Joshua Harrington - cp's OEIS frontend

A364237 a(n) is the number of non-equivalent permutations of {1,2,...,2n-1} such that no subset of consecutive terms from the permutation sums to 0 modulo 2n, where two permutations are equivalent if one can be obtained from the other by multiplying every entry with an integer relatively prime to 2n and/or reversing the permutation.

1, 1, 2, 4, 42, 504, 7492, 172480, 8639632

Offset: 1

Anna Coleman, Joshua Harrington, Maggie X. Lai, Philip Thomas, and Wing Hong Tony Wong, Jul 18 2023

If we consider all permutations of {1,2,...,2n-1} such that no subset of consecutive terms from the permutation sums to 0 modulo 2n, then the number of such permutations is given by the number of constructive orderings mentioned in A141599. For example, given the permutation 14325 that satisfies the given conditions, observe that the partial sums modulo 6, namely 1=1, 1+4=5, 1+4+3=2, 1+4+3+2=4, and 1+4+3+2+5=3, are distinct.

			When n=3, there are four permutations of {1,2,3,4,5} such that no subset of consecutive terms from the permutation sums to 0 modulo 6, namely 14325, 25314, 41352, and 52341. Note that 14325 and 52341 are equivalent by reversing the permutations. Furthermore multiplication by 5 on every entry also yields the same equivalence. Additionally, 25314 and 41352 are analogously equivalent. Hence a(3)=2.
When n=4, 6142573 and 3752416 are equivalent by reversing the permutations but not by multiplying any integer relatively prime to 8, whereas 6142573 and 2346751 are equivalent by multiplication of 3 on every entry.

Sunil K. Chebolu and Papa A. Sissokho, Zero-sum-free tuples and hyperplane arrangements, Integers 22 (2022), #A13.
Sean A. Irvine, Java program (github)

Cf. A141599.

n = 3 #the index for the sequence a(n)
orbits = {} #dictionary of permutations that are consecutive zero-sum-free
seen = [] #list of seen permutations that are consecutive zero-sum-free
a = 0 #the value of a(n)
for labeling in Permutations(range(1,2*n)):
    if labeling not in seen:
        sums = [labeling[0]]
        for i in range(1,2*n-1):
            nextsum = (labeling[i] + sums[i-1]) % (2*n)
            if any([nextsum == 0, nextsum in sums]):
                break
            sums.append(nextsum)
        if len(sums) == (2*n)-1:
            a += 1
            orbits[a] = []
            for m in [x for x in range(1,2*n) if gcd(x,2*n) == 1]:
                equiv = [(m*labeling[i]) % (2*n) for i in range(2*n-1)]
                if equiv not in orbits[a]:
                    orbits[a].append(equiv)
                seen.append(equiv)
                equiv = [equiv[2*n-2-i] for i in range(2*n-1)]
                if equiv not in orbits[a]:
                    orbits[a].append(equiv)
                seen.append(equiv)
print(f"a({n}) = {a}\n")
print("Equivalencies:")
for i in range(1,a+1):
    print(f"{i}.")
    for x in orbits[i]:
        print(x)
    print('\n')

a(8)-a(9) from Sean A. Irvine, Aug 15 2023

A346563 a(n) = n + A007978(n).

Original entry on oeis.org

3, 5, 5, 7, 7, 10, 9, 11, 11, 13, 13, 17, 15, 17, 17, 19, 19, 22, 21, 23, 23, 25, 25, 29, 27, 29, 29, 31, 31, 34, 33, 35, 35, 37, 37, 41, 39, 41, 41, 43, 43, 46, 45, 47, 47, 49, 49, 53, 51, 53, 53, 55, 55, 58, 57, 59, 59, 61, 61, 67, 63, 65, 65, 67, 67

Offset: 1

Michael Gohn, Joshua Harrington, Sophia Lebiere, Hani Samamah, Kyla Shappell, Wing Hong Tony Wong, Jul 23 2021

Beginning at n=3, a(n) represents the maximum length of consecutive numbers that are divisible by the product of their nonzero digits in base n. In particular, if n=10, the sequence of numbers that are divisible by the product of their nonzero digits is given by A055471.

