A004159 -id:A004159 - cp's OEIS frontend

A325450 Numbers k such that sum of digits (k) and sum of digits (k^2) is 9.

9, 18, 45, 90, 180, 351, 450, 900, 1800, 3510, 4500, 9000, 18000, 35100, 45000, 90000, 180000, 351000, 450000, 900000, 1800000, 3510000, 4500000, 9000000, 18000000

Offset: 1

Vincenzo Librandi, May 10 2019

A007953(A058369(n)) begins with 1, 9, 1, 9, 10, 9, 10, 10, 9, 18, ...; the 1's come from A011557, the 9's come from this sequence, the 10's come from A325451.

			a(3) = 45 because 4+5 = 9, 45^2 = 2025, and 2+0+2+5 = 9.

Cf. A004159, A007953, A325451.

Subsequence of A058369 (k and k^2 have same digit sum).

[n: n in [1..2*10^7] | &+Intseq(n^2) eq 9  and &+Intseq(n) eq 9];

Select[Range[2 10^7], Total[IntegerDigits[#]]==9&& Total[IntegerDigits[#^2]]==9&]

A325451 Numbers k such that sum of digits (k) and sum of digits (k^2) is 10.

Original entry on oeis.org

19, 46, 55, 145, 190, 361, 451, 460, 550, 1450, 1900, 3610, 4510, 4600, 5500, 14500, 19000, 20251, 36100, 45100, 46000, 55000, 145000, 190000, 202510, 361000, 451000, 460000, 550000, 1450000, 1900000, 2025100, 3610000, 4510000, 4600000, 5500000, 14500000

Offset: 1

Vincenzo Librandi, May 10 2019

Subsequence of A058369 (k and k^2 have same digit sum).

A007953(A058369(n)) begins with 1, 9, 1, 9, 10, 9, 10, 10, 9, 18, ...: the 1's come from A011557, the 9's come from A325450, and the 10's come from this sequence.

			a(4) = 145 because 1+4+5 = 10, 145^2 = 21025, and 2+1+0+2+5 = 10.

Cf. A004159, A007953, A058369, A325450.

[n: n in [1..2*10^7] | &+Intseq(n^2) eq 10  and &+Intseq(n) eq 10];

Select[Range[2 10^7], Total[IntegerDigits[#]]==10 && Total[IntegerDigits[#^2]]==10 &]

A351651 a(n) is the quotient obtained when digsum(m^2) is divided by digsum(m), with digsum = sum of digits = A007953 and m = A351650(n).

Original entry on oeis.org

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

Offset: 1

Bernard Schott, Feb 16 2022

All positive integers are terms of this sequence (see A280012).
a(n) = 1 iff m = A351650(n) is a term of A058369 \ {0}.
a(n) = digsum(n) if m = A351650(n) is a term of A061909 \ {0}.

			A351650(8) = 13, then digsum(13) = 1+3 = 4 while digsum(13^2) = digsum(169) = 1+6+9 = 16; hence, a(8) = 16/4 = 4.

Diophante, A1730 - Des chiffres à sommer pour un entier (in French).

Cf. A002283, A004159, A007953, A058369, A061909, A254066, A280012, A351650.

Select[Total[IntegerDigits[#^2]]/Total[IntegerDigits[#]]& /@ Range[300], IntegerQ] (* Amiram Eldar, Feb 16 2022 *)

lista(nn) = {my(list = List(), q); for (n=1, nn, if (denominator(q=sumdigits(n^2)/sumdigits(n))==1, listput(list, q));); Vec(list);} \\ Michel Marcus, Feb 16 2022

a(n) = A004159(A351650(n)) / A007953(A351650(n)).

More terms from Michel Marcus, Feb 16 2022

A355505 a(n) is the number of distinct cycles when iterating the function f_n(x), where f_n(x) is the sum of the digits in base n of x^2.

