A023106 -id:A023106 - cp's OEIS frontend

A256590 Base-2 Reacher numbers: numbers that are powers of the sum of their base-2 digits.

0, 1, 81, 625, 7776, 16807, 46656, 59049, 1679616, 1475789056, 6975757441, 137858491849, 576650390625, 41426511213649, 2384185791015625, 150094635296999121, 10260628712958602189, 32768000000000000000, 243569224216081305397, 655360000000000000000

Offset: 1

Jeffrey Shallit, Apr 03 2015

Named for fictional character Jack Reacher in the series of novels by Lee Child.

There are 2709 terms with 10,000 or fewer digits; a(2709) = 15402^2388. - Charles R Greathouse IV, Nov 26 2016

Lee Child, Bad Luck and Trouble, Delacorte Press, 2007. In this book, the main character, Jack Reacher, likes the number 81 because it is the square of the sum of its base-10 digits.

Hiroaki Yamanouchi and Charles R Greathouse IV, Table of n, a(n) for n = 1..388 (first 141 terms from Hiroaki Yamanouchi)
J. Shallit, Mathematics in a Jack Reacher Novel, blog post, Sep 08 2007.

Cf. A023106, A161792.

fQ[n_] := Block[{wt = DigitCount[n, 2, 1]},
Which[n <= 1, True, wt <= 1, False, True, IntegerQ@ Log[wt, n]]]; Select[Range[10^5], fQ] (* Michael De Vlieger, Apr 04 2015 *)

is(n)= n<=1 || (ispower(n,,&r) && (r==hammingweight(n) || (r^ispower(n=hammingweight(n))==n && n>1))) \\ Michel Marcus and M. F. Hasler, Apr 04 2015

list(lim)=my(v=List([0,1]),H,t); for(e=3,logint(lim\=1,3), for(b=2,min(solve(x=e,lim,x-e*log(x)/log(2)-1),sqrtnint(lim,e)), H=hammingweight(t=b^e); if(H>1 && b^valuation(H,b)==H, listput(v,t)))); Set(v) \\ Charles R Greathouse IV, Nov 26 2016

a(1) prepended and a(14)-a(20) added by Hiroaki Yamanouchi, Apr 03 2015

A381487 Numbers which are a power of their digital root.

Original entry on oeis.org

0, 1, 2, 3, 4, 5, 6, 7, 8, 9, 81, 128, 256, 512, 729, 2401, 6561, 8192, 16384, 32768, 59049, 78125, 524288, 531441, 823543, 1048576, 2097152, 4782969, 33554432, 43046721, 67108864, 134217728, 282475249, 387420489, 1220703125, 2147483648, 3486784401, 4294967296

Offset: 1

Stefano Spezia, Feb 25 2025

			a(12) = 128 is a term since 128 = 2^7 = A010888(128)^7.

Michel Marcus, Table of n, a(n) for n = 1..3347

Cf. A010888, A023106, A381491, A381492.

Digital root of k^n: A000012 (1), A153130 (2), A100401 (3), A100402 (4), A070366 (5), A100403 (6), A070403 (7), A010689 (8), A010734 (9).

A010888[n_]:=n - 9*Floor[(n-1)/9]; kmax=5*10^6; a={0,1}; For[k=2, k<=kmax, k++, If[A010888[k]!=1, If[IntegerQ[Log[A010888[k],k]], AppendTo[a,k]]]]; a

isok(k) = if ((k==0) || (k==1), return(1)); my(d=(k-1)%9+1); if (d>1, d^logint(k, d) == k); \\ Michel Marcus, Feb 26 2025

lista(nn) = my(list = List()); listput(list, 0); listput(list, 1); for (n=2, 9, for (k=1, logint(nn, n), if ((n^k-1)%9+1 == n, listput(list, n^k)););); vecsort(Vec(list)); \\ Michel Marcus, Feb 27 2025

a(n) = A381491(n)^A381492(n).

A038178 Numbers k such that k = (sum of digits of k)^(number of digits of k).

Original entry on oeis.org

0, 1, 2, 3, 4, 5, 6, 7, 8, 9, 81, 512, 2401

Offset: 1

Felice Russo

To prove completeness, consider that k^m contains more than m digits for every k >= 10 and check 1 <= k <= 9 explicitly. - Ulrich Schimke (ulrschimke(AT)aol.com)

Subset of A023106. - R. J. Mathar, Oct 20 2008

			512 is in the sequence because (5 + 1 + 2)^3 = 512.

Cf. A007953, A055642, A023106.

Select[Range[2500],#==Total[IntegerDigits[#]]^IntegerLength[#]&] (* Harvey P. Dale, Oct 26 2011 *)

A128912 Numbers m of the form (sum of digits of m)^k, k > 1.

Original entry on oeis.org

0, 1, 81, 512, 2401, 4913, 5832, 17576, 19683, 234256, 390625, 614656, 1679616, 17210368, 34012224, 52521875, 60466176, 205962976, 612220032, 8303765625, 10460353203, 24794911296, 27512614111, 52523350144, 68719476736

Offset: 1

J. M. Bergot, Apr 23 2007

Perfect powers m > 1 such that the sum of the digits of m equals one of its nontrivial roots.

Essentially a duplicate of A023106, where numbers 2 through 9 are allowed as solutions for k=1.

			234256 = 22^4 and 2+3+4+2+5+6 = 22, hence 234256 is a term.
390625 = 25^4 and 3+9+0+6+2+5 = 25, hence 390625 is a term.

