cp's OEIS Frontend

This is a front-end for the Online Encyclopedia of Integer Sequences, made by Christian Perfect. The idea is to provide OEIS entries in non-ancient HTML, and then to think about how they're presented visually. The source code is on GitHub.

Showing 1-3 of 3 results.

A102807 a(n) is the square of one plus the number consisting of n 3's.

Original entry on oeis.org

1, 16, 1156, 111556, 11115556, 1111155556, 111111555556, 11111115555556, 1111111155555556, 111111111555555556, 11111111115555555556, 1111111111155555555556, 111111111111555555555556, 11111111111115555555555556, 1111111111111155555555555556, 111111111111111555555555555556
Offset: 0

Views

Author

Ron Knott, Feb 27 2005

Keywords

Comments

Old name was: The number (333...334)^2.
An infinite sequence of squares with no zeros in base 10.
a(n) = A104265(2n) for n > 0. - Chai Wah Wu, Mar 24 2020

References

  • Albert H. Beiler, Recreations in the theory of numbers, New York, Dover, (2nd ed.) 1966. See Table 31 at p. 61.
  • Italo Ghersi, Matematica dilettevole e curiosa, pp. 111-112, Hoepli, Milano, 1967. [Vincenzo Librandi, Dec 31 2008]

Links

Crossrefs

Cf. A052041, A093137, A075415, A109344, A104265.

Programs

  • Maple
    a:= n-> (1+parse(cat(0, 3$n)))^2:
seq(a(n), n=0..20);  # Alois P. Heinz, Sep 03 2018
  • Mathematica
    Table[(10^n + 2)^2/9, {n, 0, 20}] (* Paolo Xausa, Jun 26 2024 *)

Formula

From R. J. Mathar, Jan 06 2009: (Start)
a(n) = (100^n + 4*10^n + 4)/9.
G.f.: (1 - 95*x + 490*x^2)/((1-x)*(100*x-1)*(10*x-1)). (End)
E.g.f.: exp(x)*(4 + 4*exp(9*x) + exp(99*x))/9. - Stefano Spezia, Jul 31 2024

Extensions

New name from Alois P. Heinz, Sep 03 2018

A104264 Number of n-digit squares with no zero digits.

Original entry on oeis.org

3, 6, 19, 44, 136, 376, 1061, 2985, 8431, 24009, 67983, 193359, 549697, 1563545, 4446173, 12650545, 35999714, 102439796, 291532841, 829634988, 2360947327, 6719171580, 19122499510, 54423038535, 154888366195
Offset: 1

Views

Author

Reinhard Zumkeller and Ron Knott, Feb 26 2005

Keywords

Comments

Comments from David W. Wilson, Feb 26 2005: (Start)
"There are approximately s(d) = (10^d)^(1/2) - (10^(d-1))^(1/2) d-digit squares. A random d-digit number has the probability p(d) = (9/10)^(d-1) of being zeroless (exponent d-1 as opposed to d because the first digit is not zero). So we expect p(d)s(d) zeroless d-digit squares.
"For d = 1 through 12, we get (truncating): 1, 5, 15, 44, 127, 363, 1034, 2943, 8377, 23841, 67854, 193117, ... The elements grow approximately geometrically with limit ratio (9/10)*10^(1/2) = 2.846+.
"The same naive estimate can easily be generalize to k-th powers, giving the estimate s(d) = (10^d)^(1/k) - (10^(d-1))^(1/k) for d-digit k-th powers. p(d) remains the same. The resulting estimates have ratio (9/10)*10^(1/k).
"We should expect an infinite number of zeroless k-th powers when this ratio is >= 1, which it is for k <= 21. For k >= 22, the ratio is < 1 and we should expect a finite number of zeroless k-th powers." (End)

Examples

			a(3) = #{121, 144, 169, 196, 225, 256, 289, 324, 361, 441, 484, 529, 576, 625, 676, 729, 784, 841, 961} = 19.

Crossrefs

Cf. A052041, A104265, A104266, A075415, A102807.

Programs

  • Python
    def aupton(terms):
  c, k, kk = [0 for i in range(terms)], 1, 1
  while kk < 10**terms:
    s = str(kk)
    c[len(s)-1], k, kk = c[len(s)-1] + (s.count('0')==0), k+1, kk + 2*k + 1
  return c
print(aupton(14)) # Michael S. Branicky, Mar 06 2021

Extensions

a(14)-a(18) from Donovan Johnson, Nov 05 2009
a(19)-a(21) from Donovan Johnson, Mar 23 2011
a(22)-a(25) from Donovan Johnson, Jan 29 2013

A104266 Largest n-digit square with no zero digits.

Original entry on oeis.org

9, 81, 961, 9216, 99856, 978121, 9998244, 99321156, 999887641, 9978811236, 99999515529, 999332111556, 9999995824729, 99978881115136, 999999961946176, 9999333211115556, 99999999356895225, 999978918111112681, 9999999986285964964, 99999333321111155556
Offset: 1

Views

Author

Reinhard Zumkeller, Feb 26 2005

Keywords

Comments

See Formula section for exact formula for terms whose index n is divisible by 4, and upper bounds for other cases; see Links for additional information on those other cases. - Jon E. Schoenfield, Mar 30 2015

Examples

			a(3) = Max{...., 729, 784, 841, 961} = 961.

Links

Crossrefs

Cf. A104265, A104264, A052041.

Programs

  • Maple
    f:= proc(n) local r;
  r:= floor(sqrt(10^n));
  while has(convert(r^2,base,10),0) do r:= r-1 od:
r^2
end proc:
seq(f(n),n=1..24); # Robert Israel, Mar 29 2015
  • Mathematica
    f[n_] := Block[{k = Floor[ Sqrt[10^n]]}, While[ Union[ IntegerDigits[ k^2]][[1]] == 0, k-- ]; k^2]; Table[ f[n], {n, 18}] (* Robert G. Wilson v, Mar 03 2005 *)
  • PARI
    a(n)=k=floor(sqrt(10^n));while(k,if(type(k)=="t_INT"&&vecmin(digits(k^2)), return(k^2));k--)
vector(20,n,a(n)) \\ Derek Orr, Mar 29 2015

Formula

From Jon E. Schoenfield, Mar 31 2015: (Start)
If n is divisible by 4, then a(n) = (10^(n/2) - ceiling(10^(n/4)/3))^2;
otherwise, if n is even, then a(n) < 10^(n) * (1 - (10^-((n-2)/4))* 2 / sqrt(90/1.000000000001026)) (see Links for derivation), except that a(2) = 81.
If n is odd, then a(n) ~ (floor(10^(n/2)))^2. (Although (floor(10*(n/2)))^2 gives an obvious upper bound for a(n) for all n, it seems to be a much tighter upper bound for odd values of n.) (End)

Extensions

More terms from Robert G. Wilson v, Mar 03 2005
More terms from Jon E. Schoenfield, Mar 29 2015
Showing 1-3 of 3 results.

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