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.

A333402 Numbers m such that the largest digit in the decimal expansion of 1/m is 1.

Original entry on oeis.org

1, 9, 10, 90, 99, 100, 900, 909, 990, 999, 1000, 9000, 9009, 9090, 9900, 9990, 9999, 10000, 90000, 90009, 90090, 90900, 90909, 99000, 99900, 99990, 99999, 100000, 900000, 900009, 900090, 900900, 909000, 909090, 990000, 990099, 999000, 999900, 999990, 999999, 1000000
Offset: 1

Views

Author

Bernard Schott, Mar 19 2020

Keywords

Comments

If m is a term, 10*m is also a term.
If m is a term then m has only digits {1}, {9}, {1,0} or {9,0} in its decimal representation, but this is not sufficient to be a term (see examples).
Some subsequences below (not exhaustive, see crossrefs):
m = 10^k, k >= 0, hence m is in A011557 = {1, 10, 100, 1000, 10000, ...};
m = 9*10^k, k >= 0, hence m is in A052268 = {9, 90, 900, 9000, 90000, ...};
m = 10^k-1, k >= 1, hence m is in A002283 = {9, 99, 999, 9999, 99999, ...};
m = 9*(10^k+1), k >= 1, hence m is in 9*A000533 = {99, 909, 9009, 90009, ...};
m = 9+100*(100^k-1)/11, k >= 0, hence m is in 9*A094028 = {9, 909, 90909, 9090909, ...}.

Examples

			As 1/101 = 0.009900990099..., 101 is not a term.
As 1/909 = 0.001100110011..., 909 is a term.
As 1/9099 = 0.000109902187..., 9099 is not a term.
As 1/9999 = 0.000100010001..., 9999 is also a term.

Links

Crossrefs

Cf. A333236, A333237 (similar, with 9).
Subsequences: A002283, A011557, A052268.
Subsequences: 9*A000533, 9*A094028, 9*A135577, 9*A261544, 9*A330135.

Programs

  • Mathematica
    Select[Range[10^4], Max @ RealDigits[1/#][[1]] == 1 &] (* Amiram Eldar, Mar 19 2020 *)
  • Python
    from itertools import count, islice
def A333402_gen(startvalue=1): # generator of terms >= startvalue
    for m in count(max(startvalue,1)):
        k = 1
        while k <= m:
            k *= 10
        rset = {0}
        while True:
            k, r = divmod(k, m)
            if max(str(k)) > '1':
                break
            else:
                if r in rset:
                    yield m
                    break
            rset.add(r)
            k = r
            while k <= m:
                k *= 10
A333402_list = list(islice(A333402_gen(),30)) # Chai Wah Wu, Feb 17 2022

Formula

A333236(a(n))= 1.

Extensions

More terms from Jinyuan Wang, Mar 19 2020

A269025 a(n) = Sum_{k = 0..n} 60^k.

Original entry on oeis.org

1, 61, 3661, 219661, 13179661, 790779661, 47446779661, 2846806779661, 170808406779661, 10248504406779661, 614910264406779661, 36894615864406779661, 2213676951864406779661, 132820617111864406779661, 7969237026711864406779661, 478154221602711864406779661
Offset: 0

Views

Author

Ilya Gutkovskiy, Feb 18 2016

Keywords

Comments

Partial sums of powers of 60 (A159991).
Converges in a 10-adic sense to ...762711864406779661.
More generally, the ordinary generating function for the Sum_{k = 0..n} m^k is 1/((1 - m*x)*(1 - x)). Also, Sum_{k = 0..n} m^k = (m^(n + 1) - 1)/(m - 1).

Links

Crossrefs

Cf. A159991.
Cf. similar sequences of the form (k^n-1)/(k-1): A000225 (k=2), A003462 (k=3), A002450 (k=4), A003463 (k=5), A003464 (k=6), A023000 (k=7), A023001 (k=8), A002452 (k=9), A002275 (k=10), A016123 (k=11), A016125 (k=12), A091030 (k=13), A135519 (k=14), A135518 (k=15), A131865 (k=16), A091045 (k=17), A218721 (k=18), A218722 (k=19), A064108 (k=20), A218724-A218734 (k=21..31), A132469 (k=32), A218736-A218753 (k=33..50), this sequence (k=60), A133853 (k=64), A094028 (k=100), A218723 (k=256), A261544 (k=1000).

Programs

  • Mathematica
    Table[Sum[60^k, {k, 0, n}], {n, 0, 15}]
Table[(60^(n + 1) - 1)/59, {n, 0, 15}]
LinearRecurrence[{61, -60}, {1, 61}, 15]
  • PARI
    a(n)=60^n + 60^n\59 \\ Charles R Greathouse IV, Jul 26 2016

Formula

G.f.: 1/((1 - 60*x)*(1 - x)).
a(n) = (60^(n + 1) - 1)/59 = 60^n + floor(60^n/59).
a(n+1) = 60*a(n) + 1, a(0)=1.
a(n) = Sum_{k = 0..n} A159991(k).
Sum_{n>=0} 1/a(n) = 1.016671221665660580331...
E.g.f.: exp(x)*(60*exp(59*x) - 1)/59. - Stefano Spezia, Mar 23 2023

A330135 a(n) = ((10^(n+1))^4 - 1)/9999 for n >= 0.

Original entry on oeis.org

1, 10001, 100010001, 1000100010001, 10001000100010001, 100010001000100010001, 1000100010001000100010001, 10001000100010001000100010001, 100010001000100010001000100010001
Offset: 0

Views

Author

Bernard Schott, Dec 02 2019

Keywords

Comments

This sequence was the subject of the 6th problem of the 15th British Mathematical Olympiad in 1979 (see the link BMO).
There are no prime numbers in this infinite sequence. Why?
a(0) = 1 and a(1) = 10001 = 73 * 137;
if n even = 2*k, k >= 1, then A094028(n) divides a(n);
if n odd = 2*k+1, k >= 1, then a(k) divides a(n).

Examples

			a(2) = ((10^3)^4 - 1)/9999 = 100010001 = 10101 * 9901 where 10101 = A094028(2).
a(3) = ((10^4)^4 - 1)/9999 = 1000100010001 = 10001 * 100000001 where 10001 = a(1).
From _Omar E. Pol_, Dec 04 2019: (Start)
Illustration of initial terms:
                  1;
                10001;
              100010001;
            1000100010001;
          10001000100010001;
        100010001000100010001;
      1000100010001000100010001;
    10001000100010001000100010001;
  100010001000100010001000100010001;
...
(End)

References

  • A. Gardiner, The Mathematical Olympiad Handbook: An Introduction to Problem Solving, Oxford University Press, 1997, reprinted 2011, Pb 6 pp. 68 and 201 (1979).

Links

Crossrefs

Cf. A000533 (1000...0001), A094028 (10101...101), A261544 (1001001...1001).
Cf. A131865 (similar, with 2^(n+1)).

Programs

  • Maple
    A: = seq((10^(4*n+4)-1)/9999, n=1..4);
  • Mathematica
    Table[((10^(n+1))^4 - 1)/9999, {n, 0, 8}] (* Amiram Eldar, Dec 04 2019 *)
  • PARI
    Vec(1 / ((1 - x)*(1 - 10000*x)) + O(x^11)) \\ Colin Barker, Dec 05 2019

Formula

a(n) = (10^(4*n+4) - 1)/9999 for n >= 0.
G.f.: 1 / ((1 - x)*(1 - 10000*x)). - Colin Barker, Dec 05 2019
Showing 1-3 of 3 results.

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