A061209 -id:A061209 - cp's OEIS frontend

A259313 Numbers m for which there exists a k>=2 such that m equals the average of digitsum(m^p) for p from 1 to k.

1, 9, 12, 13, 16, 19, 21, 49, 61, 67, 84, 106, 160, 191, 207, 250, 268, 373, 436, 783, 2321, 3133, 3786, 3805, 4842, 5128, 8167, 13599, 29431, 35308

Offset: 1

Pieter Post, Jun 24 2015

Digitsum = (A007953).

The 'k's are 2, 2, 4, 3, 4, 5, 7, 12, 15, 16, 19, 21, 57, 37, 38, 79, 48, 63, 72, 119, 306, 397, 469, 472, 582, 613, 927, 1461, 2926, 3449, ..., . - Robert G. Wilson v, Jul 30 2015

			Digitsum(9) is 9, digitsum(9^2) is 9. (9+9)/2 = 9. So 9 is in this sequence.
12^1 = 12, 12^2 = 144, 12^3 = 1728 and 12^4 = 20736. Digitsum(12) = 3, digitsum(144) = 9, digitsum(1728) = 18, digitsum(20736) = 18, (3+9+18+18)/4 = 12. So 12 is in this sequence.

Cf. A007953, A061910, A061209, A061210.

fQ[n_] := If[ IntegerQ@ Log10@ n, False, Block[{pwr = 2, s = Plus @@ IntegerDigits@ n}, While[s = s + Plus @@ IntegerDigits[n^pwr]; s < n*pwr, pwr++]; If[s == n*pwr, True, False]]]; k = 1; lst = {1}; While[k < 100001, If[fQ@ k, AppendTo[lst, k]]; k++]; lst (* Robert G. Wilson v, Jul 30 2015 *)

def sod(n):
    kk = 0
    while n > 0:
        kk= kk+(n%10)
        n =int(n//10)
    return kk
for c in range (2, 10**3):
    bb=0
    for a in range(1,200):
        bb=bb+sod(c**a)
        if bb==c*a:
            print (c,a)

a(21)-a(28) from Giovanni Resta, Jun 24 2015
a(1)-a(28) checked by Robert G. Wilson v, Jul 30 2015
a(29)-a(30) from Robert G. Wilson v, Jul 30 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,", ")))

A375343 Numbers which are the sixth powers of their digit sum.

Original entry on oeis.org

0, 1, 34012224, 8303765625, 24794911296, 68719476736

Offset: 1

René-Louis Clerc, Aug 12 2024

Solutions can have no more than 13 digits, since (13*9)^6 < 10^13.

			68719476736 = (6+8+7+1+9+4+7+6+7+3+6)^6 = 64^6.

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

Cf. A001014, A007953, A061209, A061210, A254000, A061211.

Cf. A055577, A152147.

for (k=0, sqrtnint(10^13,6), if (k^6 == sumdigits(k^6)^6, print1(k^6, ", ")); )

{ k : k = A007953(k)^6}.

a(n) = A055577(n)^6. - Alois P. Heinz, Aug 24 2024

A379767 Triangle read by rows: row n lists numbers which are the n-th powers of their digit sum.

Original entry on oeis.org

0, 1, 2, 3, 4, 5, 6, 7, 8, 9, 0, 1, 81, 0, 1, 512, 4913, 5832, 17576, 19683, 0, 1, 2401, 234256, 390625, 614656, 1679616, 0, 1, 17210368, 52521875, 60466176, 205962976, 0, 1, 34012224, 8303765625, 24794911296, 68719476736, 0, 1, 612220032, 10460353203, 27512614111, 52523350144, 271818611107, 1174711139837, 2207984167552, 6722988818432

Offset: 1

René-Louis Clerc, Jan 02 2025

Each row begins with 0, 1. Solutions can have no more than R(n) digits, since (R(n)*9)^n < 10^R(n), hence, for each n, there are a finite number of solutions (Property 1 and table 1 of Clerc).

