A070276 -id:A070276 - cp's OEIS frontend

A111434 Numbers k such that the sums of the digits of k, k^2 and k^3 coincide.

0, 1, 10, 100, 468, 585, 1000, 4680, 5850, 5851, 5868, 10000, 28845, 46800, 58500, 58510, 58680, 58968, 100000, 288450, 468000, 585000, 585100, 586800, 589680, 1000000, 2884500, 4680000, 5850000, 5851000, 5868000, 5896800, 10000000

Offset: 1

Giovanni Resta, Nov 21 2005

The sequence is clearly infinite, since we can add trailing zeros. Is the subset of values not ending in 0 infinite too (see A114135)?

			468 is in the sequence since 468^2 = 219024 and 468^3 = 102503232 and we have 18 = 4+6+8 = 2+1+9+0+2+4 = 1+0+2+5+0+3+2+3+2.
5851 is in the sequence because 5851, 34234201 (= 5851^2) and 200304310051 (=5851^3) all have digital sum 19.

David A. Corneth, Table of n, a(n) for n = 1..1124 (using the b-file in A114135).

Cf. A011557, A058369, A070276.

s:=proc(n) local nn: nn:=convert(n,base,10): sum(nn[j],j=1..nops(nn)): end: a:=proc(n) if s(n)=s(n^2) and s(n)=s(n^3) then n else fi end: seq(a(n),n=0..1000000); # Emeric Deutsch, May 13 2006

SumOfDig[n_]:=Apply[Plus, IntegerDigits[n]]; Do[s=SumOfDig[n]; If[s==SumOfDig[n^2] && s==SumOfDig[n^3], Print[n]], {n, 10^6}]
Select[Range[0,10000000],Length[Union[Total/@IntegerDigits[{#,#^2,#^3}]]] == 1&] (* Harvey P. Dale, Apr 26 2014 *)

b-file Corrected by David A. Corneth, Jul 22 2021

A292788 For n > 1, a(n) = least positive k, not a power of n, such that the digital sum of k in base n equals the digital sum of k^3 in base n.

Original entry on oeis.org

56953, 13, 2, 3, 20, 2, 6, 3, 8, 5, 1110, 3, 65, 8, 4, 7, 86, 9, 2374, 4, 8, 12, 114, 3, 99, 12, 135, 15, 3567, 4, 185, 15, 11, 16, 6, 19, 73, 20, 12, 5, 81, 6, 85, 23, 19, 24, 93, 7, 97, 24, 18, 27, 796, 28, 44, 7, 19, 28, 413, 4, 365, 32, 8, 31, 26, 21, 200

Offset: 2

Rémy Sigrist, Sep 23 2017

The term a(10) = 8 belongs to A070276.
For any n > 1, a(n^2) <= n.
Is this sequence defined for any n > 1 ?
Apparently, a(k) < k for any odd k > 3.
Among the first 99 999 terms, the digital sum of a(n) in base n is > n for n = 2, 12, 20, 30.
The scatterplot of the sequence shows beams on the upper part, which correspond to clusters of close points for which a(n) = k*n + (n-k-e) for some k > 0 and e in { 0, 2 }.
See also A292787 for a similar sequence involving squares instead of cubes.
The least positive k, not a power of 2, such that the hamming weight of k equals the hamming weight of k^4 is 34225258495.

			For n = 3:
- let d_3 denote the digital sum in base 3 (d_3 = A053735),
- 1 is a power of 3,
- d_3(2) = 2 and d_3(2^3) = 4,
- 3 is a power of 3,
- d_3(4) = 2 and d_3(4^3) = 4,
- d_3(5) = 3 and d_3(5^3) = 7,
- d_3(6) = 2 and d_3(6^3) = 4,
- d_3(7) = 3 and d_3(7^3) = 5,
- d_3(8) = 4 and d_3(8^3) = 8,
- 9 is a power of 3,
- d_3(10) = 2 and d_3(10^3) = 4,
- d_3(11) = 3 and d_3(11^3) = 9,
- d_3(12) = 2 and d_3(12^3) = 4,
- d_3(13) = 3 and d_3(13^3) = 3,
- hence a(3) = 13.

Rémy Sigrist, Table of n, a(n) for n = 2..10000
Rémy Sigrist, Colorized scatterplot of the sequence for n=2..100000

Cf. A053735, A070276, A292787.

With[{kk = 10^5}, Table[SelectFirst[Complement[Range[2, kk], n^Range@ Floor@ Log[n, kk]], Total@ IntegerDigits[#, n] == Total@ IntegerDigits[#^3, n] &] /. k_ /; MissingQ@ k -> -1, {n, 2, 68}]] (* Michael De Vlieger, Sep 24 2017 *)

a(n) = my (p=1); for (k=1, oo, if (k==p, p*=n, if (sumdigits(k,n) == sumdigits(k^3,n), return (k))))

A309883 Numbers k such that A003132(k^2) = A003132(k), where A003132(n) is the sum of the squares of the digits of n.

