A046459 -id:A046459 - cp's OEIS frontend

A225535 Numbers whose cubed digits sum to a cube, and have more than one nonzero digit.

168, 186, 345, 354, 435, 453, 534, 543, 618, 681, 816, 861, 1068, 1086, 1156, 1165, 1516, 1561, 1608, 1615, 1651, 1680, 1806, 1860, 3045, 3054, 3405, 3450, 3504, 3540, 4035, 4053, 4305, 4350, 4503, 4530, 5034, 5043, 5116, 5161, 5304, 5340, 5403, 5430, 5611

Offset: 1

Kevin L. Schwartz and Christian N. K. Anderson, May 10 2013

			5^3 + 6^3 + 1^3 + 1^3 = 343, which is 7^3.

Christian N. K. Anderson, Table of n, a(n) for n = 1..10000

Cf. A225534 (cubed digits sum to a prime), A197039 (square), A046459. A055012.
Cf. A165330 (cube cycle), A046197 (cubic fixed points), A000578 (cubes).
Cf. A052034 (squared digits sum to a prime), A028839, A117685.
Cf. A164882 (n such that sum of the cubes of the digits of n^3 is perfect cube). - Zak Seidov, May 21 2013

fQ[n_] := Module[{d = IntegerDigits[n]}, Count[d, 0] + 1 < Length[d] && IntegerQ[Total[d^3]^(1/3)]]; Select[Range[5611], fQ] (* T. D. Noe, May 19 2013 *)

y=rep(0,10000); len=0; x=0; library(gmp);
digcubesum<-function(x) sum(as.numeric(unlist(strsplit(as.character(as.bigz(x)),split="")))^3);
iscube<-function(x) ifelse(as.bigz(x)<2,T,all(table(as.numeric(factorize(x)))%%3==0));
nonzerodig<-function(x) sum(strsplit(as.character(x),split="")[[1]]!="0");
which(sapply(1:6000,function(x) nonzerodig(x)>1 & iscube(digcubesum(x))))

A226971 Numbers k such that the sum of digits of k^7 is equal to k.

Original entry on oeis.org

0, 1, 18, 27, 31, 34, 43, 53, 58, 68

Offset: 1

Michel Lagneau, Jun 24 2013

Only the ten integers listed have this property.

			a(3) = 18 because 18^7 = 612220032 and 6+1+2+2+2+0+0+3+2 = 18.

Cf. A046459, A055575, A055576, A055577.

Cf. A152147.

[n: n in [0..80] | &+Intseq(n^7) eq n]; // Vincenzo Librandi, Feb 23 2015

Select[Range[0, 100], #==Total[IntegerDigits[#^7]]&]

isok(k)=sumdigits(k^7)==k \\ Patrick De Geest, Dec 13 2024

[n for n in (0..70) if sum((n^7).digits()) == n] # Bruno Berselli, Feb 23 2015

A334601 Positive integers m such that sum of cubes of the digits of m, t=A055012(m), is a multiple of m (m/A055012(m) is an integer >= 1).

Original entry on oeis.org

1, 2, 3, 4, 5, 6, 7, 8, 9, 24, 27, 37, 48, 153, 370, 371, 407, 459

Offset: 1

Zak Seidov, May 07 2020 and May 12 2020

Corresponding values of t: 1, 8, 27, 64, 125, 216, 343, 512, 729, 72, 351, 370, 576, 153, 370, 371, 407, 918 (first 9 terms are all cubes).
Corresponding values of t/m: 1, 4, 9, 16, 25, 36, 49, 64, 81, 3, 13, 10, 12, 1, 1, 1, 1, 2 (first 9 terms are all squares).
The subsequence of numbers m such that sum of cubes of its digits is equal to m is A046197 \ {0}. - Bernard Schott, May 11 2020

			m = 459, t = 4^3 + 5^3 + 9^3 = 918, t/m = 2.

Cf. A005188, A046197, A046459, A055012, A055642, A101337, A306354, A306360, A306361, A329291.

Select[Range[500], Divisible[Plus @@ (IntegerDigits[#]^3), #] &] (* Amiram Eldar, May 11 2020 *)

isok(m) = my(d=digits(m)); sum(k=1, #d, d[k]^3) % m == 0; \\ Michel Marcus, May 14 2020

A350693 Number of b > 0 which permit n^3 to be written as a sum of powers of b in n parts. Each exponent c is an integer >= 0, n^3 = b^c_1 + b^c_2 + ... + b^c_n.

Original entry on oeis.org

3, 5, 8, 7, 10, 13, 17, 19, 12, 20, 16, 18, 18, 25, 25, 21, 14, 28, 31, 34, 19, 22, 29, 34, 28, 33, 29, 38, 19, 33, 30, 31, 34, 51, 44, 30, 20, 41, 38, 44, 18, 37, 42, 52, 27, 30, 37, 59, 39, 50, 28, 35, 37, 82, 64, 44, 19, 36, 27, 36, 27, 52, 85, 65, 35, 40, 29

Offset: 2

Thomas Scheuerle, Jan 12 2022

nonn

If n^3 is written in different number bases, a(n) is an upper limit for the count of number bases which allow n^3 to be written as a base-b number with a digit sum of n (generalized Dudeney numbers).

a(n) has an upper limit in the number of divisors of n^3-n. Let d be one of these divisors, then it appears that a lower limit can be found by excluding all divisors d where d+1 does not share all its prime divisors with binomial(n^3, n) (A107444).

			a(2) = 3 because 2^3 = 2^2 + 2^2 = 4^1 + 4^1 = 7^1 + 7^0.

Cf. A000005, A046459, A092517, A107444.

a(n) = sum(d=2, n^3, s=sumdigits(n^3, d); s<=n&&(n-s)%(d-1)==0); \\ Jinyuan Wang, Jan 15 2022

a(n) <= A000005(n^3-n). Conjectured to become a(n) = A000005(n^3-n), if the definition would permit negative values for b and only the absolute value of the sum needs to be equal to n^3.

More terms from Jinyuan Wang, Jan 15 2022

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A225535 Numbers whose cubed digits sum to a cube, and have more than one nonzero digit.

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A226971 Numbers k such that the sum of digits of k^7 is equal to k.

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A334601 Positive integers m such that sum of cubes of the digits of m, t=A055012(m), is a multiple of m (m/A055012(m) is an integer >= 1).

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A350693 Number of b > 0 which permit n^3 to be written as a sum of powers of b in n parts. Each exponent c is an integer >= 0, n^3 = b^c_1 + b^c_2 + ... + b^c_n.

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