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1, 4, 9, 324, 1444, 26244, 114244, 116964, 191844, 213444, 311364, 412164, 777924, 2196324, 3392964, 6441444, 11142244, 13264164, 16499844, 21734244, 23213124, 23444964, 24621444, 26193924, 27436644, 36216324, 44116164, 47361924, 61121124, 61496964, 63393444
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OFFSET
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1,2
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COMMENTS
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The sequence is infinite because the squares of numbers of the form (10^n + 14)/3, n >= 2, 38, 338, 3338, ... are 1444, 114244, 11142244, ... and have only the digits 1, 2, 4. Deleting any digit results in a number divisible by 4.
a(1) = 1 and a(3) = 9 are the only terms with an odd last digit. There are no other terms with the last digit 1, 5, 9 because the squares are odd, the tens digit must be even, and by deleting it the remaining number is odd.
The last digit of the terms cannot be 6 because the last two digits of the squares would be 16, 36, 56, 76, 96, so the penultimate digit is odd, and removing the 6 would leave an odd number. Thus, the terms a(n) with n >= 4, have the last digit 4.
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LINKS
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EXAMPLE
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324 = 18^2 is a term, since 3 divides 24, 2 divides 34 and 4 divides 32.
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MATHEMATICA
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A353729Q[n_] := Block[{d = IntegerDigits[n], i=0}, FreeQ[d, 0] && (While[++i <= Length[d] && Divisible[FromDigits[Drop[d, {i}]], d[[i]]]]; i > Length[d])];
Select[Range[15000]^2, A353729Q] (* Paolo Xausa, Feb 27 2024 *)
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PROG
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(Magma) ints:=func<n| n eq 0 select[0] else Intseq(n)>; f:=func<n, i|Seqint([ints(n)[j]: j in [1..#ints(n)]|j ne (#ints(n)-i+1)])>; [p:p in [s*s:s in [1..8000]] |not 0 in ints(p) and forall{i:i in [1..#ints(p)]|IsIntegral(f(p, i)/ints(p)[(#Intseq(p)-i+1)])}];
(PARI) isok(k) = if (issquare(k), my(d=digits(k)); if (vecmin(d), for(i=1, #d, my(list=List(d)); listpop(list, i); if (fromdigits(Vec(list)) % d[i], return(0)); ); return(1); ); ); \\ Michel Marcus, Dec 19 2023
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CROSSREFS
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KEYWORD
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nonn,base
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AUTHOR
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STATUS
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approved
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