Alex Fink - cp's OEIS frontend

A125705 Genera g such that every orientation-preserving periodic automorphism of the closed orientable surface of genus g has an invariant circle.

0, 1, 2, 3, 4, 5, 6, 7, 8, 9, 10, 12, 13, 14, 15, 17, 18, 19, 20, 21, 22, 27, 28, 30, 32, 35, 39, 42, 43, 44, 45, 48, 49, 50, 51, 60, 65, 66, 72, 73, 87, 90, 105

Offset: 1

Alex Fink, Jan 31 2007

There are no other g with g<=10000.

H. Geiges & D. Rattaggi, Periodic Automorphisms of Surfaces: Invariant Circles and Maximal Orders, Experimental Mathematics, vol. 9 (2000).

A119346 Sequence of nim-values for the game in which two players alternately cut off one inch or root two inches from a piece of string of length n. Player who runs out of string loses.

Original entry on oeis.org

0, 1, 2, 0, 1, 0, 1, 2, 0, 1, 0, 1, 2, 0, 1, 0, 1, 0, 1, 2, 0, 1, 0, 1, 2, 0, 1, 0, 1, 0, 1, 2, 0, 1, 0, 1, 2, 0, 1, 0, 1, 2, 0, 1, 0, 1, 0, 1, 2, 0, 1, 0, 1, 2, 0, 1, 0, 1, 0, 1, 2, 0, 1, 0, 1, 2, 0, 1, 0, 1, 2, 0, 1, 0, 1, 0, 1, 2, 0, 1, 0, 1, 2, 0, 1, 0, 1, 0, 1, 2, 0, 1, 0, 1, 2, 0, 1, 0, 1

Offset: 0

N. J. A. Sloane, based on email from R. K. Guy and Alex Fink, Aug 05 2006

nonn

From Michel Dekking, Feb 17 2020: (Start)
It follows from Alex Fink's remarks that (a(n)) is obtained from the sequence A276862 (removing the first 2) by mapping every 2 to 0,1 and every 3 to 0,1,2. However, the first 3 entries will be missing.
In the context of my paper "Morphic words, Beatty sequences and integer images of the Fibonacci language", this means that (a(n+3)) is obtained by decorating A006337 by the decoration delta given by delta(1) = 01, delta(2) = 012. This implies that (a(n+3)) is a morphic sequence, i.e., the letter to letter image of the fixed point of a morphism, say sigma. One obtains sigma by the 'natural' algorithm given in the "Morphic words...."-paper. In turns out that the alphabet of sigma can be chosen as {0,1,2}, and that sigma is surprisingly simple:
      sigma(0) = 01, sigma(1) = 012, sigma(2) = 01.
The letter to letter map is given by the identity. In other words, if x = 010120101... is the unique fixed point of sigma, then (a(n+3)) = x. (End)

M. Dekking, Morphic words, Beatty sequences and integer images of the Fibonacci language, Theoretical Computer Science 809, 407-417 (2020).
Alex Fink, Discussion of A119346

Cf. A003151.

To get the answers, add one to sequence A003151 and then start counting from zero, but return to zero whenever you reach a member of A003151 plus one.

Added Feb 13 2020: The simplest formula is a(n) = floor(n mod (1 + sqrt 2)). - Alex Fink (see link).

A109020 (27^n - 33^n + 1)/6.

Original entry on oeis.org

0, 1, 12, 101, 760, 5481, 38852, 273421, 1918320, 13441361, 94128892, 659020341, 4613496680, 32295539641, 226071966132, 1582513328861, 11077621999840, 77543440092321, 542804338926572, 3799631147326981, 26597420355811800, 186181949464251401, 1303273667170466212

Offset: 0

Alex Fink and R. K. Guy, Aug 18 2005

nonn

Number of incongruent integer-edged Heron triangles whose circumdiameter is the product of n distinct primes of shape 4k + 1.

Index entries for linear recurrences with constant coefficients, signature (11, -31, 21).