			Triangle begins:
  1 | 0, 1, 2, 3, 4, 5, 6, 7, 8, 9;
  2 | 0, 1, 81;
  3 | 0, 1, 512, 4913, 5832, 17576, 19683;
  4 | 0, 1, 2401, 234256, 390625, 614656, 1679616;
  5 | 0, 1, 17210368, 52521875, 60466176, 205962976;
  6 | 0, 1, 34012224, 8303765625, 24794911296, 68719476736;
  7 | 0, 1, 612220032, 10460353203, 27512614111, 52523350144, 271818611107, 1174711139837, 2207984167552, 6722988818432;
  8 | 0, 1, 20047612231936, 72301961339136, 248155780267521;
  9 | 0, 1, 3904305912313344, 45848500718449031, 150094635296999121;
  ...

René-Louis Clerc, Rows n=1...270 of triangle, flattened
René-Louis Clerc, Nombres de Niven-Harshad égaux à une puissance de la somme de leurs chiffres, 2024.

Rows 3..6 are A061209, A061210, A254000, A375343.
Row lengths are 1 + A046019(n).
Cf. A001014, A007953, A061211 (largest terms), A133509.
Cf. A152147.

R(n) = for(j=2,oo, if((j*9)^n <10^j, return(j)));
row(n) = my(L=List()); for (k=0, sqrtnint(10^R(n),n), if (k^n == sumdigits(k^n)^n, listput(L, k^n))); Vec(L)

A085754 Numbers n such that (digital sum of n)^3 = reversal of n. (Powers of 10 excluded.)

Original entry on oeis.org

215, 2150, 2385, 3194, 21500, 23850, 31940, 38691, 67571, 215000, 238500, 319400, 386910, 675710, 2150000, 2385000, 3194000, 3869100, 6757100, 21500000, 23850000, 31940000, 38691000, 67571000, 215000000, 238500000, 319400000

Offset: 1

Jason Earls, Jul 21 2003

Numbers whose digit reversals are members of A061209. - David Wasserman, Feb 09 2005

Cf. A061209.

More terms from David Wasserman, Feb 09 2005

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

A257969 Numbers m such that the sum of the digits (sod) of m, m^2, m^3, ..., m^9 are in arithmetic progression: sod(m^(k+1)) - sod(m^k) = f for k=1..8.

Original entry on oeis.org

1, 10, 100, 1000, 7972, 10000, 53941, 79720, 100000, 134242, 539410, 698614, 797200, 1000000, 1342420, 5394100, 6986140, 7525615, 7972000, 9000864, 10000000, 10057054, 13424200, 15366307, 17513566, 20602674, 23280211, 24716905, 25274655, 25665559, 32083981, 34326702, 34446204, 34534816

Offset: 1

Pieter Post, May 15 2015

All powers of 10 are terms of this sequence.
If m is a term, then so is 10*m.
Number of terms < 10^k for k >= 1: 1, 2, 3, 5, 8, 13, 20, 62.

			7972 is in the sequence, because the difference between the successive sum-of-digit values is 15:
  sod(7972) = 25;
  sod(7972^2) = 40;
  sod(7972^3) = 55;
  sod(7972^4) = 70;
  sod(7972^5) = 85;
  sod(7972^6) = 100;
  sod(7972^7) = 115;
  sod(7972^8) = 130;
  sod(7972^9) = 145;
  sod(7972^10) = 178, where the increment is no longer 15.
But there are seven numbers below 10^9 with a longer sequence (namely, 134242, 23280211, 40809168, 46485637, 59716223, 66413917, and 97134912) where sod(m^(k+1)) - sod(m^k) = f for k=1..9.
  sod(134242) = 16;
  sod(134242^2) = 40;
  sod(134242^3) = 64;
  sod(134242^4) = 88;
  sod(134242^5) = 112;
  sod(134242^6) = 136;
  sod(134242^7) = 160;
  sod(134242^8) = 184;
  sod(134242^9) = 208;
  sod(134242^10) = 232;
  sod(134242^11) = 283, where the increment is no longer 24.

