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.

Showing 1-8 of 8 results.

A238237 Numbers which when chopped into two parts with equal length, added and squared result in the same number.

Original entry on oeis.org

81, 2025, 3025, 9801, 494209, 998001, 24502500, 25502500, 52881984, 60481729, 99980001, 6049417284, 6832014336, 9048004641, 9999800001, 101558217124, 108878221089, 123448227904, 127194229449, 152344237969, 213018248521, 217930248900, 249500250000, 250500250000
Offset: 1

Views

Author

Arkadiusz Wesolowski, Feb 20 2014

Keywords

Comments

Yet another variant of the Kaprekar numbers A006886. - N. J. A. Sloane, Aug 06 2017
From Bernard Schott, Jan 21 2022: (Start)
Three subsequences:
-> {(10^m-1)^2, m >= 1} = A059988 \ {0}; see example 9801.
-> {(10^m-1)^2 * 10^(2*m) / 4, m >= 1} = A350869 \ {0}; see example 2025.
-> {(10^m+1)^2 * 10^(2*m) / 4, m >= 1} = A038544 \ {1}, see example 3025. (End)

Examples

			2025 = (20 + 25)^2, so 2025 is in the sequence.
3025 = (30 + 25)^2, so 3025 is in the sequence.
9801 = (98 + 01)^2, so 9801 is in the sequence.

Links

Crossrefs

Subsequence of A102766.
Subsequence: A350870.
Cf. A006886, A038544, A059988, A350869.
For square roots see A290449.

Programs

  • Mathematica
    Select[Range[600000]^2, EvenQ[len=IntegerLength[#]] && # == (Mod[#,10^(len/2)] + Floor[#/10^(len/2)])^2 &] (* Stefano Spezia, Jan 01 2025 *)
  • PARI
    forstep(m=1, 7, 2, p=10^((m+1)/2); for(n=10^m, 10^(m+1)-1, d=lift(Mod(n, p)); if(((n-d)/p+d)^2==n, print1(n, ", "))));

Formula

a(n) = A290449(n)^2. - Bernard Schott, Jan 20 2022

Extensions

a(12)-a(24) from Donovan Johnson, Feb 22 2014

A228103 Numbers k whose base-10 digits can be split into two parts, q and r, with k = (q-r)^2.

Original entry on oeis.org

100, 121, 6084, 10000, 10201, 82369, 132496, 1000000, 1002001, 1162084, 1201216, 1656369, 1860496, 100000000, 100020001, 123121216, 330621489, 10000000000, 10000200001, 13967221489, 113322449956, 1000000000000, 1000002000001, 1786590449956, 7438023471076
Offset: 1

Views

Author

Hans Havermann, Aug 10 2013

Keywords

Comments

q*10^m+r = (q-r)^2 = A228381^2; q,m>0; 0<=r<10^m. - Hans Havermann, Aug 21 2013

Examples

			100 = (10-0)^2.
121 = (12-1)^2.
6084 = (6-084)^2.

Links

Crossrefs

Cf. A102766, A118936, A228381.

Programs

  • Mathematica
    k=3; While[k<10^8, k++; s=k^2; d=IntegerDigits[s]; l=Length[d]; Do[a=FromDigits[Take[d,{1,i}]]; b=FromDigits[Take[d,{i+1,l}]]; If[k==Abs[a-b], w=ToString[s]; Print[StringTake[w,{1,i}], "'", StringTake[w,{i+1,l}]]], {i,l-1}]] (* Hans Havermann, Aug 10 2013, Aug 20 2013 *)

A248353 Kaprekar numbers, allowing powers of 10: n such that n=q+r and n^2=q*10^m+r, for some m >= 1, q>=0 and 0<=r<10^m.

