A175589; (k=167) A175558, A175559, A175560, A175552.

Table[Mod[Total[IntegerDigits[2^n]],n],{n,100}] (* Harvey P. Dale, Aug 12 2014 *)

Offset changed to 1 at the suggestion of Harvey P. Dale, Aug 12 2014

cp's OEIS Frontend

A004166 Sum of digits of 3^n.

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A175169 Numbers k that divide the sum of digits of 2^k.

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A175525 Numbers k that divide the sum of digits of 13^k.

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A067862 Numbers k that divide the sum of digits of 3^k.

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A175552 Numbers k such that the digit sum of 167^k is divisible by k.

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A175434 (Digit sum of 2^n) mod n.

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