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 1-9 of 9 results.

A002380 a(n) = 3^n reduced modulo 2^n.

Original entry on oeis.org

0, 1, 1, 3, 1, 19, 25, 11, 161, 227, 681, 1019, 3057, 5075, 15225, 29291, 55105, 34243, 233801, 439259, 269201, 1856179, 3471385, 6219851, 1882337, 5647011, 50495465, 17268667, 186023729, 21200275, 63600825, 1264544299, 3793632897, 7085931395
Offset: 0

Views

Author

N. J. A. Sloane

Keywords

Comments

A065554 lists the indices n such that a(n+1) = 3*a(n). - Benoit Cloitre, Apr 21 2003
a(n) = (fractional part of (3/2)^n without the decimal point)/5^n = A204544(n) / 5^n. - Michel Lagneau, Jan 25 2012

References

  • D. H. Lehmer, Guide to Tables in the Theory of Numbers. Bulletin No. 105, National Research Council, Washington, DC, 1941, p. 82.
  • S. S. Pillai, On Waring's problem, J. Indian Math. Soc., 2 (1936), 16-44.
  • N. J. A. Sloane, A Handbook of Integer Sequences, Academic Press, 1973 (includes this sequence).
  • N. J. A. Sloane and Simon Plouffe, The Encyclopedia of Integer Sequences, Academic Press, 1995 (includes this sequence).

Links

Crossrefs

Cf. A060692, A002379, A000079, A000244.
Cf. k^n mod (k-1)^n: this sequence (k=3), A064629 (k=4), A138589 (k=5), A138649 (k=6), A139786 (k=7), A138973 (k=8), A139733 (k=9).

Programs

  • Haskell
    a002380 n = 3^n `mod` 2^n  -- Reinhard Zumkeller, Jul 11 2014
  • Maple
    a:=n->3^n mod(2^n): seq(a(n), n=0..33); # Zerinvary Lajos, Feb 15 2008
  • Mathematica
    Table[ PowerMod[3, n, 2^n], {n, 0, 33}] (* Robert G. Wilson v, Dec 14 2006 *)
Table[ 3^n - 2^n * Floor[ (3/2)^n ], {n,0,33} ] (* Fred Daniel Kline, Oct 12 2017 *)
x[n_] := -(1/2) + (3/2)^n + ArcTan[Cot[(3/2)^n Pi]]/Pi;
y[n_] := 3^n - 2^n * x[n];
Array[y, 33] (* Fred Daniel Kline, Dec 21 2017 *)
  • PARI
    concat([0],vector(55,n,lift(Mod(3,2^n)^n))) \\ Joerg Arndt, Oct 14 2017

Extensions

More terms from Jason Earls, Jul 29 2001

A065560 a(n) is the smallest integer k such that floor((1+1/n)^(k+1))/floor((1+1/n)^k) = 1+1/n.

Original entry on oeis.org

2, 4, 7, 9, 12, 15, 18, 21, 25, 28, 40, 35, 39, 42, 46, 50, 54, 58, 62, 66, 70, 74, 78, 83, 87, 91, 95, 100, 104, 109, 113, 118, 122, 127, 131, 136, 141, 145, 150, 155, 159, 164, 169, 174, 179, 183, 188, 193, 198, 203, 208, 213, 218, 223, 228, 233, 238, 243, 248, 253
Offset: 2

Views

Author

Benoit Cloitre, Nov 29 2001

Keywords

Comments

a(n) is growing roughly like prime(n). a(n) < a(n+1) except for n = 12. (Is this the only exception?)
a(n) < a(n+1) except for n = 12, 108, 266, ... - Boris Gourevitch (boris(AT)pi314.net), Dec 04 2001
Conjecture: a(n)+n > prime(n).

Examples

			a(5) = 9 because 9 is the first integer satisfying floor((6/5)^(9+1))/floor((6/5)^9) = 6/5.

Links

Crossrefs

Cf. A065554, A065564.

Programs

  • PARI
    a(n) = { my(k=1, f=(n + 1)/n); while((floor(f^(k + 1))/floor(f^k)) != f, k++); k } \\ Harry J. Smith, Oct 22 2009

Formula

Asymptotic (conjectured) formula: a(n)=n*log(n)+o(log(n)).

Extensions

Terms a(53) - a(61) from Harry J. Smith, Oct 22 2009

A065564 Numbers k such that floor((4/3)^(k+1))/floor((4/3)^k) = 4/3.

Original entry on oeis.org

4, 13, 18, 20, 21, 23, 24, 34, 44, 45, 49, 56, 60, 63, 65, 66, 67, 79, 81, 83, 85, 88, 94, 102, 107, 109, 119, 125, 126, 129, 131, 132, 133, 135, 138, 139, 144, 161, 162, 164, 172, 174, 175, 190, 194, 199, 204, 217, 218, 233, 234, 240, 249, 250, 253, 255, 258
Offset: 1

Views

Author

Benoit Cloitre, Nov 30 2001

Keywords

Links

Crossrefs

Cf. A064628, A065554.

Programs

  • PARI
    isok(k) = { my(f=4/3); floor(f^(k + 1))/floor(f^k) == f } \\ Harry J. Smith, Oct 22 2009

Formula

Lim_{n->infinity} a(n)/n = 4?

Extensions

a(55)-a(57) corrected by Harry J. Smith, Oct 22 2009

A065566 Numbers k such that floor((5/4)^(k+1))/floor((5/4)^k) = 5/4.

