Arran Ireland - cp's OEIS frontend

User: Arran Ireland

Arran Ireland has authored 3 sequences.

A309889 a(n) is the maximal number of regions in the Euclidean plane made by superimposing a simple n-gon onto the resulting plane figure of a(n-1).

1, 1, 2, 10, 36

Offset: 1

Arran Ireland, Aug 21 2019

There is initially one region and the 1-gon and 2-gon are ignored, so a(1) and a(2) result in one region. Each line of the n-gon should cross as many lines as possible and avoid intersecting previous intersections.

			For n = 3 the plane is empty, so the trigon can only create 1 extra region. Thus a(3) = 2.
For n = 4 each tetragon edge intersects a maximum of 2 trigon edges, creating a total of 4 new regions. Two trigon edges intersect 2 tetragon edges, adding 4 regions, and the last trigon edge intersects all 4 tetragon edges, adding another 4 regions. Thus a(4) = 2 + 4 + 4 = 10.

Cf. A000124.

A308818 a(n) = a(a(n-1) mod n) + a(a(n-2) mod n) with a(0)=2 and a(1)=3.

Original entry on oeis.org

2, 3, 5, 7, 10, 7, 13, 15, 22, 23, 12, 6, 15, 18, 13, 25, 41, 37, 10, 22, 17, 40, 47, 40, 81, 38, 22, 53, 85, 134, 51, 29, 156, 215, 23, 47, 46, 35, 69, 98, 144, 81, 108, 116, 102, 37, 47, 37, 72, 75, 85, 104, 217, 111, 10, 15, 37, 60, 40, 147, 197, 51, 110

Offset: 0

Arran Ireland, Jun 26 2019

nonn

a(0) and a(1) are chosen to be the smallest starting numbers greater than 1 that are believed to result in a sequence that doesn't cycle.
Empirical observation of the first 10^8 terms suggests that the sequence doesn't enter a cycle.
Conjectures: (i) This sequence doesn't enter a cycle. (ii) There is an integer greater than 1 that can never appear in this sequence.

			a(2) = a(a(2-1) mod 2) + a(a(2-2) mod 2) = a(a(1) mod 2) + a(a(0) mod 2) = a(3 mod 2) + a(2 mod 2) = a(1) + a(0) = 3 + 2 = 5.

Cf. A005185, A046698, A003160.

Cf. A000027 (if a(0)=1 and a(1)=2).

a = [2, 3]
for n in range(2, 10**4 + 3):
    a.append(a[(a[n - 1] % n)] + a[(a[n - 2] % n)])
    print((n - 2), ",", a[n - 2], sep="")

A307560 a(n) = smallest m such that A307629(m) = n.

Original entry on oeis.org

0, 10, 11, 12, 13, 14, 15, 16, 17, 18, 19, 29, 39, 49, 59, 69, 79, 89, 99, 10000000000000000000, 109, 1006, 119, 100000000000000000000000, 129, 100004, 139, 1008, 149, 100000000000000000000000000000, 159, 10000000000000000000000000000000, 169, 1019, 179, 100006

Offset: 0

Arran Ireland, Apr 14 2019

			a(39) = 1039 as (1 + 0) + (1 + 3) + (1 + 9) + (0 + 3) + (0 + 9) + (3 + 9) = 39. The sums in brackets are pairs of digits of 1039. No positive integer less than 1039 has this pairwise digit sum. - _David A. Corneth_, Apr 16 2019

Arran Ireland, Table of n, a(n) for n = 0..304 (a(0) inserted by _Georg Fischer_, May 05 2019)
Arran Ireland, Python program (a(0) inserted by _Georg Fischer_, May 05 2019)
Reddit website, Integer sequence: Pairwise Combinatorial Digit Sum

Cf. A051885, A179239, A307629.

for n in [1..50] do for d in Divisors(n) do if n le 9*d*(d+1) then nd:=d+1; sdLeft:=n div d; S:=[]; for j in [1..nd-1] do if sdLeft gt 9 then S[j]:=9; else S[j]:=sdLeft-1; end if; sdLeft-:=S[j]; end for; S[nd]:=sdLeft; a:=Seqint(S); n, a; break; end if; end for; end for; // Jon E. Schoenfield, Apr 15 2019

fs[nd_, s_] := If[nd*9 < s, 0, Block[{n=10^(nd-1), f=0}, While[n < 10^nd, If[Total@ IntegerDigits@ n == s, f = n; Break[], n++]]; f]]; a[n_] := Block[{s}, Do[s = fs[d+1, n/d]; If[s > 0, Break[]], {d, Divisors[n]}]; s];  Join[{0}, Array[a,50]] (* Giovanni Resta, Apr 15 2019 *)

Let d be the smallest divisor of n for which 9*d*(d+1) >= n; then a(n) is the smallest (d+1)-digit number whose digit sum is n/d. - Jon E. Schoenfield, Apr 15 2019

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User: Arran Ireland

A309889 a(n) is the maximal number of regions in the Euclidean plane made by superimposing a simple n-gon onto the resulting plane figure of a(n-1).

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A308818 a(n) = a(a(n-1) mod n) + a(a(n-2) mod n) with a(0)=2 and a(1)=3.

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A307560 a(n) = smallest m such that A307629(m) = n.

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