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 1-4 of 4 results.

A112564 Square array, read by ascending antidiagonals, where each row is a generalized Flavius Josephus sieve (A000960).

Original entry on oeis.org

1, 1, 1, 1, 2, 1, 1, 3, 4, 1, 1, 4, 7, 6, 1, 1, 5, 13, 13, 10, 1, 1, 6, 21, 28, 19, 12, 1, 1, 7, 31, 61, 61, 27, 18, 1, 1, 8, 43, 96, 125, 88, 39, 22, 1, 1, 9, 57, 169, 241, 261, 133, 49, 30, 1, 1, 10, 73, 232, 505, 546, 421, 208, 63, 34, 1, 1, 11, 91, 361, 785, 1051, 1171, 605, 313
Offset: 0

Views

Author

Paul D. Hanna, Oct 14 2005

Keywords

Examples

			Table begins:
  1,  1,  1,   1,    1,    1,     1,     1,      1,     1, ...
  1,  2,  4,   6,   10,   12,    18,    22,     30,    34, ...
  1,  3,  7,  13,   19,   27,    39,    49,     63,    79, ...
  1,  4, 13,  28,   61,   88,   133,   208,    313,   364, ...
  1,  5, 21,  61,  125,  261,   421,   605,   1101,  1681, ...
  1,  6, 31,  96,  241,  546,  1171,  1776,   2761,  5046, ...
  1,  7, 43, 169,  505, 1051,  2527,  5083,  7729,  11635, ...
  1,  8, 57, 232,  785, 1800,  5041, 11096, 22737,  34504, ...
  1,  9, 73, 361, 1153, 3961,  8281, 20161, 43633,  95049, ...
  1, 10, 91, 460, 1981, 5950, 13951, 38080, 91081, 186130, ...
  ...

Links

Crossrefs

Cf. A002491 (row 1), A000960 (row 2), A112560 (row 3), A112561 (row 4), A112562 (row 5), A112563 (row 6), A112565 (main diagonal), A112568 (2nd diagonal), A112569 (antidiagonal sums).

Programs

  • PARI
    {T(n,k)=local(A=k,B=0,C=0);if(n==0||k==0,1, until(A==B,C=C+1;if(C%n==0,C=C+1);B=A;A=floor(A*(C+1)/C));1+A)}

A112560 Sieve performed by successive iterations of steps where step m is: keep m terms, remove the next 2 and repeat; as m = 1,2,3,.. the remaining terms form this sequence.

Original entry on oeis.org

1, 4, 13, 28, 61, 88, 133, 208, 313, 364, 541, 724, 853, 1048, 1261, 1564, 1993, 2104, 2581, 3028, 3553, 3904, 4621, 5368, 5893, 6544, 7141, 8104, 9373, 9904, 11113, 12088, 13333, 14428, 15433, 17368, 19021, 20188, 21733, 23944, 25261, 27304
Offset: 0

Views

Author

Paul D. Hanna, Oct 14 2005

Keywords

Comments

Formula: a(n) = 1 + [...[[[[n*2/1]3/2]5/4]6/5]...(k+1)/k]... where denominators k of the fractions used in the product vary over all natural numbers not congruent to 0 (mod 3); thus the product will eventually reach a maximum value of a(n).

Examples

			Sieve starts with the natural numbers:
1,2,3,4,5,6,7,8,9,10,11,12,13,14,15...
Step 1: keep 1 term, remove the next 2, repeat; giving
1,4,7,10,13,16,19,22,25,28,31,34,37,40,...
Step 2: keep 2 terms, remove the next 2, repeat; giving
1,4,13,16,25,28,37,40,49,52,61,64,73,76,...
Step 3: keep 3 terms, remove the next 2, repeat; giving
1,4,13,28,37,40,61,64,73,88,97,100,121,...
Continuing in this way, we obtain this sequence.
Using the floor function product formula:
a(2)=1+[[[[[[[(2)*2/1]3/2]5/4]6/5]8/7]9/8]11/10]12/11]=13.

Crossrefs

Cf. A073360, A112561, A112562, A112563, A112564, A112565, A112566, A112567, A112568, A112569.

Programs

  • Mathematica
    Table[1 + First@FixedPoint[{Floor[#[[1]]*(#[[2]] + 1)/#[[2]]],
      If[Mod[#[[2]] + 1, 3] == 0, #[[2]] + 2, #[[2]] + 1]} &, {n, 1},
SameTest -> (#1[[1]] == #2[[1]] &)], {n, 0, 30}] (* Birkas Gyorgy, Mar 07 2011 *)
  • PARI
    {a(n)=local(A=n,B=0,k=0); until(A==B,k=k+1;if(k%3==0,k=k+1);B=A;A=floor(A*(k+1)/k));1+A}

Formula

a(n) = 1 + 3*A073360(n).

