A055012 -id:A055012 - cp's OEIS frontend

A362953 Numbers N such that N + the sum of the cubes of its digits is again a third power.

0, 34, 352, 540, 1167, 1942, 2176, 3312, 4093, 5454, 8019, 9380, 12025, 12130, 13068, 13158, 15344, 15991, 16279, 16675, 21149, 22699, 22789, 30988, 32257, 32365, 35238, 37883, 37955, 41866, 45549, 54523, 57906, 58530, 62579, 72588, 83692, 83782, 89604, 102952

Offset: 0

Will Gosnell and M. F. Hasler, May 09 2023

			The sum of the cubes of the digits of N = 34 is 3^3 + 4^3 = 27 + 64 = 91; added to the number N itself yields  125 which is again a cube, 5^3. Therefore 34 is in this sequence.

Karl-Heinz Hofmann, Table of n, a(n) for n = 0..10000
Karl-Heinz Hofmann, Visualization of n = 0 to 11.

Cf. A000578 (the cubes), A055012 (sum of cubes of decimal digits of n).

Cf. A362954 (the same for 4th powers).

select( {is(n,p=3)=ispower(vecsum([d^p|d<-digits(n)])+n,p)}, [0..10^5])

aupto = 103000
A362953 = []
A000578 = set(cu**3 for cu in range(0, int(aupto**(1/3)+3)))
for n in range(0,aupto+1):
    if n + sum(int(digit)**3 for digit in str(n)) in A000578: A362953.append(n)
print(A362953) # Karl-Heinz Hofmann, May 24 2023

A065137 Sum of digits of n plus sum of cubes of digits of n.

Original entry on oeis.org

0, 2, 10, 30, 68, 130, 222, 350, 520, 738, 2, 4, 12, 32, 70, 132, 224, 352, 522, 740, 10, 12, 20, 40, 78, 140, 232, 360, 530, 748, 30, 32, 40, 60, 98, 160, 252, 380, 550, 768, 68, 70, 78, 98, 136, 198, 290, 418, 588, 806, 130, 132, 140, 160, 198, 260, 352, 480, 650

Offset: 0

Henry Bottomley, Oct 15 2001

			a(23) = (2+3)+(2^3+3^3) = 40.

Harvey P. Dale, Table of n, a(n) for n = 0..1000

Cf. A065138.

Table[Total[IntegerDigits[n]]+Total[IntegerDigits[n]^3],{n,0,60}] (* Harvey P. Dale, Jul 15 2017 *)

a(n) = A007953(n)+A055012(n)

A067170 Numbers n such that sum of the cubes of the distinct prime factors of n equals the sum of the cubes of the digits of n.

Original entry on oeis.org

2, 3, 5, 7, 250, 735, 2500, 25000, 250000, 1858560, 2500000, 18585600, 25000000, 91990080, 185856000, 242121642, 250000000, 919900800, 1081088775, 1390120992, 1768635648, 1858560000, 2500000000, 5435938431, 7245987840, 9199008000, 9475854336, 17996666688, 18585600000, 24214634829, 25000000000

Offset: 1

Joseph L. Pe, Feb 18 2002

If 10*m is a term (e.g. m = 25, 185856, 9199008), then 10^k * m is a term for all k >= 1. Therefore this sequence is infinite. - Amiram Eldar, Sep 28 2019

The sum of cubes of digits of a k-digit number is at most 729*k. Therefore any term with at most k digits is p-smooth where p is the largest prime < (729*k)^(1/3). - David A. Corneth, Sep 28 2019

			The prime factors of 735 are 3,5,7, the sum of whose cubes = 495 = sum of the cubes of the digits of 735; so 735 is a term of the sequence.