Original entry on oeis.org

7, 15, 17, 21, 25, 34, 52, 56, 59, 68, 74, 78, 99, 104, 111, 117, 118, 119, 124, 127, 129, 135, 136, 141, 145, 157, 162, 172, 179, 181, 184, 189, 190, 203, 204, 206, 209, 211, 212, 222, 226, 228, 245, 247, 250, 256, 258, 283, 291, 302, 315, 318, 327, 328, 331
Offset: 1

Views

Author

Benoit Cloitre, Nov 30 2001

Keywords

Comments

Also, numbers k such that (5^(k+1) mod 4^(k+1))/(5^k mod 4^k)=5, or A138589(n+1)/A138589(n) = 5. (See the Mathar link in A139768.)

Links

Crossrefs

Cf. A065554, A138589, A139768.

Programs

  • PARI
    isok(k) = { my(f=5/4); floor(f^(k + 1))/floor(f^k) == f } \\ Harry J. Smith, Oct 22 2009

Formula

Is it true that lim_{n->infinity} a(n)/n = 6?

Extensions

More terms from Jason Earls, Dec 03 2001
Edited by N. J. A. Sloane, May 24 2008

A067672 Numbers k such that floor((3/2)^(k+2))/floor((3/2)^k) = (3/2)^2.

Original entry on oeis.org

36, 40, 49, 50, 67, 72, 79, 91, 101, 102, 126, 145, 171, 178, 193, 194, 198, 204, 215, 216, 231, 236, 252, 274, 301, 304, 320, 327, 337, 350, 365, 377, 384, 385, 408, 409, 410, 411, 419, 461, 481, 485, 486, 554, 555, 567, 568, 578, 622, 697, 710, 761, 798
Offset: 1

Views

Author

Benoit Cloitre, Feb 04 2002

Keywords

Crossrefs

Cf. A065554, A002379.

A067673 Numbers k such that floor((3/2)^(k+3))/floor((3/2)^k) = (3/2)^3.

Original entry on oeis.org

49, 101, 193, 215, 384, 408, 409, 410, 485, 554, 567, 827, 828, 922, 937, 938, 939, 1002, 1021, 1022, 1023, 1115, 1132, 1188, 1238, 1310, 1314, 1315, 1316, 1317, 1409, 1419, 1424, 1506, 1520, 1616, 1629, 1754, 1773, 1817, 1848, 1873, 1881, 1882, 1883
Offset: 1

Views

Author

Benoit Cloitre, Feb 04 2002

Keywords

Crossrefs

Cf. A065554, A002379.

A067674 Numbers k such that floor((3/2)^(k+4))/floor((3/2)^k)=(3/2)^4.

Original entry on oeis.org

408, 409, 827, 937, 938, 1021, 1022, 1314, 1315, 1316, 1881, 1882, 1913, 1914, 2039, 2049, 2050, 2051, 2123, 2169, 2254, 2297, 2313, 2369, 2448, 2546, 2547, 2548, 2594, 2681, 2730, 2731, 2736, 2869, 3179, 3180, 3220, 3357, 3792, 3793, 3916, 3917, 3954
Offset: 1

Views

Author

Benoit Cloitre, Feb 04 2002

Keywords

Crossrefs

Cf. A065554, A002379.

A081723 Let b(n)=floor((3/2)^n), c(n)=floor((4/3)^n); sequence gives values of n such that b(n+1)/b(n)=3/2 and c(n+1)/c(n)=4/3.

Original entry on oeis.org

13, 24, 49, 67, 79, 88, 102, 126, 132, 162, 172, 194, 199, 204, 217, 234, 253, 255, 261, 271, 297, 320, 325, 328, 338, 351, 365, 377, 403, 411, 414, 462, 477, 482, 525, 533, 537, 541, 567, 569, 579, 601, 613, 638, 706, 740, 749, 761, 762, 773, 804, 809, 817
Offset: 1

Views

Author

Benoit Cloitre, Apr 06 2003

Keywords

Crossrefs

Cf. A065554, A065564.

Formula

It seems that a(n) is asymptotic to C*n where 12<=C<=13

A081724 Let b(n)=floor((3/2)^n), c(n)=floor((4/3)^n), d(n)=floor((5/4)^n); sequence gives values of n such that b(n+1)/b(n)=3/2, c(n+1)/c(n)=4/3 and d(n+1)/d(n)=5/4.

Original entry on oeis.org

162, 172, 204, 328, 403, 414, 809, 835, 840, 854, 1111, 1117, 1160, 1188, 1192, 1270, 1294, 1311, 1351, 1409, 1469, 1478, 1508, 1605, 1614, 1769, 1842, 1961, 2065, 2226, 2425, 2456, 2460, 2486, 2581, 2597, 2635, 2638, 2642, 2650, 2679, 2720, 2880, 2932
Offset: 1

Views

Author

Benoit Cloitre, Apr 06 2003

Keywords

Crossrefs

Cf. A065554, A065564.

Programs

  • Mathematica
    bcdQ[n_]:=Module[{b=Floor[(3/2)^n],b1=Floor[(3/2)^(n+1)],c=Floor[ (4/3)^n], c1=Floor[(4/3)^(n+1)],d=Floor[(5/4)^n],d1=Floor[(5/4)^(n+1)]}, b1/b==3/2&&c1/c==4/3&&d1/d==5/4]; Select[Range[3000],bcdQ] (* Harvey P. Dale, Jun 08 2013 *)

Formula

It seems that a(n) is asymptotic to C*n where C is around 60.
Showing 1-9 of 9 results.

Started in 1964 by Neil J. A. Sloane | Maintained by The OEIS Foundation Inc.

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