			For n=6, the least non-divisor of 6 is 4, so a(6) = 6+4 = 10. As seen in the Comments section, 55980, 55981, ..., 55989 form a sequence of length 10, where every number is divisible by the product of its nonzero digits in base n=6. Work has been done to show that 10 is the maximum length for such sequences.

Cf. A000027, A007978, A055471.

a[n_] := Module[{k = 1}, While[Divisible[n, k], k++]; n + k]; Array[a, 100] (* Amiram Eldar, Jul 23 2021 *)

a(n) = my(k=2); while(!(n % k), k++); n+k; \\ Michel Marcus, Jul 23 2021

goal = 100
these = []
n = 1
while n <= goal:
    k = 1
    while n % k == 0:
        k = k + 1
    these.append(n + k)
    n += 1
print(these)

a(2k+1) = 2k+3.

a(2k) >= 2k+3.

A346537 Squares that are divisible by the product of their nonzero digits.

Original entry on oeis.org

1, 4, 9, 36, 100, 144, 400, 900, 1024, 1296, 2304, 2500, 2916, 3600, 10000, 11664, 12100, 14400, 22500, 32400, 40000, 41616, 78400, 82944, 90000, 102400, 110224, 121104, 122500, 129600, 152100, 176400, 186624, 200704, 202500, 219024, 230400, 250000, 260100, 291600

Offset: 1

Michael Gohn, Joshua Harrington, Sophia Lebiere, Hani Samamah, Kyla Shappell, Wing Hong Tony Wong, Jul 23 2021

			For the perfect square 1024 = 32^2 the product of its nonzero digits is 8 which divides 1024.

Jean-Marie De Koninck and Florian Luca, Positive integers divisible by the product of their nonzero digits, Port. Math. 64 (2007) 75-85. (This proof for upper bounds contains an error. See the paper below.)
Jean-Marie De Koninck and Florian Luca, Corrigendum to "Positive integers divisible by the product of their nonzero digits", Portugaliae Math. 64 (2007), 1: 75-85, Port. Math. 74 (2017), 169-170.

Intersection of A000290 and A055471.

Select[Range[500]^2, Divisible[#, Times @@ Select[IntegerDigits[#], #1 > 0 &]] &] (* Amiram Eldar, Jul 23 2021 *)

isok(m) = issquare(m) && !(m % vecprod(select(x->(x>0), digits(m))));
lista(nn) = for (m=1, nn, if (isok(m^2), print1(m^2, ", "))); \\ Michel Marcus, Jul 23 2021

from math import prod
def nzpd(n): return prod([int(d) for d in str(n) if d != '0'])
def ok(sqr): return sqr > 0 and sqr%nzpd(sqr) == 0
print(list(filter(ok, (i*i for i in range(541))))) # Michael S. Branicky, Jul 23 2021

A339999 Squares that are divisible by both the sum of their digits and the product of their nonzero digits.

Original entry on oeis.org

1, 4, 9, 36, 100, 144, 400, 900, 1296, 2304, 2916, 3600, 10000, 11664, 12100, 14400, 22500, 32400, 40000, 41616, 82944, 90000, 121104, 122500, 129600, 152100, 176400, 186624, 202500, 219024, 230400, 260100, 291600, 360000, 419904, 435600, 504100

Offset: 1

Michael Gohn, Joshua Harrington, Sophia Lebiere, Hani Samamah, Kyla Shappell, Wing Hong Tony Wong, Jul 23 2021

			For the perfect square 144 = 12^2, the sum of its digits is 9, which divides 144, and the product of its nonzero digits is 16, which also divides 144 so 144 is a term of the sequence.