Table[(2*7^n-3*3^n+1)/6,{n,0,30}] (* or *) LinearRecurrence[{11,-31,21},{0,1,12},30] (* Harvey P. Dale, Aug 10 2012 *)

a(0)=0, a(1)=1, a(2)=12, a(n)=11*a(n-1)-31*a(n-2)+21*a(n-3) -- From Harvey P. Dale, Aug 10 2012

G.f.: (-x^2-x)/(21*x^3-31*x^2+11*x-1) -- From Harvey P. Dale, Aug 10 2012

A109021 (27^n - 63^n + 4)/6.

Original entry on oeis.org

0, 0, 8, 88, 720, 5360, 38488, 272328, 1915040, 13431520, 94099368, 658931768, 4613230960, 32294742480, 226069574648, 1582506154408, 11077600476480, 77543375522240, 542804145216328, 3799630566196248, 26597418612419600, 186181944234074800, 1303273651479936408

Offset: 0

Alex Fink and R. K. Guy, Aug 18 2005

Number of incongruent integer-edged Heron triangles whose circumdiameter is the product of n distinct primes of shape 4k + 1 and which are not right-angled.

Index entries for linear recurrences with constant coefficients, signature (11, -31, 21).

Cf. A016212, A109020, A003462.

Table[(2*7^n-6*3^n+4)/6,{n,0,30}] (* or *) LinearRecurrence[{11,-31,21},{0,0,8},30] (* Harvey P. Dale, Jan 30 2013 *)

a(n) = 8*A016212(n-2).
(0)=0, a(1)=0, a(2)=8, a(n)=11*a(n-1)-31*a(n-2)+21*a(n-3). - Harvey P. Dale, Jan 30 2013
G.f.: -8*x^2 / ( (x-1)*(3*x-1)*(7*x-1) ). - R. J. Mathar, Feb 10 2016

A083417 Primitive recursive function r(z, r(s, r(s, r(s, p_2)))) at (n, 0).

Original entry on oeis.org

0, 1, 2, 1, 0, 5, 2, 3, 3, 2, 2, 3, 4, 1, 8, 5, 4, 2, 2, 3, 3, 2, 2, 7, 2, 9, 5, 2, 12, 9, 7, 5, 4, 2, 2, 3, 4, 1, 8, 5, 4, 2, 2, 3, 3, 2, 2, 15, 8, 5, 1, 43, 20, 13, 10, 3, 14, 7, 3, 11, 8, 3, 8, 5, 4, 2, 2, 3, 4, 1, 24, 13, 5, 4, 2, 11, 4, 5, 5, 4, 1, 13, 6, 5, 5, 4, 2, 7, 5, 3, 1, 3, 3, 2, 2, 31, 14, 10, 3

Offset: 0

Alex Fink, Jun 08 2003

nonn

S. Wolfram, A New Kind of Science, 2001, p. 908.

Charles R Greathouse IV, Table of n, a(n) for n = 0..10000
S. Wolfram, A New Kind of Science, pages 907-908.

A253099 gives locations of zeros.

z := x -> 0: s := x -> (1 + op(1, x)): p := x -> subs(q = x, y -> op(q, y)): c := x -> subs(q = x, y -> eval((op(1, q))([(seq(op(i, q), i = 2..nops(q)))(y)]))): r := x -> subs(q = x, y -> eval(`if`(op(1, y) = 0, (op(1, q))([op(2, y)]), (op(2, q))([r(q)([op(1, y) - 1, op(2, y)]), op(1, y) - 1, op(2, y)])))): seq(r([z, r([s, r([s, r([s, p(2)])])])])([i, 0]), i = 0..109);

(* Mathematica code from New Kind of Science, p. 908, added by N. J. A. Sloane, Feb 17 2015 *)
F = Fold[Fold[
     2^Ceiling[Log[2, Ceiling[(#1 + 2)/(#2 + 2)]]] (#2 + 2) -
       2 - #1 &, #2, Range[#1]] &, 0, Range[#]] &
Table[F[n], {n, 0, 98}]

f(x,y)=(y+2)<Charles R Greathouse IV, Jan 25 2012

A057336 1) Write n in binary; 2) Find run lengths of this expression; 3) Replace these as follows: 1 -> 0, 2 -> 010, 3 -> 01010, 4 -> 0101010...; 4) Remove final 0 and append an initial 1; 5) The term a(n) is the number with the obtained Zeckendorf expression.