Original entry on oeis.org

2, 5, 3, 4, 4, 7, 4, 3, 4, 6, 4, 7, 4, 8, 6, 3, 3, 7, 5, 7, 9, 7, 4, 6, 4, 7, 5, 9, 5, 12, 7, 3, 9, 5, 8, 9, 5, 10, 9, 6, 4, 16, 8, 9, 8, 7, 5, 7, 9, 7, 7, 8, 4, 9, 8, 8, 11, 9, 4, 14, 7, 13, 11, 3, 8, 16, 7, 6, 9, 16, 8, 8, 5, 9, 9, 11, 13, 17, 7, 6, 6, 7, 5, 17, 6, 15, 11, 9, 4

Offset: 2

Wouter Zandsteeg, Jul 04 2022

The trajectory from every starting point will enter a cycle given a sufficient number of iterations of f_n.
To determine a(n), only starting points 0, 1, 2, ..., n^2 have to be checked for cycles. Larger starting points will always lead to a cycle reached from one (or more) of 0, 1, 2, ..., n^2.
From Iain Fox, Jul 09 2022: (Start)
Since f_n(x) <= (n-1)*(1 + floor(2*log_n(x))), only numbers less than the largest zero of (n-1)*(1 + floor(2*log_n(x))) - x need to be checked.
The value mentioned above is less than or equal to (2-2n)*W_{-1}(log(n)/((2-2n)*sqrt(n)))/log(n) where W_k(x) is the k-th branch of the Lambert W function. (End)

			a(8) = 4 because there are 4 cycles for n = 8:
  f_8(0) = 0 (since 0^2 = 0 = 0_8, with digit sum 0),
  f_8(1) = 1 (1^2 = 1 = 1_8, with digit sum 1),
  f_8(4) = 2 (4^2 = 20_8) and f_8(2) = 4 (2^2 = 4_8), and
  f_8(7) = 7 (7^2 = 61_8, with digit sum 7).

Iain Fox, Table of n, a(n) for n = 2..10000

Cf. A004159, A061903, A159918.

a(n) = my(d=1); while(d<=logint(((n-1)*d)^2, n)+1, d++); my(l=(n-1)*(d-1)+1, x=vector(l, i, l-i), y=[], z=[], j, c=0); for(i=1, #x, y=setunion(y, z); j=x[i]; z=[]; while(!setsearch(z, j), if(setsearch(y, j), next(2)); z=setunion(z, [j]); j=sumdigits(j^2, n)); c++); c \\ Iain Fox, Jul 09 2022

A369956 a(n) is the least integer m such that the sum of the digits of m^2 is k+n where k is the number of digits of n.

Original entry on oeis.org

1, 101, 11, 2, 149, 32, 4, 12, 3, 8, 106, 16, 7, 103, 13, 108, 24, 17, 1019, 124, 43, 1013, 113, 67, 114, 63, 10024, 1024, 133, 83, 1067, 167, 1044, 264, 314, 10087, 1303, 313, 10093, 1183, 707, 1374, 1333, 836, 10343, 1667, 100264, 10714, 2236, 10386, 3114

Offset: 0

Zhining Yang, Feb 06 2024

			a(5)=32 because 32 is the least integer with 2 digits and 32^2=1024 and 1+0+2+4=2+5.

Zhining Yang, Table of n, a(n) for n = 0..152

Cf. A004159, A369953, A369955.

Table[SelectFirst[Range@200000,Total[IntegerDigits[#^2]]==n+Length@IntegerDigits@#&],{n,0,50}]

A061806 Numbers n such that the iterative cycle: n -> sum of digits of n^2 has only three distinct elements.

Original entry on oeis.org

4, 5, 17, 26, 28, 32, 37, 40, 43, 44, 49, 50, 53, 62, 63, 64, 67, 73, 74, 76, 77, 82, 83, 86, 87, 88, 89, 91, 92, 93, 94, 97, 98, 107, 109, 113, 114, 116, 117, 118, 122, 124, 125, 126, 127, 128, 133, 137, 141, 143, 149, 154, 157, 158, 161, 164, 166, 167, 169, 170

Offset: 1

Asher Auel, May 17 2001

			4 -> 1+6 = 7 -> 4+9 = 13 -> 1+6+9 = 16 -> 2+5+6 = 13, thus {7,13,16} are the only distinct elements of the iterative cycle of 4.

Cf. A007953, A004159, A061903 - A061910.

A061905 The iterative cycle: n -> sum of digits of n^2 has only two distinct elements.