H. G. Grundman, Sequences of consecutive n-Niven numbers, Fibonacci Quarterly (1994) 32 (2): 174-175.
Jean-Marie De Koninck and Florian Luca, Positive integers divisible by the product of their nonzero digits, Port. Math. 64 (2007) 75-85. (This proof for upper bounds contains an error. See the paper below.)
Jean-Marie De Koninck and Florian Luca, Corrigendum to "Positive integers divisible by the product of their nonzero digits", Portugaliae Math. 64 (2007), 1: 75-85, Port. Math. 74 (2017), 169-170.

Intersection of A000290, A005349 and A055471.

Cf. A118547, A342262, A342650.

Select[Range[720]^2, And @@ Divisible[#, {Plus @@ (d = IntegerDigits[#]), Times @@ Select[d, #1 > 0 &]}] &] (* Amiram Eldar, Jul 23 2021 *)

from math import prod
def sumd(n): return sum(map(int, str(n)))
def nzpd(n): return prod([int(d) for d in str(n) if d != '0'])
def ok(sqr): return sqr > 0 and sqr%sumd(sqr) == 0 and sqr%nzpd(sqr) == 0
print(list(filter(ok, (i*i for i in range(1001)))))
# Michael S. Branicky, Jul 23 2021

A346657 Numbers that are not divisible by the product of their nonzero digits.

Original entry on oeis.org

13, 14, 16, 17, 18, 19, 21, 22, 23, 25, 26, 27, 28, 29, 31, 32, 33, 34, 35, 37, 38, 39, 41, 42, 43, 44, 45, 46, 47, 48, 49, 51, 52, 53, 54, 55, 56, 57, 58, 59, 61, 62, 63, 64, 65, 66, 67, 68, 69, 71, 72, 73, 74, 75, 76, 77, 78, 79, 81, 82, 83, 84, 85, 86, 87

Offset: 1

Michael Gohn, Joshua Harrington, Sophia Lebiere, Hani Samamah, Kyla Shappell, Wing Hong Tony Wong, Jul 27 2021

			The product of the nonzero digits of 42 is 4*2 = 8, which does not divide 42.

Complement of A055471.

Select[Range[100], !Divisible[#, Times @@ Select[IntegerDigits[#], #1 > 0 &]] &] (* Amiram Eldar, Jul 27 2021 *)

isok(k) = k % vecprod(select(x->(x>0), digits(k))); \\ Michel Marcus, Jul 28 2021

from math import prod
def ok(n): return n > 0 and n%prod([int(d) for d in str(n) if d != '0'])
print(list(filter(ok, range(88)))) # Michael S. Branicky, Jul 27 2021

A317244 For n >= 3, smallest prime number N such that for every prime p >= N, every element in Z_p can be expressed as a sum of two n-gonal numbers mod p, without allowing zero as a summand.

Original entry on oeis.org

11, 11, 11, 11, 11, 11, 11, 11, 11, 11, 11, 13, 11, 11, 11, 11, 11, 11, 11, 23, 11, 11, 13, 29, 11, 11, 11, 11, 11, 11, 11, 37, 11, 13, 11, 11, 11, 11, 23, 11, 11, 11, 11, 47, 13, 11, 29, 53, 11, 11, 11, 11, 11, 11, 11, 13, 11, 23, 11, 61, 11, 11, 37, 11, 11, 11, 13, 71, 11, 29, 11, 73, 11, 11, 11, 11, 23, 13, 11, 83, 11, 11, 11, 89, 11, 11, 47, 11, 13, 11, 11, 11, 29, 37, 53, 23, 11

Offset: 3

Theresa Baren, James Hammer, Joshua Harrington, Ziyu Liu, Sean E. Rainville, Melea Roman, Hongkwon V. Yi, Jul 24 2018

nonn

Joshua Harrington, Lenny Jones, and Alicia Lamarche, Representing Integers as the Sum of Two Squares in the Ring Z_n, Journal of Integer Sequences, Vol. 17 (2014), Article 14.7.4.
Bernard M. Moore and H. Joseph Straight, Pythagorean triples in multiplicative groups of prime power order, Pi Mu Epsilon Journal, vol. 14, no. 3, 2015, pp. 191-198.