David A. Corneth, Table of n, a(n) for n = 1..11790 (first 260 terms from Giovanni Resta, terms < 10^34)

Cf. A005064, A006753, A055012, A067184.

f[n_] := Module[{a, l, t, r}, a = FactorInteger[n]; l = Length[a]; t = Table[a[[i]][[1]], {i, 1, l}]; r = Sum[(t[[i]])^3, {i, 1, l}]]; g[n_] := Module[{b, m, s}, b = IntegerDigits[n]; m = Length[b]; s = Sum[(b[[i]])^3, {i, 1, m}]]; Select[Range[2, 10^6], f[ # ] == g[ # ] &]

sd(n) = my(d=digits(n)); sum(k=1, #d, d[k]^3); \\ A055012
sp(n) = my(f=factor(n)); sum(k=1, #f~, f[k,1]^3); \\ A005064
isok(n) = sp(n) == sd(n); \\ Michel Marcus, Sep 28 2019

a(10)-a(14) from Amiram Eldar, Sep 28 2019
a(15)-a(18) from Michel Marcus, Sep 28 2019
a(20)-a(29) from David A. Corneth, Sep 28 2019
Missing a(19) from Giovanni Resta, Sep 28 2019

A072082 Numbers divisible by the cube of the sum of their digits in base 10.

Original entry on oeis.org

1, 10, 100, 200, 500, 512, 1000, 2000, 2401, 4913, 5000, 5103, 5120, 5832, 10000, 10206, 11000, 11200, 11664, 13122, 14000, 17576, 19000, 19683, 20000, 20412, 21141, 23000, 23328, 24010, 28000, 29160, 32000, 37000, 39366, 40000, 40824

Offset: 1

Labos Elemer, Jun 14 2002

If k is a term, then 10 * k is a term. There are an infinite number of terms that are not divisible by 10. The numbers m = 24 * 10^(294 * k - 292) +1 are divisible by 7^3 = digsum(m)^3. Also, the numbers s = 491 * 10^(4624 * k - 4623) + 3, k >= 1, are divisible by 17^3 = digsum(s)^3. - Marius A. Burtea, Mar 18 2020

			k=98415: sumdigits(98415)=27, q=98415=5*27*27*27.

Donovan Johnson, Table of n, a(n) for n = 1..1000

Cf. A055012, A005349, A003634, A072081, A072083.

[k:k in [1..41000]| k mod &+Intseq(k)^3 eq 0]; // Marius A. Burtea, Mar 18 2020

sud[x_] := Apply[Plus, IntegerDigits[x]] Do[s=sud[n]^3; If[IntegerQ[n/s], Print[n]], {n, 1, 10000}]
Select[Range[50000],Divisible[#,Total[IntegerDigits[#]]^3]&] (* Harvey P. Dale, Mar 22 2016 *)

is(n)=n%sumdigits(n)^3==0 \\ Charles R Greathouse IV, Mar 19 2020

A072884 3rd-order digital invariants: the sum of the cubes of the digits of n equals some number k and the sum of the cubes of the digits of k equals n.

Original entry on oeis.org

1, 136, 153, 244, 370, 371, 407, 919, 1459

Offset: 1

Robert G. Wilson v and Harvey P. Dale, Aug 09 2002

			136 is included because 1^3 + 3^3 + 6^3 = 244 and 2^3 + 4^3 + 4^3 = 136.
244 is included because 2^3 + 4^3 + 4^3 = 136 and 1^3 + 3^6 + 6^3 = 244.

J.-M. De Koninck and A. Mercier, 1001 Problèmes en Théorie Classique Des Nombres, Problem 257 pp. 41; 185 Ellipses Paris 2004.
David Wells, The Penguin Dictionary of Curious and Interesting Numbers, Revised Edition, London, England, 1997, pp. 124-125.

Cf. A072409.

f[n_] := Apply[Plus, IntegerDigits[Apply[Plus, IntegerDigits[n]^3]]^3]; Select[ Range[10^7], f[ # ] == # &]
Select[Range[10000], Plus@@(IntegerDigits[Plus@@(IntegerDigits[ # ]^3)]^3)== #&]

k such that f(f(k)) = k, where f(k) = A055012(k). - Lekraj Beedassy, Sep 10 2004

A165551 Sum of cube of digits is sum of digits of cube.