Original entry on oeis.org

1, 9, 10, 45, 55, 99, 100, 297, 703, 999, 1000, 2223, 2728, 4879, 4950, 5050, 5292, 7272, 7777, 9999, 10000, 17344, 22222, 38962, 77778, 82656, 95121, 99999, 100000, 142857, 148149, 181819, 187110, 208495, 318682, 329967, 351352, 356643, 390313, 461539
Offset: 1

Views

Author

Reinhard Zumkeller, Oct 05 2014

Keywords

Comments

Powers of 10 were excluded in Kaprekar's original definition (A006886), see also A045913.

Links

Crossrefs

Cf. A006886 (subsequence), A045913, A053816, A011557, A102766.

Programs

  • Haskell
    a248353 n = a248353_list !! (n-1)
a248353_list = filter k [1..] where
   k x = elem x $ map (uncurry (+)) $
         takeWhile ((> 0) . fst) $ map (divMod (x ^ 2)) a011557_list

Formula

a(n) = sqrt(A102766(n)).

A350870 Numbers k = x.y such that x.y = (x+y)^2, when x and y have the same number of digits, "." means concatenation, and y may not begin with 0.

Original entry on oeis.org

81, 2025, 3025, 494209, 24502500, 25502500, 52881984, 60481729, 6049417284, 6832014336, 101558217124, 108878221089, 123448227904, 127194229449, 152344237969, 213018248521, 217930248900, 249500250000, 250500250000, 284270248900, 289940248521, 371718237969, 413908229449, 420744227904
Offset: 1

Views

Author

Bernard Schott, Jan 20 2022

Keywords

Comments

Problem proposed on French site Diophante (see link).
We have to solve Diophantine equation (x+y)^2 = x*10^m + y where m = length(x) = length(y).
Squares of a variant of Kaprekar numbers (A045913).
Number of solutions with 2*m digits for m >= 1: 1, 2, 1, 4, 2, 21, ...
Compare with A347541 where x*y divides x.y, when x and y have the same number of digits, "." means concatenation, and y may not begin with 0.

Examples

			81 = (8+1)^2, hence 81 is a term.
3025 = (30+25)^2, hence 3025 is another term.

Links

Crossrefs

Equals A238237 \ A350918.
Subsequence of A102766.
Subsequences: A038544, A350869.
Cf. A045913, A347541.

Programs

  • PARI
    upto(n) = {i = 4; i2 = i^2; res = List(); while(i2 <= n, i++; i2 = i^2; if(#digits(i2) % 2 == 1, i = sqrtint(10^(#digits(i2))) + 1; i2 = i^2; ); if(is(i2), listput(res, i2) ); ); res }
is(n) = { my(d = digits(n), d1, d2, frd2); if(#d % 2 == 1, return(0)); d1 = vector(#d \ 2, i, d[i]); d2 = vector(#d \ 2, i, d[i + #d \ 2]); frd2 = fromdigits(d2); 10^(#d \ 2 - 1) <= frd2 && (fromdigits(d1) + frd2)^2 == n } \\ David A. Corneth, Jan 21 2022

Formula

a(n) = A045913(n+1)^2.

A104113 Numbers which when chopped into one, two or more parts, added and squared result in the same number.

Original entry on oeis.org

0, 1, 81, 100, 1296, 2025, 3025, 6724, 8281, 9801, 10000, 55225, 88209, 136161, 136900, 143641, 171396, 431649, 455625, 494209, 571536, 627264, 826281, 842724, 893025, 929296, 980100, 982081, 998001, 1000000, 1679616, 2896804, 3175524, 4941729, 7441984
Offset: 1

Views

Author

Bodo Zinser, Mar 05 2005

Keywords

Comments

Every term is congruent to 0 or 1 modulo 9. - Andrea Tarantini, Sep 27 2021

Examples

			1296 is a term since (1+29+6)^2 = 36^2 = 1296.

Links

Crossrefs

Cf. A038206, A102766.

