A023110 Squares which remain squares when the last digit is removed.

0, 1, 4, 9, 16, 49, 169, 256, 361, 1444, 3249, 18496, 64009, 237169, 364816, 519841, 2079364, 4678569, 26666896, 92294449, 341991049, 526060096, 749609641, 2998438564, 6746486769, 38453641216, 133088524969, 493150849009, 758578289296, 1080936581761

Offset: 1

Author

David W. Wilson

Keywords

• nonn
• base

Comments

This A023110 = A031149^2 is the base 10 version of A001541^2 = A055792 (base 2), A001075^2 = A055793 (base 3), A004275^2 = A055808 (base 4), A204520^2 = A055812 (base 5), A204518^2 = A055851 (base 6), A204516^2 = A055859 (base 7), A204514^2 = A055872 (base 8) and A204502^2 = A204503 (base 9). - M. F. Hasler, Sep 28 2014

For the first 4 terms the square has only one digit. It is understood that deleting this digit yields 0. - Colin Barker, Dec 31 2017

References

• R. K. Guy, Neg and Reg, preprint, Jan 2012.

Links

• Jon E. Schoenfield, Table of n, a(n) for n = 1..70 (terms 1..38 from David W. Wilson, terms 39..40 from Robert G. Wilson v, terms 41..67 from Dmitry Petukhov)
• M. F. Hasler, Truncated squares, OEIS wiki, Jan 16 2012
• Joshua Stucky, Pell's Equation and Truncated Squares, Number Theory Seminar, Kansas State University, Feb 19 2018.
• Index to sequences related to truncating digits of squares.

Crossrefs

Cf. A023111.

Cf. A031150, A053784, A031149, A055792, A055793, A055808, A055812, A055851, A055859, A055872.

Cf. A001541, A001075, A004275, A204520, A204518, A204516, A204514, A204502, A204503.

Programs

Maple

count:= 1: A[1]:= 0:
for n from 0 while count < 35 do
  for t in [1,4,6,9] do
    if issqr(10*n^2+t) then
       count:= count+1;
       A[count]:= 10*n^2+t;
    fi
  od
od:
seq(A[i],i=1..count); # Robert Israel, Sep 28 2014

Mathematica

fQ[n_] := IntegerQ@ Sqrt@ Quotient[n^2, 10]; Select[ Range@ 1000000, fQ]^2 (* Robert G. Wilson v, Jan 15 2011 *)

PARI

for(n=0,1e7, issquare(n^2\10) & print1(n^2",")) \\  M. F. Hasler, Jan 16 2012

Formula

Appears to satisfy a(n)=1444*a(n-7)+a(n-14)-76*sqrt(a(n-7)*a(n-14)) for n >= 16. For n = 15, 14, 13, ... this would require a(1) = 16, a(0) = 49, a(-1) = 169, ... - Henry Bottomley, May 08 2001; edited by Robert Israel, Sep 28 2014

a(n) = A031149(n)^2. - M. F. Hasler, Sep 28 2014

Conjectures from Colin Barker, Dec 31 2017: (Start)

G.f.: x^2*(1 + 4*x + 9*x^2 + 16*x^3 + 49*x^4 + 169*x^5 + 256*x^6 - 1082*x^7 - 4328*x^8 - 9738*x^9 - 4592*x^10 - 6698*x^11 - 6698*x^12 - 4592*x^13 + 361*x^14 + 1444*x^15 + 3249*x^16 + 256*x^17 + 169*x^18 + 49*x^19 + 16*x^20) / ((1 - x)*(1 + x + x^2 + x^3 + x^4 + x^5 + x^6)*(1 - 1442*x^7 + x^14)).

a(n) = 1443*a(n-7) - 1443*a(n-14) + a(n-21) for n>22.

(End)

Extensions

More terms from M. F. Hasler, Jan 16 2012