A073361 -id:A073361 - cp's OEIS frontend

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.

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

Paul D. Hanna, Oct 14 2005

nonn

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).

			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.

Cf. A073361, A112560, A112562, A112563, A112564, A112565, A112566, A112567, A112568, A112569.

{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}

a(n) = 1 + 4*A073361(n).

A073362 Nested floor product of n and fractions (k+1)/k for all k>0 (mod 5), divided by 5.

Original entry on oeis.org

1, 6, 19, 48, 109, 234, 355, 552, 1009, 1518, 2371, 3804, 4141, 6342, 8803, 12096, 14389, 18438, 24043, 27720, 36397, 45366, 60499, 75876, 80137, 97566, 114931, 140892, 166321, 205926, 218587, 266664, 292429, 342006, 394651, 477336, 481429

Offset: 1

Paul D. Hanna, Jul 29 2002

			a(1)=1 since (1/5)[[[[1(2/1)](3/2)](4/3)](5/4)]=1

Cf. A073359, A073360, A073361.

f[n_] := Block[{k = 1, p = n}, While[q = Floor[p*(k + 1)/k]; q != p, p = q; k++; If[ Mod[k, 5] == 0, k++ ]]; p/5]; Table[ f[n], {n, 1, 37}] (* Robert G. Wilson v *)

a(n)=(1/5)[...[[[[n(2/1)](3/2)](4/3)](5/4)](7/6)]...(k+1)/k]..., k>0 (mod 5), where [x] = floor of x; this infinite nested floor product will eventually level-off at a(n).

More terms from Robert G. Wilson v, Dec 27 2003

A073363 Nested floor product of n and fractions (k+1)/k for all k>0 (mod 6), divided by 6.

Original entry on oeis.org

1, 7, 28, 84, 175, 421, 847, 1288, 1939, 3780, 5656, 9247, 15148, 22099, 25375, 39676, 54607, 75208, 90559, 129360, 166321, 209832, 240268, 320719, 399595, 536956, 672672, 816733, 906444, 1115275, 1321741, 1595832, 1908088, 2323944

Offset: 1

Paul D. Hanna, Jul 29 2002

When is a(n) not divisible by 7?

			a(1)=1 since (1/6)[[[[1(2/1)](3/2)](4/3)](5/4)](6/5)]=1

Cf. A073359, A073360, A073361, A073362.

a(n)=(1/6)[...[[[[[[n(2/1)](3/2)](4/3)](5/4)](6/5)](8/7]...(k+1)/k]..., k>0 (mod 6), where [x] = floor of x; this infinite nested floor product will eventually level-off at a(n).

cp's OEIS Frontend

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.

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A073362 Nested floor product of n and fractions (k+1)/k for all k>0 (mod 5), divided by 5.

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A073363 Nested floor product of n and fractions (k+1)/k for all k>0 (mod 6), divided by 6.

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