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 1-5 of 5 results.

A096022 Numbers that are congruent to {15, 27, 39, 51} mod 60.

Original entry on oeis.org

15, 27, 39, 51, 75, 87, 99, 111, 135, 147, 159, 171, 195, 207, 219, 231, 255, 267, 279, 291, 315, 327, 339, 351, 375, 387, 399, 411, 435, 447, 459, 471, 495, 507, 519, 531, 555, 567, 579, 591, 615, 627, 639, 651, 675, 687, 699, 711, 735, 747, 759, 771, 795
Offset: 1

Views

Author

Klaus Brockhaus, Jun 15 2004

Keywords

Comments

Numbers n such that (n+j) mod (2+j) = 1 for j from 0 to 2 and (n+3) mod 5 <> 1.
This is one of a family of sequences which are defined (or could be defined) according to the same scheme: Numbers n such that (n+j) mod (2+j) = 1 for j from 0 to k-1 and (n+k) mod (2+k) <> 1. We have A007310 for k = 1, A017629 for k = 2, this one (A096022) for k = 3, A096023 for k = 5, A096024 for k = 6, A096025 for k = 7, A096026 for k = 9, A096027 for k = 11. Remarkably these sequences are empty for k = 4, 8, 10, ... (i.e., if k+1 is a term of A080765).
Numbers n such that n mod 12 = 3 and n mod 60 <> 3.
Subsequence of A017557: 12n+3.

Examples

			51 mod 2 = 52 mod 3 = 53 mod 4 = 1 and 54 mod 5 = 4, hence 51 is in the sequence; 3 mod 2 = 4 mod 3 = 5 mod 4 = 6 mod 5 = 1, hence 3 is not in the sequence.

Links

Crossrefs

Cf. A007310, A017557, A017629, A080765, A096023, A096024, A096025, A096026, A096027.

Programs

  • Magma
    [ n : n in [1..1500] | n mod 60 in [15, 27, 39, 51] ] // Vincenzo Librandi, Mar 24 2011
  • Maple
    A096022:=n->3*(10*n-3-I^(2*n)-(1-I)*I^(-n)-(1+I)*I^n)/2: seq(A096022(n), n=1..80); # Wesley Ivan Hurt, Jun 04 2016
  • Mathematica
    Table[3*(10n-3-I^(2n)-(1-I)*I^(-n)-(1+I)*I^n)/2, {n, 80}] (* Wesley Ivan Hurt, Jun 04 2016 *)
  • PARI
    {k=3;m=800;for(n=1,m,j=0;b=1;while(b&&j

Formula

G.f.: 3*x*(5+4*x+4*x^2+4*x^3+3*x^4) / ( (1+x)*(x^2+1)*(x-1)^2 ). - R. J. Mathar, Oct 08 2011
From Wesley Ivan Hurt, Jun 04 2016: (Start)
a(n) = a(n-1) + a(n-4) - a(n-5) for n>5.
a(n) = 3*(10*n-3-i^(2*n)-(1-i)*i^(-n)-(1+i)*i^n)/2 where i=sqrt(-1). (End)
E.g.f.: 3*(3 + sin(x) - cos(x) + (5*x - 1)*sinh(x) - (2 - 5*x)*cosh(x)). - Ilya Gutkovskiy, Jun 05 2016

Extensions

New definition from Ralf Stephan, Dec 01 2004

A096023 Numbers congruent to {63, 123, 183, 243, 303, 363} mod 420.

Original entry on oeis.org

63, 123, 183, 243, 303, 363, 483, 543, 603, 663, 723, 783, 903, 963, 1023, 1083, 1143, 1203, 1323, 1383, 1443, 1503, 1563, 1623, 1743, 1803, 1863, 1923, 1983, 2043, 2163, 2223, 2283, 2343, 2403, 2463, 2583, 2643, 2703, 2763, 2823, 2883, 3003, 3063, 3123
Offset: 1

Views

Author

Klaus Brockhaus, Jun 15 2004

Keywords

Comments

Numbers n such that (n+j) mod (2+j) = 1 for j from 0 to 4 and (n+5) mod 7 <> 1.
Numbers n such that n mod 60 = 3 and n mod 420 <> 3.

