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.

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A046156 Limit set for operation of repeatedly replacing a number with the sum of the cubes of its digits.

Original entry on oeis.org

0, 1, 55, 133, 136, 153, 160, 217, 244, 250, 352, 370, 371, 407, 919, 1459
Offset: 1

Views

Author

Richard C. Schroeppel

Keywords

Comments

Range of A165330; A165330(a(n))=a(n); A165331(a(n))=0. - Reinhard Zumkeller, Sep 17 2009

Crossrefs

Programs

  • Mathematica
    lst = {}; k = 0; While[k < 1500, a = NestWhile[Plus @@ (IntegerDigits@ #^3) &, k, Unequal, All]; If[FreeQ[lst, a], AppendTo[lst, a]]; k++]; Sort@ lst (* Robert G. Wilson v, Jan 19 2006, revised Jan 03 2015 *)
    Table[Nest[Total[IntegerDigits[#]^3]&,n,30],{n,0,1500}]//Union (* Harvey P. Dale, Aug 04 2018 *)

A165333 Numbers that eventually reach the fixed point 370 under "x -> sum of cubes of digits of x" (see A055012).

Original entry on oeis.org

7, 19, 34, 37, 43, 58, 67, 70, 73, 76, 85, 88, 91, 109, 118, 124, 139, 142, 145, 148, 154, 157, 166, 169, 175, 178, 181, 184, 187, 190, 193, 196, 214, 223, 226, 232, 241, 247, 259, 262, 268, 274, 277, 286, 295, 304, 307, 319, 322, 334, 340, 343, 346, 355, 358
Offset: 1

Views

Author

Reinhard Zumkeller, Sep 17 2009

Keywords

Comments

A165330(a(n)) = 370;
Subsequence of A031179 and of A016777; a(n) mod 3 = 1.

Examples

			a(3)=34: 34 -> 3^3+4^3=91 -> 9^3+1=730 -> 7^3+3^3+0=370.
		

Crossrefs

Programs

  • Mathematica
    f[n_] := Plus@@(IntegerDigits[n]^3); Trajectory[n_] := Most[NestWhileList[f, n, UnsameQ ,All]]; Select[Range[358], Last[Trajectory[#]] == 370&] (* Ant King, May 24 2013 *)

A165334 Numbers that eventually reach the fixed point 371 under "x -> sum of cubes of digits of x" (see A055012).

Original entry on oeis.org

2, 5, 8, 11, 14, 17, 20, 23, 26, 29, 32, 35, 38, 41, 44, 50, 53, 56, 59, 62, 65, 68, 71, 80, 83, 86, 92, 95, 101, 104, 107, 110, 113, 116, 119, 122, 125, 128, 131, 134, 137, 140, 143, 146, 149, 152, 155, 158, 161, 164, 167, 170, 173, 176, 179, 182, 185, 188, 191
Offset: 1

Views

Author

Reinhard Zumkeller, Sep 17 2009

Keywords

Comments

A165330(a(n)) = 371;
Subsequence of A031179;
complement of A165335 with respect to A016789; a(n) mod 3 = 2.

Examples

			a(10)=29: 29 -> 2^3+9^3=737 -> 2*7^3+3^3=713 -> 7^3+1+3^3=371.
		

Crossrefs

Programs

  • Mathematica
    f[n_] := Plus@@(IntegerDigits[n]^3); Trajectory[n_] := Most[NestWhileList[f, n, UnsameQ ,All]]; Select[Range[191], Last[Trajectory[#]]==371 &] (* Ant King, May 24 2013 *)

A165335 Numbers that eventually reach the fixed point 407 under "x -> sum of cubes of digits of x" (see A055012).

Original entry on oeis.org

47, 74, 77, 89, 98, 407, 449, 470, 494, 578, 587, 668, 686, 704, 707, 740, 758, 770, 785, 788, 809, 857, 866, 875, 878, 887, 890, 908, 944, 980, 1124, 1139, 1142, 1148, 1157, 1175, 1178, 1184, 1187, 1193, 1214, 1241, 1319, 1367, 1376, 1391, 1412, 1418, 1421
Offset: 1

Views

Author

Reinhard Zumkeller, Sep 17 2009

Keywords

Comments

A165330(a(n)) = 407.

