This is a front-end for the Online Encyclopedia of Integer Sequences, made by Christian Perfect. The idea is to provide OEIS entries in non-ancient HTML, and then to think about how they're presented visually. The source code is on GitHub.
A109408=Select[Table[n^3, {n, 200}], PrimeQ[Plus@@IntegerDigits[ # ]]&];
Select[Total[IntegerDigits[#]]&/@(Range[1000]^3),PrimeQ] (* Harvey P. Dale, Jul 16 2017 *)
36^3 = 46656. DigitSum(46656) = 27 (also a cube). So, 36 is a member of this sequence.
isok(n) = ispower(sumdigits(n^3), 3); \\ Michel Marcus, Feb 09 2014
The table T(n, k) begins 0 0 0 0 0 0 0 0 ... 0 1 2 3 4 5 6 7 ... 0 2 4 6 8 1 3 5 ... 0 3 6 9 3 6 9 3 ... 0 4 8 3 7 2 6 10 ... 0 5 1 6 2 7 3 8 ... 0 6 3 9 6 3 9 6 ... 0 7 5 3 10 8 6 13 ... ...
T[n_,k_]:=Total[IntegerDigits[n*k]]; Table[T[n-k,k],{n,0,12},{k,0,n}]//Flatten
T(n, k) = sumdigits(n*k);
s=0; Do[s=s+Apply[Plus, IntegerDigits[n^3]]; Print[s], {n, 1, 128}]
nxt[{n_,a_}]:={n+1,a+Total[IntegerDigits[(n+1)^3]]}; NestList[nxt,{1,1},60][[;;,2]] (* Harvey P. Dale, Aug 30 2025 *)
162 is a term because s(162) = 9, s(162^2) = 18, s(162^3) = 27, s(162^4) = 54 and 9 + 18 + 27 = 54.
Select[Range[0, 20000], Sum[i*(DigitCount[ # ][[i]] + DigitCount[ #^2][[i]] + DigitCount[ #^3][[i]]), {i, 1, 9}] == Sum[i*DigitCount[ #^4][[i]], {i, 1, 9}] &] (* Stefan Steinerberger, May 04 2006 *)
s[n_] := Plus @@ IntegerDigits@n; Select[ Range[0, 16217], s@# + s[ #^2] + s[ #^3] == s[ #^4] &] (* Robert G. Wilson v, May 04 2006 *)
Parallelize[While[True,If[Total[IntegerDigits[n]]+Total[IntegerDigits[n^2]]+Total[IntegerDigits[n^3]]==Total[IntegerDigits[n^4]],Print[n]];n++];n] (* J.W.L. (Jan) Eerland, Dec 25 2021 *)
is(n)=my(s=sumdigits); s(n)+s(n^2)+s(n^3) == s(n^4) \\ Anders Hellström, Sep 16 2015
select(isA118470(n)={sumdigits(n)+sumdigits(n^2)+sumdigits(n^3) == sumdigits(n^4)}, [0..1000]) \\ J.W.L. (Jan) Eerland, Dec 25 2021
def sd(n): return sum(map(int, str(n))) def ok(n): return sd(n) + sd(n**2) + sd(n**3) == sd(n**4) print([k for k in range(20000) if ok(k)]) # Michael S. Branicky, Dec 25 2021
Total[IntegerDigits[#]] & /@ (Prime[Range[5000000]]^3) // Union (* The program generates the first 39 terms of the sequence. To generate more, increase the Range constant. *) (* Harvey P. Dale, Sep 19 2021 *)
list(maxx)={v=List();n=2; while(n<=maxx,q=n^3;summ=sumdigits(q);
if(setsearch(v,summ)<1,listput(v,summ));n=nextprime(n+1));vecsort(v,,8) ;} \\ Bill McEachen, Jan 29 2014
8 is in the sequence because 8^3 = 512 and 5 + 1 + 2 = 8, and 8/8 = 1.
Select[Range[1000], Divisible[Plus@@IntegerDigits[#^3], #] &] (* Alonso del Arte, Jul 07 2019 *)
isok(n) = !(sumdigits(n^3) % n); \\ Michel Marcus, Jul 07 2019
The digit sum of 7978887 is 54, the digit sum of 7978887^3 mod 10^7 = 1110103 is 7. 54-7=47, which means that randomly finding numbers for which the sum of digits of x^3 is smaller than the sum of digits of x is easiest when choosing x with the given seven ending digits (that is, of course, depending on the preferred size of x). Heuristically it can be conjectured that there are infinitely many numbers x for which the digit sum of x^3 is smaller than the digit sum of x.
a(n)={r=0; i=10^n; for(x=1, i-1, s1=sumdigits(x); s2=sumdigits(x^3%i); d=s2-s1; if(d
/* The following program finds the first 69 terms, and also many of the larger terms, correctly. The 70th term will be incorrect because of the truncation to 10^4*(9/10)=9000 record numbers for each number of digits; if the truncation is increased to 10^5*(9/10) records (n=5 at the beginning), the first 237 terms give a(n) with corresponding maxima. (Further restrictions on ending digits find optimal solutions for a higher number of digits.) */
{n=4; d=10^n; v=vector(d*9/10); print("keeping "#v" record numbers"); w=vector(d*9); for(x=1, d-1, if(x%10, y=sumdigits((x^3)%d); z=sumdigits(x); v[9*(x\10)+(x%10)]=x+d*(y-z))); v=vecsort(v); while(d<10^250, for(i=1, #v, for(j=0, 9, x=d*j+v[i]%d; y=sumdigits((x^3)%(d*10)); z=sumdigits(x); w[10*(i-1)+j+1]=x+d*10*(y-z))); d*=10; n++; w=vecsort(w); v=vecextract(w,Str("1.."#v)); print(n" digits - record differences: "vecextract(v,"1..5")\d" / record mod "10"^"n": "v[1]%d))}
For n=42, 42^3 = 74088 has sum of digits 27 so a(42) = (27 - 42)/3 = -5.
#include#include int32_t sumDigits(int64_t num) { int32_t sum = 0; while (num > 0) { sum += num % 10; num /= 10; } return sum; } int main() { for (int64_t i=0; i<10000; ++i) { int64_t num = i * i * i; int32_t sum = sumDigits(num); printf("%ld, ", (sum - i)/3); } return 0; }
read("transforms"):
A372924 := proc(n)
(digsum(n^3)-n)/3 ;
end proc:
seq(A372924(n),n=0..80) ; # R. J. Mathar, Jul 03 2024
Table[(DigitSum[n^3] - n)/3, {n, 0, 100}] (* Paolo Xausa, Jul 03 2024 *)
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