Examples

			63 mod 2 = 64 mod 3 = 65 mod 4 = 66 mod 5 = 67 mod 6 = 1 and 68 mod 7 = 5, hence 63 is in the sequence.

Links

Crossrefs

Cf. A007310, A017629, A096022, A096024, A096025, A096026, A096027.
Cf. A047391 (see MAGMA code). - Bruno Berselli, Mar 25 2011

Programs

  • Magma
    [ n : n in [1..3500] | n mod 420 in [63, 123, 183, 243, 303, 363] ] // Vincenzo Librandi, Mar 24 2011
  • Magma
    /* Alternatively:*/ &cat[ [ 60*n+3, 60*n+63 ]: n in [1..52] | n mod 7 in [1,3,5] ]; // Bruno Berselli, Mar 25 2011
  • Maple
    A096023:=n->420*floor(n/6)+[63, 123, 183, 243, 303, 363][(n mod 6)+1]: seq(A096023(n), n=0..80); # Wesley Ivan Hurt, Jul 22 2016
  • Mathematica
    Select[Range[0, 5*10^3], MemberQ[{63, 123, 183, 243, 303, 363}, Mod[#, 420]] &] (* Wesley Ivan Hurt, Jul 22 2016 *)
  • PARI
    {k=5;m=3150;for(n=1,m,j=0;b=1;while(b&&j

Formula

G.f.: 3*x*(21+20*x+20*x^2+20*x^3+20*x^4+20*x^5+19*x^6) / ( (1+x)*(1+x+x^2)*(x^2-x+1)*(x-1)^2 ). - R. J. Mathar, Oct 08 2011
From Wesley Ivan Hurt, Jul 22 2016: (Start)
a(n) = a(n-1) + a(n-6) - a(n-7) for n>7; a(n) = a(n-6) + 420 for n>6.
a(n) = (210*n - 96 - 30*cos(n*Pi/3) - 30*cos(2*n*Pi/3) - 15*cos(n*Pi) + 30*sqrt(3)*sin(n*Pi/3) + 10*sqrt(3)*sin(2*n*Pi/3))/3.
a(6k) = 420k-57, a(6k-1) = 420k-117, a(6k-2) = 420k-177, a(6k-3) = 420k-237, a(6k-4) = 420k-297, a(6k-5) = 420k-357. (End)

Extensions

New definition from Ralf Stephan, Dec 01 2004

A096024 Numbers n such that (n+j) mod (2+j) = 1 for j from 0 to 5 and (n+6) mod 8 <> 1.

Original entry on oeis.org

423, 1263, 2103, 2943, 3783, 4623, 5463, 6303, 7143, 7983, 8823, 9663, 10503, 11343, 12183, 13023, 13863, 14703, 15543, 16383, 17223, 18063, 18903, 19743, 20583, 21423, 22263, 23103, 23943, 24783, 25623, 26463, 27303, 28143, 28983, 29823
Offset: 1

Views

Author

Klaus Brockhaus, Jun 15 2004

Keywords

Comments

Numbers n such that n mod 840 = 423.

Examples

			423 mod 2 = 424 mod 3 = 425 mod 4 = 426 mod 5 = 427 mod 6 = 428 mod 7 = 1 and 429 mod 8 = 5, hence 423 is in the sequence.

Links

Crossrefs

Cf. A007310, A017629, A096022, A096023, A096025, A096026, A096027.

Programs

  • Magma
    [n: n in [1..30000] | forall{j: j in [0..5] | IsOne((n+j) mod (2+j)) and (n+6) mod 8 ne 1}]; // Bruno Berselli, Apr 11 2013
  • PARI
    {k=6;m=30000;for(n=1,m,j=0;b=1;while(b&&j

Formula

a(n) = 2*a(n-1)-a(n-2). G.f.: 3*x*(139*x+141) / (x-1)^2. - Colin Barker, Apr 11 2013
a(n) = 840*n-417. [Bruno Berselli, Apr 11 2013]

A096025 Numbers n such that (n+j) mod (2+j) = 1 for j from 0 to 6 and (n+7) mod 9 <> 1.