Original entry on oeis.org

0, 1, 2, 10, 20, 31, 100, 103, 123, 200, 203, 301, 302, 310, 1000, 1003, 1030, 1230, 1302, 1312, 1321, 2000, 2003, 2030, 2312, 3001, 3002, 3010, 3020, 3032, 3100, 3112, 3211, 3213, 3221, 10000, 10003, 10030, 10033, 10232, 10300, 10303, 11223, 12033

Offset: 0

Rémy Sigrist, Sep 21 2009

			For n=31, n^3=29791, sum of digits of cube is 2+9+7+9+1=28, sum of cube of digits is 3^3+1=27+1=28, thus 31 appears in the sequence.

Harvey P. Dale, Table of n, a(n) for n = 0..1000

Cf. A007953, A055012, A165550.

Select[Range[0,13000],Total[IntegerDigits[#^3]]==Total[IntegerDigits[ #]^3]&] (* Harvey P. Dale, Nov 13 2011 *)

cubdigsum(n) = local(s=0);while(n,s=s+(n%10)^3;n=n\10);return(s)
for(n=0,100000, if(cubdigsum(n)==sumdigits(n^3), print1(n, ",") ) )

A169662 Numbers divisible by the sum of their digits, and by the sum of their digits squared, by the sum of their digits cubed and by the sum of 4th powers of their digits.

Original entry on oeis.org

1, 10, 100, 110, 111, 1000, 1010, 1011, 1100, 1101, 1110, 2000, 5000, 10000, 10010, 10011, 10100, 10101, 10110, 11000, 11001, 11010, 11100, 20000, 50000, 55000, 100000, 100010, 100011, 100100, 100101, 100110, 101000, 101001, 101010, 101100

Offset: 1

Michel Lagneau, Apr 05 2010

The numbers such that all digits are nonzero are rare (see the subsequence A176194).

			1121211 is a term since 1^4 + 1^4 + 2^4 + 1^4 + 2^4 + 1^4 + 1^4 = 37 and 1121211 = 37*30303 ; 1^3 + 1^3 + 2^3 + 1^3 + 2^3 + 1^3 + 1^3 = 21 and 1121211 = 21*53391 ; 1^2 + 1^2 + 2^2 + 1^2 + 2^2 + 1^2 + 1^2 = 13 and 1121211 = 13* 86247 ; 1 + 1 + 2 + 1 + 2 + 1 + 1 = 9 and 1121211 = 9*124579.

Amiram Eldar, Table of n, a(n) for n = 1..10000

Intersection of A005349, A034087, A034088 and A169665.

Cf. A072081, A117562, A176194.

isA169662 := proc(n)
        dgs := convert(n,base,10) ;
        if (n mod ( add(d,d=dgs) ) = 0)  and (n mod (add(d^2,d=dgs) )) =0 and (n mod (add(d^3,d=dgs))) =0 and (n mod (add(d^4,d=dgs))) = 0 then
                true;
        else
                false;
        end if;
end proc:
for i from 1 to 110000 do
        if isA169662(i) then
                printf("%d,",i) ;
        end if;
end do: # R. J. Mathar, Nov 07 2011

q[n_] := And @@ Divisible[n, Plus @@@ Transpose @ Map[#^Range[4] &, IntegerDigits[n]]]; Select[Range[10^5], q] (* Amiram Eldar, Jan 31 2021 *)

{n : A007953(n)|n and A003132(n)|n and A055012(n)| n and A055013(n)| n}.

A176194 Numbers with no zero digits divisible by the sum of the k-th powers of their digits, for each k = 1,2,3,4.