Original entry on oeis.org

1, 2, 4, 6, 3, 7, 12, 17, 10, 5, 9, 19, 11, 20, 33, 46, 28, 16, 27, 14, 8, 15, 25, 51, 31, 18, 30, 53, 32, 54, 88, 122, 75, 45, 74, 43, 26, 44, 72, 38, 23, 13, 22, 40, 24, 41, 67, 135, 83, 50, 82, 48, 29, 49, 80, 140, 86, 52, 85, 142, 87, 143, 232, 321, 198, 121, 197, 119

Offset: 1

Alex Fink, Aug 27 2000

nonn

A permutation of the positive integers.

			a(24) = 51 because: 1) 24 in binary is 11000 2) the run lengths are 2, 3 3) 01001010 4) 10100101 5) the Zeckendorf expression of 51 is 10100101 because 51 = 34 + 13 + 3 + 1

Inverse of A057337.

More terms from David W. Wilson, May 12 2001

A057337 1) Write the Zeckendorf expression of n; 2) Remove initial 1 and append a final 0; 3) Replace numbers in this as follows: 0 -> 1, 010 -> 2, 01010 -> 3, 0101010 -> 4...; 4) Find a binary number with run lengths from step 3 (starting with 1); 5) The term a(n) is the decimal equivalent of this binary number.

Original entry on oeis.org

1, 2, 5, 3, 10, 4, 6, 21, 11, 9, 13, 7, 42, 20, 22, 18, 8, 26, 12, 14, 85, 43, 41, 45, 23, 37, 19, 17, 53, 27, 25, 29, 15, 170, 84, 86, 82, 40, 90, 44, 46, 74, 36, 38, 34, 16, 106, 52, 54, 50, 24, 58, 28, 30, 341, 171, 169, 173, 87, 165, 83, 81, 181, 91, 89, 93, 47, 149, 75

Offset: 1

Alex Fink, Aug 27 2000

A permutation of the positive integers

			a(18) = 26 because: 1) the Zeckendorf expression for 18 is 101000 (13 + 5) 2) this becomes 010000 3) 010 -> 2, 0 -> 1, 0 -> 1, 0 -> 1 4) 2 ones, 1 zero, 1 one, 1 zero: 11010 5) the binary number 11010 in decimal is 26.

Inverse of A057336.

More terms from David W. Wilson, May 12 2001

A067462 a(n) = (1! + 2! + ... + (n-1)!) mod n.

Original entry on oeis.org

0, 1, 0, 1, 3, 3, 5, 1, 0, 3, 0, 9, 9, 5, 3, 9, 12, 9, 8, 13, 12, 11, 20, 9, 13, 9, 9, 5, 16, 3, 1, 25, 0, 29, 33, 9, 4, 27, 9, 33, 3, 33, 15, 33, 18, 43, 17, 9, 47, 13, 12, 9, 12, 9, 33, 33, 27, 45, 27, 33, 21, 1, 54, 25, 48, 33, 64, 29, 66, 33, 67, 9, 54, 41, 63, 65, 33, 9, 19, 73, 63

Offset: 1

Alex Fink, Feb 24 2002

A.H.M. Smeets, Table of n, a(n) for n = 1..20000 (first 1000 terms from T. D. Noe)

Cf. A049782.

