A061912 -id:A061912 - cp's OEIS frontend

A061910 Positive numbers k such that sum of digits of k^2 is a square.

1, 2, 3, 6, 9, 10, 11, 12, 13, 14, 15, 18, 20, 21, 22, 23, 30, 31, 39, 41, 45, 48, 51, 58, 59, 60, 67, 68, 76, 77, 85, 86, 90, 94, 95, 100, 101, 102, 103, 104, 105, 110, 111, 112, 113, 120, 121, 122, 130, 131, 139, 140, 148, 150, 157, 158, 166, 175, 176, 180, 184, 185

Offset: 1

Asher Auel, May 17 2001

			6^2 = 36 and 3+6 = 9 is a square. 13^2 = 169 and 1+6+9 = 16 is a square.

Bruno Berselli, Table of n, a(n) for n = 1..1000
MathStackExchange, Numbers n such that the digit sum of n2 is a square, 2015-2016.

Cf. A007953, A004159, A053057, A061909, A061911, A061912.

Sequence A293832 gives the start of the first run of n consecutive values.

[ n: n in [1..185] | IsSquare(&+Intseq(n^2)) ];  // Bruno Berselli, Jul 29 2011

readlib(issqr): f := []: for n from 1 to 200 do if issqr(convert(convert(n^2,base,10),`+`)) then f := [op(f), n] fi; od; f;

Select[Range[185], IntegerQ[Sqrt[Total[IntegerDigits[#^2]]]] &] (* Jayanta Basu, May 06 2013 *)

is(n)=n=eval(Vec(Str(n^2)));issquare(sum(i=1,#n,n[i])) \\ Charles R Greathouse IV, Jul 29 2011

select( is_A061910(n)=issquare(sumdigits(n^2)), [0..199]) \\ Includes the initial 0. - M. F. Hasler, Oct 16 2017

from gmpy2 import is_square
A061910 = [n for n in range(1,10**3) if is_square(sum(int(d) for d in str(n*n)))] # Chai Wah Wu, Sep 03 2014

A067075 a(n) is the smallest number m such that the sum of the digits of m^3 is equal to n^3.

Original entry on oeis.org

0, 1, 2, 27, 1192, 341075, 3848163483, 2064403725539899

Offset: 0

Amarnath Murthy, Jan 05 2002

If n = 6*k, a(n) <= A002283(n^3/18). For example, a(6) = 3848163483 <= A002283(6^3/18) = 999999999999. - Seiichi Manyama, Aug 12 2017
a(n) >= ceiling(A051885(n^3)^(1/3)). For example a(7) >= ceiling(A051885(7^3)^(1/3)) = ceiling((2*10^38-1)^(1/3)) = 5848035476426 - David A. Corneth, Aug 23 2018
From Zhining Yang, Jun 20 2024: (Start)
a(8) <= 99995999799995999999999.
a(9) <= 999699989999999949999999999999999.
a(10) <= 199999999929999999999949999999999999999999999.
(End)

			a(3) = 27 as 27^3 = 19683 is the smallest cube whose digit sum = 27 = 3^3.

Cf. A051885, A061912, A067074. Subsequence of A067177.

Do[k = 1; While[Plus @@ IntegerDigits[k^3] != n^3, k++ ]; Print[k], {n, 1, 6}] (* Ryan Propper, Jul 07 2005 *)

a(n) = my(k=0); while (sumdigits(k^3) != n^3, k++); k; \\ Seiichi Manyama, Aug 12 2017

Corrected and extended by Ryan Propper, Jul 07 2005
a(0)=0 prepended by Seiichi Manyama, Aug 12 2017
a(7) from Zhining Yang, Jun 20 2024

A280012 a(n) = least positive integer k such that sumdigits(k^2) = n*sumdigits(k).