Examples

			a(4)=89: 89 -> 8^3+9^3=1241 -> 1+2^3+4^3+1=74 -> 7^3+4^3=407.
		

Crossrefs

Subsequence of A031179.

Programs

  • Mathematica
    f[n_] := Plus@@(IntegerDigits[n]^3); Trajectory[n_] := Most[NestWhileList[f,n,UnsameQ,All]]; Select[Range[1421], Last[Trajectory[#]]==407 &] (* Ant King, May 24 2013 *)
    Select[Range[1500],FixedPoint[Total[IntegerDigits[#]^3]&,#,100]==407&] (* Harvey P. Dale, Apr 17 2020 *)

Formula

Complement of A165334 with respect to A016789; a(n) mod 3 = 2.

A225534 Numbers whose sum of cubed digits is prime.

Original entry on oeis.org

11, 101, 110, 111, 113, 115, 122, 124, 128, 131, 139, 142, 146, 148, 151, 155, 164, 166, 182, 184, 193, 199, 212, 214, 218, 221, 223, 227, 232, 236, 238, 241, 245, 254, 256, 263, 265, 269, 272, 278, 281, 283, 287, 289, 296, 298, 311, 319, 322, 326, 328, 335
Offset: 1

Views

Author

Keywords

Comments

Note that 11 is the only two-digit number in the sequence.
a(n) ~ n. For 414 < n < 10000, 6.38*n - 528 provides an estimate of a(n) to within 6%.

Examples

			139 is in the sequence because 1^3 + 3^3 + 9^3 = 757, which is prime.
		

Crossrefs

Programs

  • Mathematica
    Select[Range[350],PrimeQ[Total[IntegerDigits[#]^3]]&] (* Harvey P. Dale, Mar 16 2016 *)
  • R
    digcubesum<-function(x) sum(as.numeric(strsplit(as.character(x),split="")[[1]])^3); library(gmp);
    which(sapply(1:1000,function(x) isprime(digcubesum(x))>0))

A194025 Number of fixed points under iteration of sum of cubes of digits in base b.

Original entry on oeis.org

1, 2, 9, 3, 4, 7, 6, 8, 5, 8, 5, 5, 3, 3, 24, 3, 2, 9, 2, 3, 16, 5, 2, 20, 2, 2, 7, 9, 3, 14, 2, 6, 8, 4, 10, 12, 2, 8, 8, 7, 2, 12, 4, 5, 17, 5, 4, 27, 6, 5, 10, 4, 2, 11, 9, 5, 9, 6, 3, 25, 5, 6, 24, 5, 4, 17, 5, 5, 9, 10, 1, 15, 4, 3, 13, 3, 5, 19, 4, 13, 7
Offset: 2

Views

Author

Martin Renner, Aug 22 2011

Keywords

Comments

If b >= 2 and n >= 2*b^3, then S(n,3,b) < n. For each positive integer n, there is a positive integer m such that S^m(n,3,b) < 2*b^3. (Grundman/Teeple, 2001, Lemma 8 and Corollary 9.)
From Christian N. K. Anderson, May 23 2013: (Start)
1 is considered a fixed point in all bases, 0 is not.
In order for a number with d digits in base n to be a fixed point, it must satisfy the condition d*(n-1)^3 < n^d, which can only occur when d < 4 for n > 2. Because all binary numbers are "happy" (become 1 under iteration), there are no fixed points with more than 4 digits in any base. It can further be demonstrated that all 4-digit solutions begin with 1 in base n.
Unlike the number of fixed points under iteration of sum of squares of digits (A193583), this sequence contains many even numbers, and its histogram converges to a smooth distribution (approximately gamma(2.64,2.8); see "histogram" in links). (End)

Examples

			In the decimal system all integers go to (1); (153); (370); (371); (407) or (55, 250,133); (136, 244); (160, 217, 352); (919, 1459) under the iteration of sum of cubes of digits, hence there are five fixed points, two 2-cycles and two 3-cycles. Therefore a(10) = 5.
		

Crossrefs

Solutions for a(10): A046197.
Largest of the a(n) fixed points: A226026.
Related sequences for sum of squared digits: A193583, A209242.

