A023052 -id:A023052 - cp's OEIS frontend

A134703 Powerful numbers (2b): a sum of nonnegative powers of its digits.

0, 1, 2, 3, 4, 5, 6, 7, 8, 9, 24, 43, 63, 89, 132, 135, 153, 175, 209, 224, 226, 254, 258, 262, 263, 264, 267, 283, 308, 332, 333, 334, 347, 357, 370, 371, 372, 373, 374, 375, 376, 377, 378, 379, 407, 445, 463, 472, 518, 538, 598, 629, 635, 653, 675, 730, 731, 732, 733, 734

Offset: 1

David W. Wilson, Sep 05 2009

Here 0 digits may be used, with the convention that 0^0 = 1. Of course 0^1 = 0, so one is free to use the 0 digit to get an extra 1, or not.

			43 = 4^2 + 3^3; 254 = 2^7 + 5^3 + 4^0 = 128 + 125 + 1.
209 = 2^7 + 0^1 + 9^2.
732 = 7^0 + 3^6 + 2^1.

David Wilson, Table of n, a(n) for n = 1..10000
Index entries for sequences related to powerful numbers

Cf. A001694, A005934, A005188, A003321, A014576, A023052, A046074.

Different from A007532 and A061862, which are variations.

If n = d_1 d_2 ... d_k in decimal then there are integers m_1 m_2 ... m_k >= 0 such that n = d_1^m_1 + ... + d_k^m_k.

A061862 Powerful numbers (2a): a sum of nonnegative powers of its digits.

Original entry on oeis.org

0, 1, 2, 3, 4, 5, 6, 7, 8, 9, 24, 43, 63, 89, 132, 135, 153, 175, 209, 224, 226, 254, 258, 262, 263, 264, 267, 283, 332, 333, 334, 347, 357, 370, 371, 372, 373, 374, 375, 376, 377, 378, 379, 407, 445, 463, 472, 518, 538, 598, 629, 635, 653, 675, 730, 731, 732

Offset: 1

Erich Friedman, Jun 23 2001

Zero digits cannot be used in the sum. - N. J. A. Sloane, Aug 31 2009

More precisely, digits 0 do not contribute to the sum, in contrast to A134703 where it is allowed to use 0^0 = 1. - M. F. Hasler, Nov 21 2019

			43 = 4^2 + 3^3; 254 = 2^7 + 5^3 + 4^0 = 128 + 125 + 1.
209 = 2^7 + 9^2.
732 = 7^0 + 3^6 + 2^1.

Cf. A001694, A005934, A005188, A003321, A014576, A023052, A046074.

Different from A007532 and A134703, which are variations.

a061862 n = a061862_list !! (n-1)
a061862_list = filter f [0..] where
   f x = g x 0 where
     g 0 v = v == x
     g u v = if d <= 1 then g u' (v + d) else v <= x && h 1
             where h p = p <= x && (g u' (v + p) || h (p * d))
                   (u', d) = divMod u 10
-- Reinhard Zumkeller, Jun 02 2013

f[ n_ ] := Module[ {}, a=IntegerDigits[ n ]; e=g[ Length[ a ] ]; MemberQ[ Map[ Apply[ Plus, a^# ] &, e ], n ] ] g[ n_ ] := Map[ Take[ Table[ 0, {n} ]~Join~#, -n ] &, IntegerDigits[ Range[ 10^n ], 10 ] ] For[ n=0, n >= 0, n++, If[ f[ n ], Print[ n ] ] ]

If n = d_1 d_2 ... d_k in decimal then there are integers m_1 m_2 ... m_k >= 0 such that n = d_1^m_1 + ... + d_k^m_k.

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

Kevin L. Schwartz and Christian N. K. Anderson, May 24 2013

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.

			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.

Christian N. K. Anderson, Table of n, a(n) for n = 2..250

Cf. A226064 (greatest fixed point).
Cf. A052455 (fixed points in base 10).
Cf. A023052, A046074, A046197.

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

A226064 Largest fixed point in base n for the sum of the fourth power of its digits.

Original entry on oeis.org

1, 1, 243, 419, 1, 1, 273, 1824, 9474, 10657, 1, 8194, 1, 53314, 47314, 36354, 1, 246049, 53808, 378690, 170768, 185027, 1, 247507, 1, 1002324, 722739, 278179, 301299, 334194, 1004643, 959859, 1, 1538803, 1798450, 1, 4168450, 2841074, 1, 1877793, 5556355

Offset: 2

Kevin L. Schwartz and Christian N. K. Anderson, May 24 2013

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.

