A058369 -id:A058369 - cp's OEIS frontend

A176760 Numbers k such that k^2 and k^4 have the same sum of digits.

0, 1, 3, 10, 17, 19, 27, 30, 57, 93, 100, 170, 190, 219, 267, 270, 300, 314, 327, 359, 387, 417, 423, 424, 570, 685, 693, 807, 828, 891, 917, 930, 963, 1000, 1207, 1223, 1317, 1333, 1673, 1693, 1700, 1827, 1864, 1900, 1917, 2141, 2190, 2202, 2213, 2364, 2367

Offset: 1

Ulrich Krug (leuchtfeuer37(AT)gmx.de), Apr 25 2010

Let sod(n) := digital sum of n (A007953); here we have sod(n^2) = sod(n^4).
Trivial cases:
(I) Powers of 10, as sod((10^k)^2) = sod((10^k)^4) = 1.
(II) If N is a term of sequence, then so is 10 * N.

			sod(3^2) = sod(9) = 9 = sod(81) = sod(3^4), so 3 is a term.
sod(17^2) = sod(289) = 19 = sod(83521) = sod(17^4), so 17 is a term.

Hans Schubart, Einfuehrung in die klassische und moderne Zahlentheorie, Vieweg, Braunschweig, 1974.

Cf. A000290, A000583, A058369, A176012, A176111, A176465.

Select[Range[0,2000],Total[IntegerDigits[#^2]]==Total[IntegerDigits[#^4]]&]  (* Harvey P. Dale, Jan 19 2011 *)

Edited by D. S. McNeil, Nov 21 2010

a(43)-a(51) from Jason Yuen, Oct 13 2024

A357756 a(n) is the least k > 0 such that A007953(nk) equals A007953((nk)^2), where A007953 is the sum of the digits.

Original entry on oeis.org

1, 1, 5, 3, 25, 2, 3, 27, 62, 1, 1, 5, 15, 27, 128, 3, 31, 17, 1, 1, 5, 9, 9, 2, 75, 4, 18, 7, 64, 5, 3, 16, 56, 3, 85, 17, 5, 27, 5, 9, 25, 9, 45, 13, 27, 1, 1, 27, 66, 54, 2, 9, 9, 18, 22, 1, 32, 15, 25, 135, 3, 18, 8, 3, 28, 9, 3, 43, 47, 72, 27, 8, 25, 126, 27

Offset: 0

Thomas Scheuerle, Oct 12 2022

A task in the German competition "Bundeswettbewerb Mathematik 2021" was to prove that for each positive integer n there exists a k such that A007953(n*k) = A007953((n*k)^2).
One of the proposed proofs uses the argument that numbers of the form m = (10^x-1)*(10^y) will have the desired property A007953(m) = A007953(m^2). Thus we need to prove that we can find for all n a k, x and y such that n*k = (10^x-1)*(10^y). Let n be of the form b*2^c*5^d with b odd and not divisible by 5, then we know that y = max(c, d). From Euler's totient theorem we know that 10^x-1 will be divisible by e if x = A000010(e) where A000010 is Euler's totient function. See the formula section for the corresponding resulting k.
a(n) will never be divisible by 10.
If n is divisible by 3 but not by 9, then a(n) is divisible by 3. - Robert Israel, Oct 13 2022

Johann Beurich, Wie kann man das beweisen? (Bundeswettbewerb Mathematik 2021), YouTube video, 2021 (in German).
Bundeswettbewerb Mathematik 2021, Die Aufgaben_und_Loesungen der 2. Runde 2021 (in German). The proof and solutions.