Original entry on oeis.org

0, 1, 10, 35, 100, 152, 350, 377, 452, 539, 709, 1000, 1299, 1398, 1439, 1519, 1520, 1569, 1591, 1679, 1965, 2599, 2838, 3332, 3500, 3598, 3770, 4520, 4586, 4754, 4854, 5390, 5501, 5835, 5857, 6388, 6595, 6735, 6861, 6951, 7090, 7349, 7887, 8395, 9795, 10000, 10056, 10159, 10389, 11055, 11091, 12990, 12999

Offset: 1

Antonio Roldán, Aug 21 2019

If k is in the sequence, then so are k*10^r, r >= 1.

			377^2 = 142129, A003132(377) = 3^2 + 7^2 + 7^2 = 107, A003132(142129) = 1^2 + 4^2 + 2^2 + 1^2 + 2^2 + 9^2 = 107.

Robert Israel, Table of n, a(n) for n = 1..10000

Cf. A003132, A058369, A070276, A174460, A176670, A178213, A307735, A309884.

[0] cat [k:k in [1..13000]| &+[c^2: c in Intseq(k)] eq &+[c^2: c in Intseq(k^2)]]; // Marius A. Burtea, Aug 24 2019

filter:= proc(n) local t;
  add(t^2, t = convert(n,base,10)) = add(t^2, t = convert(n^2,base,10))
end proc:
select(filter, [$0..20000]); # Robert Israel, Apr 30 2023

digSum[n_] := Total[IntegerDigits[n]^2]; Select[Range[0, 13000], digSum[#] == digSum[#^2] &] (* Amiram Eldar, Aug 22 2019 *)

for(i = 0, 30000, if(norml2(digits(i^2)) == norml2(digits(i)), print1(i, ", ")))

def A003132(n):
    s = 0
    while n > 0:
        s, n = s+(n%10)**2, n//10
    return s
n, a = 0, 0
while n < 50:
    if A003132(a) == A003132(a*a):
        n = n+1
        print(n,a)
    a = a+1 # A.H.M. Smeets, Aug 23 2019

A204324 Numbers k such that A007953(k) >= A007953(k^3), where A007953 = digital sum in base 10.

Original entry on oeis.org

0, 1, 8, 10, 80, 100, 171, 378, 468, 487, 577, 585, 586, 587, 684, 800, 1000, 1710, 3780, 4680, 4870, 4877, 5770, 5850, 5851, 5860, 5868, 5870, 6840, 8000, 10000, 15877, 17100, 28845, 28847, 28885, 28887, 37800, 46800, 46877, 48700, 48770, 48784, 49468

Offset: 1

M. F. Hasler, Jan 14 2012

When k is in the sequence, then 10*k is in the sequence, too.

Cf. A064399, A064209, A064210.

Select[Range[0,51000],Total[IntegerDigits[#]]>=Total[IntegerDigits[#^3]]&] (* Harvey P. Dale, Jul 05 2025 *)

for(n=0,1e6, A007953(n)>=A007953(n^3)&print1(n","))

A204324 = A064209 union A070276.

A266315 Numbers n such that 2*n and n^3 have the same digit sum.

Original entry on oeis.org

0, 171, 324, 378, 468, 684, 1710, 3240, 3780, 4680, 6840, 17100, 28845, 29241, 32400, 34884, 37800, 46800, 46944, 48924, 68400, 69174, 84348, 171000, 242424, 288450, 292410, 324000, 348840, 378000, 467424, 468000, 469440, 489240, 493794, 684000, 691740

Offset: 1

Vincenzo Librandi, Jan 01 2016

			171 is in the sequence because 171^3 = 5000211 and 2*171 = 342 have the same digit sum: 9.

G. C. Greubel, Table of n, a(n) for n = 1..160

Cf. A000578, A005843, A070276, A260906.

[n: n in [0..2*10^6] | &+Intseq(2*n) eq &+Intseq(n^3)];

Select[Range[0, 7 10^5], Total[IntegerDigits[2 #]] == Total[IntegerDigits[#^3]] &]

isok(n) = sumdigits(2*n)==sumdigits(n^3); \\ Michel Marcus, Jan 02 2016

A309884 Numbers k such that A003132(k^3) = A003132(k), where A003132(n) is the sum of the squares of the digits of n.

Original entry on oeis.org

0, 1, 10, 74, 100, 740, 1000, 3488, 7400, 10000, 23658, 30868, 34880, 47508, 48517, 52187, 58947, 59468, 67685, 68058, 74000, 76814, 78368, 78845, 84878, 100000, 108478, 145877, 149217, 163871, 179685, 186884, 188647, 218977, 219878, 236580, 238758, 248967, 278638, 292597, 308680

Offset: 1

Antonio Roldán, Aug 21 2019

If k is in the sequence, then so are k*10^r, with r >= 1.

