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 1-6 of 6 results.

A067499 Powers of 2 with digit sum also a power of 2.

Original entry on oeis.org

1, 2, 4, 8, 512, 68719476736, 38685626227668133590597632, 95780971304118053647396689196894323976171195136475136, 25108406941546723055343157692830665664409421777856138051584, 1606938044258990275541962092341162602522202993782792835301376
Offset: 1

Views

Author

Amarnath Murthy, Apr 19 2001

Keywords

Comments

Question is the sequence finite or infinite?

Examples

			512 = 2^9 and 5+1+2 = 8 = 2^3. 68719476736 = 2^36, sum of digits = 64 = 2^6.
		

Crossrefs

Subsequence of A000079, A005349.
Cf. A095412.

Programs

  • Maple
    with(numtheory): pow2 := [2^i$ i=1..2000]: for n from 1 to 1000 do L1 := convert(2^n, base, 10): if member(sum(L1[i], i=1..nops(L1)), pow2) then printf(`%d,`,2^n) fi: od: # James Sellers, Apr 19 2001
  • Mathematica
    Select[Table[2^n,{n,0,250}],IntegerQ[Log[2,Total[IntegerDigits[#]]]] &] (* Jayanta Basu, May 19 2013 *)
    Module[{nn=200,pw2},pw2=2^Range[0,nn];Select[pw2,MemberQ[pw2,Total[ IntegerDigits[ #]]]&]] (* Harvey P. Dale, Dec 01 2022 *)

Formula

a(n) = 2^k with digit sum a(n) = 2^r.
{ A000079 } intersect { A005349 }.

Extensions

More terms from James Sellers, Apr 19 2001

A072081 Numbers divisible by the square of the sum of their digits in base 10.

Original entry on oeis.org

1, 10, 20, 50, 81, 100, 112, 162, 200, 243, 324, 392, 400, 405, 500, 512, 605, 648, 810, 972, 1000, 1053, 1100, 1120, 1134, 1183, 1215, 1296, 1400, 1620, 1701, 1900, 1944, 2000, 2025, 2106, 2156, 2240, 2268, 2300, 2401, 2430, 2511, 2592, 2704, 2800, 2916
Offset: 1

Views

Author

Labos Elemer, Jun 14 2002

Keywords

Comments

If k is a term, then 10 * k is a term. There are an infinite number of terms that are not divisible by 10. The numbers m = 24 * 10^(42 * k - 40) +1, k >= 1, are divisible by 7^2 = digsum(m)^2. Also, the numbers s = 491 * 10^(42 * k - 8) + 3, k >= 1, are divisible by 17^2 = digsum(s)^2. - Marius A. Burtea, Mar 19 2020
The numbers 2^A095412(n), n >= 5, are terms. - Marius A. Burtea, Apr 02 2020

Examples

			k=9477, sumdigits(9477)=27, q=9477=27*27*13.
		

Crossrefs

Programs

  • Magma
    [k:k in [1..3000]| k mod &+Intseq(k)^2 eq 0]; // Marius A. Burtea, Mar 19 2020
    
  • Mathematica
    sud[x_] := Apply[Plus, IntegerDigits[x]] Do[s=sud[n]^2; If[IntegerQ[n/s], Print[n]], {n, 1, 10000}]
    Select[Range[3000],Divisible[#,Total[IntegerDigits[#]]^2]&] (* Harvey P. Dale, May 04 2011 *)
  • PARI
    for(n=1,10^4,s=sumdigits(n);if(!(n%s^2),print1(n,", "))) \\ Derek Orr, Apr 29 2015
    
  • Python
    def ok(n): return n and n%sum(di for di in map(int, str(n)))**2 == 0
    print([k for k in range(3000) if ok(k)]) # Michael S. Branicky, Jan 10 2025

A072083 Numbers divisible by the 4th power of the sum of their digits in base 10.