Programs

  • Maple
    S:=proc(n,p,b) local Q,k,N,z; Q:=[n]; for k from 1 do N:=Q[k]; z:=convert(sum(N['i']^p,'i'=1..nops(N)),base,b); if not member(z,Q) then Q:=[op(Q),z]; else Q:=[op(Q),z]; break; fi; od; return Q; end:
    a:=proc(b) local F,i,A,Q,B,C; A:=[]: for i from 1 to 2*b^3 do Q:=S(convert(i,base,b),3,b); A:={op(A),Q[nops(Q)]}; od: F:={}: for i from 1 while nops(A)>0 do B:=S(A[1],3,b); C:=[seq(B[i],i=1..nops(B)-1)]: if nops(C)=1 then F:={op(F),op(C)}: fi: A:=A minus {op(B)}; od: return(nops(F)); end:
    # Martin Renner, Aug 24 2011
  • R
    #See A226026 for an optimized version
    inbase=function(n, b) { x=c(); while(n>=b) { x=c(n%%b, x); n=floor(n/b) }; c(n, x) }; yn=rep(NA, 30)
    for(b in 2:30) yn[b]=sum(sapply(1:(2*b^3), function(x) sum(inbase(x, b)^3))==1:(2*b^3)); yn # Christian N. K. Anderson, Jun 08 2013
  • Sage
    def A194025(n):
        # inefficient but straightforward
        return len([i for i in (1..2*n**3) if i==sum(d**3 for d in i.digits(base=n))]) # D. S. McNeil, Aug 23 2011
    

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

Original entry on oeis.org

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

Views

Author

Keywords

Examples

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

Crossrefs

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

Programs

  • Mathematica
    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 *)
  • R
    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))))

A255668 Number of perfect digital invariants of order n, i.e., numbers equal to the sum of n-th powers of their digits.

Original entry on oeis.org

1, 10, 2, 6, 5, 8, 3, 7, 5, 6, 3, 10, 2, 3, 3, 2, 4, 6, 2, 6, 3, 4, 2, 7, 5, 10, 2, 9, 2, 9, 2, 6, 3, 5, 3, 6, 3, 5, 5, 7, 2, 2, 4, 9, 6, 9, 5, 7, 2, 3, 2, 4, 2, 3, 6, 4, 5, 4, 2, 4, 4, 4, 3, 7, 3, 6, 3, 4, 3, 3, 4, 3, 4, 5, 3, 4, 5, 5, 3, 3, 2, 3, 2, 4, 3, 8, 3, 5, 2, 7, 3
Offset: 0

Views

Author

M. F. Hasler, Apr 14 2015

Keywords

Comments

Row lengths of the table A252648.
For a number with d digits, the sum of n-th powers cannot exceed d*9^n, but the number is not less than 10^(d-1). Therefore there is only a finite number of possible perfect digital invariants for any n, the largest of which has at most d* digits, where d* = 1+(n*log(9)+log d*)/log(10).

Examples

			a(0)=1 because 1 is the only number equal to the sum of 0th powers of its digits.
a(1)=10 because { 0, 1, ... 9 } are the only numbers equal to the sum of their digits (taken to the power 1).
a(2)=2 because 0 and 1 are the only numbers equal to the sum of the squares of their digits.
a(3)=6 because { 0, 1, 153, 370, 371, 407 } is the set of all numbers equal to the sum of the 3rd powers of their digits, cf. A046197.
For more examples, see the table A252648.
		

Crossrefs

Programs

  • Mathematica
    Reap@ For[n = 0, n < 6, n++, Sow@ Length@ Select[Range[0, 10^(n + 1)], Plus @@ (IntegerDigits[#]^n) == # &]] // Flatten // Rest (* Michael De Vlieger, Apr 14 2015 *)

Formula

a(n) >= 2 for all n > 0, since 0 and 1 are digital invariants for any power n > 0.

Extensions

a(10)-a(90) from Don Knuth, Sep 09 2015

A226026 Maximum fixed points under iteration of sum of cubes of digits in base n.