A317254 a(n) is the smallest integer such that for all s >= a(n), there are at least n-1 different partitions of s into n parts, namely {x_{11},x_{12},...,x_{1n}}, {x_{21},x_{22},...,x_{2n}},..., and {x_{n-1,1},x_{n-1,2},...,x_{n-1,n}}, such that the products of every set are equal.

Original entry on oeis.org

19, 23, 23, 26, 27, 29, 31, 32, 35, 36, 38, 40, 42, 44, 45, 47, 49, 50, 52, 53, 54, 55, 57, 58, 59, 61, 62, 63, 64, 66, 67, 69, 70, 71, 73, 74, 75, 77, 78, 80, 81, 82, 84, 85, 86, 87, 88, 89, 90, 91, 93, 94, 95, 96, 97, 99, 100, 101

Offset: 3

Byungchul Cha, Adam Claman, Joshua Harrington, Ziyu Liu, Barbara Maldonado, Alexander M. Miller, Ann Palma, Wing Hong Tony Wong, Hongkwon V. Yi, Jul 25 2018

			a(3)=19. From s=19 onward, there are at least 2 different partitions of s into 3 parts with equal products:
s=19: {12,4,3} & {9,8,2}:
  12 + 4 + 3 = 9 + 8 + 2 =  19;
  12 * 4 * 3 = 9 * 8 * 2 = 144;
s=20: {15,3,2} & {10,9,1}:
  15 + 3 + 2 = 10 + 9 + 1 = 20;
  15 * 3 * 2 = 10 * 9 * 1 = 90;
s=21: {16,3,2} & {12,8,1}:
  16 + 3 + 2 = 12 + 8 + 1 = 21;
  16 * 3 * 2 = 12 * 8 * 1 = 96.

Byungchul Cha et al., An Investigation on Partitions with Equal Products, arXiv:1811.07451 [math.NT], 2018.
John B. Kelly, Partitions with equal products, Proc. Amer. Math. Soc. 15 (1964), 987-990.

Cf. A060277, A316945, A316946.

Do[maxsumnotwork = 0;  Do[intpart = IntegerPartitions[sum, {n}];   prod = Table[Times @@ intpart[[i]], {i, Length[intpart]}];   prodtally = Tally[prod];   repeatprod = Select[prodtally, #[[2]] >= n - 1 &];   If[repeatprod == {}, maxsumnotwork = sum], {sum, 12, 200}];  Print[n, " ", maxsumnotwork + 1], {n, 3, 60}]

A316945 A triple of positive integers (n,p,k) is admissible if there exist at least two different multisets of k positive integers, {x_1,x_2,...,x_k} and {y_1,y_2,...,y_k}, such that x_1+x_2+...+x_k=y_1+y_2+...+y_k = n and x_1x_2...x_k = y_1y_2...y_k = p. For each n, let A(n)={k:(n,p,k) is admissible for some p}, and let a(n) = |A(n)|.

Original entry on oeis.org

0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 1, 2, 4, 4, 6, 7, 7, 9, 10, 11, 12, 13, 14, 15, 16, 17, 18, 19, 20, 21, 22, 23, 24, 25, 26, 27, 28, 29, 30, 31, 32, 33, 34, 35, 36, 37, 38, 39, 40, 41, 42, 43, 44, 45, 46, 47, 48, 49, 50, 51, 52, 53, 54, 55, 56, 57, 58, 59, 60, 61, 62, 63, 64, 65, 66, 67, 68, 69, 70, 71, 72, 73, 74

Offset: 1

Byungchul Cha, Adam Claman, Joshua Harrington, Ziyu Liu, Barbara Maldonado, Alexander M. Miller, Ann Palma, Wing Hong Tony Wong, Hongkwon V. Yi, Jul 20 2018

John Conway proposed an interesting math puzzle in the 1960s, which is now generally known as "Conway's wizard problem." Here is the problem.
Last night I sat behind two wizards on a bus and overheard the following:
Blue Wizard: I have a positive integer number of children, whose ages are positive integers. The sum of their ages is the number of this bus, while the product is my own age.
Red Wizard: How interesting! Perhaps if you told me your age and the number of your children, I could work out their individual ages?
Blue Wizard: No, you could not.
Red Wizard: Aha! At last, I know how old you are!
Apparently the Red Wizard had been trying to determine the Blue Wizard's age for some time. Now, what was the number of the bus?
This problem posed by Conway looks at different multisets that correspond to the same ordered triple, which motivated the study of this sequence.

