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.

Previous Showing 11-17 of 17 results.

A077502 Smallest square greater than the previous term and beginning with n.

Original entry on oeis.org

0, 1, 25, 36, 49, 529, 625, 729, 841, 900, 1024, 1156, 1225, 1369, 1444, 1521, 1600, 1764, 1849, 1936, 2025, 2116, 2209, 2304, 2401, 2500, 2601, 2704, 2809, 2916, 3025, 3136, 3249, 3364, 3481, 35344, 36100, 37249, 38025, 39204, 40000, 41209, 42025, 43264
Offset: 0

Views

Author

Amarnath Murthy, Nov 08 2002

Keywords

Crossrefs

Cf. A018796, A272676 (the square roots).

Extensions

Corrected and extended by Ray Chandler, Aug 11 2003
Added initial 0. - N. J. A. Sloane, May 21 2016

A272680 Smallest square that begins with n (in binary).

Original entry on oeis.org

0, 1, 4, 25, 4, 81, 25, 121, 16, 9, 81, 361, 25, 441, 225, 121, 16, 1089, 36, 625, 81, 169, 361, 1521, 49, 25, 841, 441, 225, 3721, 121, 2025, 64, 529, 1089, 4489, 36, 2401, 1225, 625, 81, 5329, 169, 2809, 11449, 361, 5929, 1521, 6241, 49, 100, 6561, 841, 6889
Offset: 0

Views

Author

N. J. A. Sloane, May 22 2016

Keywords

Examples

			a(10)=81, because 81 = 9^2 = 1010001_2 begins with 1010 = 10_2.
		

References

  • Allan C. Wechsler, posting to math-fun mailing list May 22 2016.

Crossrefs

Programs

  • Python
    from gmpy2 import isqrt
    def A272680(n):
        if n == 0:
            return 0
        else:
            d, nd = 1, n
            while True:
                x = (isqrt(nd-1)+1)**2
                if x < nd+d:
                    return int(x)
                d *= 2
                nd *= 2 # Chai Wah Wu, May 22 2016

Extensions

More terms from Chai Wah Wu, May 22 2016

A272681 Smallest binary square that begins with the binary expansion of n.

Original entry on oeis.org

0, 1, 100, 11001, 100, 1010001, 11001, 1111001, 10000, 1001, 1010001, 101101001, 11001, 110111001, 11100001, 1111001, 10000, 10001000001, 100100, 1001110001, 1010001, 10101001, 101101001, 10111110001, 110001, 11001, 1101001001, 110111001, 11100001, 111010001001
Offset: 0

Views

Author

N. J. A. Sloane, May 22 2016

Keywords

Examples

			a(10)=1010001 = 81_10, because 1010001_2 begins with 1010 = 10_2.
		

References

  • Allan C. Wechsler, posting to math-fun mailing list May 22 2016.

Crossrefs

Programs

  • Python
    from gmpy2 import isqrt
    def A272681(n):
        if n == 0:
            return 0
        else:
            d, nd = 1, n
            while True:
                x = (isqrt(nd-1)+1)**2
                if x < nd+d:
                    return int(bin(x)[2:])
                d *= 2
                nd *= 2 # Chai Wah Wu, May 22 2016

Extensions

More terms from Chai Wah Wu, May 22 2016

A272676 Smallest number greater than previous term such that a(n)^2 begins with n.

Original entry on oeis.org

0, 1, 5, 6, 7, 23, 25, 27, 29, 30, 32, 34, 35, 37, 38, 39, 40, 42, 43, 44, 45, 46, 47, 48, 49, 50, 51, 52, 53, 54, 55, 56, 57, 58, 59, 188, 190, 193, 195, 198, 200, 203, 205, 208
Offset: 0

Views

Author

Keywords

Examples

			a(4) = 7 because 7^2 = 49 begins with 4. (Without the monotonic condition, we could have used 4 - this would give A018851.)
		

Crossrefs

Cf. A077502 (the squares), A018851, A018796.

A272678 Smallest number m such that A272677(m) = n.

