A359344 -id:A359344 - cp's OEIS frontend

A359342 Least pandigital square with n digits.

1026753849, 10057482369, 100549873216, 1000574082369, 10000938205476, 100005740082369, 1000000973875264, 10000057400082369, 100000030347218596, 1000000574000082369, 10000000365759287524, 100000005740000082369, 1000000003751486308921, 10000000057400000082369

Offset: 10

Martin Renner, Dec 27 2022

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

For number of digits n >= 11, every second term is of the form 10...05740...082369 with (n-1)/2 - 3 zeros after the leading 1 and (n-1)/2 - 5 zeros after the middle three digits 547. This term is 10...0287^2 with (n-1)/2 - 3 zeros after the leading 1. This is the case since (10^m + 287)^2 = 10^(2*m) + 574*10^m + 82369 with m = (n-1)/2 and n >= 11 odd, and is the first n-digit square containing all digits from 0 to 9.

Cf. A225218, A359343, A359344.

a:=proc(n::posint) local s, k, K: if n<10 then s:=NULL: else for k from ceil(sqrt(10^(n-1))) to floor(sqrt(10^n)) 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);

from math import isqrt
def c(n): return len(set(str(n))) == 10
def a(n): return next((k*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, 24)]) # Michael S. Branicky, Dec 27 2022

a(n) = 10^(n-1) + 574*10^((n-1)/2) + 82369 for n >= 11 odd.

A359345 Roots of largest pandigital squares with n digits.

Original entry on oeis.org

99066, 315996, 999366, 3162039, 9999629, 31622524, 99999629, 316227746, 999999629, 3162277591, 9999999629, 31622776461, 99999999629, 316227765995, 999999999629, 3162277660155, 9999999999629, 31622776601681, 99999999999629, 316227766016811, 999999999999629

Offset: 10

Martin Renner, Dec 27 2022

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

Cf. A359343, A359344.

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 for k in range(ub, lb-1, -1) if c(k*k)), None)
print([a(n) for n in range(10, 31)]) # Michael S. Branicky, Dec 27 2022

a(n) = sqrt(A359344(n)).

a(n) = (10^(n/2-3) - 1)*10^3 + 629 for n >= 14 even.

cp's OEIS Frontend

A359342 Least pandigital square with n digits.

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