Matthew Ryan - cp's OEIS frontend

User: Matthew Ryan

Matthew Ryan has authored 4 sequences.

A330405 a(1) = 0; thereafter a(n) = (a(n-1)^2+1) mod n.

0, 1, 2, 1, 2, 5, 5, 2, 5, 6, 4, 5, 0, 1, 2, 5, 9, 10, 6, 17, 17, 4, 17, 2, 5, 0, 1, 2, 5, 26, 26, 5, 26, 31, 17, 2, 5, 26, 14, 37, 17, 38, 26, 17, 20, 33, 9, 34, 30, 1, 2, 5, 26, 29, 17, 10, 44, 23, 58, 5, 26, 57, 37, 26, 27, 4, 17, 18, 49, 22, 59, 26, 20, 31, 62, 45, 24

Offset: 1

Matthew Ryan, Dec 12 2019

Does the value 0 appear infinitely many times? - Rémy Sigrist, Dec 16 2019
From Michael De Vlieger, Jan 26 2020: (Start)
Observations based on a(n) for 1 <= n <= 300000:
The value 0 appears at indices n = {1, 13, 26, 89, 205, 530, 2041, 276205, ...}.
The value 1 appears at indices n = {2, 4, 14, 27, 50, 90, 99, 175, 188, 206, 531, 2042, 5445, 6845, 7200, 18225, 24389, 25215, 37538, 46875, 48672, 53066, 79527, 93900, 147875, 176267, 186576, 196025, 254457, 276206, ...}. Let M be the indices in a(n) where 1 appears.
The subsequence {1, 2, 5, 26} appears with the first term at index n = 27, and apparently for all subsequent indices listed n M.
The subsequence {1, 2, 5, 26, 677} appears with the first term at index n = 2042, and apparently for all subsequent indices listed n M.
A stable next term in the subsequence S = {1, 2, 5, 26, 677} is not yet apparent, given 300000 terms of a(n). (End)

			a(1) = 0; a(2) = (0^2+1) mod 2 = 1; a(3) = (1^2+1) mod 2 = 2.

Rémy Sigrist, Table of n, a(n) for n = 1..10000

Cf. A003095, A064434.

Nest[Append[#1, Mod[#1[[#2 - 1]]^2 + 1, #2]] & @@ {#, Length@ # + 1} &, {0}, 76] (* Michael De Vlieger, Dec 16 2019 *)

v=0; for (n=1, 77, print1 (v=(v^2+1)%n", ")) \\ Rémy Sigrist, Dec 16 2019

a(1) = 0; a(n) = (a(n-1)^2+1) mod n.

A272041 Smallest integer that can be expressed as the sum of n primes in at least n distinct ways.

Original entry on oeis.org

2, 10, 15, 18, 19, 22, 25, 27, 29, 32, 34, 36, 39, 42, 44, 46, 49, 51, 53, 55, 58, 60, 63, 65, 67, 69, 72, 74, 76, 78, 80, 83, 85, 87, 90, 92, 94, 96, 98, 100, 102, 105, 107, 109, 111, 113, 115, 117, 120, 122, 124, 126, 128, 131, 133, 135, 137, 139, 141, 143

Offset: 1

Matthew Ryan, Apr 21 2016

nonn

Initial terms found by exhaustive search in Excel.

			The sequence is defined here as starting at n=1 to avoid the term a(0). Even though there cannot be exactly zero ways to add zero primes, there is always at least one way to add 0 primes to get any n (i.e., the sum of itself for any nonprime or (1+..+1) for any prime), and zero would be the lowest such number.
Sum of 1 prime in 1 way: 2.
Sum of 2 primes in 2 ways: 3+7 = 5+5 = 10.
Sum of 3 primes in 3 ways: 3+5+7 = 5+5+5 = 2+2+11 = 15.
Sum of 4 primes in 4 ways: 2+2+3+11 = 2+2+7+7 = 3+3+5+7 = 3+5+5+5 = 18.
Sum of 60 primes in 61 ways, e.g.: 57*2 + 3 + 7 + 19 = 37*2 + 23*3 = 143. - _Lars Blomberg_, Jul 18 2017

Lars Blomberg and Giovanni Resta, Table of n, a(n) for n = 1..5000 (first 99 terms from Lars Blomberg)

Cf. A054845, A054859.

a[n_] := Block[{k = 1}, While[Length@ Quiet@ IntegerPartitions[k,{n}, Prime@ Range@ PrimePi@ k, n] < n, k++]; k]; Array[a, 50]

a(36)-a(60) from Lars Blomberg, Jul 18 2017

A257925 a(n) = (n^2 - n + 1)*(n^2 + n - 1).

