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A344570
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Number of pairs of n-digit squares such that the final (n-1) digits of the first square coincide with the initial (n-1) digits of the second.
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2
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0, 4, 6, 7, 10, 5, 10, 4, 15, 0, 13, 5, 16, 3, 57, 2, 8, 2, 5, 1, 119, 1, 13, 8, 39, 5, 55, 2, 53, 7, 12, 7, 76, 1, 193, 2, 21, 2, 59, 11, 35, 15, 42, 7, 541, 7, 17, 20, 37, 1, 233, 3, 32, 2, 373, 19, 65, 0, 15, 16, 181, 15, 8637, 5, 175, 15
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OFFSET
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1,2
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COMMENTS
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For n=2, certain numbers (16 and 64) appear in more than one pair. No such numbers have been observed up to n=9, but so far there is no proof of this property.
a(10)=0 was found by Andrew Howroyd. Does this sequence contain infinitely many nonzero terms?
The sequence is inspired by a problem from the 2020 Polish Juniors' Mathematics Olympiad. It is problem 1: 'Is there a six-digit positive integer such that any two consecutive digits form a perfect square?'
Are there any other terms such that a(n) = 0 besides n=1 and n=10? - Chai Wah Wu, May 26 2021
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LINKS
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EXAMPLE
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For n=2: (81,16), (16,64), (36,64), (64,49).
For n=3: (144,441), (196,961), (225,256), (625,256), (484,841), (784,841).
For n=4: (3136,1369), (4624,6241), (5184,1849), (5476,4761), (7396,3969), (7921,9216), (9409,4096).
For n=20: (64764644930975528100, 47646449309755281001) is the only pair. - Andrew Howroyd, May 23 2021
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PROG
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(PARI) a(n)={sum(k=sqrtint(10^(n-1))+1, sqrtint(10^n-1), my(t=k^2*10%10^n); t>=10^(n-1) && sqrtint(t+9)^2\10==t\10)} \\ Andrew Howroyd, May 24 2021
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CROSSREFS
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KEYWORD
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nonn,base,more
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AUTHOR
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EXTENSIONS
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STATUS
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approved
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