Original entry on oeis.org

7, 8, 13, 14, 16, 19, 22, 23, 24, 25, 27, 29, 31, 33, 34, 35, 36, 38, 41, 42, 46, 47, 52, 54, 55, 56, 57, 58, 59, 61, 65, 66, 68, 69, 70, 71, 72, 75, 78, 79, 80, 81, 84, 85, 95, 96, 99, 103, 104, 106, 108, 112, 115, 119, 121, 123, 129, 130, 131, 132, 135, 138, 139, 140

Offset: 1

Asher Auel, May 17 2001

It seems that {10,1}, {13,16} and {9,18} are the only iterative cycles with 2 distinct elements.

			7 -> 4+9 = 13 -> 1+6+9 = 16 -> 2+5+6 = 13, thus only {13,16} are contained in the iterative cycle of 7. 24 -> 5+7+6 = 18 -> 3+2+4 = 9 -> 8+1 = 9, thus {18,9} are the only elements of the iterative cycle of 24.

Cf. A007953, A004159, A061903 - A061910.

A239937 Numbers k such that DigitSum(k^2) > DigitSum((k+1)^2).

Original entry on oeis.org

3, 7, 8, 9, 14, 17, 19, 24, 26, 28, 29, 31, 33, 34, 37, 38, 39, 43, 44, 47, 48, 53, 54, 57, 59, 63, 64, 67, 69, 70, 74, 77, 78, 79, 83, 84, 87, 88, 89, 93, 94, 97, 98, 99, 104, 107, 109, 114, 117, 118, 119, 122, 124, 126, 128, 129, 133, 134, 137, 138, 141, 143

Offset: 1

Oliver Bel, Mar 29 2014

			For k=3, we have DigitSum(3^2) = 9 > 7 = DigitSum(4^2).

Cf. A007953 (sum of digits of n), A004159 (sum of digits of n^2).

lis = Table[Total[IntegerDigits[n^2, 10]], {n, 1, 100}];
Flatten[Position[Greater @@@ Partition[lis, 2, 1], True]]
Position[Partition[Total[IntegerDigits[#]]&/@(Range[150]^2),2,1],?(#[[1]]> #[[2]]&),1,Heads->False]//Flatten (* _Harvey P. Dale, Feb 19 2020 *)

isok(k) = sumdigits(k^2) > sumdigits((k+1)^2); \\ Michel Marcus, Jul 03 2021

More terms from Jon E. Schoenfield, Mar 29 2014

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

A362069 Numbers k such that k+digitsum(k^2) is a square.

Original entry on oeis.org

0, 17, 62, 71, 117, 125, 197, 206, 296, 297, 305, 413, 414, 557, 558, 692, 702, 711, 863, 864, 872, 873, 1062, 1070, 1071, 1268, 1493, 1502, 1727, 1736, 1737, 1745, 1998, 2006, 2267, 2276, 2285, 2564, 2565, 2573, 2879, 2888, 2889, 3221, 3222

Offset: 1

Alexandru Petrescu, May 17 2023

Conjecture: there are infinitely many pairs of consecutive terms. Example: (296,297); (413,414); (863,864).

			k=17 is a term because k^2=289 and 17+2+8+9=36=6^2.

Cf. A004159, A066564.

Select[Range[0, 3300], IntegerQ[Sqrt[# + Plus @@ IntegerDigits[#^2]]] &] (* Amiram Eldar, May 18 2023 *)

PARI
```
isok(k)=issquare(k+sumdigits(k^2))
```

cp's OEIS Frontend

A325450 Numbers k such that sum of digits (k) and sum of digits (k^2) is 9.

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A325451 Numbers k such that sum of digits (k) and sum of digits (k^2) is 10.

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A351651 a(n) is the quotient obtained when digsum(m^2) is divided by digsum(m), with digsum = sum of digits = A007953 and m = A351650(n).

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A355505 a(n) is the number of distinct cycles when iterating the function f_n(x), where f_n(x) is the sum of the digits in base n of x^2.

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A369956 a(n) is the least integer m such that the sum of the digits of m^2 is k+n where k is the number of digits of n.

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A061806 Numbers n such that the iterative cycle: n -> sum of digits of n^2 has only three distinct elements.

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A061905 The iterative cycle: n -> sum of digits of n^2 has only two distinct elements.

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A239937 Numbers k such that DigitSum(k^2) > DigitSum((k+1)^2).

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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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