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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A359343 Square roots of least pandigital squares with n digits.

Original entry on oeis.org

32043, 100287, 317096, 1000287, 3162426, 10000287, 31622792, 100000287, 316227814, 1000000287, 3162277718, 10000000287, 31622776661, 100000000287, 316227766026, 1000000000287, 3162277660177, 10000000000287, 31622776601685, 100000000000287, 316227766016843
Offset: 10

Views

Author

Martin Renner, Dec 27 2022

Keywords

Comments

Pandigital squares are perfect squares containing each digit from 0 to 9 at least once.

Crossrefs

Programs

  • Maple
    f:= proc(n); local k;
      for k from ceil(10^((n-1)/2)) do
        if convert(convert(k^2,base,10),set) = {$0..9} then return k fi
      od
    end proc:
    map(f, [$10..30]); # Robert Israel, Dec 29 2022
  • Python
    from math import isqrt
    def c(n): return len(set(str(n))) == 10
    def a(n): return next((k for k in range(isqrt(10**(n-1))+1, isqrt(10**n-1)+1) if c(k*k)), None)
    print([a(n) for n in range(10, 31)]) # Michael S. Branicky, Dec 27 2022

Formula

a(n) = sqrt(A359342(n)).
If n is odd, a(n) = 10^((n-1)/2) + 287. - Robert Israel, Dec 29 2022
a(n) = 10^((n-1)/2) + O(1). - Charles R Greathouse IV, Dec 30 2022

A359344 Largest pandigital square with n digits.

Original entry on oeis.org

9814072356, 99853472016, 998732401956, 9998490637521, 99992580137641, 999984024130576, 9999925800137641, 99999987340240516, 999999258000137641, 9999999562540763281, 99999992580000137641, 999999991102375684521, 9999999925800000137641, 99999999986188478340025
Offset: 10

Views

Author

Martin Renner, Dec 27 2022

Keywords

Comments

Pandigital squares are perfect squares containing each digit from 0 to 9 at least once.
For number of digits n >= 14, every second term is of the form 9...92580...0137641 with n/2 - 3 leading nines and n/2 - 6 zeros after the middle three digits 258. This term is 9...9629^2 with n/2 - 3 leading nines. This is the case since ((10^m - 1)*10^3 + 629)^2 = 10^(2*m+6) - 2*10^(m+6) + 10^(m+6) + 258*10^(m+3) + 10^6 - 1258*10^3 + 395641 = (10^m - 1)*10^(m+6) + 258*10^(m+3) + 137641 with m = (n-6)/2 and n >= 14 even, and is the last n-digit square containing all digits from 0 to 9.

Crossrefs

Programs

  • Maple
    a:=proc(n::posint) local s, k, K: if n<10 then s:=NULL: else for k from floor(sqrt(10^n)) to ceil(sqrt(10^(n-1))) by -1 do K:=convert(k^2,base,10); if nops({op(K)})=10 then s:=k^2: break: fi: od: fi: return s; end:
    seq(a(n),n=10..30);
  • Python
    from math import isqrt
    def c(n): return len(set(str(n))) == 10
    def a(n):
        ub, lb = isqrt(10**n-1), isqrt(10**(n-1)) if n&1 else isqrt(10**(n-1))+1
        return next((k*k for k in range(ub, lb-1, -1) if c(k*k)), None)
    print([a(n) for n in range(10, 24)]) # Michael S. Branicky, Dec 27 2022

Formula

a(n) = (10^(n/2-3)-1)*10^(n/2+3) + 258*10^(n/2) + 137641 for n >= 14 even.
Showing 1-2 of 2 results.