cp's OEIS Frontend

This is a front-end for the Online Encyclopedia of Integer Sequences, made by Christian Perfect. The idea is to provide OEIS entries in non-ancient HTML, and then to think about how they're presented visually. The source code is on GitHub.

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A348303 a(n) is the largest n-digit number whose square has a digital sum equal to A348300(n).

Original entry on oeis.org

7, 83, 937, 9417, 94863, 987917, 9893887, 99483667, 994927133, 9486778167, 99497231067, 999949483667, 9892825177313
Offset: 1

Views

Author

Keywords

Crossrefs

Programs

  • Mathematica
    Array[#1 + Position[#2, Max[#2]][[-1, -1]] - 1 & @@ {#1, Map[Total@ IntegerDigits[#^2] &, Range[#1, #2]]} & @@ {10^(# - 1), 10^# - 1} &, 8] (* Michael De Vlieger, Oct 12 2021 *)

Extensions

a(11) from Chai Wah Wu, Nov 18 2021
a(12)-a(13) from Martin Ehrenstein, Nov 20 2021

A351650 Integers m such that digsum(m) divides digsum(m^2) where digsum = sum of digits = A007953.

Original entry on oeis.org

1, 2, 3, 9, 10, 11, 12, 13, 18, 19, 20, 21, 22, 24, 27, 30, 31, 33, 36, 42, 45, 46, 54, 55, 63, 72, 74, 81, 90, 92, 99, 100, 101, 102, 103, 108, 110, 111, 112, 113, 117, 120, 121, 122, 123, 126, 128, 130, 132, 135, 144, 145, 153, 162, 171, 180, 189, 190, 191, 198
Offset: 1

Views

Author

Bernard Schott, Feb 16 2022

Keywords

Comments

This is a generalization of a problem proposed by French site Diophante in link.
The smallest term k such that the corresponding quotient = n is A280012(n).
The quotient is 1 iff m is in A058369 \ {0}.
If k is in A061909, then digsum(k^2) = digsum(k)^2.
If k is a term, 10*k is also a term.
There are infinitely many m such that both m and m+1 are in the sequence, for example subsequence A002283 \ {0}.
Corresponding quotients are in A351651.

Examples

			digit sum of 42 = 4+2 = 6; then 42^2 = 1764, digit sum of 1764 = 1+7+6+4 = 18; as 6 divides 18, 42 is a term.
		

Crossrefs

Programs

  • Mathematica
    Select[Range[200], Divisible[Total[IntegerDigits[#^2]], Total[IntegerDigits[#]]] &] (* Amiram Eldar, Feb 16 2022 *)
  • PARI
    is(n)=sumdigits(n^2)%sumdigits(n) == 0 \\ David A. Corneth, Feb 16 2022
    
  • Python
    def sd(n): return sum(map(int, str(n)))
    def ok(n): return sd(n**2)%sd(n) == 0
    print([m for m in range(1, 200) if ok(m)]) # Michael S. Branicky, Feb 16 2022

Formula

A004159(a(n)) = A007953(a(n)) * A351651(n).

Extensions

More terms from David A. Corneth, Feb 16 2022

A369953 a(n) is the least integer k such that the sum of the digits of k^2 is 9*n.

Original entry on oeis.org

0, 3, 24, 63, 264, 1374, 3114, 8937, 60663, 94863, 545793, 1989417, 5477133, 20736417, 82395387, 260191833, 706399164, 2428989417, 9380293167, 28105157886, 99497231067, 538479339417, 1974763271886, 4472135831667, 14106593458167, 62441868958167, 244744764757083, 836594274358167
Offset: 0

Views

Author

Zhining Yang, Feb 06 2024

Keywords

Comments

3|a(n).

Examples

			a(3)=63 because k=63 is the least integer k such that the sum of the digits of k^2 = 3969 is 9*3 = 27 (3+9+6+9 = 27).
		