Programs

  • Mathematica
    Join[{0},Select[Select[Range@3000^2,Mod[#,9]<2&],(n=#;MemberQ[(Total/@(FromDigits/@#&/@Union[DeleteCases[SplitBy[#,#==-1&],{-1}]&/@(Insert[IntegerDigits@n,-1,#]&/@(List/@#&/@Rest@Subsets[Range@IntegerLength@n]))]))^2,#])&]] (* Giorgos Kalogeropoulos, Oct 28 2021 *)
  • Python
    def expr(t, d): # can you express target t with digits d, only adding +'s
    if t < 0: return False
    if t == int(d): return True
    return any(expr(t-int(d[:i]), d[i:]) for i in range(1, len(d)))
def aupto(limit):
    alst, k, k2 = [], 0, 0
    while k2 <= limit:
        if expr(k, str(k2)):
            alst.append(k2)
        k, k2 = k+1, k2 + 2*k + 1
    return alst
print(aupto(7500000)) # Michael S. Branicky, Sep 27 2021

Formula

a(n) = A038206(n)^2. - Andrea Tarantini, Sep 27 2021

Extensions

a(30) and beyond from Mehrad Mahmoudian, Dec 16 2019

A384538 Positive integers k >= 10 for which for every pair of nonempty substrings that concatenate to give k one substring divides the other.

Original entry on oeis.org

10, 11, 12, 13, 14, 15, 16, 17, 18, 19, 20, 21, 22, 24, 26, 28, 30, 31, 33, 36, 39, 40, 41, 42, 44, 48, 50, 51, 55, 60, 61, 62, 63, 66, 70, 71, 77, 80, 81, 82, 84, 88, 90, 91, 93, 99, 100, 101, 102, 105, 110, 111, 120, 121, 122, 123, 124, 126, 130, 131, 140, 141
Offset: 1

Views

Author

Felix Huber, Jun 09 2025

Keywords

Examples

			324 is a term because 3 divides 24 and 4 divides 32.
105 is a term because 1 divides 05 = 5 and 5 divides 10.
2500 is a term because 2 divides 500. 25 divides 00 = 0 and 250 divides 0.
104 is not a term: Although 1 divides 04 = 4, 4 does not divide 10.

Links

Crossrefs

Supersequence of A384539.
Cf. A102766, A228103.

Programs

  • Maple
    A384538:=proc(n)
    option remember;
    local i,j,k,p,m,q,L;
    if n=1 then
        10
    else
        for k from procname(n-1)+1 do
            L:=ListTools:-Reverse(convert(k,'base',10));
            m:=length(k)-1;
            for j to m do
                p:=parse(cat(seq(L[i],i=1..j)));
                q:=k-p*10^(m+1-length(p));
                if q mod p<>0 and p mod q<>0 then
                    break
            	 elif j=m then
                    return k
                fi
            od
        od
    fi;
end proc;	
seq(A384538(n),n=1..62);
  • PARI
    isok(k) = my(nb=logint(k, 10), d=10); for (i=1, nb, my(sa = k%d, sb=k\d); if ((sa % sb) && (sb % sa), return(0)); d *= 10;); return(1); \\ Michel Marcus, Jun 19 2025
  • Python
    def c(k, m): return (k > 0 and m%k == 0) or (m > 0 and k%m == 0)
def ok(n):
    s = str(n)
    return n > 9 and all(c(int(s[:i]), int(s[i:])) for i in range(1, len(s)))
print([k for k in range(150) if ok(k)]) # Michael S. Branicky, Jun 17 2025

A384539 Zeroless positive integers k for which for every pair of nonempty substrings that concatenate to give k one substring divides the other.