Original entry on oeis.org

1, 10, 100, 1000, 2000, 2401, 5000, 10000, 13122, 20000, 24010, 50000, 100000, 110000, 131220, 140000, 190000, 200000, 230000, 234256, 240100, 280000, 320000, 370000, 390625, 400221, 410000, 460000, 500000, 512000, 550000, 614656, 640000
Offset: 1

Views

Author

Labos Elemer, Jun 14 2002

Keywords

Comments

If k is a term, then 10*k is a term. There are an infinite number of terms that are not divisible by 10. The numbers m = 24 * 10^(294*k - 292) + 1, k = 7*a - 6, a >= 1, are divisible by 7^4 = digsum(m)^4. Also, the numbers s = 491 * 10^(4624*k - 4623) + 3, k = 17*u - 11, u >= 1, are divisible by 17^4 = digsum(s)^4. - Marius A. Burtea, Mar 19 2020
The numbers 2^A095412(n), n >= 6, are terms. - Marius A. Burtea, Apr 02 2020

Examples

			k=614656: sumdigits(614656)=28, q=1, since k=28*28*28*28.
		

Crossrefs

Programs

  • Magma
    [k:k in [1..640000]| k mod &+Intseq(k)^4 eq 0]; // Marius A. Burtea, Mar 19 2020
    
  • Mathematica
    sud[x_] := Apply[Plus, IntegerDigits[x]] Do[s=sud[n]^4; If[IntegerQ[n/s], Print[n]], {n, 1, 10000}]
    Select[Range[700000],Divisible[#,Total[IntegerDigits[ #]]^4]&] (* Harvey P. Dale, Jun 28 2011 *)
  • PARI
    isok(m) = (m % sumdigits(m)^4) == 0; \\ Michel Marcus, Apr 02 2020

A358633 a(n) is the smallest k > 1 such that the sum of digits of n^k is a power of n (or -1 if no such k exists).

Original entry on oeis.org

2, 2, 2, 18, 8, 7, 4, 3, 2, 2, 45741764, 4216, 32, 537, 39, 44, 3, 3, 1187, 13, 67, 4
Offset: 1

Views

Author

Jon E. Schoenfield, Nov 24 2022

Keywords

Comments

a(11), if it is not -1, seems likely to exceed 4*10^6.
Additional known terms: a(12)..a(22) = {4216, 32, 537, 39, 44, 3, 3, 1187, 13, 67, 4}; a(24)..a(28) = {88, 4, 3, 3, 4}; a(30) = 399, a(31) = 7, a(33)..a(55) = {159, 7, 5, 4, 191, 188, 228, 13, 389332, 236, 7, 11543, 6, 5, 302, 292, 15405, 788, 337, 18213, 7, 6, 21248}; a(57)..a(60) = {413, 7, 25683, 1044}; a(62) = 476.
a(10^m) = 2 for all m >= 0, since the sum of digits of (10^m)^2 is 1.
If n is not divisible by 10, then a(n) tends to be fairly close to a number x such that the number of digits of n^x is (2/9)*n^j for some positive integer j, i.e., log_10(n^x) ~ (2/9)*n^j, so a(n) ~ (2/9)*n^j/log_10(n) for some integer j. E.g., a(12) = 4216 ~ 4269.90... = (2/9)*12^4/log_10(12). For n = 11, such numbers x are (2/9)*11^j/log_10(11) = 0.213389... * 11^j, which, for j = 1..7, round to 2, 26, 284, 3124, 34366, 378032, and 4158357. (By exhaustive search, a(n) > 10^5 (or a(n) = -1) for n = 23, 29, 32, 56, and 61, and (if a(11) != -1) a(11) > 4*10^5, so a(11) seems very likely to be either in the general vicinity of 4.16*10^6 or > 4.5*10^7.)
a(32) = 4950773; a(61) = 1722427. - Martin Ehrenstein, Nov 25 2022
3*10^7 < a(23) <= 555650815. - Martin Ehrenstein, Nov 28 2022

Examples

			The sum of digits of 1^2 = 1 is 1, which is a power of 1, so a(1) = 2.
The sum of digits of 2^2 = 4 is 4, which is a power of 2, so a(2) = 2.
For k = 2..17, the sum of digits of 4^k is {7, 10, 13, 7, 19, 22, 25, 19, 31, 25, 37, 40, 43, 37, 58, 61}, none of which is a power of 4, but the sum of digits of 4^18 = 68719476736 is 6+8+7+1+9+4+7+6+7+3+6 = 64 = 4^3, so a(4) = 18.
		