			74^3 = 405224, A003132(74) = 7^2 + 4^2 = 65, A003132(405224) = 4^2 + 0^2 + 5^2 + 2^2 + 2^2 + 4^2 = 65.

Cf. A003132, A058369, A070276, A174460, A176670, A178213, A307735, A309883.

[0] cat [k:k in [1..310000]| &+[c^2: c in Intseq(k)] eq &+[c^2: c in Intseq(k^3)]]; // Marius A. Burtea, Aug 26 2019

digSum[n_] := Total[IntegerDigits[n]^2]; Select[Range[0, 310000], digSum[#] == digSum[#^3] &] (* Amiram Eldar, Aug 22 2019 *)

for(i = 0, 400000, if(norml2(digits(i^3)) == norml2(digits(i)), print1(i, ", ")))

A321881 Numbers whose sum and product of digits are cubes.

Original entry on oeis.org

0, 1, 8, 10, 80, 100, 107, 170, 206, 260, 305, 350, 404, 440, 503, 530, 602, 620, 701, 710, 800, 999, 1000, 1007, 1016, 1025, 1034, 1043, 1052, 1061, 1070, 1106, 1124, 1142, 1160, 1205, 1214, 1241, 1250, 1304, 1340, 1403, 1412, 1421, 1430, 1502, 1520, 1601, 1610, 1700

Offset: 1

Enrique Navarrete, Nov 20 2018

The first numbers in the sequence that are cubes themselves are 0,1,8,1000,8000.
a(22)=999 is the only term up to n=120 related to the cube 27 (the previous ones relate to 0,1,8).
Also, a(22)=999 is the first term that has more than one digit and consists of a single repeated digit; the next ones are 11111111 and 333333333.

			93111111111111111 (15 ones) is in the sequence since the sum and the product of the digits is 27 (a cube).
333 is not in the sequence since the product of the digits is 27 but the sum is 9 (not a cube).

Robert Israel, Table of n, a(n) for n = 1..10000

Cf. A007953, A046031, A062398, A070276, A059094, A237767.

[n:n in [0..2000]| IsPower((&+Intseq(n)), 3) and IsPower((&*Intseq(n)), 3)] // Marius A. Burtea, Jan 21 2019

filter:= proc(n) local L;
  L:= convert(n,base,10);
  simplify(convert(L,`+`)^(1/3))::integer and
  simplify(convert(L,`*`)^(1/3))::integer;
end proc:
select(filter, [$0..1000]); # Robert Israel, Jan 21 2019

cubeQ[n_] := IntegerQ[Surd[n, 3]]; aQ[n_] := cubeQ[Plus @@ IntegerDigits[n]] &&
cubeQ[Times @@ IntegerDigits[n]]; Select[Range[0, 3000], aQ] (* Amiram Eldar, Nov 20 2018 *)

isok(n) = my(d=digits(n)); ispower(vecsum(d), 3) && ispower(vecprod(d), 3); \\ Michel Marcus, Nov 29 2018

A124667 Prime numbers p such that the sum of the digits of p equals the sums of the digits of p^3.

Original entry on oeis.org

487, 577, 4877, 5851, 15877, 467587, 496187, 697967, 5889959, 8194787, 14596991, 17978887, 27698887, 47959487, 58590487, 58678903, 59489371, 79492771, 79897897, 79932871, 109148887, 109696969, 145969757, 227799577, 276857947

Offset: 1

Tanya Khovanova, Dec 23 2006

			487^3 = 115501303 -- the sum of the digits of 487 and 115501303 is the same and is equal to 19.

An equivalent sequence for squares is A058370 = Primes p such that p and p^2 have same digit sum. This sequence is prime subsequence of A070276 = Sum of digits of n equals the sum of digits of n^3.

Select[Range[10000000], PrimeQ[ # ] && Plus @@ IntegerDigits[ # ] == Plus @@ IntegerDigits[ #^3] &]
Select[Prime[Range[151*10^5]],Total[IntegerDigits[#]]==Total[ IntegerDigits[ #^3]]&] (* Harvey P. Dale, Feb 17 2018 *)

More terms from Olaf Voß, Feb 11 2008

cp's OEIS Frontend

A111434 Numbers k such that the sums of the digits of k, k^2 and k^3 coincide.

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A292788 For n > 1, a(n) = least positive k, not a power of n, such that the digital sum of k in base n equals the digital sum of k^3 in base n.

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A309883 Numbers k such that A003132(k^2) = A003132(k), where A003132(n) is the sum of the squares of the digits of n.

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A204324 Numbers k such that A007953(k) >= A007953(k^3), where A007953 = digital sum in base 10.

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A266315 Numbers n such that 2*n and n^3 have the same digit sum.

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A309884 Numbers k such that A003132(k^3) = A003132(k), where A003132(n) is the sum of the squares of the digits of n.

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A321881 Numbers whose sum and product of digits are cubes.

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