Original entry on oeis.org

1, 2, 3, 13, 113, 1113, 11113, 211113, 101011113, 1101111211, 110101111211

Offset: 1

M. F. Hasler, Apr 14 2017

a(n) exists for any n, since sum_{i=0..n-1} 10^(2^i-1) is an integer with the required property, having n digits 1, with its square having n digits 1 at positions 2^i-1 (n>=i>=1), and n(n-1)/2 digits 2 at positions 2^i+2^j-1 (n>=i>j>=0 i.e. at positions 1<=k<2^(n+1) for k in A099628).

a(12) <= 21201101101122, a(13) <= 10101010101101122. - Giovanni Resta, Apr 15 2017

Les Reid, Problem #12, Challenge Problem Archive, Missouri State University Math Department, Academic year 2013-2014.

Cf. A007953, A061912, A099628.

a(n)=for(k=1,9e9,sumdigits(k^2)==n*sumdigits(k)&&return(k))

a(10)-a(11) from Giovanni Resta, Apr 15 2017

A067072 a(n) = smallest square m^2 such that the sum of the digits of m^2 is equal to n^2.

Original entry on oeis.org

0, 1, 4, 9, 169, 4489, 69696, 2778889, 277788889, 8998988769, 2699997789889, 898999897988929, 498999778899898896, 969988797999759789889, 1679898987989978888999689

Offset: 0

Amarnath Murthy, Jan 05 2002

			a(5) = 4489 = 67^2 as it is the smallest square whose digit sum = 25 = 5^2.

Cf. A061912, A067074.

a(n) = A061912(n)^2.

More terms from Jason Earls, Jan 09 2002
a(11)-a(14) from Donovan Johnson, Jul 10 2012
a(0)=0 prepended by Seiichi Manyama, Aug 12 2017

A286650 a(n) is the smallest number m such that the sum of the digits of m^4 is equal to n^4.

Original entry on oeis.org

0, 1, 11, 1434, 1269681358

Offset: 0

Seiichi Manyama, Aug 15 2017

			a(2) = 11 as 11^4 = 14641 is the smallest fourth power whose digit sum = 16 = 2^4.

Cf. A061912, A067075, A069648, A291145.

Cf. A000583 (n^4), A055565 (sum of digits of n^4).

{a(n) = my(k=0); while(sumdigits(k^4) != n^4, k++); k}

A061911 Square root of the sum of the digits of k^2 when this sum is a square.

Original entry on oeis.org

1, 2, 3, 3, 3, 1, 2, 3, 4, 4, 3, 3, 2, 3, 4, 4, 3, 4, 3, 4, 3, 3, 3, 4, 4, 3, 5, 4, 5, 5, 4, 5, 3, 5, 4, 1, 2, 3, 4, 4, 3, 2, 3, 4, 5, 3, 4, 5, 4, 4, 4, 4, 4, 3, 5, 5, 5, 4, 5, 3, 5, 4, 5, 5, 2, 3, 4, 4, 3, 4, 5, 4, 5, 4, 4, 5, 4, 4, 4, 3, 5, 5, 6, 4, 5, 5, 5, 5, 5, 5, 3, 4, 4, 4, 5, 3, 4, 3, 5, 4, 5, 4, 5, 4, 3

Offset: 1

Asher Auel, May 17 2001

			6^2 = 36 and 3+6 = 9 is a square, thus 3 is in the sequence. 13^2 = 169 and 1+6+9 = 16 is a square, thus 4 is in the sequence.

Cf. A007953, A004159, A061909, A061910, A061912.

readlib(issqr): f := []: for n from 1 to 200 do if issqr(convert(convert(n^2,base,10),`+`)) then f := [op(f),sqrt(convert(convert(n^2,base,10),`+`))] fi; od; f;

Select[Table[Sqrt[Total[IntegerDigits[n^2]]],{n,350}],IntegerQ] (* Jayanta Basu, May 06 2013 *)

a(n) = sqrt(A004159(A061910(n))) = sqrt(A007953((A061910(n))^2)). - Zak Seidov, Jul 04 2012

A328364 a(n) is the smallest number m such that the sum of the digits of m^5 is equal to n^5.

Original entry on oeis.org

0, 1, 47, 13174539

Offset: 0

Seiichi Manyama, Oct 14 2019

			a(2) = 47 as 47^5 = 229345007 is the smallest fifth power whose digit sum = 32 = 2^5.

Cf. A061912, A069648, A067075, A286650, A328374.