			The fixed points in base 8 are {1,16,17,256,257,272,273}, because in base 8, these are written as {1,20,21,400,401,420,421} and 1^4 = 1, 2^4 + 0^4 = 16, 2^4 + 1^4 = 17, 4^4 + 0^4 + 0^4 = 256, etc. The largest of these is 273 = a(8).

Christian N. K. Anderson, Table of n, a(n) for n = 2..250
Christian N. K. Anderson, Table of base, largest fixed point, number of fixed points, and a list of all fixed points in base 10 and base n for n = 1..250

Cf. A226063 (number of fixed points).
Cf. A052455 (fixed points in base 10).
Cf. A023052, A046074, A046197.

for(b in 2:50) {
    fp=c()
    for(w in 1:b-1) for(x in 1:b-1) if((v1=w^4+x^4)<=(v2=w*b^3+x*b^2))
        for(y in 1:b-1) if((u1=v1+y^4)<=(u2=v2+y*b) & u1+b^4>u2+b-1) {
            z=which(u1+(1:b-1)^4==u2+(1:b-1))-1
            if(length(z)) fp=c(fp,u2+z)
        }
    cat("Base",b,":",fp[-1],"\n")
}

A257814 Numbers n such that k times the sum of the digits (d) to the power k equal n, so n=k*sum(d^k), for some positive integer k, where k is smaller than sum(d^k).

Original entry on oeis.org

2, 3, 4, 5, 6, 7, 8, 9, 50, 298, 130004, 484950, 3242940, 4264064, 5560625, 36550290, 47746195, 111971979, 129833998, 9865843497, 46793077740, 767609367921, 4432743262896, 42744572298532, 77186414790914, 99320211963544, 99335229415136, 456385296642870

Offset: 1

Pieter Post, May 10 2015

The first nine terms are trivial, but then the terms become scarce. The exponent k must be less than the "sum of the digits" raised to the k-th power, otherwise there will be infinitely many terms containing 1's and 0's, like 11000= 5500*(1^5500+1^5500+0^5500+0^5500+0^5500). It appears this sequence is finite, because there is a resemblance with the Armstrong numbers (A005188).

			50 = 2*(5^2+0^2);
484950 = 5*(4^5+8^5+4^5+9^5+5^5+0^5).

Chai Wah Wu, Table of n, a(n) for n = 1..54 (terms < 10^32, n = 1..53 from Giovanni Resta)

Cf. A005188, A023052, A257768.

sdk(d, k) = sum(j=1, #d, d[j]^k);
isok(n) = {d = digits(n); k = 1; while ((val=k*sdk(d,k)) != n, k++; if (val > n, return (0))); k < sdk(d,k);} \\ Michel Marcus, May 30 2015

def mod(n,a):
    kk = 0
    while n > 0:
        kk= kk+(n%10)**a
        n =int(n//10)
    return kk
for a in range (1, 10):
    for c in range (1, 10**7):
        if c==a*mod(c,a) and a

a(16)-a(28) from Giovanni Resta, May 10 2015

A258484 Numbers m such that m equals a fixed number raised to the powers of the digits.

Original entry on oeis.org

1, 10, 12, 100, 101, 111, 1000, 1010, 1033, 1100, 2112, 4624, 10000, 10001, 11101, 20102, 31301, 100000, 100010, 100011, 100100, 100101, 100110, 101000, 101001, 101010, 101100, 101110, 101111, 101121, 110000, 110001, 110010, 110100, 110110, 110111, 111000

Offset: 1

Pieter Post, May 31 2015

Let m = abcde... and z is a fixed radix -> m = z^a +z^b +z^c +z^d +z^e...

A number m made of k ones and h zeros is a member if m-h is divisible by k. Several other large members exist, including 12095925296900865188 (base = 113) and 115330163577499130079377256005 (base = 1500). - Giovanni Resta, Jun 01 2015

			12 = 3^1 + 3^2;
31301 = 25^3 + 25^1 + 25^3 + 25^0 + 25^1;
595968 = 4^5 + 4^9 + 4^5 + 4^9 + 4^6 + 4^8;
13177388 = 7^1 + 7^3 + 7^1 + 7^7 + 7^7 + 7^3 + 7^8 + 7^8.