Original entry on oeis.org

0, 2, 5, 35, 296, 2600, 25317, 251416, 2504474, 25010000, 250044723, 2500100000, 25000316228, 250002000003, 2500004472137, 25000010000000, 250000044721361, 2500000141421358, 25000000316227767
Offset: 0

Views

Author

Keywords

Comments

Given n, this is the smallest number m with the property that the smallest square beginning with m has n more digits than n.
a(n) >= 25*10^(n-3). Conjecture: a(n)/(25*10^(n-3)) -> 1 as n -> oo. - Chai Wah Wu, May 21 2016
For odd n > 2, it seems that a(n) is about 25 * 10^(n-3) + 10^(floor((n-1)/2)), although a(13) breaks that pattern. - David A. Corneth, May 22 2016
Except for n = 1 and 13, a(n) appears to be approximately equal to either 25*10^(n-3)+sqrt(10^(n-1)) (for n = 0, 2, 3, 5, 6, 9, 11, 12, 15, 18, ... ) or 25*10^(n-3)+sqrt(2*10^(n-1)) (for n = 4, 7, 8, 14, 16, 17, ...). For n = 1, a(n) is approximately 25*10^(n-3)+sqrt(3*10^(n-1)) and for n = 13, a(n) is about equal to 25*10^(n-3)+sqrt(4*10^(n-1)). Conjecture: a(n) is always approximately to 25*10^(n-3)+sqrt(k*10^(n-1)) for some small integer k > 0. - Chai Wah Wu, May 22 2016
Using the above conjecture as a guide, upper bounds for a(n) can be computed (see file in links) which coincide with a(n) for n <= 19. - Chai Wah Wu, May 23 2016

Examples

			The smallest square beginning with 5 is 529, which has two more digits than 5, and corresponds to a(2) = 5.
		

Crossrefs

Extensions

a(6)-a(8) from Chai Wah Wu, May 21 2016
a(9)-a(10), a(15)-a(18) and corrected a(12) from Chai Wah Wu, May 22 2016
a(11)-a(14) from David A. Corneth, May 22 2016

A309123 a(1) = 50 and for any n > 0, a(n+1)^2 is the smallest square that begins with a(n).

Original entry on oeis.org

50, 71, 267, 517, 2274, 1508, 3884, 1971, 444, 667, 817, 286, 535, 732, 856, 2926, 541, 736, 858, 293, 542, 233, 483, 695, 834, 2888, 16995, 41225, 20304, 4506, 6713, 2591, 5091, 22564, 47502, 21795, 46686, 21607, 46484, 215602, 46433, 68142, 261041, 510922
Offset: 1

Views

Author

Rémy Sigrist, Jul 13 2019

Keywords

Comments

This sequence is similar to A308055.
The initial value (50) seems to be the first one for which the iteration of A018796 diverges; there are neither duplicates nor squares among the first 79000 terms.

Examples

			The first terms, alongside the square of a(n+1), are:
  n   a(n)  a(n+1)^2
  --  ----  --------
   1    50      5041
   2    71     71289
   3   267    267289
   4   517   5171076
   5  2274   2274064
   6  1508  15085456
   7  3884   3884841
   8  1971    197136
   9   444    444889
  10   667    667489
  11   817     81796
  12   286    286225
  13   535    535824
  14   732    732736
  15   856   8561476
  16  2926    292681
		

Crossrefs

Programs

  • PARI
    See Links section.

Formula

a(1) = 50 and then a(n+1) = A018796(a(n)) for n > 0.

A111443 Least square beginning with the digit reversal of n^2.

Original entry on oeis.org

1, 4, 9, 61009, 529, 63001, 9409, 4624, 1849, 1, 121, 441, 961, 69169, 522729, 652864, 982081, 423801, 16384, 4, 144, 484, 925444, 675684, 5262436, 676, 927369, 487204, 14884, 9, 169, 42016324, 9801, 651117289, 5221225, 692110864, 96314596
Offset: 1

Views

Author

Amarnath Murthy, Aug 03 2005

Keywords

Comments

a(10n) = a(n). a(10^n) = 1. a(k) = k^2 if k^2 is a palindromic square. (Comment corrected by David Wasserman, Jan 21 2009)

Examples

			a(8) = 4624 = 68^2 beginning with 46 the digit reversal of 64 = 8^2.
		

Formula

a(n) = A018796(A004086(n^2)). [From David Wasserman, Jan 21 2009]

Extensions

Corrected and extended by David Wasserman, Jan 21 2009
Previous Showing 11-17 of 17 results.