Cf. A061209, A115518, A257784, A258722.

fQ[n_] := Block[{g}, g[x_] := Power[x, #] & /@ Range@ 9; Length@ DeleteDuplicates@ Differences[Total[IntegerDigits@ #] & /@ g@ n] == 1]; Select[Range@ 1000000, fQ] (* Michael De Vlieger, Jun 12 2015 *)
Select[Range[35*10^6],Length[Union[Differences[Total/@IntegerDigits[ #^Range[9]]]]] ==1&] (* Harvey P. Dale, Aug 23 2017 *)

isok(n) = {my(osod = sumdigits(n^2)); my(f = osod - sumdigits(n)); for (k=3, 9, my(nsod = sumdigits(n^k)); if (nsod - osod != f, return (0)); osod = nsod;); return (1);} \\ Michel Marcus, May 28 2015

{m : sod(m^(k+1)) - sod(m^k) = f for k=1..8}.

Corrected and extended by Harvey P. Dale, Aug 23 2017

Edited by Jon E. Schoenfield, Mar 01 2022

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, ", ")))

A385158 Numbers k such that the sum of the digits of k is a number that appears as a substring of k, and every nonzero digit of k appears in that sum.

Original entry on oeis.org

1, 2, 3, 4, 5, 6, 7, 8, 9, 10, 20, 30, 40, 50, 60, 70, 80, 90, 100, 199, 200, 300, 400, 500, 600, 700, 800, 900, 919, 1000, 1188, 1818, 1881, 1909, 1990, 2000, 2999, 3000, 4000, 5000, 6000, 7000, 8000, 8118, 8181, 9000, 9019, 9190, 9299, 9929, 10000

Offset: 1

Rivka Maryles, Jun 19 2025

			1818 is in the sequence because 1 + 8 + 1 + 8 = 18, and 18 appears within the number. Also, all nonzero digits (1 and 8) are found in the digit sum (18).

Cf. A052018, A007953, A031286 (numbers containing a given substring), A070939 (numbers equal to the sum of their digits concatenated), A061209 (numbers containing their digit sum).

filter:= proc(k) local L,s,S,nL,nS,i;
  L:= convert(k,base,10);
  nL:= nops(L);
  s:= convert(L,`+`);
  S:= convert(s,base,10);
  nS:= nops(S);
  (convert(L,set) minus {0} = convert(S,set) minus {0}) and member(S, [seq(L[i..i+nS-1],i=1..nL-nS+1)])
end proc:
select(filter, [$1..10000]); # Robert Israel, Jul 23 2025

isok[n_] := Module[{digits, sum, sumStr},
  digits = IntegerDigits[n];
  sum = Total[digits];
  sumStr = ToString[sum];
  StringContainsQ[ToString[n], sumStr] &&
    AllTrue[DeleteCases[digits, 0],
      DigitCount[sum, 10, #] > 0 &]
];
Select[Range[9999], isok]

def ok(n):
    digits = str(n)
    digit_sum_str = str(sum(map(int, digits)))
    return digit_sum_str in digits and all(d in digit_sum_str for d in set(digits) - {'0'})
print([k for k in range(1, 10001) if ok(k)])

cp's OEIS Frontend

A259313 Numbers m for which there exists a k>=2 such that m equals the average of digitsum(m^p) for p from 1 to k.

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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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A375343 Numbers which are the sixth powers of their digit sum.

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A379767 Triangle read by rows: row n lists numbers which are the n-th powers of their digit sum.

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A085754 Numbers n such that (digital sum of n)^3 = reversal of n. (Powers of 10 excluded.)

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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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A257969 Numbers m such that the sum of the digits (sod) of m, m^2, m^3, ..., m^9 are in arithmetic progression: sod(m^(k+1)) - sod(m^k) = f for k=1..8.

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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.

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