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.

A381226 a(n) is the number of distinct positive integers that can be obtained by starting with n!, and optionally applying the operations square root, floor, and ceiling, in any order.

Original entry on oeis.org

1, 2, 4, 6, 7, 8, 8, 9, 10, 10, 10, 11, 12, 12, 12, 12, 12, 13, 13, 13, 14, 14, 14, 14, 14, 14, 14, 14, 15, 15, 15, 15, 15, 15, 16, 16, 16, 16, 16, 16, 16, 16, 16, 16, 16, 16, 16, 17, 17, 17, 17, 17, 17, 17, 17, 17, 17, 18, 18, 18, 18, 18, 18, 18, 18, 18, 18, 18
Offset: 1

Views

Author

N. J. A. Sloane, Feb 24 2025

Keywords

Comments

This sequence, A381227, and A381228 arose in connection with the problem of showing that every positive integer can be represented using a single 4. Hans Havermann has pointed out that A139004 is related to this question and has many references. - N. J. A. Sloane, Feb 25 2025

Examples

			For n = 8, 8! = 40320; sqrt(40320) = 200.798..., floor and ceiling give 200 and 201. Sqrt(200) = 14.142..., and floor and ceiling give 14 and 15. From 14 we get 3 and 4; from 3 we get 1 and 2. 15 and 4 give nothing more. In all, we get a(8) = 9 different numbers: 40320, 200, 201, 14, 15, 3, 4, 1, 2.
Note that at each step, we must consider three "parents": if x was a term at the previous step, we get floor(sqrt(x)), sqrt(x), and ceiling(sqrt(x)) as potential parents at the next step.
		

Crossrefs

Motivated by trying to understand A000319.

Programs

  • PARI
    f(n) = my(t); if(n<4, [1..n], t=sqrtint(n); if(issquare(n), concat(f(t), n), Set(concat([f(t), f(t+1), [n]]))));
    a(n) = #f(n!); \\ Jinyuan Wang, Feb 25 2025

Extensions

More terms from Jinyuan Wang, Feb 25 2025

A381227 Irregular triangle read by rows: row n lists the A381226(n) numbers constructed in the definition of A381226, in increasing order.

Original entry on oeis.org

1, 1, 2, 1, 2, 3, 6, 1, 2, 3, 4, 5, 24, 1, 2, 3, 4, 10, 11, 120, 1, 2, 3, 5, 6, 26, 27, 720, 1, 2, 3, 8, 9, 70, 71, 5040, 1, 2, 3, 4, 14, 15, 200, 201, 40320, 1, 2, 3, 4, 5, 24, 25, 602, 603, 362880, 1, 2, 3, 6, 7, 43, 44, 1904, 1905, 3628800, 1, 2, 3, 8, 9, 79, 80, 6317, 6318, 39916800
Offset: 1

Views

Author

N. J. A. Sloane, Feb 25 2025

Keywords

Examples

			Triangle begins:
   1;
   1, 2;
   1, 2, 3, 6;
   1, 2, 3, 4,  5, 24;
   1, 2, 3, 4, 10, 11, 120;
   1, 2, 3, 5,  6, 26,  27,  720;
   1, 2, 3, 8,  9, 70,  71, 5040;
   1, 2, 3, 4, 14, 15, 200,  201, 40320;
   1, 2, 3, 4,  5, 24,  25,  602,   603,  362880;
   1, 2, 3, 6,  7, 43,  44, 1904,  1905, 3628800;
   ...
		

Crossrefs

A381228 Smallest k such that n appears in row k of the triangle in A381227, or -1 if n never appears in A381227.

Original entry on oeis.org

1, 2, 3, 4, 4, 3, 10, 7, 7, 5, 5, 12, 12, 8, 8, 13, 13, 21, 36, 22, 22, 37, 14, 4, 9, 6, 6, 39, 39, 24, 24, 15, 15, 69, 41, 41, 25, 25, 42, 42, 72, 72, 10, 10, 43, 16, 16, 74, 128, 44, 44, 75, 130, 76, 76, 27, 27, 77, 77, 134, 134, 78, 46, 46, 17, 17, 79, 79, 28
Offset: 1

Views

Author

N. J. A. Sloane, Feb 25 2025

Keywords

Comments

Conjecture: Every positive integer appears in A381227.

Examples

			14 first appears in row 8 of A381227, so a(14) = 8.
		

Crossrefs

Extensions

More terms from Jinyuan Wang, Feb 25 2025

A139003 Number of operations A000142 (i.e., x!) or A000196 (i.e., floor(sqrt(x))) needed to get n, starting with 3.

Original entry on oeis.org

1, 2, 0, 20, 4, 1, 14, 17, 31, 6, 26, 41, 35, 20, 31, 31, 19, 28, 27, 38, 21, 33, 21, 21, 26, 3, 51, 38, 28, 26, 20, 35, 36, 36, 13, 23, 27, 62, 45, 50, 45, 40, 9, 15, 31, 8, 32, 52, 36, 13, 68, 69, 57, 33, 54, 36, 46, 34, 49, 63, 56, 68, 14, 63, 23, 33, 36, 47, 43, 16, 38, 66, 38
Offset: 1

Views

Author

M. F. Hasler, Apr 09 2008

Keywords

Comments

Knuth conjectured that any number can be obtained in that way, starting from 4.
This seems also to be true using 3 as the starting value. Since 3 is the minimal possible choice, this variant could be considered to be more natural.
To ensure the sequence is well-defined, define a(n)=-1 if it is not possible to get n in the given way.
See A139004 for references and links.
In fact a single 2 is enough to get any positive integer, if Knuth's conjecture that one 4 is enough is true. From 2, (((-tan(2.))!)!)! = 5.592..., then floor, factorial gets 120, then sqrt, sqrt gives 3.162..., and floor gives 3, or negate, floor, negate gives 4. - N. J. A. Sloane, Feb 26 2025

Examples

			Representing the operation x -> floor(sqrt(x)) by "s" and x -> x! by "f",
we have:
a(1) = 1 since 1 = s3 is clearly the shortest way to obtain 1 from 3.
a(2) = 2 since 2 = sf3 is clearly the shortest way to obtain 2 from 3.
a(3) = 0 since no operation is required to get 3 which is there at the beginning.
a(5) = 4 since 5 = ssff3 is the shortest way to obtain 5 from 3.
a(6) = 1 since 6 = f3 is certainly the shortest way to get 6 from 3.
a(4) = 20 = 7+9+a(5) since 4 = ssssssfsssssssffssff3 = floor(35!^(1/2^6)), 35 = floor((5!)!^(1/2^7)).
		

Crossrefs

Cf. A139004.

Programs

  • PARI
    A139003( n, S=Set(3), LIM=10^5 )={ for( i=0,LIM, setsearch( S, n) & return(i); S=setunion( S, setunion( Set( vector( #S, j, sqrtint(eval(S[j])))), Set( vector( #S, j, if( LIM > j=eval(S[j]), j!))))))}

Formula

a(3) = 0; a(n) = min { a(k)+1 ; n^2 <= k < (n+1)^2 or k! = n }

Extensions

a(9)-a(11) from Max Alekseyev, Nov 03 2008
Corrected formula, added terms from a(12) onward. - Jon E. Schoenfield, Nov 17 2008, Nov 19 2008
Comments and example edited by Jon E. Schoenfield, Sep 15 2013
Showing 1-4 of 4 results.