Original entry on oeis.org

1, 111, 1121211, 11243232, 12132432, 12413232, 22331232, 23111352, 23411232, 24113232, 41223312, 42131232, 44662464, 111111111, 112452144, 114251424, 135964224, 211412544, 246134592, 313212312, 332131212, 382941675, 416283624, 442114512, 523173456, 671635575, 979652772

Offset: 1

Michel Lagneau, Apr 11 2010

For the numbers divisible by the sum of k-th powers of digits including 0, see A169662. The numbers such that the digits are > 0 are rare.

			For n = 246134592 we obtain :
2^4 + 4^4 + 6^4 + 1^4 + 3^4 + 4^4 + 5^4 + 9^4 + 2^4 = 9108, and 246134592 = 9108*27024 ;
2^3 + 4^3 + 6^3 + 1^3 + 3^3 + 4^3 + 5^3 + 9^3 + 2^3 = 1242, and 246134592 = 1242*198176 ;
2^2 + 4^2 + 6^2 + 1^2 + 3^2 + 4^2 + 5^2 + 9^2 + 2^2 = 192, and 246134592 = 192*1281951 ;
2 + 4 + 6 + 1 + 3 + 4 + 5 + 9 + 2 = 36, and 246134592 = 36*6837072.

Amiram Eldar, Table of n, a(n) for n = 1..464

Intersection of A052382 and A169662.

Cf. A034087, A072081, A117562, A034087, A072081, A117562, A034088, A034088.

with(numtheory):for n from 2 to 500000000 do:l:=evalf(floor(ilog10(n))+1):n0:=n:s1:=0:s2:=0:s3:=0:s4:=0:p:=1:for m from 1 to l do:q:=n0:u:=irem(q,10):v:=iquo(q,10):n0:=v :s1:=s1+u:p:=p*u:s2:=s2+u^2:s3:=s3+u^3:s4:=s4+u^4: od:if irem(n,s1)=0 and irem(n,s2)=0 and irem(n,s3)=0 and irem(n,s4)=0 and p<>0 then print(n):else fi:od:

A007953 (n)|n and A003132(n)|n and A055012 (n)| n and A055013 (n)| n and all digits < > 0.

a(1)-a(2) and more terms add by Amiram Eldar, Apr 20 2023

A182111 Number of iterations of the map n -> sum of the cubes of the decimal digits of n.

Original entry on oeis.org

1, 7, 3, 6, 6, 10, 6, 6, 4, 1, 8, 5, 5, 6, 10, 3, 8, 2, 2, 7, 5, 4, 7, 3, 3, 8, 2, 4, 3, 3, 5, 7, 6, 3, 6, 6, 1, 8, 6, 6, 6, 3, 3, 7, 5, 5, 1, 6, 4, 6, 10, 3, 6, 5, 3, 5, 5, 8, 10, 10, 3, 8, 6, 5, 5, 6, 7, 11, 6, 6, 8, 2, 1, 1, 5, 7, 7, 8, 4, 6, 2, 4, 8, 6, 8

Offset: 1

Michel Lagneau, Apr 12 2012

a(n) is the number of times you obtain the sums of cubes of digits of n before reaching a fixed point (last number of the cycle).

			a(3) = 3 because :
3^3  = 27 -> 2^3 + 7^3 = 351;
351 -> 3^3 + 5^3 + 1^3 = 153;
153 -> 1^3+5^3+3^3 = 153 is the end because this number is already in the trajectory. Hence we obtain the map : 3 -> 27 -> 351 -> 153 with 3 iterations.

Cf. A000578, A055012, A152077, A160862, A165331.

a:= proc(n) local k, m, s; m:= n; s:= {};
      for k from 0 do
        m:= add(i^3, i=convert(m, base, 10));
        if m in s then return k fi;
        s:= s union {m}
      od
    end:
seq(a(n), n=1..85);  # Alois P. Heinz, Mar 01 2018

A225808 Values (Sum_{1<=i<=k} x_i)^2 = Sum_{1<=i<=k} x_i^3 for 1 <= x_1 <= x_2 <=...<= x_k ordered lexicographically according to (x1, x2,..., xk).

