A307588 -id:A307588 - cp's OEIS frontend

A307371 Numbers k such that the digits of sqrt(k) begin with k.

0, 1, 98, 99, 100, 9998, 9999, 10000, 999998, 999999, 1000000, 99999998, 99999999, 100000000, 9999999998, 9999999999, 10000000000, 999999999998, 999999999999, 1000000000000, 99999999999998, 99999999999999, 100000000000000, 9999999999999998, 9999999999999999

Offset: 1

Dmitry Kamenetsky, Apr 17 2019

From Chai Wah Wu, Jan 17 2020: (Start)
Theorem: A number n is a term if and only if n is 0, 1, 10^(2m), 10^(2m)-1 or 10^(2m)-2 for some m >= 1.
Proof: k <= sqrt(k)*10^m < k+1. For m = 0, the only solutions are 0 and 1. For m > 0, k^2 <= k*10^(2m) < (k+1)^2. This is equivalent to k <= 10^2m < k + 2 + 1/k, i.e., 10^(2m)-2-1/k < k <= 10^(2m). Thus the only solutions for k are 10^(2m), 10^(2m)-1 and 10^(2m)-2. (End)

			sqrt(9998) = 99.989..., which begins with "9998", so 9998 is in the sequence.

Dmitry Kamenetsky, Java program to compute terms

Cf. A307588, A307600.

A307371_list = [0, 1, 98, 99, 100, 9998]
for _ in range(100):
    A307371_list.append(101*A307371_list[-3]-100*A307371_list[-6]) # Chai Wah Wu, Jan 18 2020

From Chai Wah Wu, Jan 17 2020: (Start)
a(n) = 101*a(n-3) - 100*a(n-6) for n > 6.
G.f.: x^2*(100*x^4 - x^3 + 99*x^2 + 98*x + 1)/(100*x^6 - 101*x^3 + 1). (End)

a(12)-a(25) from Jon E. Schoenfield, May 01 2019

A307600 Numbers k such that the digits of k^(1/4) begin with k.

Original entry on oeis.org

0, 1, 21, 463, 464, 9999, 10000, 215443, 4641588, 99999999, 100000000, 2154434689, 2154434690, 46415888335, 46415888336, 999999999999, 1000000000000, 21544346900318, 464158883361277, 9999999999999999, 10000000000000000, 215443469003188371, 215443469003188372

Offset: 1

Dmitry Kamenetsky, Apr 17 2019

Program is in A307371.
From Bernard Schott, May 01 2019: (Start)
There are two nontrivial families in this sequence:
1st: 21, 215443, 2154434689, 2154434690, 21544346900318, ...
2nd: 463, 464, 4641588, 46415888335, 46415888336, ... (End)
From Jon E. Schoenfield, May 04 2019: (Start)
For each number k such that the digits of k^(1/m) begin with k, we have, for each m >= 2, floor(k^(1/m) * 10^d) = k for some integer d, so k^(1/m) * 10^d ~= k; solving for k gives k ~= 10^(d*m/(m-1)).
In the m=4 case (this sequence), this gives k ~= 10^(d*4/3) so, as d is incremented by 1, 10^(d*4/3) increases by a factor of 10^(4/3) = 10000^1/3 = 21.5443469...:
.
   d |     10^(d*4/3)
  ---+---------------------
   0 |             1
   1 |            21.544...
   2 |           464.158...
   3 |         10000
   4 |        215443.469...
   5 |       4641588.833...
   6 |     100000000
   7 |    2154434690.031...
   8 |   46415888336.127...
   9 | 1000000000000
.
Each nonnegative integer d corresponds to one or two terms in the sequence. Letting j = floor(10000^(d/3)), j is necessarily a term; j-1 is also a term iff (j-1)^(1/4)*10^d < j. This inequality is satisified
   for d == 1 (mod 3) at d = 7, 13, 16, 34, 37, ...;
   for d == 2 (mod 3) at d = 2, 8, 20, 29, 32, 35, ...;
   and at every d == 0 (mod 3).
(The sequence contains no other terms than numbers k of the form j or j-1 where j = floor(10000^(d/3)) for some nonnegative integer d.)
(End)

			215443^(1/4) = 21.544335..., which begins with "215443", so 215443 is in the sequence.

Chai Wah Wu, Table of n, a(n) for n = 1..1172

Cf. A307371, A307588.

Cf. A052211 (analog for 4th power instead of 1/4).

a(12)-a(23) from Jon E. Schoenfield, May 01 2019

A331492 Numbers k such that the digits of k^(1/5) begin with k.

