cp's OEIS Frontend

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.

Showing 1-3 of 3 results.

A231688 a(n) = Sum_{i=0..n} digsum(i)^3, where digsum(i) = A007953(i).

Original entry on oeis.org

0, 1, 9, 36, 100, 225, 441, 784, 1296, 2025, 2026, 2034, 2061, 2125, 2250, 2466, 2809, 3321, 4050, 5050, 5058, 5085, 5149, 5274, 5490, 5833, 6345, 7074, 8074, 9405, 9432, 9496, 9621, 9837, 10180, 10692, 11421, 12421, 13752, 15480, 15544, 15669, 15885, 16228, 16740, 17469, 18469, 19800, 21528, 23725, 23850, 24066, 24409, 24921, 25650, 26650, 27981, 29709, 31906, 34650
Offset: 0

Views

Author

N. J. A. Sloane, Nov 13 2013

Keywords

References

  • Grabner, P. J.; Kirschenhofer, P.; Prodinger, H.; Tichy, R. F.; On the moments of the sum-of-digits function. Applications of Fibonacci numbers, Vol. 5 (St. Andrews, 1992), 263-271, Kluwer Acad. Publ., Dordrecht, 1993.

Crossrefs

Partial sums of A118880.

Programs

  • Maple
    See A037123.
  • Mathematica
    Accumulate[Table[Total[IntegerDigits[n]]^3,{n,0,60}]] (* Harvey P. Dale, Aug 06 2021 *)
  • PARI
    a(n) = sum(i=0, n, sumdigits(i)^3); \\ Michel Marcus, Jan 07 2017

A182128 Number of iterations of the map n -> (sum of the decimal digits of n)^3 before reaching the last number of the cycle.

Original entry on oeis.org

0, 0, 2, 3, 3, 2, 3, 3, 1, 2, 1, 2, 3, 3, 2, 3, 3, 1, 2, 2, 2, 3, 3, 2, 3, 3, 1, 2, 2, 2, 3, 3, 2, 3, 3, 1, 2, 2, 2, 2, 3, 2, 3, 3, 1, 2, 2, 2, 2, 3, 2, 3, 3, 1, 2, 2, 2, 2, 3, 2, 3, 3, 1, 2, 2, 2, 2, 3, 2, 2, 3, 1, 2, 2, 2, 2, 3, 2, 2, 3, 1, 2, 2, 2, 2, 3, 2
Offset: 0

Views

Author

Michel Lagneau, Apr 13 2012

Keywords

Comments

a(n) is the number of times that the cube of the sum of the digits must be calculated before reaching the last number of the cycle.

Examples

			0 is in the sequence twice because 0 -> 0 and 1 -> 1;
a(3) = 3:
    3 ->       3^3 = 27;
   27 ->   (2+7)^3 = 729;
  729 -> (7+2+9)^3 = 18^3 = 5832 is the end of the map because 5832 -> (5+8+3+2)^3 = 18^3 is already in the trajectory. Hence we obtain the map: 3 -> 27 -> 729 -> 5832 with 3 iterations.
		

Crossrefs

Programs

  • Maple
    A182128 := proc(n)
            local traj ,c;
            traj := n ;
            c := [n] ;
            while true do
                    traj := A118880(traj) ;
                    if member(traj,c) then
                            return nops(c)-1 ;
                    end if;
                    c := [op(c),traj] ;
            end do:
    end proc:
    seq(A182128(n),n=0..80) ; # R. J. Mathar, Jul 08 2012

A162840 Numbers k such that the cube of the sum of digits of k equals the product of digits of k.

Original entry on oeis.org

0, 1, 666666, 1377789, 1377798, 1377879, 1377897, 1377978, 1377987, 1378779, 1378797, 1378977, 1379778, 1379787, 1379877, 1387779, 1387797, 1387977, 1389777, 1397778, 1397787, 1397877, 1398777, 1555888, 1558588, 1558858
Offset: 0

Views

Author

Boris Hostnik (megpplus(AT)siol.net), Jul 14 2009

Keywords

Examples

			666666 is in the sequence because (1) cubed sum of its digits is (6+6+6+6+6+6)^3 = 46656, (2) the product of its digits is 6*6*6*6*6*6=46656; 46656=46656.
		

Crossrefs

Cf. A007954, A118880, A117720, A034710. - R. J. Mathar, Jul 19 2009

Programs

  • Maple
    A007953 := proc(n) add(d,d=convert(n,base,10)) ; end: A007954 := proc(n) mul(d,d=convert(n,base,10)) ; end: A118880 := proc(n) (A007953(n))^3; end: for n from 1 to 2000000 do if A118880(n) = A007954(n) then printf("%d,\n",n) ; fi; od: # R. J. Mathar, Jul 19 2009
  • Mathematica
    Select[Range[0,156*10^4],Total[IntegerDigits[#]]^3==Times@@IntegerDigits[#]&] (* Harvey P. Dale, Jul 07 2022 *)

Formula

{n: A118880(n)=A007954(n)}. - R. J. Mathar, Jul 19 2009
Showing 1-3 of 3 results.