A130181 -id:A130181 - cp's OEIS frontend

A130179 Largest k such that k <= 81(number of digits of k^n)(number of digits of k^(n+1)).

2268, 7776, 18954, 35397, 56376, 85050, 119556, 159894, 209952, 267300, 331047, 402084, 479520, 570807, 670032, 777195, 892296, 1015335, 1146312, 1285227, 1432080, 1586871, 1749600, 1932498, 2125035, 2312712, 2522340, 2741607

Offset: 1

Klaus Brockhaus, May 20 2007

a(n) is an upper bound for A130181(n) and all the more so for A126783(n); apparently even A130181(n) < a(n)/4.
All terms are divisible by 81; the quotients a(n)/81 are in A130085.
For some n (18, 34, 35, 38, 42, 58, 59, ...) the line y = x and the graph of the staircase function y = 81*(number of digits of x^n)*(number of digits of x^(n+1)) intersect twice; this possibility has to be taken into account by the program.

			Let D(n,k) = 81*(number of digits of k^n)*(number of digits of k^(n+1)).
D(2,k) > k for k = 1..4641, D(2,k) = 7776 for k = 4642..9999, D(2,k) < k for k >= 10000, hence a(2) = 7776.
D(18,k) > k for k = 1..885866, D(18,k) = 997272 for k = 885867..999999, D(18,k) = 1015335 for k = 1000000..1128837, D(18,k) < k for k >= 1128838, hence a(18) = 1015335.

Klaus Brockhaus, Table of n, a(n) for n=1..100

Cf. A126783, A130181, A130085.

{for(n=1, 28, s=30*n; k=s; while(k<81*length(Str(k^n))*length(Str(k^(n+1))), k+=s); r=0; g=0; k-=s; b=1; while(b, p=81*length(Str(k^n))*length(Str(k^(n+1))); if(rr, b=0, g=h)); k++); print1(g, ","))}

A126783 Smallest k > 1 such that (sum of digits of k^n)*(sum of digits of k^(n+1)) = k, or 0 if no such k exists.

Original entry on oeis.org

80, 80, 70, 3905, 4004, 700, 19278, 32761, 5600, 8100, 24940, 10600, 56330, 68040, 81760, 149705, 116180, 126360, 123580, 0, 65500, 311003, 205030, 114400, 454951, 317350, 312170, 296270, 359380, 332750, 699785, 723338, 498150, 499130, 901368

Offset: 1

Paolo P. Lava and Giorgio Balzarotti, Apr 17 2007

For each n there is an upper bound (see A130179) for values of k such that (sum of digits of k^n)*(sum of digits of k^(n+1)) = k, hence the number of such k is finite, possibly zero, (see A130180) and if the number is not zero there is a largest one (see A130181).

			For n = 2 the smallest such k is 80: 80^2 = 6400 and 6+4+0+0 = 10; 80^3 = 512000 and 5+1+2+0+0+0 = 8; 10*8 = 80. Hence a(2) = 80.
For n = 3 the smallest such k is 70: 70^3 = 343000 and 3+4+3+0+0+0 = 10; 70^4 = 24010000 and 2+4+0+1+0+0+0+0 = 7; 10*7 = 70. Hence a(3) = 70.

Klaus Brockhaus, Table of n, a(n) for n=1..54

Cf. A130179, A130180, A130181.

P:=proc(n) local a,i,j,k,w,x; for a from 1 by 1 to n do for i from 1 by 1 to n*n do w:=0;k:=i^a;j:=0;x:=i^(a+1); while k>0 do w:=w+k-(trunc(k/10)*10); k:=trunc(k/10); od; while x>0 do j:=j+x-(trunc(x/10)*10); x:=trunc(x/10); od; if (i=w*j and i>1) then print(i); break; fi; od; od; end: P(1000);

Edited and a(17) to a(35) added by Klaus Brockhaus, May 14 2007

A130180 Number of numbers k > 1 such that (sum of digits of k^n)*(sum of digits of k^(n+1)) = k.

Original entry on oeis.org

5, 3, 12, 2, 6, 8, 4, 1, 13, 8, 7, 14, 8, 3, 9, 1, 5, 12, 4, 0, 13, 4, 7, 7, 1, 4, 7, 2, 5, 8, 2, 4, 8, 7, 1, 10, 5, 2, 8, 4, 2, 10, 2, 6, 10, 2, 3, 6, 2, 4, 4, 2, 3, 9, 2, 3, 8, 1, 3, 8, 5, 3, 6, 4, 6, 8, 4, 3, 10, 0, 1, 6, 3, 6, 6, 4, 2, 7, 2, 1

Offset: 1

Klaus Brockhaus, May 14 2007

			80, 1036, 1215 are the only numbers k > 1 such that (sum of digits of k^2)*(sum of digits of k^3) = k, hence a(2) = 3.

Lars Blomberg, Table of n, a(n) for n = 1..116

Cf. A126783 (smallest k), A130179 (upper bound), A130181 (largest k).

a(55)-a(80) and b-file from Lars Blomberg, Dec 11 2011

A357080 Numbers k such that the sum of the digits of k multiplied by the sum of the digits of k^2 equals k.

Original entry on oeis.org

0, 1, 80, 162, 243, 476, 486

Offset: 1

Tanya Khovanova, Sep 10 2022

Suppose k has m digits, then the sum of the digits of k multiplied by the sum of the digits of k^2 is bounded by 9m times 9*(2m), which equals 162m^2. On the other hand, k is greater than 10^(m-1), which grows much faster than 162m^2. It follows that k can't have more than 4 digits.

			The sum of the digits of 80 is 8, the sum of the digits of 80^2 = 6400 is 10. The number 80 itself is 8*10. Thus, 80 is in this sequence.

Cf. A007953, A004159, A130181.

Select[Range[100000], # == Total[IntegerDigits[#]] Total[IntegerDigits[#^2]] &]

isok(k) = k == sumdigits(k)*sumdigits(k^2); \\ Michel Marcus, Sep 11 2022

def sd(n): return sum(map(int, str(n)))
def ok(n): return sd(n) * sd(n*n) == n
print([k for k in range(10**5) if ok(k)]) # Michael S. Branicky, Sep 11 2022

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A130179 Largest k such that k <= 81*(number of digits of k^n)*(number of digits of k^(n+1)).

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A126783 Smallest k > 1 such that (sum of digits of k^n)*(sum of digits of k^(n+1)) = k, or 0 if no such k exists.

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A130180 Number of numbers k > 1 such that (sum of digits of k^n)*(sum of digits of k^(n+1)) = k.

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Crossrefs

Extensions

A357080 Numbers k such that the sum of the digits of k multiplied by the sum of the digits of k^2 equals k.

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Author

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Mathematica

PARI

Python

A130179 Largest k such that k <= 81(number of digits of k^n)(number of digits of k^(n+1)).