			For n = 3, the only partitions of 3 are {3}, {1,2}, and {1,1,1}. Hence, there is no admissible triple (3,p,k).
For n = 4, the only partitions of 4 are {4}, {1,3}, {2,2},{1,1,2}, and {1,1,1,1}. Hence, there is no admissible triple (4,p,k).
For n = 12, the only admissible triple (12,p,k) is when p = 48 and k = 4. This is achieved by the following multisets: {1,3,4,4} and {2,2,2,6}. Thus a(12) = 1.

Jay Bennett, Riddle of the week #34: Two wizards ride a bus, Popular Mechanics. Hearst Communications, Inc., 4 Aug. 2017. 12 Jun. 2018 Accessed.
John B. Kelly, Partitions with equal products, Proc. Amer. Math. Soc. 15 (1964), 987-990.
Tanya Khovanova, Conway's Wizards, arXiv:1210.5460 [math.HO], 2012.

Cf. A060277, A316946.

Table[Count[
  Table[intpart = IntegerPartitions[sum, {n}];
   DuplicateFreeQ[
    Table[Product[intpart[[i]][[j]], {j, n}], {i,
      Length[intpart]}]], {n, sum}], False], {sum,50}]

a(n) = 0 for 1 <= n <= 11, a(12) = 1, a(13) = 2, a(14) = 4, a(15) = 4, a(16) = 6, a(17) = 7, a(18) = 7, and a(n) = n - 10 for n >= 19.

A316946 A triple of positive integers (n,p,k) is admissible if there exist at least two different multisets of k positive integers, {x_1,x_2,...,x_k} and {y_1,y_2,...,y_k}, such that x_1+x_2+...+x_k = y_1+y_2+...+y_k = n and x_1x_2...x_k = y_1y_2...y_k = p. For each n, let A(n) = {p:(n,p,k) is admissible for some k}, and let a(n) = |A(n)|.

Original entry on oeis.org

0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 1, 2, 5, 6, 10, 14, 19, 26, 33, 43, 54, 68, 87, 106, 129, 157, 187, 226, 269, 319, 378, 445, 521, 610, 712, 825, 952, 1099, 1261, 1443, 1655, 1889, 2148, 2440, 2769, 3135, 3542, 4000, 4494, 5049, 5661, 6346, 7099, 7938, 8857, 9862, 10972, 12190, 13532, 15000, 16611, 18366

Offset: 1

Byungchul Cha, Adam Claman, Joshua Harrington, Ziyu Liu, Barbara Maldonado, Alexander M. Miller, Ann Palma, Wing Hong Tony Wong, Hongkwon V. Yi, Jul 20 2018

nonn

John Conway proposed an interesting math puzzle in the 1960s, which is now generally known as the "Conway's wizard problem." Here is the problem.
Last night I sat behind two wizards on a bus and overheard the following:
Blue Wizard: I have a positive integer number of children, whose ages are positive integers. The sum of their ages is the number of this bus, while the product is my own age.
Red Wizard: How interesting! Perhaps if you told me your age and the number of your children, I could work out their individual ages?
Blue Wizard: No, you could not.
Red Wizard: Aha! At last, I know how old you are!
Apparently the Red Wizard had been trying to determine the Blue Wizard's age for some time. Now, what was the number of the bus?
This problem posed by Conway looks at different multisets that correspond to the same ordered triple, which motivated the study of this sequence.

			a(15) = 6 since A(15) = {36,40,48,72,96,144}:
p = 36 [9, 2, 2, 1, 1], [6, 6, 1, 1, 1]
p = 40 [10, 2, 2, 1], [8, 5, 1, 1]
p = 48 [6, 2, 2, 2, 1, 1, 1], [4, 4, 3, 1, 1, 1, 1]
p = 72 [9, 2, 2, 2], [8, 3, 3, 1], [6, 6, 2, 1],
p = 96 [8, 3, 2, 2], [6, 4, 4, 1], [6, 2, 2, 2, 2, 1], [4, 4, 3, 2, 1, 1]
p = 144 [6, 3, 2, 2, 2], [4, 4, 3, 3, 1].