Crossrefs

Programs

  • Mathematica
    n=1;lst={};For[k=0,k<10^8,k+=3,If[Total[IntegerDigits[k^2]]==9*n,AppendTo[lst,k];n++]];lst
  • PARI
    a(n) = my(k=0); while(sumdigits(k^2) != 9*n, k+=3); k; \\ Michel Marcus, Feb 17 2024
  • Python
    n=1
    lst=[]
    for k in range(0,10**8,3):
        if sum(int(d) for d in str(k*k))==9*n:
            lst.append(k)
            n=n+1
    print(lst)
    

Formula

a(n) = A067179(4n).

Extensions

a(19)-a(27) from Zhao Hui Du, Feb 09 2024

A370522 a(n) is the least n-digit number whose square has the maximum sum of digits (A348300(n)).

Original entry on oeis.org

7, 83, 836, 8937, 94863, 987917, 9893887, 99477133, 994927133, 9380293167, 99497231067, 926174913167, 9892825177313, 89324067192437, 943291047332683, 9949874270443813, 83066231922477313, 707106074079263583, 9429681807356492126, 94180040294109027313, 888142995231510436417, 8882505274864168010583
Offset: 1

Views

Author

Zhining Yang, Feb 21 2024

Keywords

Comments

a(n) is the last n-digit term in A067179.
As the last two of the only nine known numbers whose square has a digit mean above 8.25 (see A164841), there is a high probability that a(30)=314610537013606681884298837387 and a(31)=9984988582817657883693383344833.

Examples

			a(3) = 836 because among all 3-digit numbers, 836 is the smallest whose square 698896 has the maximum sum of digits, 46 = A348300(3).
		

Crossrefs

Programs

  • Mathematica
    A348300={13,31,46,63,81,97,112,130,148,162,180};
    A370522[n_]:=Do[If[Total@IntegerDigits[k^2]==A348300[[n]],Return[k];],{k,10^(n-1),10^n-1}];
    Table[A370522[n],{n,8}]
  • Python
    def A370522(n):
        A348300=[0,13,31,46,63,81,97,112,130,148,162,180]
        for k in range(10**(n-1), 10**n):
            if sum(int(d) for d in str(k**2))==A348300[n]:
                return(k)
    print([A370522(n) for n in range(1,9)])

Extensions

a(11)-a(24) from Zhao Hui Du, Feb 23 2024

A056528 Sum of digits of square of sum of digits of square.

Original entry on oeis.org

1, 7, 9, 13, 13, 9, 16, 1, 9, 1, 7, 9, 13, 13, 9, 16, 10, 9, 1, 7, 9, 13, 13, 9, 16, 10, 9, 10, 16, 9, 13, 13, 9, 16, 1, 9, 10, 16, 9, 13, 13, 9, 16, 10, 9, 1, 16, 9, 13, 13, 9, 16, 10, 9, 1, 16, 9, 13, 13, 9, 16, 10, 18, 10, 16, 9, 13, 13, 9, 16, 1, 9, 10, 16, 9, 13, 13, 9, 16, 1, 9
Offset: 1

Views

Author

Henry Bottomley, Jun 19 2000

Keywords

Examples

			a(2)=7 because sum of digits of square of 2 is 4 and sum of digits of square of 4 is 1+6=7
		

Crossrefs

Cf. A004159 for sum of digits of square, A056020 where iteration settles to 1, A056020 where iteration settles to 9, A056527 where iteration settles to 13 and 16. See also A056529.

Programs

  • Mathematica
    Array[Total[IntegerDigits[(Total[IntegerDigits[#^2]])^2]]&,90] (* Harvey P. Dale, Jan 17 2012 *)
    Table[Nest[Total[IntegerDigits[#^2]]&,n,2],{n,90}] (* Harvey P. Dale, Mar 07 2018 *)

Formula

a(n)=A004159(A004159(n))

A056529 Number of iterations of sum of digits of square to reach 1, 9, 13 or 16.

