A130680 -id:A130680 - cp's OEIS frontend

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

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

A257768 Numbers m such that for some power k, m is the sum of d + d^k as d runs through the digits of m.

Original entry on oeis.org

12, 18, 30, 90, 666, 870, 960, 1998, 7816, 42648, 119394, 302034, 360522, 1741752, 12051036, 909341082, 931186956, 1136424308, 1145082306, 8390370196, 49388550660, 52927388760, 100552730520, 41845367362266, 51671446297908, 245917854035004, 607628544623816, 858683110606660, 4023730658941192

Offset: 1

Pieter Post, May 07 2015

The power k of most terms in this sequence is equal to or one more or one less than the number of digits in the term. One exception is 302034: 302034 = 3^9 + 0^9 + 2^9 + 0^9 + 3^9 + 4^9 + 3+0+2+0+3+4.

			666 = (6+6+6) + (6^3 + 6^3 + 6^3).
7816 = (7+8+1+6) + (7^4 + 8^4 + 1^4 + 6^4).
360522 = (3+6+0+5+2+2) + (3^7 + 6^7 + 0^7 + 5^7 + 2^7 + 2^7).

Cf. A115518, A130680, A111434.

mmax:= 10:  # to get all terms < 10^mmax
Res:= NULL:
score:= (c,p) -> add(c[i+1]*(i+i^p), i=0..9):
for m from 2 to mmax do
comps:= convert(map(`-`,combinat:-composition(10+m,10),[1$10]),list):
for c in comps do
  cL:= [seq(i$c[i+1], i=0..9)];
  if max(c[3..-1]) = 0 then slim:= 0 else slim:= 10^m fi;
  for p from 1 do
    s:= score(c,p);
    L:= sort(convert(s,base,10));
    if L = cL then Res:= Res,s; break fi;
    if s >= slim then break fi;
  od:
od:
od:
sort([Res]); # Robert Israel, May 08 2015

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

a(14)-a(29) from Giovanni Resta, May 08 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

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

A368939 Numbers k such that the sum of the digits times the sum of the fourth powers of the digits equals k.

Original entry on oeis.org

0, 1, 182380, 444992

Offset: 1

René-Louis Clerc, Jan 10 2024

There are exactly 4 such numbers (Property 16 of Clerc).

			182380 = (1+8+2+3+8)*(1^4 + 8^4 + 2^4 + 3^4 + 8^4) = 22*8290.

René-Louis Clerc, Quelques nombres de Niven-Harshad particuliers, 2023.

Cf. A115518, A257766, A061209, A061210, A254000, A130680, A366507, A366512.

Select[Range[0,10^7],#==Total[IntegerDigits[#]]*Total[IntegerDigits[#]^4]&] (* James C. McMahon, Jan 11 2024 *)

niven14(k) = my(d=digits(k)); vecsum(d)*sum(i=1, #d, d[i]^4) == k;
for(k=1,10^7,if(niven14(k)==1,print1(k,", ")))

A257786 Numbers n such that the square root of the sum of the digits times the sum of the digits of n in some power equal n.

Original entry on oeis.org

0, 1, 27, 376, 13131, 234595324075, 54377519037479592374299, 8326623359858152426050700, 1513868951125582592290131113769528

Offset: 1

Pieter Post, May 08 2015

It appears that this sequence is finite.

			376 = sqrt(3+7+6)*(3^2+7^2+6^2).
13131 = sqrt(1+3+1+3+1)*(1^7+3^7+1^7+3^7+1^7).

Cf. A028839, A061209, A115518, A130680.

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:
        k= kk+(n%10)
        n =int(n//10)
    return kk
for a in range (1, 10):
    for c in range (1, 10**8):
        if c**2==sod(c)*moda(c,a)**2:
            print (a,c, sod(c),moda(c,a))

a(6) from Giovanni Resta, May 09 2015

a(7)-a(9) from Chai Wah Wu, Nov 29 2015

A370250 Numbers k such that the sum of the digits times the square of the sum of the fourth powers of the digits equals k.

Original entry on oeis.org

0, 1, 5873656512, 7253758561, 29961747275

Offset: 1

René-Louis Clerc, Feb 13 2024

There are exactly 5 such numbers (Property 17 of Clerc).

			7253758561 = (7+2+5+3+7+5+8+5+6+1)*(7^4 + 2^4 + 5^4 + 3^4 + 7^4 + 5^4 + 8^4 + 5^4 + 6^4 + 1^4)^2 = 49*148035889 = 7253758561.

René-Louis Clerc, Quelques nombres de Niven-Harshad particuliers, 2023.

Cf. A368939, A115518, A257766, A061209, A061210, A254000, A130680, A366507, A366512.

niven142(k) = my(d=digits(k)); vecsum(d)*sum(i=1, #d, d[i]^4)^2 == k;
for(k=0,10^12,if(niven142(k)==1,print1(k, ", ")))

cp's OEIS Frontend

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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A257768 Numbers m such that for some power k, m is the sum of d + d^k as d runs through the digits of m.

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

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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.

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A368939 Numbers k such that the sum of the digits times the sum of the fourth powers of the digits equals k.

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A257786 Numbers n such that the square root of the sum of the digits times the sum of the digits of n in some power equal n.

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