Crossrefs

Cf. A066005 (sum of digits of 11^n).

Programs

  • PARI
    isok(k,n) = my(s=sumdigits(n^k), x, y); if (s==1, return(2)); (s==n) || ((ispower(s,,&x)) && ((x==n) || (ispower(n,,&y) && (y==x))));
    a(n) = my(k=2); while (!isok(k,n), k++); k; \\ Michel Marcus, Nov 25 2022

Extensions

a(11)-a(22) from Martin Ehrenstein, Nov 26 2022

A333474 Numbers k such that 2^k + 1 is divisible by the sum of its decimal digits.

Original entry on oeis.org

0, 1, 2, 3, 9, 98, 135, 200, 665, 1782, 4230, 4521, 6815, 17010, 19635, 30338, 31365, 35427, 49555, 96619, 102897, 157850, 193734, 273050, 393225, 449217, 477333, 483310, 493350, 534465, 661815, 918918, 947925, 1050858, 1114690, 1134945, 1204686, 1350990, 1428105
Offset: 1

Views

Author

Daniel Starodubtsev, Mar 23 2020

Keywords

Comments

Numbers k such that A000051(k) is in A005349.

Examples

			9 is in the sequence, because 2^9 + 1 = 513 is divisible by 5 + 1 + 3.
		

Crossrefs

Programs

  • PARI
    isok(k) = my(x=2^k+1); !(x % sumdigits(x)); \\ Michel Marcus, Mar 23 2020
  • Python
    print([i for i in range(5000) if (2**i+1)%sum([int(i) for i in str(2**i+1)]) == 0])
    

Extensions

More terms from Giovanni Resta, Mar 23 2020

A358667 T(n,k) is the k-th integer j > 1 such that the sum of digits of n^j is a power of n (or -1 if no such k-th integer exists); table read by downward antidiagonals.

Original entry on oeis.org

2, 3, 2, 4, 3, 2, 5, 9, 3, 18, 6, 36, 4, 88, 8, 7, 85, 5, 97, 208, 7, 8, 176, 9, 100, 977, 8, 4, 9, 194, 10, 1521, 1007, 9, 11, 3, 10, 200, 11, 6034, 4938, 10, 4433, 12, 2, 11, 375, 13, 6052, 24709, 13, 30810, 125, 18, 2, 12, 1517, 16, 96867, 24733, 51, 216613, 1014, 1503, 3
Offset: 1

Views

Author

Jon E. Schoenfield, Nov 25 2022

Keywords

Comments

T(11,1) is unknown at this time.

Examples

			Table begins:
.
   n\k|   1    2     3     4      5     6     7     8    9    10   11 ...
   ---+------------------------------------------------------------------
    1 |   2    3     4     5      6     7     8     9   10    11   12 ...
    2 |   2    3     9    36     85   176   194   200  375  1517  ...
    3 |   2    3     4     5      9    10    11    13   16   ...
    4 |  18   88    97   100   1521  6034  6052 96867  ...
    5 |   8  208   977  1007   4938 24709 24733   ...
    6 |   7    8     9    10     13    51   ...
    7 |   4   11  4433 30810 216613   ...
    8 |   3   12   125  1014    ...
    9 |   2   18  1503   ...
   10 |   2    3   ...
   11 |   ?  ...
  ... | ...
		

Crossrefs

Formula

T(n,1) = A358633(n).
T(1,k) = k+1.
T(2,k) = A095412(k+2).
T(3,k) = A118872(k+2).
Showing 1-6 of 6 results.