Original entry on oeis.org

0, 3, 1, 2, 2, 1, 1, 2, 0, 1, 3, 1, 0, 1, 1, 0, 3, 1, 2, 3, 1, 1, 1, 2, 1, 3, 2, 3, 1, 1, 1, 2, 2, 1, 2, 2, 3, 1, 1, 2, 1, 2, 2, 3, 1, 2, 1, 1, 2, 2, 1, 1, 3, 2, 2, 1, 2, 1, 1, 1, 1, 3, 3, 3, 1, 2, 2, 1, 2, 1, 2, 2, 3, 2, 2, 2, 2, 2, 1, 2, 2, 3, 2, 2, 1, 2, 3, 2, 3, 1, 3, 2, 3, 2, 1, 2, 2, 3, 2, 1, 3, 1, 1, 1, 1
Offset: 1

Views

Author

Henry Bottomley, Jun 19 2000

Keywords

Examples

			a(2)=3 because successive iterations give 2 -> 4 -> 7 -> 13 -> 16 -> 13 etc.
		

Crossrefs

Cf. A004159 for sum of digits of square, A056528 for two iterations, A056020 where iteration settles to 1, A056020 where iteration settles to 9, A056527 where iteration settles to 13 and 16.

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

Views

Author

Asher Auel, May 17 2001

Keywords

Examples

			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.
		

Crossrefs

Programs

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

Formula

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

A117685 Squares for which the sum of the digits are cubes.

Original entry on oeis.org

0, 1, 100, 3969, 7569, 8649, 10000, 12996, 13689, 15876, 19881, 33489, 34596, 36864, 42849, 45369, 46656, 47961, 49284, 51984, 54756, 56169, 59049, 66564, 71289, 74529, 76176, 77841, 79524, 82944, 84681, 86436, 88209, 91809, 93636, 95481, 97344, 99225
Offset: 0

Views

Author

Luc Stevens (lms022(AT)yahoo.com), Apr 12 2006

Keywords

Examples

			8649 is in the sequence because it is a square and the sum of the digits 8+6+4+9=27 is a cube.
		

Crossrefs

Programs

  • Mathematica
    Select[Range[0,400]^2,IntegerQ[Power[Total[IntegerDigits[#]], (3)^-1]]&] (* Harvey P. Dale, Jan 28 2012 *)

Extensions

Corrected (a(18) inserted) and extended by Harvey P. Dale, Jan 28 2012

A147523 a(n) = number of n-digit terms in A058369.

Original entry on oeis.org

3, 8, 37, 195, 1159, 6951, 42421, 261430, 1631123, 10238173, 64536808
Offset: 1

Views

Author

Zak Seidov, Apr 25 2009

Keywords

Examples

			a(1)=3: 0,1,9; a(2)=8: 10,18,19,45,46,55,90,99.
		

Crossrefs

A058369 Numbers n such that n and n^2 have same digit sum. A004159 Sum of digits of n^2. A007953 Digital sum (i.e. sum of digits) of n.

Extensions

a(9) from Donovan Johnson, Dec 08 2009
a(10)-a(11) from Donovan Johnson, Jul 31 2011

A153751 Numbers k such that there are 15 digits in k^2 and for each factor f of 15 (1,3,5) the sum of digit groupings of size f is a square.

Original entry on oeis.org

10000000, 10000001, 10000002, 10000003, 10000004, 10000005, 10000010, 10000011, 10000012, 10000013, 10000020, 10000021, 10000022, 10000030, 10000031, 10000200, 10000300, 10011003, 10022000, 10035990, 10042440
Offset: 1

Views

Author

Doug Bell, Dec 31 2008

Keywords

Comments

This sequence is a subsequence of both A153745 and A061910.
The last term is a(2782) = 31616301. - Giovanni Resta, Jun 06 2015

Examples

			10000011^2 = 100000220000121;
1+0+0+0+0+0+2+2+0+0+0+0+1+2+1 = 9 = 3^2;
100+000+220+000+121 = 441 = 21^2;
10000+02200+00121 = 12321 = 111^2.
		

Crossrefs

Programs

  • Mathematica
    Select[Range[10^7,31622776],AllTrue[{Sqrt[Total[IntegerDigits[#^2]]],Sqrt[Total[ FromDigits/@ Partition[IntegerDigits[#^2],3]]],Sqrt[Total[FromDigits/@Partition[IntegerDigits[#^2],5]]]},IntegerQ]&] (* Harvey P. Dale, Apr 11 2023 *)
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