Giovanni Resta, Table of n, a(n) for n = 1..956 (terms < 10^12)
Giovanni Resta, Table of a(1..956) values and corresponding bases

Cf. A005188, A023052, A257784, A257766, A257787, A257814.

okQ[v_] := Block[{b, d=IntegerDigits@ v, y, t}, t = Last@ Tally@ Sort@d; b = Floor[ (v/t[[2]]) ^ (1/t[[1]])]; While[(y = Total[b^d]) > v, b--]; v==y]; Select[Range[10^5],okQ] (* Giovanni Resta, Jun 01 2015 *)

for(n=1,10^5,d=digits(n);for(m=1,n,s=sum(i=1,#d,m^d[i]);if(s==n,print1(n,", ");break);if(s>n,break))) \\ Derek Orr, Jun 12 2015

def moda(n,a,m):
    kk = 0
    while n > 0:
        na=int(n%m)
        kk= kk+a**na
        n =int(n//m)
    return kk
for c in range (1, 10**8):
    for a in range (1,20):
        if  c==moda(c,a,10):
            print (a,c)

More terms from Giovanni Resta, Jun 01 2015

A302649 Numbers that are the sum of some fixed power of the digits of their ten's complement.

Original entry on oeis.org

5, 8, 14, 3953, 33626, 89843301, 71341793655800, 245916794707565, 19429639306542698, 36106092555634673, 1818632037625982420, 4099389352522800257, 51096092690519702666, 1361788669288181208317, 80939622935362328928524, 3061856409269150191916609

Offset: 1

Paolo P. Lava, Apr 11 2018

No other term up to 3*10^12. - Giovanni Resta, Apr 12 2018

			(10 - 5) = 5 and 5^1 = 5;
(10 - 8) = 2 and 2^3 = 8;
(100 - 14) = 86 and 8^1 + 6^1 = 14;
(10000 - 3953) = 6047 and 6^4 + 0^4 + 4^4 + 7^4 = 3953;
(100000 - 33626) = 66374 and 6^5 + 6^5 + 3^5 + 7^5 + 4^5 = 33626;
(100000000 - 89843301) = 10156699 and 1^8 + 0^8 + 1^8 + 5^8 + 6^8 + 6^8 + 9^8 + 9^8 = 89843301.

Chai Wah Wu, Table of n, a(n) for n = 1..22

Cf. A023052, A261433, A302651.

with(numtheory): P:=proc(q) local a,b,i,j,k,n;
for n from 1 to q do a:=convert(10^(ilog10(n)+1)-n,base,10);
b:=convert(a,`+`); j:=1; i:=0; while n>b do
if i=b then break; else i:=b; j:=j+1; b:=add(a[k]^j,k=1..nops(a)); fi; od;
if n=b then print(n); fi; od; end: P(10^9);

a(7)-a(16) from Chai Wah Wu, Jun 06 2018

A192636 Powerful sums of two powerful numbers.

Original entry on oeis.org

8, 9, 16, 25, 32, 36, 64, 72, 81, 100, 108, 121, 125, 128, 144, 169, 196, 200, 216, 225, 243, 256, 288, 289, 324, 343, 361, 392, 400, 432, 441, 484, 500, 512, 576, 625, 648, 675, 676, 729, 784, 800, 841, 864, 900, 961, 968, 972, 1000, 1024, 1089, 1125, 1152, 1156, 1225

Offset: 1

Charles R Greathouse IV, Jul 06 2011

nonn

Browning & Valckenborgh conjecture that a(n) ~ kn^2 with k approximately 0.139485255. See their Conjecture 1 and equation (14). Their Theorems 1 and 2 establish upper and lower asymptotic bounds.

Amiram Eldar, Table of n, a(n) for n = 1..10000 (terms 1..5000 from Charles R Greathouse IV)
Tim D. Browning and K. Van Valckenborgh, Sums of three squareful numbers, Experimental Mathematics, Vol. 21, No. 2 (2012), pp. 204-211; arXiv preprint, arXiv:1106.4472 [math.NT], 2011.

Subsequence of A001694 and of A076871.

Cf. A001694, A007532, A005934, A005188, A003321, A014576, A023052, A046074, A013929, A076871, A143813. - Jonathan Vos Post, Jul 10 2011

With[{m = 1225}, pow = Select[Range[m], # == 1 || Min[FactorInteger[#][[;; , 2]]] > 1 &]; Intersection[pow, Plus @@@ Tuples[pow, {2}]]] (* Amiram Eldar, Feb 12 2023 *)

isPowerful(n)=if(n>3,vecmin(factor(n)[,2])>1,n==1)
sumset(a,b)={
  my(c=vectorsmall(#a*#b));
  for(i=1,#a,
    for(j=1,#b,
      c[(i-1)*#b+j]=a[i]+b[j]
    )
  );
  vecsort(c,,8)
}; selfsum(a)={
  my(c=vectorsmall(binomial(#a+1,2)),k);
  for(i=1,#a,
    for(j=i,#a,
      c[k++]=a[i]+a[j]
    )
  );
  vecsort(c,,8)
};
list(lim)={
  my(v=select(isPowerful, vector(floor(lim),i,i)));
  select(n->n<=lim && isPowerful(n), Vec(selfsum(v)))
};

Numbers k such that there exists some a, b, c with A001694(a) + A001694(b) = k = A001694(c).

