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 11-15 of 15 results.

A062939 Numbers k that, when expressed in base 6 and then interpreted in base 9, give a multiple of k.

Original entry on oeis.org

0, 1, 2, 3, 4, 5, 51, 54, 105, 306, 324, 630, 2646, 6711, 8998, 9003, 19847, 29513, 30127, 30132, 67662, 71267, 314751, 314928, 405972, 427602, 1009394, 1347704, 1888506, 1889568, 2321838, 2840097, 5383299, 6056364, 7143622, 8086224, 11331036, 11337408, 14382561
Offset: 1

Views

Author

Erich Friedman, Jul 21 2001

Keywords

Examples

			51 in base 6 is 123, which interpreted in base 9 is 102 = 2*51.
		

Crossrefs

Extensions

More terms from Naohiro Nomoto, Aug 06 2001
Offset changed to 1 and a(33)-a(39) from Georg Fischer, Mar 13 2023

A062942 Numbers k that, when expressed in base 6 and then interpreted in base 10, give a multiple of k.

Original entry on oeis.org

0, 1, 2, 3, 4, 5, 308, 4920, 11284, 11914, 144393, 195453, 518659, 866358, 925148, 1010765, 1172718, 1369865, 2141968, 2557924, 4287428, 4296908, 6064590, 8219190, 15347544, 16891738, 18409156, 18532263, 21880744, 23693054, 25724568, 25781448, 88115915, 93066844
Offset: 1

Views

Author

Erich Friedman, Jul 21 2001

Keywords

Comments

Zero followed by A032546. [From R. J. Mathar, Oct 02 2008]

Examples

			308 in base 6 is 1232, which interpreted in base 10 is 1232 = 4*308.
		

Crossrefs

Extensions

More terms from Naohiro Nomoto, Aug 06 2001
Offset changed to 1 and a(29)-a(34) from Georg Fischer, Mar 13 2023

A331841 When expressed in base 2 and then interpreted in base 5, is a multiple of the original number.

Original entry on oeis.org

0, 1, 3, 6, 9, 10, 18, 21, 27, 30, 54, 57, 60, 63, 89, 90, 108, 114, 126, 130, 178, 180, 189, 228, 300, 356, 378, 390, 630, 712, 780, 900, 1170, 1299, 1300, 1890, 1953, 2340, 2370, 2730, 3510, 3900, 3906, 4740, 7020, 7110, 7410, 7800, 8100, 8190, 9261, 11700
Offset: 1

Views

Author

Dimiter Skordev, Jan 29 2020

Keywords

Examples

			30 = 11110_2; 11110_5 = 780 = 26*30.
		

Crossrefs

Cf. (with base 2 and b): A062845 (b=3), A062846 (b=4), A062847 (b=6), A062848 (b=7), A062849 (b=8), A062850 (b=9), A032533 (b=10).

Programs

  • Magma
    [0] cat [k:k in [1..12000]|Seqint(Intseq(Seqint(Intseq(k, 2))), 5) mod k eq 0]; // Marius A. Burtea, Jan 29 2020
  • Mathematica
    Prepend[Select[Range[12000], Divisible[FromDigits[IntegerDigits[#, 2], 5], #] &], 0] (* Amiram Eldar, Jan 29 2020 *)
  • PARI
    isok(n) = (n == 0) || (fromdigits(digits(n, 2), 5) % n) == 0; \\ Michel Marcus, Jan 29 2020
    

A387069 Numbers whose representation in base b are their representation in base b+1 with a "0" added at the end for some b.

Original entry on oeis.org

10, 12, 273, 546, 582, 11808655, 34503520, 35555155, 54813136653716, 1066338805156287287067, 1124161332414632881704, 2305867155177711644802, 2306166776784312535170, 5744341611556736174883, 2705154287309969771123575182312
Offset: 1

Views

Author

Keywords

Comments

The corresponding b's are: 2, 2, 3, 3, 3, 5, 5, 5, 7, 9, 9, 9, 9, 9, 11, ...
From Sean A. Irvine, Aug 18 2025: (Start)
Numbers k such that k = Sum_{j>=1} d(j) * b^j = Sum_{j>=0} d(j+1) * (b+1)^j where 0 <= d(j) < b.
For k to be a solution in base b, we require Sum d(j) = 0 (mod b) and Sum_{j>=2} (-1)^j*d(j) = 2*d(1) (mod b+1). (End)

Examples

			a(1) = 10, is 1010 in base 2 and 101 in base 3.
a(2) = 12, is 1100 in base 2 and 110 in base 3.
a(3) = 273, is 101010 in base 3 and 10101 in base 4.
		

Crossrefs

Programs

  • PARI
    isok(k) = for (b=2, k-1, if (digits(k, b) == concat(digits(k, b+1), 0), return(b));); \\ Michel Marcus, Aug 16 2025

Extensions

a(9)-a(15) from Sean A. Irvine, Aug 18 2025

A330435 a(n) is the least k >= n that written in base n and then interpreted in base n+1 is a multiple of k.

Original entry on oeis.org

5, 53, 1060, 697, 14027, 7830448093388, 278269, 7794416, 1784625167675, 217659538, 7299226328, 58429863516468189, 2265720635440119410, 11301046374119, 5483279396166772909558757483, 9796440127236265192879361141313874782657110677228434, 24212615127434834, 1506888944866952574
Offset: 2

Views

Author

Giovanni Resta, Dec 14 2019

Keywords

Comments

Theorem: a(n) > n^(n*log(2)). Proof: if k=(d_m...d_2d_1d_0)n, and K=(d_m...d_2d_1d_0){n+1} then K <= k*(1+1/n)^m. Suppose k=a(n). Then K >= 2*k, hence (1+1/n)^m >= 2, and therefore m >= log(2)/log(1+1/n) > n*log(2). Since d_m can be assumed to be >= 1, we may assume that k >= n^m, and this yields k > n^(n*log(2)). - Dimiter Skordev, Mar 14 2020

Examples

			a(5) = 697 = (10242)_5 and (10242)_6 = 1394 = 2 * 697.
a(7) = 7830448093388 = (1435505542406624)_7, which when interpreted in base 8 is equal to 7 * 7830448093388.
		

Crossrefs

Programs

  • Mathematica
    a[n_] := Block[{k = n}, While[ Mod[ FromDigits[ IntegerDigits[k, n], n + 1], k] > 0, k++]; k]; a /@ Range[2, 6]
  • Python
    def BaseUp(n,b):
        up, b1 = 0, 1
        while n > 0:
            up, b1, n = up+(n%b)*b1, b1*(b+1), n//b
        return up
    n = 2
    while n < 20:
        k = n
        while BaseUp(k,n)%k != 0:
            k = k+1
        print(n,k)
        n = n+1 # A.H.M. Smeets, Mar 31 2020
Previous Showing 11-15 of 15 results.