A112562 Sieve performed by successive iterations of steps where step m is: keep m terms, remove the next 4 and repeat; as m = 1,2,3,.. the remaining terms form this sequence.

Original entry on oeis.org

1, 6, 31, 96, 241, 546, 1171, 1776, 2761, 5046, 7591, 11856, 19021, 20706, 31711, 44016, 60481, 71946, 92191, 120216, 138601, 181986, 226831, 302496, 379381, 400686, 487831, 574656, 704461, 831606, 1029631, 1092936, 1333321, 1462146
Offset: 0

Views

Author

Paul D. Hanna, Oct 14 2005

Keywords

Comments

Formula: a(n) = 1 + [..[[[[n*2/1]3/2]4/3]5/4]...(k+1)/k]...] where denominators k of the fractions used in the product vary over all natural numbers not congruent to 0 (mod 5); thus the product will eventually reach a maximum value of a(n).

Examples

			Sieve starts with the natural numbers:
1,2,3,4,5,6,7,8,9,10,11,12,13,14,15,...
Step 1: keep 1 term, remove the next 4, repeat; giving
1,6,11,16,21,26,31,36,41,46,51,56,61,66,...
Step 2: keep 2 terms, remove the next 4, repeat; giving
1,6,31,36,61,66,91,96,121,126,151,156,...
Step 3: keep 3 terms, remove the next 4, repeat; giving
1,6,31,96,121,126,211,216,241,306,331,...
Continuing in this way, we obtain this sequence.
Using the floor function product formula:
a(2) = 1 + [..[(2)*2/1]*3/2]*4/3]*6/5]*7/6]*8/7]*10/9]*
11/10]*12/11]*14/13]*15/14]*16/15]*18/17]*19/18]*20/19] = 21.

Crossrefs

Cf. A073362, A112560, A112561, A112563, A112564, A112565, A112566, A112567, A112568, A112569.

Programs

  • PARI
    {a(n)=local(A=n,B=0,k=0); until(A==B,k=k+1;if(k%5==0,k=k+1);B=A;A=floor(A*(k+1)/k));1+A}

Formula

a(n) = 1 + 5*A073362(n).

A112563 Sieve performed by successive iterations of steps where step m is: keep m terms, remove the next 5 and repeat; as m = 1,2,3,.. the remaining terms form this sequence.

Original entry on oeis.org

1, 7, 43, 169, 505, 1051, 2527, 5083, 7729, 11635, 22681, 33937, 55483, 90889, 132595, 152251, 238057, 327643, 451249, 543355, 776161, 997927, 1258993, 1441609, 1924315, 2397571, 3221737, 4036033, 4900399, 5438665, 6691651
Offset: 0

Views

Author

Paul D. Hanna, Oct 14 2005

Keywords

Comments

Formula: a(n) = 1 + [..[[[[n*2/1]3/2]4/3]5/4]...(k+1)/k]...] where denominators k of the fractions used in the product vary over all natural numbers not congruent to 0 (mod 6); thus the product will eventually reach a maximum value of a(n).

Examples

			Sieve starts with the natural numbers:
1,2,3,4,5,6,7,8,9,10,11,12,13,14,15,...
Step 1: keep 1 term, remove the next 5, repeat; giving
1,7,13,19,25,31,37,43,49,55,61,67,73,79,...
Step 2: keep 2 terms, remove the next 5, repeat; giving
1,7,43,49,85,91,127,133,169,175,211,217,...
Step 3: keep 3 terms, remove the next 5, repeat; giving
1,7,43,169,175,211,337,343,379,505,511,547,...
Continuing in this way, we obtain this sequence.
Using the floor function product formula:
a(2) = 1+[..[(2)*2/1]*3/2]*4/3]*5/4]*7/6]*8/7]*9/8]*10/9]*
12/11]*13/12]*14/13]*15/14]*17/16]*18/17]*19/18]*20/19]*
22/21]*23/22]*24/23]*25/24]*27/26]*28/27]*29/28]*30/29] =43.

Crossrefs

Cf. A073363, A112560, A112561, A112562, A112564, A112565, A112566, A112567, A112568, A112569.

Programs

  • PARI
    {a(n)=local(A=n,B=0,k=0); until(A==B,k=k+1;if(k%6==0,k=k+1);B=A;A=floor(A*(k+1)/k));1+A}

Formula

a(n) = 1 + 6*A073363(n).
Showing 1-4 of 4 results.

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