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

1,3

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)

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 *)

Prog

(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(b<n, for(i=0, 9, if(b==1||(s\(b\10)-i)%2, my(k=i*b+s); if(k<=n&&(b==1||k^2\(b\10)*11\10%2), if(i>0&&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

Crossrefs

Cf. A030141, A030151, A030152, A030153.

Sequence in context: A068879 A030152 A030288 * A122541 A133743 A235484

Adjacent sequences: A030151 A030152 A030153 * A030155 A030156 A030157

Keyword

nonn,base

Author

Patrick De Geest

Extensions

Offset changed by David A. Corneth, Aug 05 2018

Status

approved