A061209 -id:A061209 - cp's OEIS frontend

A061210 Numbers which are the fourth powers of their digit sum.

0, 1, 2401, 234256, 390625, 614656, 1679616

Offset: 1

Amarnath Murthy, Apr 21 2001

It can be shown that 1679616 = 36^4 is the largest such number.

			614656 = ( 6+1+4+6+5+6)^4 =28^4.

Amarnath Murthy, The largest and the smallest m-th power whose digit sum is the m-th root. (To be published)
Alfred S. Posamentier, Math Charmers, Tantalizing Tidbits for the Mind, Prometheus Books, NY, 2003, page 36.

Cf. A061209 (with cubes), A061211.

Cf. A046000, A076090, A046017; A252648 and references there.

Select[Range[0,17*10^5],#==Total[IntegerDigits[#]]^4&] (* Harvey P. Dale, Sep 22 2019 *)

isok(n) = n == sumdigits(n)^4; \\ Michel Marcus, Jan 22 2015

Corrected by Ulrich Schimke, Feb 11 2002

Initial 0 added by M. F. Hasler, Apr 12 2015

A061211 Largest number m such that m is the n-th power of the sum of its digits.

Original entry on oeis.org

9, 81, 19683, 1679616, 205962976, 68719476736, 6722988818432, 248155780267521, 150094635296999121, 480682838924478847449, 23316389970546096340992, 2518170116818978404827136, 13695791164569918553628942336, 4219782742781494680756610809856

Offset: 1

Amarnath Murthy, Apr 21 2001

Clearly m = 1 always works, so a(n) exists for all n. - Farideh Firoozbakht, Nov 23 2007

105 is the smallest number n such that a(n)=1. This means that if n<105 there exists at least one number m greater than 1 such that m is the n-th power of the sum of its digits while 1 is the only number m such that m is the 105th power of the sum of its digits. A133509 gives n such that a(n) = 1. - Farideh Firoozbakht, Nov 23 2007

			a(3) = 19683 = 27^3 and no bigger number can have this property. (This has been established in the Murthy reference.)
a(4) = 1679616 = (1+6+7+9+6+1+6)^4 = 36^4.

Amarnath Murthy, The largest and the smallest m-th power whose digits sum /product is its m-th root. To appear in Smarandache Notions Journal.
Amarnath Murthy, e-book, "Ideas on Smarandache Notions", manuscript.

T. D. Noe, Table of n, a(n) for n = 1..105

Cf. A061209, A061210, A046000, A076090, A046017.

meanDigit = 9/2; translate = 900; upperm[1] = translate;
upperm[n_] := Exp[-ProductLog[-1, -Log[10]/(meanDigit*n)]] + translate;
a[n_] := (For[max = m = 1, m <= upperm[n], m++, If[m == Total[ IntegerDigits[ m^n ] ], max = m]]; max^n);
Array[a, 14] (* Jean-François Alcover, Jan 09 2018 *)

More terms from Ulrich Schimke, Feb 11 2002

Edited by N. J. A. Sloane at the suggestion of Farideh Firoozbakht, Dec 04 2007

A023106 a(n) is a power of the sum of its digits.

Original entry on oeis.org

0, 1, 2, 3, 4, 5, 6, 7, 8, 9, 81, 512, 2401, 4913, 5832, 17576, 19683, 234256, 390625, 614656, 1679616, 17210368, 34012224, 52521875, 60466176, 205962976, 612220032, 8303765625, 10460353203, 24794911296, 27512614111, 52523350144, 68719476736

Offset: 0

David W. Wilson

Base-10 Reacher numbers: named for the character Jack Reacher in the series of books by Lee Child. Reacher likes the number 81 because it is the square of the sum of its base-10 digits. - Jeffrey Shallit, Apr 03 2015

Contains A061209 and A061210 and all terms of A061211. See A252648 for numbers which are the sum of some power of their digits. - M. F. Hasler, Apr 13 2015

			2401 is an element because 2401 = 7^4 is a power of its digit sum 7.

Alfred S. Posamentier, Math Charmers, Tantalizing Tidbits for the Mind, Prometheus Books, NY, 2003, page 36.

David W. Wilson, Table of n, a(n) for n = 0..1137
Jeffrey Shallit, Mathematics in a Jack Reacher Novel, blog post, September 8 2007.