Cf. A000010, A007953, A051628, A058369, A060284, A132740, A178505.

f:= proc(n) local k;
   for k from 1 do if sd(n*k) = sd((n*k)^2) then return k fi od
end proc:
map(f, [$1..100]); # Robert Israel, Oct 13 2022

a(n) = {my(k = 1); while(sumdigits(n*k)!=sumdigits((n*k)^2),k++);k}

def sd(n): return sum(map(int, str(n)))
def a(n):
    k = 1
    while not sd(n*k) == sd((n*k)**2): k += 1
    return k
print([a(n) for n in range(75)]) # Michael S. Branicky, Oct 13 2022

a(A058369(n)) = 1.
a(a(n)) <= n.
a(n) <= A132740(n)*A060284(A132740(n))*10^A051628(n)/n.
or a(n) <= (10^A000010(A132740(n))-1)*10^A051628(n)/n.

A374024 Integers k such that digsum(k) = digsum(k^2) = p, where p is prime and digsum(i) = A007953(i).

Original entry on oeis.org

199, 289, 379, 388, 496, 559, 568, 595, 739, 775, 838, 955, 1099, 1189, 1198, 1468, 1495, 1585, 1738, 1747, 1765, 1792, 1855, 1990, 2098, 2494, 2665, 2881, 2890, 3169, 3196, 3259, 3349, 3466, 3493, 3745, 3790, 3880, 4249, 4519, 4735, 4951, 4960, 5149, 5482

Offset: 1

Gonzalo Martínez, Jul 05 2024

Subsequence of A058369.
If k is a term, then digsum(k) = 19, 37 or 73, for k < 10^9.
If k is an integer such that digsum(k) = digsum(k^2) = p, with p prime, then p == 1 (mod 9) (A061237).
This sequence has infinitely many terms of the form 1999...9 (A067272). If p is a prime with p == 1 (mod 9), i.e., p = 9m + 1 for some m, then t = 2*10^m - 1 = 1999...9, i.e., 1 followed by m 9's, is in this sequence since digsum(t) = 9m + 1 = p and t^2 = 39...960...01, where there are (m - 1) 9's and (m - 1) 0's, so digsum(t^2) = 3 + 9*(m - 1) + 6 + 1 = 9m + 1 = p. Dirichlet's theorem guarantees the existence of infinitely many primes of the form 9w + 1 and hence infinitely many terms of this sequence.
2*10^m - 1 is the least number with digit sum 9*m + 1.  Since the next prime congruent to 1 (mod 9) after 73 is 109 = 9*12 + 1, the first term with digit sum other than 19, 37 or 73 is 2*10^12 - 1. - Robert Israel, Jul 07 2024

			199 is a term, because its digital sum is 1 + 9 + 9 = 19 and 199^2 = 39601, whose digital sum is 3 + 9 + 6 + 0 + 1 = 19, which is prime.

Cf. A007953, A058369, A061237, A067272, A254066.

ds:= n -> convert(convert(n,base,10),`+`):
filter:= proc(n) local p;
  p:= ds(n);
  isprime(p) and ds(n^2) = p
end proc:
select(filter, [seq(i,i=1..1000, 9)]); # Robert Israel, Jul 05 2024

Select[Range[5490],PrimeQ[dg=DigitSum[#]]&&(dg==DigitSum[#^2])&] (* Stefano Spezia, Jul 05 2024 *)

isok(k) = my(s=sumdigits(k)); isprime(s) && (s==sumdigits(k^2)); \\ Michel Marcus, Jul 06 2024

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A176760 Numbers k such that k^2 and k^4 have the same sum of digits.

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A357756 a(n) is the least k > 0 such that A007953(nk) equals A007953((nk)^2), where A007953 is the sum of the digits.

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A374024 Integers k such that digsum(k) = digsum(k^2) = p, where p is prime and digsum(i) = A007953(i).

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cp's OEIS Frontend

A176760 Numbers k such that k^2 and k^4 have the same sum of digits.

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Mathematica

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A357756 a(n) is the least k > 0 such that A007953(n*k) equals A007953((n*k)^2), where A007953 is the sum of the digits.

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Formula

A374024 Integers k such that digsum(k) = digsum(k^2) = p, where p is prime and digsum(i) = A007953(i).

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Maple

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PARI

A357756 a(n) is the least k > 0 such that A007953(nk) equals A007953((nk)^2), where A007953 is the sum of the digits.