Mod[ #[[1]], #[[2]]]&/@Transpose[{Rest[FoldList[Plus, 0, Factorial[Range[90]]]], Range[90]}]
Table[Mod[Total[Range[n-1]!],n],{n,90}] (* Harvey P. Dale, Jan 23 2023 *)

a(n) = my(s=0, p=1); for (i=1, n-1, f = Mod(i, n); p*=f; s += p); lift(s); \\ Michel Marcus, Aug 19 2019

n = 0
while n < 200:
    n,f,fs,i = n+1,1,0,1
    while i < n:
        f,i = (f*i)%n,i+1
        fs = (fs+f)%n
    print(n,fs) # A.H.M. Smeets, Aug 19 2019

A061342 Period of the stationary component of the pattern which a row of n cells becomes in Conway's Game of Life.

Original entry on oeis.org

1, 1, 2, 1, 2, 1, 1, 1, 2, 15, 2, 1, 2, 1, 1, 2, 1, 1, 1, 1, 2, 1, 1, 1, 2, 1, 1, 1, 2, 2, 2, 1, 2, 1, 1, 1, 2, 1, 1, 1, 3, 2, 1, 1, 2, 1, 2, 2, 1, 2, 2, 1, 2, 2, 2, 2, 2, 1, 1, 2, 2, 2, 2, 1, 2, 2, 2, 2, 2, 2, 2, 2, 2, 2, 2, 2, 2, 2, 2, 2, 2, 2, 1, 2, 2, 2, 2

Offset: 1

Alex Fink, Jun 06 2001

nonn

For n = 56, the pattern produces gliders and thus does not oscillate in the truest sense of the word. For n = 57, no gliders are produced and a(57) = 2. There exists an N such that for all n >= N, the row of n alive cells produces gliders and thus does not oscillate. - Nathaniel Johnston, Jun 08 2009

			a(10) = 15 because a line of 10 cells becomes the pentadecathlon.
a(56) = 2 because a line of 56 cells becomes 4 gliders (which we ignore) and a remaining pattern of period 2. - _Eric M. Schmidt_, May 24 2014

Eric M. Schmidt, Table of n, a(n) for n = 1..1000
LifeWiki, One-cell-thick pattern

Cf. A152389.

More general definition and sequence extended by Eric M. Schmidt, May 24 2014

A073833 Numerators of b(n) where b(1) = 1, b(i) = b(i-1) + 1/b(i-1).

Original entry on oeis.org

1, 2, 5, 29, 941, 969581, 1014556267661, 1099331737522548368039021, 1280590510388959061548230114212510564911731118541, 1726999038066943724857508638586386504281539279376091034086485112150121338989977841573308941492781

Offset: 1

Alex Fink, Aug 12 2002

a(n) is also the numerator of the fractional chromatic number of the Mycielski graph M_n. - Eric W. Weisstein, Mar 05 2011
It appears that lim_{n->infinity} (1/n)*exp(2*(b(n)^2-2n)) = c1 = 0.57...... - Benoit Cloitre, Oct 16 2002
c1 = 0.574810274671785...; see A232975. - Jon E. Schoenfield, Nov 30 2013
b(n)^2 = t/2 + u + (u - 1/2)/t + (-u^2 + 2*u - 11/12)/t^2 + (4*u^3/3 - 5*u^2 + 17*u/3 - 65/36)/t^3 + ... where t = 4*n, u = (log n)/2 + c, and c = -0.2768576248625765389364372...; see A233770. - Jon E. Schoenfield, Dec 15 2013
a(n) is also the numerator of b(n) where b(0) = b(1) = 1 and b(n) = (b(n-1)^2 + b(n-2)^2) / b(n-2) for n > 1 where the denominator of b(n) is partial products of A073834. - Michael Somos, Aug 16 2014
a(n) is also the numerator of b(n) where b(1) = 1 and b(2) = 2 and b(n) = b(n-2) + b(n-1) - (b(n-2)^2/b(n-1)) for n > 2. This has a geometric interpretation: One can prove, given two half lines starting at the center of a series of concentric circles, and a set of triangles each defined by the intersections of the two half lines with any given circle and one of the intersections of the rays with the next circle, that if the circles have radii specified by b(n), the triangle areas are all equal. - Sjoerd C. de Vries, Aug 13 2015

			1, 2, 5/2, 29/10, 941/290, 969581/272890, 1014556267661/264588959090, 1099331737522548368039021/268440386798659418988490, ...

