cp's OEIS Frontend

This is a front-end for the Online Encyclopedia of Integer Sequences, made by Christian Perfect. The idea is to provide OEIS entries in non-ancient HTML, and then to think about how they're presented visually. The source code is on GitHub.

Showing 1-8 of 8 results.

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

Views

Author

T. D. Noe, Dec 03 2010

Keywords

Comments

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

Crossrefs

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.

Programs

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

Extensions

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

Views

Author

Shyam Sunder Gupta and Amarnath Murthy, Feb 16 2002

Keywords

Comments

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

Examples

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

Crossrefs

Cf. A004166, A175169, A175435.

Programs

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

Extensions

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

Views

Author

N. J. A. Sloane, Dec 03 2010

Keywords

Comments

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

Links

Crossrefs

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.

Programs

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

Extensions

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

A067864 Numbers k such that k divides the sum of digits of 6^k.

Original entry on oeis.org

1, 3, 9, 18, 90
Offset: 1

Views

Author

Amarnath Murthy and Shyam Sunder Gupta, Feb 16 2002

Keywords

Comments

Conjecture: the sequence is finite, since (sum of the digits of 6^k)/k -> log_10(6)*4.5 ~ 3.50168 as k->infinity (this is also a conjecture). - Robert Gerbicz, May 08 2008
Next term, if it exists, exceeds 100000. - Sean A. Irvine, Apr 05 2010
The keyword "hard" refers to the difficulty of finding, or disproving the existence of, the next term. - N. J. A. Sloane, Nov 09 2024

Examples

			a(1)=3, so 3 divides the sum of digits of 6^3 (i.e., 2 + 1 + 6 = 9).

Crossrefs

Cf. A000400, A066002, A175457, A175169, A067862, A067863.

Programs

  • Mathematica
    Select[Range[100],Mod[Total[IntegerDigits[6^#]],#]==0&] (* Harvey P. Dale, Nov 09 2024 *)

A175435 (Digit sum of 3^n) mod n.

Original entry on oeis.org

0, 1, 0, 1, 4, 0, 4, 2, 0, 7, 5, 6, 1, 3, 6, 11, 10, 9, 17, 5, 6, 1, 8, 6, 13, 11, 0, 16, 14, 3, 19, 31, 6, 31, 11, 9, 16, 14, 3, 10, 17, 6, 22, 2, 9, 16, 32, 21, 10, 44, 15, 13, 29, 0, 35, 34, 51, 10, 58, 39, 13, 29, 27, 16, 5, 51, 28, 17, 15, 4, 38, 18, 7
Offset: 1

Views

Author

N. J. A. Sloane, Dec 03 2010

Keywords

Examples

			a(5) = 4 because digsum(3^5) mod 5 = digsum(243) mod 5 = 9 mod 5 = 4.

Links

Crossrefs

Cf. A000244, A004166, A067862; A000079, A001370, A175434, A175169.

Programs

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

Views

Author

N. J. A. Sloane, Dec 03 2010

Keywords

Examples

			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

Crossrefs

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.

Programs

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

Extensions

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

A175512 (Digit sum of 7^n) mod n.

Original entry on oeis.org

0, 1, 1, 3, 2, 4, 4, 7, 1, 3, 5, 1, 0, 2, 4, 4, 7, 1, 3, 16, 19, 9, 16, 1, 22, 8, 10, 21, 5, 28, 22, 7, 28, 6, 17, 28, 4, 16, 19, 22, 25, 10, 40, 43, 28, 3, 34, 40, 31, 34, 13, 40, 8, 1, 49, 43, 10, 9, 34, 19, 22, 8, 19, 40, 52, 64, 0, 66, 37, 49, 52, 55, 58, 70, 37, 49, 11, 64, 6, 25, 28, 67, 43, 55, 40, 20, 64, 13, 8, 28, 49, 34, 82, 0, 79, 82, 85, 61, 73, 67
Offset: 1

Views

Author

N. J. A. Sloane, Dec 03 2010

Keywords

Crossrefs

Cf. A000420, A066003, A067863; A000079, A001370, A175434, A175169.

Programs

  • Mathematica
    Table[Mod[Plus@@IntegerDigits[7^n],n],{n,150}] (* Harvey P. Dale, Dec 16 2010 *)

A220365 a(n) is conjectured to be the largest power k for which k divides the sum of digits of n^k.

Original entry on oeis.org

1, 70, 486, 35, 10, 90, 805, 5218, 243, 1, 35, 1494, 469004, 1045, 288, 116, 7, 195, 29, 70, 16710, 23, 2, 1017, 28, 58, 162, 166, 209, 486, 205, 106, 1206, 2053, 37120
Offset: 1

Views

Author

Robert G. Wilson v, Dec 12 2012

Keywords

Comments

a(36) >= 423378.
Please consult the argument in A067863 for the reason that it is believed that all individual such sequences (all k's which divide b^k) terminate.

Examples

			a(2) = 70 since the sum of digits of 2^70 is divisible by 70 and it is believed that there does not exist any larger exponent which satisfies this criterion.

Crossrefs

Numbers n such that n divides the sum of digits of k^n: A175169 (k=2), A067862 (k=3), A067864 (k=6), A067863 (k=7), A062933 (k=8), A062927 (k=9), A175525 (k=13), A175589 (k=21), A220364 (k=36), A175552 (k=167).

Programs

  • Mathematica
    For any individual base, b, fQ[n_] := Mod[Plus @@ IntegerDigits[b^n], n] == 0; k = 1; lst = {}; While[k < 100001, If[ fQ@ k, AppendTo[lst, k]; Print[k]]; k++]; lst

Formula

If a(n) = k, then a(10*n) = k.

Extensions

Definition and example corrected by Giovanni Resta, Dec 14 2012
Showing 1-8 of 8 results.

Started in 1964 by Neil J. A. Sloane | Maintained by The OEIS Foundation Inc.

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