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 1-6 of 6 results.

A112565 Main diagonal of square table A112564 of generalized Flavius Josephus sieves.

Original entry on oeis.org

1, 2, 7, 28, 125, 546, 2527, 11096, 43633, 186130, 809831, 3423432, 14022373, 58574894, 250708291, 1038612976, 4353755777, 18333324162, 74663736859, 311293807040, 1286700247561, 4917768055222, 20458840039199, 83985256000824
Offset: 0

Views

Author

Paul D. Hanna, Oct 14 2005

Keywords

Comments

Appears to be a self-convolution of an integer sequence (A112567).

Links

Crossrefs

Cf. A112564, A112566, A112567, A112568.

Programs

  • PARI
    {a(n)=local(A=n,B=0,C=0);if(n==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)}
  • PARI
    A002491(n) = local(a, b); a = n; b = n - 1; while (b > 1, a = b*ceil(a/b); b--); a;
T(n, k) = local(A = k, C = 1, q, d, x); if (n*k == 0, return(1)); if (n == 1, return(A002491(k + 1))); while (q = A\C, d = A%C; x = d\q + 1; A += x*(n - 1)*(A\C); C += x*n); 1 + A; \\ David Wasserman, Jun 25 2009

Formula

a(n) = 1 + n*A112566(n) for n >= 0.

Extensions

More terms from David Wasserman, Jun 25 2009

A112567 Self-convolution equals A112565.

Original entry on oeis.org

1, 1, 3, 11, 47, 193, 869, 3583, 12399, 51287, 217551, 867797, 3274727, 13099145, 55568089, 216651145, 878354151, 3609915051, 13596245715, 55617915013, 220140920491, 676614665133, 2955537029341, 11835711831601, 48642769948139
Offset: 0

Views

Author

Paul D. Hanna, Oct 14 2005

Keywords

Comments

A112565 is the main diagonal of square table A112564 of generalized Flavius Josephus sieves. It is not yet certain that this sequence consists entirely of integers.
The first non-integral term is a(84) = 102317017376376435483208037493436891797489491884241/2. [From David Wasserman, Jul 02 2009]

Links

Crossrefs

Cf. A112564, A112565, A112566.

Extensions

More terms from David Wasserman, Jul 02 2009

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).

A112561 Sieve performed by successive iterations of steps where step m is: keep m terms, remove the next 3 and repeat; as m = 1,2,3,.. the remaining terms form this sequence.

Original entry on oeis.org

1, 5, 21, 61, 125, 261, 421, 605, 1101, 1681, 2525, 2781, 4201, 5645, 6741, 9541, 11765, 13701, 17641, 21305, 27981, 29401, 37265, 43521, 51541, 59945, 65781, 78121, 89345, 99981, 121381, 124445, 144321, 173041, 189965, 212361, 229381
Offset: 0

Views

Author

Paul D. Hanna, Oct 14 2005

Keywords

Comments

Formula: a(n) = 1 + [..[[[[n*2/1]3/2]4/3]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 4); 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 3, repeat; giving
1,5,9,13,17,21,25,29,33,37,41,45,49,53,57,61,65,...
Step 2: keep 2 terms, remove the next 3, repeat; giving
1,5,21,25,41,45,61,65,81,85,101,105,121,125,141,...
Step 3: keep 3 terms, remove the next 3, repeat; giving
1,5,21,61,65,81,121,125,141,181,185,201,241,245,261,...
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. A073361, A112560, A112562, 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%4==0,k=k+1);B=A;A=floor(A*(k+1)/k));1+A}

Formula

a(n) = 1 + 4*A073361(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 1-6 of 6 results.

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