Cf. A001597 (perfect powers), A007953 (sum of digits).

{m=10^5; z=10^11; v=[]; for(n=0, m, k=2; while((p=n^k)<=z, s=sumdigits(p); if(n==s, v=concat(v, p)); k++)); v=vecsort(v); print(v)} \\ Klaus Brockhaus, Apr 24 2007, edited by M. F. Hasler, Apr 14 2015

is(n)=ispower(n)&&(1M. F. Hasler, Apr 14 2015

Edited, corrected and extended by Klaus Brockhaus, Apr 24 2007

Definition simplified and initial terms 0, 1 added by M. F. Hasler, Apr 14 2015

A283034 Numbers k such that k = (sum of digits of k)^(last digit of k).

Original entry on oeis.org

1, 4913, 19683, 52521875, 24794911296, 68719476736, 271818611107, 1174711139837

Offset: 1

Shmelev Aleksei, Feb 27 2017

The check must be done up to 10^22 (then for 23 digits in a number max result can be (23*10)^9 = 4, 6*10^20 < 10^22).

			1 = 1^1,
4913 = (4+9+1+3)^3,
19683 = (1+9+6+8+3)^3,
52521875 = (5+2+5+2+1+8+7+5)^5.

Cf. A007953, A010879, A023106.

Union[Reap[nd=1; Sow[1]; While[Ceiling[(10^(nd-1))^(1/9)] <= 9 nd, Do[ Do[v = s^e; If[Mod[v, 10] == e && Plus @@ IntegerDigits@ v == s, Sow[v]], {s, Ceiling[ (10^(nd-1))^(1/e)], Min[ Floor[10^(nd/e)], 9 nd]}], {e, 2, 9}]; nd++]][[2, 1]]] (* all terms, Giovanni Resta, Feb 27 2017 *)

Sub calcul()
   Sheets("Result").Select
   Range("A1").Select
   For i = 1 To 10000000
      Sum = 0
      For k = 1 To Len(i)
         Sum = Sum + Mid(i, k, 1)
      Next
      If Sum ^Mid(i, len(i), 1)= i Then
         ActiveCell.Value = i
         ActiveCell.Offset(1, 0).Select
      End If
   Next
End Sub

a(5)-a(8) from Giovanni Resta, Feb 27 2017

A368245 Numbers k such that there is a positive integer r for which k^(1/r) = digsum(k) - r.

Original entry on oeis.org

4, 25, 64, 125, 196, 216, 289, 343, 2744, 3375, 4096, 3518743761, 13537086546263552, 15633814156853823

Offset: 1

Stefano Spezia, Jan 23 2024

Some other terms: 50714860157241037295616, 188031682201497672618081, 4817904819828488880132096, 13214788658781797667045376, 45587487211290846582931833112449, 112410921330388974282595778471993, 282429536481000000000000000000000000, 21936950640377856000000000000000000000. - Chai Wah Wu, Jan 30 2024

			The terms 4, 25, 64, 196, and 289 are obtained for r = 2:
  4^(1/2) = sqrt(4) = 2 = digsum(4) - 2;
  25^(1/2) = sqrt(25) = 5 = digsum(25) - 2 = 7 - 2;
  64^(1/2) = sqrt(64) = 8 = digsum(64) - 2 = 10 - 2;
  196^(1/2) = sqrt(196) = 14 = digsum(196) - 2 = 16 - 2;
  289^(1/2) = sqrt(289) = 17 = digsum(289) - 2 = 19 - 2.
The terms 125, 216, 343, 2744, 3375, and 4096 are obtained for r = 3:
  125^(1/3) = 5 = digsum(125) - 3 = 8 - 3;
  216^(1/3) = 6 = digsum(216) - 3 = 9 - 3;
  343^(1/3) = 7 = digsum(343) - 3 = 10 - 7;
  2744^(1/3) = 14 = digsum(2744) - 3 = 17 - 3;
  3375^(1/3) = 15 = digsum(3375) - 3 = 18 - 3;
  4096^(1/3) = 16 = digsum(4096) - 3 = 19 - 3.
The term 3518743761 is obtained for r = 6:
  3518743761^(1/6) = 39 = digsum(3518743761) - 6 = 45 - 6.

Cf. A001597 (supersequence), A007953, A029837.

Cf. A023106, A128912.

min = 0; max = 10^8; list=Union@ Flatten@ Table[ n^expo, {expo, Prime@ Range@ PrimePi@ Log2@ max}, {n, Floor[1 + min^(1/expo)], max^(1/expo)}]; (* A001597 *)
a = {};  For[k = 1, k <=Length[list], k++, For[r = 2, r <= Ceiling[Log2[Part[list,k]]], r++, If[Part[list,k]^(1/r) == Total[IntegerDigits[Part[list,k]]]-r, AppendTo[a, Part[list,k]]]]]; a

a(13)-a(14) from Chai Wah Wu, Jan 30 2024

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A256590 Base-2 Reacher numbers: numbers that are powers of the sum of their base-2 digits.

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A381487 Numbers which are a power of their digital root.

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A038178 Numbers k such that k = (sum of digits of k)^(number of digits of k).

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A128912 Numbers m of the form (sum of digits of m)^k, k > 1.

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A283034 Numbers k such that k = (sum of digits of k)^(last digit of k).

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A368245 Numbers k such that there is a positive integer r for which k^(1/r) = digsum(k) - r.

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