Original entry on oeis.org

843, 1683, 3363, 4203, 5883, 6723, 8403, 9243, 10923, 11763, 13443, 14283, 15963, 16803, 18483, 19323, 21003, 21843, 23523, 24363, 26043, 26883, 28563, 29403, 31083, 31923, 33603, 34443, 36123, 36963, 38643, 39483, 41163, 42003, 43683
Offset: 1

Views

Author

Klaus Brockhaus, Jun 15 2004

Keywords

Comments

Numbers n such that n mod 840 = 3 and n mod 2520 <> 3.

Examples

			843 mod 2 = 844 mod 3 = 845 mod 4 = 846 mod 5 = 847 mod 6 = 848 mod 7 = 849 mod 8 = 1 and 850 mod 9 = 4, hence 843 is in the sequence.

Links

Crossrefs

Cf. A007310, A017629, A096022, A096023, A096024, A096026, A096027.

Programs

  • Magma
    [n: n in [1..44000] | forall{j: j in [0..6] | IsOne((n+j) mod (2+j)) and (n+7) mod 9 ne 1}]; // Bruno Berselli, Apr 11 2013
  • Mathematica
    LinearRecurrence[{1,1,-1},{843,1683,3363},40] (* Harvey P. Dale, Nov 22 2015 *)
  • PARI
    {k=7;m=44000;for(n=1,m,j=0;b=1;while(b&&j

Formula

a(n) = -3*(209+70*(-1)^n-420*n). a(n) = a(n-1)+a(n-2)-a(n-3). G.f.: 3*x*(279*x^2+280*x+281) / ((x-1)^2*(x+1)). - Colin Barker, Apr 11 2013

A096027 Numbers k such that (k+j) mod (2+j) = 1 for j from 0 to 10 and (k+11) mod 13 <> 1.

Original entry on oeis.org

27723, 55443, 83163, 110883, 138603, 166323, 194043, 221763, 249483, 277203, 304923, 332643, 388083, 415803, 443523, 471243, 498963, 526683, 554403, 582123, 609843, 637563, 665283, 693003, 748443, 776163, 803883, 831603, 859323, 887043
Offset: 1

Views

Author

Klaus Brockhaus, Jun 15 2004

Keywords

Comments

Numbers k such that k mod 27720 = 3 and k mod 360360 <> 3.

Examples

			27723 mod 2 = 27724 mod 3 = 27725 mod 4 = 27726 mod 5 = 27727 mod 6 = 27728 mod 7 = 27729 mod 8 = 27730 mod 9 = 27731 mod 10 = 27731 mod 11 = 27731 mod 12 = 1 and 27732 mod 13 = 3, hence 27723 is in the sequence.

Links

Crossrefs

Cf. A007310, A017629, A096022, A096023, A096024, A096025, A096026.

Programs

  • Magma
    [n: n in [1..900000] | forall{j: j in [0..10] | IsOne((n+j) mod (2+j)) and (n+11) mod 13 ne 1}]; // Bruno Berselli, Apr 11 2013
  • PARI
    {k=11;m=900000;for(n=1,m,j=0;b=1;while(b&&j

Formula

G.f.: 3*x*(9239*x^12 +9240*x^11 +9240*x^10 +9240*x^9 +9240*x^8 +9240*x^7 +9240*x^6 +9240*x^5 +9240*x^4 +9240*x^3 +9240*x^2 +9240*x +9241) / ((x -1)^2*(x +1)*(x^2 -x +1)*(x^2 +1)*(x^2 +x +1)*(x^4 -x^2 +1)). - Colin Barker, Apr 11 2013
Showing 1-5 of 5 results.

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

Content is available under The OEIS End-User License Agreement.