Original entry on oeis.org

0, 1, 17, 315, 316, 5623, 99999, 100000, 1778279, 31622776, 562341324, 562341325, 9999999999, 10000000000, 177827941003, 3162277660168, 56234132519034, 999999999999999, 1000000000000000, 17782794100389227, 17782794100389228, 316227766016837932, 316227766016837933

Offset: 1

Chai Wah Wu, Jan 18 2020

The following algorithm will generate all numbers k such that the digits of k^(1/b) begins with k: For each integer m >= 0, compute r = floor(10^(b*m/(b-1))). Let s <= r be the largest integer >= 0 such that (r-s)*10^(b*m) < (r-s+1)^b. Then r, r-1, ... r-s are such numbers k and there are no other such numbers.

			5623^(1/5) = 5.6233305990931... which starts with the digits 5623, so 5623 is in the sequence.

Chai Wah Wu, Table of n, a(n) for n = 1..1140

Cf. A307371, A307588, A307600.

A331493 Numbers k such that the digits of k^(1/6) begin with k.

Original entry on oeis.org

0, 1, 15, 250, 251, 3980, 3981, 63095, 999999, 1000000, 15848931, 251188642, 251188643, 3981071705, 63095734447, 63095734448, 999999999999, 1000000000000, 15848931924610, 15848931924611, 251188643150957, 251188643150958, 3981071705534972, 63095734448019324

Offset: 1

Chai Wah Wu, Jan 18 2020

			63095^(1/6) =  6.309561203942..., which starts with the digits 63095, so 63095 is a term.

Chai Wah Wu, Table of n, a(n) for n = 1..1141

Cf. A307371, A307588, A307600, A331492.

A331494 Numbers k such that the digits of k^(1/7) begin with k.

Original entry on oeis.org

0, 1, 14, 215, 3162, 46415, 681291, 681292, 9999999, 10000000, 146779926, 2154434689, 2154434690, 31622776601, 464158883361, 6812920690579, 99999999999999, 100000000000000, 1467799267622069, 21544346900318837, 316227766016837933, 4641588833612778892, 68129206905796128549

Offset: 1

Chai Wah Wu, Jan 18 2020

			681291^(1/7) = 6.8129191633529... which starts with the digits 681291, so 681291 is a term.

Chai Wah Wu, Table of n, a(n) for n = 1..1120

Cf. A307371, A307588, A307600, A331492, A331493.

A331495 Numbers k such that the digits of k^(1/8) begin with k.

Original entry on oeis.org

0, 1, 13, 192, 193, 2682, 37275, 517947, 7196856, 99999999, 100000000, 1389495494, 19306977288, 268269579527, 3727593720314, 51794746792311, 51794746792312, 719685673001151, 719685673001152, 9999999999999999, 10000000000000000, 138949549437313763, 1930697728883250166

Offset: 1

Chai Wah Wu, Jan 18 2020

			7196856^(1/8) = 7.1968566387600... which starts with the digits 7196856, so 7196856 is a term.

Chai Wah Wu, Table of n, a(n) for n = 1..1111

Cf. A307371, A307588, A307600, A331492, A331493, A331494.

A331496 Numbers k such that the digits of k^(1/9) begin with k.

Original entry on oeis.org

0, 1, 13, 177, 2371, 31622, 421696, 5623413, 74989420, 999999999, 1000000000, 13335214321, 177827941003, 2371373705661, 31622776601683, 421696503428582, 5623413251903490, 74989420933245582, 999999999999999999, 1000000000000000000, 13335214321633240256, 177827941003892280122

Offset: 1

Chai Wah Wu, Jan 18 2020

			13335214321^(1/9) = 13.33521432156287... which starts with the digits 13335214321, so 13335214321 is a term.

Chai Wah Wu, Table of n, a(n) for n = 1..1088

Cf. A307371, A307588, A307600, A331492, A331493, A331494, A331495.

A331497 Numbers k such that the digits of k^(1/10) begin with k.

Original entry on oeis.org

0, 1, 12, 166, 2154, 27825, 359381, 4641588, 59948424, 59948425, 774263682, 9999999999, 10000000000, 129154966501, 1668100537199, 1668100537200, 21544346900318, 278255940220712, 3593813663804627, 46415888336127788, 599484250318941014, 7742636826811270597, 99999999999999999999

Offset: 1

Chai Wah Wu, Jan 18 2020

			59948425^(1/10) = 5.994842502870... which starts with the digits 59948425 so 59948425 is a term.

Chai Wah Wu, Table of n, a(n) for n = 1..1085

Cf. A307371, A307588, A307600, A331492, A331493, A331494, A331495, A331496,.

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A307371 Numbers k such that the digits of sqrt(k) begin with k.

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A307600 Numbers k such that the digits of k^(1/4) begin with k.

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A331492 Numbers k such that the digits of k^(1/5) begin with k.

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A331493 Numbers k such that the digits of k^(1/6) begin with k.

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A331494 Numbers k such that the digits of k^(1/7) begin with k.

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A331495 Numbers k such that the digits of k^(1/8) begin with k.

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A331496 Numbers k such that the digits of k^(1/9) begin with k.

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A331497 Numbers k such that the digits of k^(1/10) begin with k.

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