H. L. Montgomery, Ten Lectures on the Interface Between Analytic Number Theory and Harmonic Analysis, Amer. Math. Soc., 1996, p. 187.
D. J. Newman, A Problem Seminar, Springer; see Problem #60.
J. H. Silverman, The arithmetic of dynamical systems, Springer, 2007, see p. 113 Table 3.1.

Elijah M. Kin, Table of n, a(n) for n = 1..13
Sjoerd C. de Vries, Mathematica file illustrating geometric application of the sequence
Steven Finch, Popa's "Recurrent Sequences" and Reciprocity, arXiv:2412.11806 [math.CA], 2024. See p. 16.
Clark Kimberling, Polynomials associated with reciprocation, JIS 12 (2009) 09.3.4.
Eric Weisstein's World of Mathematics, Fractional Chromatic Number
Eric Weisstein's World of Mathematics, Mycielski Graph

See A073834 for denominators. See A232975 for c1; see A233770 for c.

f[n_]:=n+1/n;Prepend[Numerator[NestList[f,2,9]],1] (* Vladimir Joseph Stephan Orlovsky, Nov 19 2010 *)
Numerator[NestList[# + 1/# &, 1, 10]] (* Eric W. Weisstein, Mar 05 2001 *)
a[ n_] := If[ n<1, 0, If[ n<3, n, With[{x = a[n-2]^2, y = a[n-1]}, y y + x y - x x]]]; (* Michael Somos, Aug 16 2014 *)
Numerator@RecurrenceTable[{b[n] == b[-2 + n] - b[-2 + n]^2/b[-1 + n] + b[-1 + n], b[1] == 1,
   b[2] == 2}, b, {n, 1, 10}] (* Sjoerd C. de Vries, Aug 13 2015 *)

{a(n) = if( n<1, 0, if( n<3, n, my(x = a(n-2)^2, y = a(n-1)); y^2 + x*y -x^2))}; /* Michael Somos, Mar 05 2012 */

a(n) = a(n-1)^2 + A073834(n-1)^2; A073834(n) = a(n-1) * A073834(n-1). - Franklin T. Adams-Watters, Aug 04 2008

0 = a(n)^2*(a(n+1) - a(n)^2) - (a(n+2) - a(n+1)^2) for all n > 0. - Michael Somos, Aug 16 2014

cp's OEIS Frontend

User: Alex Fink

A125705 Genera g such that every orientation-preserving periodic automorphism of the closed orientable surface of genus g has an invariant circle.

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A119346 Sequence of nim-values for the game in which two players alternately cut off one inch or root two inches from a piece of string of length n. Player who runs out of string loses.

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Formula

A109020 (2*7^n - 3*3^n + 1)/6.

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Mathematica

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A109021 (2*7^n - 6*3^n + 4)/6.

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Mathematica

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A083417 Primitive recursive function r(z, r(s, r(s, r(s, p_2)))) at (n, 0).

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Maple

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A057336 1) Write n in binary; 2) Find run lengths of this expression; 3) Replace these as follows: 1 -> 0, 2 -> 010, 3 -> 01010, 4 -> 0101010...; 4) Remove final 0 and append an initial 1; 5) The term a(n) is the number with the obtained Zeckendorf expression.

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Extensions

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A067462 a(n) = (1! + 2! + ... + (n-1)!) mod n.

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Mathematica

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Python

A061342 Period of the stationary component of the pattern which a row of n cells becomes in Conway's Game of Life.

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Extensions

A109020 (27^n - 33^n + 1)/6.

A109021 (27^n - 63^n + 4)/6.