Cf. A061209, A061210, A061211, A252648.

fQ[n_] := Block[{b = Plus @@ IntegerDigits[n]}, If[b > 1, IntegerQ[ Log[b, n]] ]]; Take[ Select[ Union[ Flatten[ Table[n^m, {n, 55}, {m, 9}]]], fQ[ # ] &], 31] (* Robert G. Wilson v, Jan 28 2005 *)
Join[{0,1},Select[Range[0,1700000],IntegerQ[Log[Total[IntegerDigits[#]],#]]&]//Quiet] (* The program generates the first 21 terms of the sequence. *) (* Harvey P. Dale, Mar 30 2024 *)

is(n)={n<10||(!(n%s=sumdigits(n))&&s>1&&n==s^round(log(n)/log(s)))} \\ M. F. Hasler, Apr 13 2015

import math
def is_valid(n): dsum = sum(map(int, str(n))); return dsum ** int(round(math.log(n, dsum))) == n if dsum > 1 else n < 2
# Victor Dumbrava, May 02 2018

A257784 Numbers n such that the sum of the digits squared times the sum of the digits of n to some power equals n.

Original entry on oeis.org

0, 1, 512, 2511, 4913, 5832, 17576, 19683, 24624, 32144, 37000, 111616, 382360, 415000, 420224, 2219400, 14041600, 16328000, 19300032, 30681423, 39203125, 62025728, 78535423, 186836625, 214292000, 432265248, 1120141312, 3479669440, 18529084125, 25342447725

Offset: 1

Pieter Post, May 08 2015

When the power is 1 the numbers are the cubes of their digit sum (A061209).

There are no 2-digit and 18-digit terms. - Chai Wah Wu, Jan 11 2016

			For power 2: 24624 = (2+4+6+2+4)^2*(2^2+4^2+6^2+2^2+4^2).
For power 3: 111616 = (1+1+1+6+1+6)^2*(1^3+1^3+1^3+6^3+1^3+6^3).

Giovanni Resta and Chai Wah Wu, Table of n, a(n) for n = 1..80 n = 1..43 from Giovanni Resta

Cf. A061209, A115518, A130680.

# WARNING: this prints numbers in the sequence, but not in increasing order.
def moda(n,a):
    kk = 0
    while n > 0:
        kk= kk+(n%10)**a
        n =int(n//10)
    return kk
def sod(n):
    kk = 0
    while n > 0:
        kk= kk+(n%10)
        n =int(n//10)
    return kk
for a in range (1, 10):
    for c in range (1, 10**8):
        if c==sod(c)**2*moda(c,a):
            print(c, end=",")

a(16)-a(30) from Giovanni Resta, May 09 2015

A254000 Numbers equal to the fifth powers of the sums of their digits.

Original entry on oeis.org

0, 1, 17210368, 52521875, 60466176, 205962976

Offset: 1

Michal Paulovic, Jan 21 2015

			205962976 = 46^5 = (2 + 0 + 5 + 9 + 6 + 2 + 9 + 7 + 6)^5.

Alfred S. Posamentier, Math Charmers, Tantalizing Tidbits for the Mind, Prometheus Books, NY, 2003, page 36.

Cf. A061209 (with cubes), A061210 (with 4th powers), A061211.

Select[Range@ 210000000, Plus @@ IntegerDigits@ # ^ 5 == # &] (* Michael De Vlieger, Feb 25 2015 *)

lista(nn) = {for (n=0, nn, if (n^5 == sumdigits(n^5)^5, print1(n^5, ", ")););} \\ Michel Marcus, Feb 23 2015

a(n) = A055576(n)^5. - Michel Marcus, Feb 23 2015

A366507 Numbers k such that the sum of the digits of k times the square of the sum of the digits cubed of k equals k.

Original entry on oeis.org

1, 4147200, 12743163, 21147075, 39143552, 52921472, 156754936, 205889445, 233935967

Offset: 1

René-Louis Clerc, Oct 11 2023

There are exactly 9 such numbers (Property 13 of Clerc).

			4147200 = (4+1+4+7+2)*(4^3+1+4^3+7^3+2^3)^2 = 18*230400.

