A030154 Squares such that in n and sqrt(n) the parity of digits alternates.

0, 1, 4, 9, 16, 25, 36, 49, 81, 256, 529, 729, 1296, 4761, 5476, 6561, 9216, 16129, 32761, 34969, 87616, 763876, 5414929, 5612161, 7414729, 7436529, 7634169, 14561856, 21058921, 34503876, 43072969, 43414921, 45252529, 69272329

Offset: 1

Author

Patrick De Geest

Keywords

• nonn
• base

Comments

The more digits there are in n, the lower the likelihood that the parity of n's digits will strictly alternate. Thus, the terms of the sequence become increasingly rare as n gets larger. - Harvey P. Dale, Aug 05 2018

For n > 3 the last digit of a(n) isn't 0 or 4. - David A. Corneth, Aug 05 2018

Links

• Andrew Howroyd, Table of n, a(n) for n = 1..1226 (first 101 terms from Harvey P. Dale, terms 102..223 from David A. Corneth)

Crossrefs

Cf. A030141, A030151, A030152, A030153.

Programs

Mathematica

pdaQ[n_]:=Module[{a=Mod[IntegerDigits[n],2],b=Mod[IntegerDigits[ Sqrt[ n]],2]},Length[ Split[a]] ==IntegerLength[n]&&Length[Split[b]]== IntegerLength[ Sqrt[n]]]; Join[{0},Select[Range[8500]^2,pdaQ]] (* Harvey P. Dale, Aug 05 2018 *)

PARI

alternating(n)={my(v=digits(n)%2);0==#select(i->v[i]==v[i-1],[2..#v])}
{ for(n=0, 10^5, if(alternating(n^2) && alternating(n), print1(n^2, ", "))) } \\ Andrew Howroyd, Aug 05 2018

PARI

\\ for larger n: requires alternating function above
upto(n)={local(R=List([0])); my(recurse(s,b)=if(b0&&alternating(k^2\b), listput(R, k)); self()(k, 10*b)))))); recurse(0,1); listsort(R); Vec(R)}
apply(n->n^2, upto(sqrtint(10^12))) \\ Andrew Howroyd, Aug 05 2018

Extensions

Offset changed by David A. Corneth, Aug 05 2018