Original entry on oeis.org

1, 9, 16, 36, 81, 81, 100, 144, 256, 169, 225, 324, 361, 625, 144, 256, 324, 441, 324, 361, 441, 625, 256, 576, 729, 784, 576, 729, 900, 961, 1089, 1296, 484, 625, 784, 900, 484, 441, 576, 729, 784, 900, 1089, 1089, 1156, 1369, 625, 784, 729, 900, 1089, 1369, 1296, 1600, 900, 961, 1089

Offset: 1

Charles R Greathouse IV, Jimmy Zotos, and Balarka Sen, Jul 29 2013

a(n) <= k^4 where k is the size of the ordered tuple (x_1, x_2,..., x_k).

This sequence is closed under multiplication, that is, if m and n are in this sequence, so is m*n.

			1;
9, 16;
36, 81;
81, 100, 144, 256;
169, 225, 324, 361, 625;
144, 256, 324, 441, 324, 361, 441, 625, 256, 576, 729, 784, 576, 729, 900, 961, 1089, 1296;
484, 625, 784, 900, 484, 441, 576, 729, 784, 900, 1089, 1089, 1156, 1369, 625, 784, 729, 900, 1089, 1369, 1296, 1600, 900, 961, 1089, 1600, 1296, 1600, 2025, 2401;

Balarka Sen, Rows n = 1..10 of irregular triangle, flattened
Edward Barbeau and Samer Seraj, Sum of cubes is square of sum, arXiv:1306.5257 [math.NT], 2013.
John Mason, Generalising 'sums of cubes equal to squares of sums', The Mathematical Gazette 85:502 (2001), pp. 50-58.
Alasdair McAndrew, A cute result relating to sums of cubes
David Pagni, 82.27 An interesting number fact, The Mathematical Gazette 82:494 (1998), pp. 271-273.
Balarka Sen, Table of rows, n = 1..10
W. R. Utz, The Diophantine Equation (x_1 + x_2 + ... + x_n)^2 = x_1^3 + x_2^3 + ... + x_n^3, Fibonacci Quarterly 15:1 (1977), pp. 14, 16. Part 1, part 2.

Cf. A158649, A055012, A118881.

row[n_] := Reap[Module[{v, m}, v = Table[1, {n}]; m = n^(4/3); While[ v[[-1]] < m, v[[1]]++; If[v[[1]] > m, For[i = 2, i <= m, i++, If[v[[i]] < m, v[[i]]++; For[j = 1, j <= i - 1, j++, v[[j]] = v[[i]]]; Break[]]]]; If[Total[v^3] == Total[v]^2, Sow[Total[v]^2]]]]][[2, 1]];
Array[row, 7] // Flatten (* Jean-François Alcover, Feb 23 2019, from PARI *)

row(n)=my(v=vector(n,i,1),N=n^(4/3)); while(v[#v]N,for(i=2, N,if(v[i]

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A362953 Numbers N such that N + the sum of the cubes of its digits is again a third power.

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A065137 Sum of digits of n plus sum of cubes of digits of n.

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A067170 Numbers n such that sum of the cubes of the distinct prime factors of n equals the sum of the cubes of the digits of n.

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A072082 Numbers divisible by the cube of the sum of their digits in base 10.

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A072884 3rd-order digital invariants: the sum of the cubes of the digits of n equals some number k and the sum of the cubes of the digits of k equals n.

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A165551 Sum of cube of digits is sum of digits of cube.

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A169662 Numbers divisible by the sum of their digits, and by the sum of their digits squared, by the sum of their digits cubed and by the sum of 4th powers of their digits.

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A176194 Numbers with no zero digits divisible by the sum of the k-th powers of their digits, for each k = 1,2,3,4.

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