René-Louis Clerc, Quelques nombres de Niven-Harshad particuliers, pp. 1-17, hal-04235744, 2023.
René-Louis Clerc, Nombres S+P, maxSP, minSP et |P-S|, hal-04507547 [math.nt], 2024. (In French)

Cf. A115518, A257766, A061209, A061210, A254000, A130680.

niven12()={for(a=0,9,for(b=0,9,for(c=0,9,for(d=0,9,for(e=0,9,for(f=0,9,for(g=0,9,for(h=0,9,for(i=0,9,for(j=0,9,if((a+b+c+d+e+f+g+h+i+j)*(a^3+b^3+c^3+d^3+e^3+f^3+g^3+h^3+i^3+j^3)^2==1000000000*a+100000000*b+10000000*c+1000000*d+100000*e+10000*f+1000*g+100*h+10*i+j,print1(1000000000*a+100000000*b+10000000*c+1000000*d+100000*e+10000*f+1000*g+100*h+10*i+j,";"))))))))))))}

isok(k) = my(d=digits(k)); vecsum(d)*sum(i=1, #d, d[i]^3)^2 == k; \\ Michel Marcus, Oct 12 2023

A366512 Numbers k such that the square of the sum of the digits times the sum of the cubes of the digits equals k.

Original entry on oeis.org

1, 32144, 37000, 111616, 382360

Offset: 1

René-Louis Clerc, Oct 11 2023

There are exactly 5 such numbers (Property 14 of Clerc).

			32144 = ((3+2+1+4+4)^2)*(3^3 + 2^3 + 1^3 + 4^3 + 4^3) = 196*164.

René-Louis Clerc, Quelques nombres de Niven-Harshad particuliers, pp. 1-17, hal-04235744, 2023.
René-Louis Clerc, Nombres S+P, maxSP, minSP et |P-S|, hal-04507547 [math.nt], 2024. (In French)

Cf. A115518, A257766, A061209, A061210, A254000, A130680.

Select[Range[10^6], #1 == Total[#2]^2*Total[#2^3] & @@ {#, IntegerDigits[#]} &] (* Michael De Vlieger, Mar 25 2024 *)

niven23()={for(a=0,9,for(b=0,9,for(c=0,9,for(d=0,9,for(e=0,9,for(f=0,9,for(g=0,9,for(h=0,9,if((a+b+c+d+e+f+g+h)^2*(a^3+b^3+c^3+d^3+e^3+f^3+g^3+h^3)==10000000*a+1000000*b+100000*c+10000*d+1000*e+100*f+10*g+h,print1(10000000*a+1000000*b+100000*c+10000*d+1000*e+100*f+10*g+h,", "))))))))))}

isok(k) = my(d=digits(k)); vecsum(d)^2*sum(i=1, #d, d[i]^3) == k; \\ Michel Marcus, Oct 12 2023

A236220 Cubes which when divided by their digital sum, become square.

Original entry on oeis.org

1, 8, 512, 4913, 5832, 17576, 19683, 64000, 125000, 314432, 421875, 1000000, 1124864, 1259712, 1404928, 3176523, 8000000, 91125000, 130323843, 191102976, 274625000, 348913664, 512000000, 791453125

Offset: 1

K. D. Bajpai, Jan 20 2014

The term digital sum represents: sum of the digits of a number e.g. the digital sum of 4913 is (4+9+1+3 = 17).

			19683 is in the sequence because 19683 divided by its digital sum (1+9+6+8+3 = 27) gives 729 which is a square: 729 = 27^2.
314432 is in the sequence because 314432 divided by its digital sum (3+1+4+4+3+2 = 17) gives 18496 which is a square: 18496 = 136^2.

K. D. Bajpai and Giovanni Resta, Table of n, a(n) for n = 1..1000 (first 110 terms from K. D. Bajpai)

Cf. A000290 (squares), A007953 (digital sum), A061209.

with(StringTools):KD := proc() local a,b,d,e; a:=n^3; b:=add( i,i = convert((a), base,10))(a); d:=a/b; if d=floor(d) then e:=evalf(d^(1/2)); if e=floor(e)then RETURN (a);fi;fi; end: seq(KD(), n=1..3000);

Select[Range[10^4]^3, IntegerQ@ Sqrt[#/Total[ IntegerDigits@#]] &] (* Giovanni Resta, Jan 20 2014 *)

i2d(x) = if(x==0, return([0])); v=[]; while(x>0, y=x%10; x\=10; v=concat(y, v)); v
digsum(n) = d=i2d(n); sum(i=1, #d, d[i])
s=[]; for(n=1, 1000, c=n^3; if(issquare(c/digsum(c)), s=concat(s, c))); s \\ Colin Barker, Jan 20 2014

A257787 Numbers n such that the sum of the digits of n to some power divided by the sum of the digits equal n.