Original entry on oeis.org

11, 12, 13, 14, 15, 16, 17, 18, 19, 21, 22, 24, 26, 28, 31, 33, 36, 39, 41, 42, 44, 48, 51, 55, 61, 62, 63, 66, 71, 77, 81, 82, 84, 88, 91, 93, 99, 111, 121, 122, 123, 124, 126, 131, 141, 142, 147, 151, 153, 155, 161, 162, 164, 168, 171, 181, 182, 183, 186, 189
Offset: 1

Views

Author

Felix Huber, Jun 09 2025

Keywords

Comments

It is conjectured that this sequence is finite and contains 132 terms.
From David A. Corneth, Jun 19 2025: (Start)
A term with at least 3 digits cannot end in 14. Else we can split it in 10*k + 1 and 4 where k >= 1. So we'd need 4 | 10*k + 1. A contradiction.
If a term with at least 5 digits ends in 89 then it is 100*k + 89 where k >= 1000.
We'd need 9 | 1000*k + 8, 89 | 100*k so 89 | k. This largely restrict the possibilities for k. (End)
a(133) > 10^11, if it exists. - Michael S. Branicky, Jun 18 2025
a(133) > 10^25, if it exists. - David A. Corneth, Jun 19 2025

Examples

			168 is a term because 1 divides 68 and 8 divides 16.
4284 is a term because 4 divides 284, 42 divides 84 and 4 divides 428.
222222 is a term because 2 divides 22222, 22 divides 2222, 222 divides 222 and vice versa.

Links

Crossrefs

Subsequence of A052382.
Subsequence of A384538.
Cf. A102766, A228103.

Programs

  • Maple
    A384539:=proc(n)
    option remember;
    local i,j,k,p,m,q,L;
    if n=1 then
        11
    else
        for k from procname(n-1)+1 do
            L:=ListTools:-Reverse(convert(k,'base',10));
            if not member(0,L) then
	        m:=length(k)-1;
	        for j to m do
	            p:=parse(cat(seq(L[i],i=1..j)));
	            q:=k-p*10^(m+1-length(p));
	            if max(p,q) mod min(p,q)<>0 then
	                break
	            elif j=m then
	                return k
	            fi
	        od
	    fi
        od
    fi;
end proc;	
seq(A384539(n),n=1..60);
  • Python
    def c(k, m): return m%k == 0 or k%m == 0
def ok(n):
    s = str(n)
    return n > 9 and "0" not in s and all(c(int(s[:i]), int(s[i:])) for i in range(1, len(s)))
print([k for k in range(200) if ok(k)]) # Michael S. Branicky, Jun 18 2025

A379591 Numbers k whose base-10 digits can be split into two parts, q and r, with k = (|q-r|)^3.

Original entry on oeis.org

1000, 456533, 474552, 1000000, 69426531, 1000000000, 1000000000000, 1000000000000000, 1000000000000000000, 1000000000000000000000, 1000000000000000000000000, 1000000000000000000000000000, 1000000000000000000000000000000, 1000000000000000000000000000000000
Offset: 1

Views

Author

Travis Vasquez, Dec 26 2024

Keywords

Examples

			1000 = (|10-00|)^3.
456533 = (|456-533|)^3.
1000000 = (|100-0000|)^3.

Crossrefs

Cf. A102766, A228103.

Programs

  • Maple
    filter:= proc(x) local m;
for m from 1 to ilog10(x) do
  if x = abs((x mod 10^m) - floor(x/10^m))^3 then return true fi
od;
false
end proc:
select(filter, [seq(i^3,i=1..10^7)]); # Robert Israel, Jan 01 2025
  • PARI
    lista(nn) = my(d, m, q, s, t=1, v=List([])); while(t1&&gcd(f[1], (t+1)/f[1])==1, d=[x[1]^x[2]|x<-f[2]~]; forvec(x=vector(#d, i, [0, 1]), s=lift(chinese(chinese(vector(#d, i, Mod((-1)^x[i], d[i]))), Mod(0, (t+1)/f[1]))); q=(s^3-s)/(t+1); if(q>0&&s+q=s&&q-sx<=nn, v)); \\ Jinyuan Wang, Jan 03 2025

Extensions

a(11)-a(12) from Michael S. Branicky, Jan 01 2025
a(13)-a(14) from Jinyuan Wang, Jan 03 2025
Showing 1-8 of 8 results.

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