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A269588
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Numbers n such that n^2 ends with the digits of n reversed (A004086(n)).
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2
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1, 5, 6, 963, 9867, 65766, 69714, 6317056, 90899553, 169605719, 4270981082, 96528287587, 465454256742, 692153612536, 182921919071841, 655785969669834, 650700037578750084, 125631041500927357539, 673774165549097456624, 16719041449406813636569
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OFFSET
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1,2
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COMMENTS
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a(29)>10^32 (if it exists)
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LINKS
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EXAMPLE
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6317056^2 = 39905196507136 which ends with 6507136, so 6317056 is a term.
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MATHEMATICA
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Select[Range[10^7], Function[k, Take[IntegerDigits[#^2], -Length@ k] == Reverse@ k]@ IntegerDigits@ # &] (* Michael De Vlieger, Mar 04 2016 *)
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PROG
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(PARI) isA269588(n)=dn = digits(n); rn = subst(Polrev(dn), x, 10); nbd = #dn; (n^2 - rn) % 10^nbd == 0; \\ Michel Marcus, Mar 01 2016
(PARI) \\ printA269588len(d) prints all terms of the sequence with d digits
rev(n) = eval(concat(Vecrev(Str(n))));
{ printA269588len(d) = my(l, u, n); l=ceil(d/2); u=floor(d/2); for(y=0, 10^l-1, n=rev(y^2 % 10^u)*10^l+y; if(#Str(n)==d && Mod(n, 10^d)^2==rev(n), print(n)); ); }
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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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EXTENSIONS
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a(28) from Dieter Beckerle, Jun 09 2016
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STATUS
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approved
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