Original entry on oeis.org

1, 2, 3, 4, 5, 6, 7, 8, 9, 37, 48, 415, 231591, 3829377463694454, 56407086228259246207394322684

Offset: 1

Pieter Post, May 08 2015

The first nine terms are trivial, but then the terms become very rare. It appears that this sequence is finite.

			37 = (3^3+7^3)/(3+7).
231591 = (2^7+3^7+1^7+5^7+9^7+1^7)/(2+3+1+5+9+1).

Cf. A061209, A115518, A111434, A114135, A130680, A257784, A257768.

def moda(n,a):
    kk = 0
    while n > 0:
        kk= kk+(n%10)**a
        n =int(n//10)
    return kk
def sod(n):
    kk = 0
    while n > 0:
        kk= kk+(n%10)
        n =int(n//10)
    return kk
for a in range (1, 10):
    for c in range (1, 10**6):
        if c*sod(c)==moda(c, a):
            print (a,c, moda(c,a),sod(c))

a(14) from Giovanni Resta, May 09 2015

a(15) from Chai Wah Wu, Nov 30 2015

A257860 Numbers n such that a digit of n to the power k plus the sum of the other digits of n equals n, where k is a positive integer.

Original entry on oeis.org

1, 89, 132, 264, 518, 739, 2407, 6579, 8200, 8201, 8202, 8203, 8204, 8205, 8206, 8207, 8208, 8209, 32780, 32781, 32782, 32783, 32784, 32785, 32786, 32787, 32788, 32789, 59060, 59061, 59062, 59063, 59064, 59065, 59066, 59067, 59068, 59069, 78145, 524300, 524301, 524302, 524303, 524304, 524305, 524306, 524307, 524308, 524309, 531459, 823567, 2097178

Offset: 1

Pieter Post, May 11 2015

There are numbers that come in groups of 10, like 8200, 32780 and 524300. But there are also a few stand-alone numbers. Like 531459 (=5+3+1+4+5+9^6).

It is easy to generate large terms in the sequence, for example, 9^104+409 and 9^1047+4561 are the smallest terms with 100 and 1000 digits, respectively. - Giovanni Resta, May 12 2015

			89 is in the sequence because 89 = 8+9^2.
2407 is in the sequence because 2407 = 2+4+0+7^4.
8202 is in the sequence because 8202 = 8+ 2^13 +0+2, also 8202 = 8+2+0+2^13.

Cf. A061209, A130680, A257766, A257768, A257784,

Cf. A007953.

import Data.List (nub); import Data.List.Ordered (member)
a257860 n = a257860_list !! (n-1)
a257860_list = 1 : filter f [1..] where
   f x = any (\d -> member (x - q + d) $ ps d) $ filter (> 1) $ nub ds
         where q = sum ds; ds = (map (read . return) . show) x
   ps x = iterate (* x) (x ^ 2)
-- Reinhard Zumkeller, May 12 2015

def sod(n):
    kk = 0
    while n > 0:
        kk= kk+(n%10)
        n =int(n//10)
    return kk
for i in range (1,10**7):
    for j in range(1,len(str(i))+1):
        k=(i//(10**(j-1)))%10
        for m in range (2,30):
            if i==sod(i)+k**m-k:
                print (i)

One more term and some missing data added by Reinhard Zumkeller, May 12 2015

cp's OEIS Frontend

A061210 Numbers which are the fourth powers of their digit sum.

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A061211 Largest number m such that m is the n-th power of the sum of its digits.

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A023106 a(n) is a power of the sum of its digits.

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A257784 Numbers n such that the sum of the digits squared times the sum of the digits of n to some power equals n.

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A254000 Numbers equal to the fifth powers of the sums of their digits.

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A366507 Numbers k such that the sum of the digits of k times the square of the sum of the digits cubed of k equals k.

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A366512 Numbers k such that the square of the sum of the digits times the sum of the cubes of the digits equals k.

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