Jay Bennett, Riddle of the week #34: Two wizards ride a bus, Popular Mechanics. Hearst Communications, Inc., 4 Aug. 2017. 12 Jun. 2018 Accessed.
John B. Kelly, Partitions with equal products, Proc. Amer. Math. Soc. 15 (1964), 987-990.

Cf. A060277, A316945.

Do[repeats = {};  Do[intpart = IntegerPartitions[sum, {n}];   prod = Tally[Table[Times @@ intpart[[i]], {i, Length[intpart]}]];   repeatprod = Select[prod, #[[2]] > 1 &];   If[repeatprod != {},    repeats = Join[repeats, Transpose[repeatprod][[1]]]], {n, 3,    sum - 8}]; output = DeleteDuplicates[repeats];  Print[sum, " ", Length[output]], {sum, 12, 100}]

A290322 Sum modulo n of all units u in Z/nZ such that Phi(5,u) is a unit, where Phi is the cyclotomic polynomial.

Original entry on oeis.org

1, 0, 0, 4, 0, 0, 0, 0, 9, 1, 0, 0, 0, 3, 0, 0, 0, 0, 8, 0, 12, 0, 0, 20, 0, 0, 0, 0, 18, 1, 0, 24, 0, 14, 0, 0, 0, 0, 16, 1, 0, 0, 24, 9, 0, 0, 0, 0, 45, 0, 0, 0, 0, 14, 0, 0, 0, 0, 36, 1, 32, 0, 0, 13, 24, 0, 0, 0, 14, 1, 0, 0, 0, 15, 0, 28, 0, 0, 32, 0, 42, 0

Offset: 2

Joshua Harrington, Michael Mueller, Jordan Lenchitz, Tristan Phillips, Madison Wellen, Eric Jovinelly, Jul 27 2017

Conjecture: If n is divisible by 5 then a(n) > 0. - Robert Israel, Jan 23 2024

Robert Israel, Table of n, a(n) for n = 2..10000

Cf. A058026, A289460.

with(numtheory): m:=5: for n from 2 to 100 do S:={}: for a from 1 to n-1 do if gcd(a,n)=1 and gcd(cyclotomic(m,a),n)=1 then S:={op(S),a}: fi: od: print(sum(op(i,S),i=1..nops(S)) mod n): od:

Table[Mod[Total@ Select[Range[n - 1], CoprimeQ[#, n] && CoprimeQ[Cyclotomic[5, #], n] &], n], {n, 83}] (* Michael De Vlieger, Jul 29 2017 *)

a(n) = sum(k=0, n-1, k*((gcd(n, k)==1) && (gcd(n, polcyclo(5, k))==1))) % n; \\ Michel Marcus, Jul 29 2017

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User: Joshua Harrington

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A346563 a(n) = n + A007978(n).

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A346537 Squares that are divisible by the product of their nonzero digits.

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A339999 Squares that are divisible by both the sum of their digits and the product of their nonzero digits.

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A346657 Numbers that are not divisible by the product of their nonzero digits.

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A317244 For n >= 3, smallest prime number N such that for every prime p >= N, every element in Z_p can be expressed as a sum of two n-gonal numbers mod p, without allowing zero as a summand.

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A317254 a(n) is the smallest integer such that for all s >= a(n), there are at least n-1 different partitions of s into n parts, namely {x_{11},x_{12},...,x_{1n}}, {x_{21},x_{22},...,x_{2n}},..., and {x_{n-1,1},x_{n-1,2},...,x_{n-1,n}}, such that the products of every set are equal.

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