Cf. A000584 (n^5), A055566 (sum of digits of n^5).

{a(n) = my(k=0); while(sumdigits(k^5) != n^5, k++); k}

A280798 a(n) is the smallest integer m such that sumdigits(m^2) = 4^n.

Original entry on oeis.org

1, 2, 13, 16667

Offset: 0

Michel Marcus, Jan 08 2017

See the Mathematical Reflections link for a proof that a(n) exists for all n.
a(4) > 300*10^6.
From Jon E. Schoenfield, Jan 08 2017: (Start)
As shown below, 264575131106460 <= a(4) <= 13663784168010583.
The maximal sum of the last three digits of a square is 25, which occurs only when those digits are 889 (e.g., 83^2 = 6889), so since sumdigits(a(4)^2) = 256, the sum of the digits that precede the last three must be at least 256 - 25 = 231, thus a(4)^2 >= 69999999999999999999999999889 (call this number j), so a(4) >= ceiling(sqrt(j)) = 264575131106460. (This bound could easily be improved; e.g., since j is not a square, m^2 cannot be less than the smallest number greater than j whose digit sum is 256 and whose last three digits are 889, i.e., m^2 >= 78999999999999999999999999889, so m >= 281069386451104.)
The last digit of a square m^2 is maximized (at 9) iff the last digit of m is 3 or 7. The sum of the last two digits of m^2 is maximized (at 17) iff the last two digits are 89, which occurs iff the last two digits of m are 33, 83, 17, or 67. For k >= 3, it appears that the sum of the last k digits of m^2 is maximized (at 9k-2) iff the last k digits are all 9s except for the two 8s immediately before the final 9, which occurs iff the k-digit suffix of m takes one of eight values, as shown in the table below; the line extending upward from each suffix of more than one digit connects it to the suffix from which it inherits all but its first digit.
.
k   k-digit suffixes of m that maximize sumdigits(m^2 mod 10^k)
= ===============================================================
1               3                               7
               / \                             / \
              /   \                           /   \
             /     \                         /     \
            /       \                       /       \
           /         \                     /         \
          /           \                   /           \
         /             \                 /             \
2      33              83              17              67
        |\              |\              |\              |\
3     833 333         583 083         417 917         167 667
        |\              |\              |\              |\
4    1833 6833       0583 5583       9417 4417       8167 3167
        |\              |\              |\              |\
5   41833 91833     10583 60583     89417 39417     58167 08167
        |\              |\              |\              |\
6  041833 541833   010583 510583   989417 489417   958167 458167
        |\              |\              |\              |\
7 5041833 0041833 8010583 3010583 1989417 6989417 4958167 9958167
        |\              |\              |\              |\
.     ... ...         ... ...         ... ...         ... ...
.
Since numbers m ending in these suffixes have squares m^2 such that sumdigits(m^2 mod 10^k) is maximized, their full digit sums sumdigits(m^2) tend to be larger than those of nearby numbers with other suffixes. A search over a range of prefixes using the 10-digit suffix 4168010583 found that m = 13663784168010583 has sumdigits(m^2) = 256, which yields an upper bound for a(4). (End)

			a(1)=2 since 2^2=4 with sum of digits 4.

Mathematical Reflections, Solution to Problem J307, Issue 5, 2015, p. 1.

Cf. A000302, A007953, A061912.

a(n) = my(k=1); while (sumdigits(k^2) != 4^n, k++); k;

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A061910 Positive numbers k such that sum of digits of k^2 is a square.

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A067075 a(n) is the smallest number m such that the sum of the digits of m^3 is equal to n^3.

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A280012 a(n) = least positive integer k such that sumdigits(k^2) = n*sumdigits(k).

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A067072 a(n) = smallest square m^2 such that the sum of the digits of m^2 is equal to n^2.

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A286650 a(n) is the smallest number m such that the sum of the digits of m^4 is equal to n^4.

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A061911 Square root of the sum of the digits of k^2 when this sum is a square.

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A328364 a(n) is the smallest number m such that the sum of the digits of m^5 is equal to n^5.

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A280798 a(n) is the smallest integer m such that sumdigits(m^2) = 4^n.

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