Corrected (on the advice of Donovan Johnson) by Charles R Greathouse IV, Sep 25 2012

A302651 Numbers that are the product of some fixed power of the digits of their ten's complement.

Original entry on oeis.org

5, 8, 81, 2016, 2205

Offset: 1

Paolo P. Lava, Apr 11 2018

No other terms up to 10^40. - Giovanni Resta Apr 12 2018

No other terms up to 10^51 - Chai Wah Wu, Jun 06 2018

			(10 - 5) = 5 and 5^1 = 5;
(10 - 8) = 2 and 2^3 = 8;
(100 - 81) = 19 and 1^2 * 9^2 = 81;
(10000 - 2016) = 7984 and 7^1 * 9^1 * 8^1 * 4^1 = 2016;
(10000 - 2205) = 7795 and 7^1 * 7^1 * 9^1 * 5^1 = 2205;

Cf. A023052, A261433, A302649.

with(numtheory): P:=proc(q) local a,b,i,j,k,n;
for n from 1 to q do a:=convert(10^(ilog10(n)+1)-n,base,10);
b:=convert(a,`*`); j:=1; i:=0; while n>b do
if i=b then break; else i:=b; j:=j+1; b:=add(a[k]^j,k=1..nops(a)); fi; od;
if n=b then print(n); fi; od; end: P(10^9);

A307417 Numbers that can be expressed in a base in such a way that the sum of cubes of their digits in this base equals the original number.

Original entry on oeis.org

8, 9, 16, 17, 27, 28, 29, 35, 43, 54, 55, 62, 64, 65, 72, 91, 92, 99, 118, 125, 126, 127, 128, 133, 134, 152, 153, 189, 190, 216, 217, 224, 243, 244, 250, 251, 280, 307, 341, 342, 343, 344, 351, 370, 371, 407, 432, 433, 468, 469, 512, 513, 514, 520, 539, 559

Offset: 1

César Eliud Lozada, Apr 07 2019

There are infinitely many such numbers (proof in the second Johnson link).

			a(1) = 8 = [2, 0] (base 4) =  2^3 + 0^3
a(2) = 9 = [2, 1] (base 4) =  2^3 + 1^3
a(3) = 16 = [2, 2] (base 7) =  2^3 + 2^3
a(4) = 17 = [1, 2, 2] (base 3) =  1^3 + 2^3 + 2^3

César Eliud Lozada, Table of n, a(n) for n = 1..114
Allan Wm. Johnson Jr., Crux Mathematicorum, Vol. 5, No. 1, January 1979, problem 407, 16.
Allan Wm. Johnson Jr., Crux Mathematicorum, Vol. 5, No. 9, November 1979, solution to problem 407, 273-277.

Cf. A005188, A023052, A046197.

sqn:= []; lis:=[];
for n to 1000 do
  b := 2;
  while b < n do #needs to be adjusted
    q := convert(n, base, b);
    s := convert(map(proc (X) options operator, arrow; X^3 end proc, q), `+`);
    if evalb(s = n) then
      sqn := [op(sqn), n];
      lis := [op(lis), [n, b, ListTools[Reverse](q)]];
      break
    end if;
    b := b+1
  end do
end do;
lis := lis; #list of decompositions [number, base, conversion]
sqn := sqn; #sequence

cp's OEIS Frontend

A134703 Powerful numbers (2b): a sum of nonnegative powers of its digits.

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A061862 Powerful numbers (2a): a sum of nonnegative powers of its digits.

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

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R

A226064 Largest fixed point in base n for the sum of the fourth power of its digits.

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R

A257814 Numbers n such that k times the sum of the digits (d) to the power k equal n, so n=k*sum(d^k), for some positive integer k, where k is smaller than sum(d^k).

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A258484 Numbers m such that m equals a fixed number raised to the powers of the digits.

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Mathematica

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A302649 Numbers that are the sum of some fixed power of the digits of their ten's complement.

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Maple

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A192636 Powerful sums of two powerful numbers.

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