Original entry on oeis.org

1, 17, 62, 118, 251, 250, 433, 1052, 407, 1280, 2002, 1968, 793, 3052, 5614, 1456, 5337, 5939, 2413, 5615, 20217, 11648, 11080, 31024, 5425, 1737, 28027, 26846, 17451, 33535, 10261, 64019, 23552, 44937, 30086, 84870, 17353, 55243, 48824, 108936, 58618, 87977
Offset: 2

Views

Author

Keywords

Comments

1 is considered a fixed point in all bases, 0 is not.
a(n)=1 iff A194025(n)=1.
In order for a number with d digits in base n to be a fixed point, it must satisfy the condition d*(n-1)^32. Because all binary numbers are "happy" (become 1 under iteration), there are no fixed points with more than 4 digits in any base.
Furthermore, 4-digit solutions of the form x0mm or xmmm (where m is n-1) represent extreme values of sum of cubed digits, and so 4-digit numbers can only be solutions if xn^3+n^2-1<=2n^3+x^3. For x=2 this reduces to n<=3, so any 4-digit solution must begin with 1 in bases above 3.

Examples

			In base 5, the numbers 1, 28 and 118 are written as 1, 103, and 433. The sum of the cubes of their digits are 1, 1+0^3+3^3=28, and 4^3+3^3+3^3=118. There are no other solutions, so a(5)=118.
		

Crossrefs

Number of fixed points in base n: A194025.
All fixed points in base 10: A046197.

Programs

  • R
    inbase=function(n,b) { x=c(); while(n>=b) { x=c(n%%b,x); n=floor(n/b) }; c(n,x) }
    yfp=vector("list",100)
    for(b in 2:100) { fp=c()
        for(w in 0:1) for(x in 1:b-1) for(y in 1:b-1) if((u1=w^3+x^3+y^3)<=(u2=w*b^3+x*b^2+y*b) & u1+b^3>u2+b-1)
            if(length((z=which((1:b-1)*((1:b-1)^2-1)==u2-u1)-1))) fp=c(fp,u2+z)
        yfp[[b]]=fp[-1]
        cat("Base",b,":",fp,"\n")
    }

A226063 Number of fixed points in base n for the sum of the fourth power of its digits.

Original entry on oeis.org

1, 1, 3, 4, 1, 1, 7, 3, 4, 3, 1, 2, 1, 7, 2, 2, 1, 4, 2, 6, 2, 3, 1, 3, 1, 11, 3, 3, 2, 2, 7, 4, 1, 4, 3, 1, 3, 4, 1, 2, 2, 2, 3, 4, 2, 2, 1, 2, 1, 2, 1, 2, 4, 3, 3, 2, 2, 1, 3, 2, 5, 2, 9, 2, 1, 2, 1, 1, 3, 2, 2, 1, 2, 5, 1, 5, 5, 4, 2, 5, 3, 2, 2, 3, 3, 1, 2
Offset: 2

Views

Author

Keywords

Comments

All fixed points in base n have at most 5 digits. Proof: In order to be a fixed point, a number with d digits in base n must meet the condition n^d <= d*(n-1)^4, which is only possible for d < 5.
For 5-digit numbers vwxyz in base n, only numbers where v*n^4 + n^3 - 1 <= v^4 + 3*(n-1)^4 or v*n^4 + n^4 - 1 <= v^4 + 4*(n-1)^4 are possible fixed points. v <= 2 for n <= 250.

Examples

			For a(8)=7, the solutions are {1,16,17,256,257,272,273}. In base 8, these are written as {1, 20, 21, 400, 401, 420, 421}. Because 1^4 = 1, 2^4 + 0^4 = 16, 2^4 + 1^4 = 17, 4^4 + 0^4 + 0^4 = 256, etc., these are the fixed points in base 8.
		

Crossrefs

Cf. A226064 (greatest fixed point).
Cf. A052455 (fixed points in base 10).

Programs

  • R
    inbase=function(n,b) { x=c(); while(n>=b) { x=c(n%%b,x); n=floor(n/b) }; c(n,x) }
    yn=rep(NA,20)
    for(b in 2:20) yn[b]=sum(sapply(1:(1.5*b^4),function(x) sum(inbase(x,b)^4))==1:(1.5*b^4)); yn
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