Original entry on oeis.org

1, 15, 77, 247, 609, 1271, 2365, 4047, 6497, 9919, 14541, 20615, 28417, 38247, 50429, 65311, 83265, 104687, 129997, 159639, 194081, 233815, 279357, 331247, 390049, 456351, 530765, 613927, 706497, 809159, 922621

Offset: 1

Matthew Ryan, Apr 17 2016

Subsequence of a(m,n)=(m^2 + n).(n^2 + m)/(m - n)^3 with m=n-1. Q N4 of the 2012 International Mathematical Olympiad paper poses the problem of proving more than 500 solutions exist below 2012 for the equation: a(m,n).(m - n)^3=(m^2 + n).(n^2 + m). Such solutions a(m,n) were called 'Friendly'. If m=2k-1 and n=k-1, solutions of the form a=4k-3 for some integer k, satisfy this requirement although others do exist for other (m,n) pairs e.g. if (m,n)=(1,2), a(m,n)=15.
If m=n-2, a(n)=(n^2 - 3*n + 4)*(n^2 + n - 2)/8. This is the sequence A176145 [t*(t-3)*(t^2-7*t+14)/8] with t=n+2.
Satisfies a linear recurrence having signature (5, -10, 10, -5, 1). - Harvey P. Dale, Apr 18 2019

			For n=1, a(1) = 1;
For n=2, a(2) = 15;
For n=3, a(3) = 77.

International Mathematical Olympiad 2012, Number Theory Question 4

Cf. A002061, A028387, A176145.

Table[(n^2-n+1)(n^2+n-1),{n,40}] (* Harvey P. Dale, Apr 18 2019 *)

a(n) = (n^2 - n + 1)*(n^2 + n - 1); \\ Michel Marcus, Apr 17 2016

a(n) = (n^2 - n + 1)*(n^2 + n - 1).

a(n) = A002061(n)*A028387(n-1). - Michel Marcus, Apr 17 2016

A106529 Numbers having k prime factors (counted with multiplicity), the largest of which is the k-th prime.

Original entry on oeis.org

2, 6, 9, 20, 30, 45, 50, 56, 75, 84, 125, 126, 140, 176, 189, 196, 210, 264, 294, 315, 350, 396, 416, 440, 441, 490, 525, 594, 616, 624, 660, 686, 735, 875, 891, 924, 936, 968, 990, 1029, 1040, 1088, 1100, 1225, 1386, 1404, 1452, 1456, 1485, 1540, 1560

Offset: 1

Matthew Ryan (mattryan1994(AT)hotmail.com), May 30 2005

nonn

It seems that the ratio between successive terms tends to 1 as n increases, meaning perhaps that most numbers are in this sequence.
The number of terms that have the k-th prime as their largest prime factor is A000984(k), the k-th central binomial coefficient. E.g., 6 and 9 are the A000984(2)=2 terms in {a(n)} that have prime(2)=3 as their largest prime factor.
The sequence contains the positive integers m such that the rank of the partition B(m) = 0. For m >= 2, B(m) is defined as the partition obtained by taking the prime decomposition of m and replacing each prime factor p with its index i (i.e., i-th prime = p); also B(1) = the empty partition. For example, B(350) = B(2*5^2*7) = [1,3,3,4]. B is a bijection between the positive integers and the set of all partitions. The rank of a partition P is the largest part of P minus the number of parts of P. - Emeric Deutsch, May 09 2015
Also Heinz numbers of balanced partitions, counted by A047993. The Heinz number of an integer partition (y_1,...,y_k) is prime(y_1)*...*prime(y_k). - Gus Wiseman, Feb 08 2021

			a(7)=50 because 50=2*5*5 is, for k=3, the product of k primes, the largest of which is the k-th prime, and 50 is the 7th such number.

Alois P. Heinz, Table of n, a(n) for n = 1..10000

Cf. A000984.
A001222 counts prime factors.
A056239 adds up prime indices.
A061395 selects maximum prime index.
A112798 lists the prime indices of each positive integer.
Other balance-related sequences:
- A010054 counts balanced strict partitions.
- A047993 counts balanced partitions.
- A090858 counts partitions of rank 1.
- A098124 counts balanced compositions.
- A340596 counts co-balanced factorizations.
- A340598 counts balanced set partitions.
- A340599 counts alt-balanced factorizations.
- A340600 counts unlabeled balanced multiset partitions.
- A340653 counts balanced factorizations.
Cf. A006141, A064174, A096401, A117409, A168659, A324522, A340654, A340655, A340656, A340657.

with(numtheory): a := proc (n) options operator, arrow: pi(max(factorset(n)))-bigomega(n) end proc: A := {}: for i from 2 to 1600 do if a(i) = 0 then A := `union`(A, {i}) else  end if end do: A; # Emeric Deutsch, May 09 2015

Select[Range@ 1560, PrimePi@ FactorInteger[#][[-1, 1]] == PrimeOmega@ # &] (* Michael De Vlieger, May 09 2015 *)

For all terms, A001222(a(n)) = A061395(a(n)). - Gus Wiseman, Feb 08 2021

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User: Matthew Ryan

A330405 a(1) = 0; thereafter a(n) = (a(n-1)^2+1) mod n.

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A272041 Smallest integer that can be expressed as the sum of n primes in at least n distinct ways.

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A257925 a(n) = (n^2 - n + 1)*(n^2 + n - 1).

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A106529 Numbers having k prime factors (counted with multiplicity), the largest of which is the k-th prime.

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