A348491 Positive numbers whose square starts and ends with exactly one 9.

3, 97, 303, 307, 313, 953, 957, 963, 967, 973, 977, 983, 987, 993, 3003, 3007, 3013, 3017, 3023, 3027, 3033, 3037, 3043, 3047, 3053, 3057, 3063, 3067, 3073, 3077, 3083, 3087, 3093, 3097, 3103, 3107, 3113, 3117, 3123, 3127, 3133, 3137, 3143, 9487, 9493, 9497, 9503, 9507, 9513, 9517

Offset: 1

Author

Bernard Schott, Nov 02 2021

Keywords

• nonn
• base
• easy

Comments

When a square ends with 9, it ends with only one 9.

From Marius A. Burtea, Nov 02 2021 : (Start)

The sequence is infinite because the numbers 303, 3003, 30003, ..., 3*(10^k + 1), k >= 2, are terms with squares 91809, 9018009, 900180009, 90001800009, ... 9*(10^(2*k) + 2*10^k + 1), k >= 2.

Numbers 97, 967, 9667, 96667, 966667, ..., (29*10^n + 1) / 3, k >= 1, are terms and have no digits 0, because their squares are 9409, 935089, 93450889, 9344508889, 934445088889, ...

Also 963, 9663, 96663, 966663, 9666663, 96666663, ... (29*10^k - 11) / 3, k >= 2, are terms and have no digits 0, because their squares are 927369, 93373569, 9343735569, 934437355569, 93444373555569, 9344443735555569, ... (End)

Examples

			97^2 = 9409, hence 97 is a term.
997^2 = 994009, hence  997 is not a term.
		

Crossrefs

Subsequence of A305719, A063226, and A045863.

Cf. A017377, A045863, A273374 (squares ending with 9).

Similar to: A348487 (k=1), A348488 (k=4), A348489 (k=5), A348490 (k=6), this sequence (k=9).

Programs

Magma

[3] cat [n:n in [4..9600]|Intseq(n*n)[1] eq 9 and Intseq(n*n)[#Intseq(n*n)] eq 9]; // Marius A. Burtea, Nov 02 2021

Mathematica

Join[{3}, Select[Range[10, 10^4], (d = IntegerDigits[#^2])[[1]] == d[[-1]] == 9 && d[[2]] != 9 &]] (* Amiram Eldar, Nov 02 2021 *)

PARI

isok(k) = my(d=digits(sqr(k))); (d[1]==9) && (d[#d]==9) && if (#d>2, (d[2]!=9) && (d[#d-1]!=9), 1); \\ Michel Marcus, Nov 03 2021

PARI

list(lim)=my(v=List([3])); for(d=2, 2*#digits(lim\=1), my(s=sqrtint(9*10^(d-1)-1)+1); s+=[3,2,1,0,3,2,1,0,5,4][s%10+1]; forstep(n=s, min(sqrtint(10^d-10^(d-2)-1), lim), if(s%10==3, [4,6], [6,4]), listput(v, n))); Vec(v) \\ Charles R Greathouse IV, Nov 03 2021

Python

from itertools import count, takewhile
def ok(n):
  s = str(n*n); return len(s.rstrip("9")) == len(s.lstrip("9")) == len(s)-1
def aupto(N):
  r = takewhile(lambda x: x<=N, (10*i+d for i in count(0) for d in [3, 7]))
  return [k for k in r if ok(k)]
print(aupto(9517)) # Michael S. Branicky, Nov 02 2021