Matthijs Coster - cp's OEIS frontend

A252496 Numbers k such that arctan(1/k) = arctan(1/x) - arctan(1/y) for some integers 0

3, 7, 8, 13, 17, 18, 21, 30, 31, 32, 41, 43, 46, 47, 50, 55, 57, 68, 72, 73, 75, 76, 83, 91, 93, 98, 99, 100, 105, 111, 112, 117, 119, 122, 123, 128, 129, 132, 133, 142, 144, 155, 157, 162, 172, 173, 174, 177, 182, 183, 185, 187, 189, 192, 193, 200, 203, 211

Offset: 1

Matthijs Coster, Dec 17 2014

nonn

arctan(1/a(n)) = arctan(1/x) - arctan(1/y) for some integers x and y where 0 < x < y < a(n). We use the formula tan(a+b) = (tan a + tan b)/(1 - tan a.tan b) which implies that 1/a(n) = (1/x - 1/y)/(1+1/(xy)) or a(n) = (xy+1)/(y-x) = x + (x^2+1)/(y-x). So we look for divisors of x^2+1.

			8 is in the sequence since arctan(1/8) = arctan(1/3) - arctan(1/5)

Robert Israel, Table of n, a(n) for n = 1..10000

N:= 1000: # to get all terms <= N
A:= {}:
for x from 1 to N/2 do
   ds:= select(d -> (d <= x and d >= (x^2+1)/(N-x)), numtheory:-divisors(x^2+1));
   A:= A union map(d -> x + (x^2+1)/d, ds);
od:
A;
# if using Maple 11 or earlier, uncomment the next line
# sort(convert(A,list));
# Robert Israel, Dec 19 2014

S = []
bound = 50
for b in range(1, bound-1):
    bb = b*b+1
    for d in divisors(bb):
        if (2*b < d) & (d-b < 2*bound):
            c = d-b
            a = (b*c-1)/(b+c)
            S.append((c, b, a))
S.sort()
print(S)

A248979 Numbers n such that 11 is not a divisor of A002805(11*n).

Original entry on oeis.org

0, 33, 77, 110, 847, 880, 924, 957, 1210, 1243, 1287, 1320, 9328, 9372, 9416, 9702, 9768, 10538, 10582, 10626, 14201, 14223, 102608, 102641, 102685, 102718, 103136, 103158, 116413, 116457, 116501, 156255, 156277, 1128688, 1128721, 1128765, 1128798, 1129073

Offset: 1

Matthijs Coster, Oct 18 2014

nonn

For other primes after a few exceptions it seems that all denominators of harmonic numbers are divisible by that prime. For 11 there are many more exceptions. Maybe infinitely many?

			33 is in the sequence since H(33) = p/q and 11 is not a divisor of q. Here H(n) = Sum_{i=1..n} 1/i.
Of course if H(33) has no denominator with a factor 11 the same is true for 34, 35, ..., 43.

Cf. A002805.

lista(nn) = {forstep (n=0, nn, 11, if (denominator(sum(k=2,n,1/k)) % 11, print1(n, ", ")););} \\ Michel Marcus, Oct 19 2014

n = 10000
sum11 = 0
resu = [0]
for i in range(11, n, 11):
    D = (1 / i).partial_fraction_decomposition()[1]
    sum11 += sum(v for v in D if 11.divides(v.denominator()))
    if sum11 >= 1:
        sum11 -= 1
    if sum11 == 0:
        resu.append(i)
resu

A208893 First number in smaller of two distinct sets of n consecutive numbers with identical corresponding prime signatures.

Original entry on oeis.org

2, 6, 11, 68, 130, 88, 658, 1339, 1603, 1603, 2485, 47261, 7058, 7058, 69858, 69858, 62279, 62279

Offset: 1

Matthijs Coster, Mar 03 2012

A208892 is the main sequence for this entry.

A208892 gives both sets, A208894 the first number in the larger set.

a(11)-a(18) from T. D. Noe, Mar 13 2012

A208894 First number in larger of two distinct sets of n consecutive numbers with identical corresponding prime signatures.

Original entry on oeis.org

3, 10, 17, 76, 138, 232, 1090, 2491, 3763, 3763, 15589, 90461, 133058, 133058, 566658, 566658, 1601927, 1601927

Offset: 1

Matthijs Coster, Mar 03 2012

A208892 is the main sequence for this entry.

A208892 gives both sets, A208893 the first number in the smaller set.

a(11)-a(18) from T. D. Noe, Mar 13 2012

A208892 Two distinct sets of n consecutive numbers with identical corresponding prime signatures.

Original entry on oeis.org

2, 3, 6, 7, 10, 11, 11, 12, 13, 17, 18, 19, 68, 69, 70, 71, 76, 77, 78, 79, 130, 131, 132, 133, 134, 138, 139, 140, 141, 142, 88, 89, 90, 91, 92, 93, 232, 233, 234, 235, 236, 237, 658, 659, 660, 661, 662, 663, 664, 1090, 1091, 1092, 1093, 1094, 1095, 1096

Offset: 1

Matthijs Coster, Mar 03 2012

The sets are chosen to minimize the larger of the two start values, so for example, (6,7) and (10,11) would be chosen over (5,6) and (13,14) since 10 is less than 13.
The sequence can be read as follows:
2,                           3,
6,7,                         10,11,
11,12,13,                    17,18,19,
68,69,70,71,                 76,77,78,79,
130,131,132,133,134,         138,139,140,141,142
88,89,90,91,92,93,           232,233,234,235,236,237
658,659,660,661,662,663,664, 1090,1091,1092,1093,1094,1095,1096
etc

			At n=4, the consecutive numbers 68, 69, 70, and 71 have the same prime signature as 76, 77, 78, and 79.  Both have the form p^2*q, p*q, p*q*r, and p, respectively.

Matthijs Coster, improved Sage program for computing A208892

Cf. A208893, A208894.

Cf. A046523 (smallest number with same prime signature as n).

A160559 Minimal covering numbers.

Original entry on oeis.org

12, 80, 90, 210, 280, 378, 448, 1386, 1650, 2200, 2464, 5346, 9750, 11264, 11466, 13000, 14994, 18954, 20384, 23166, 26656, 27846, 30294, 31122, 33150, 33858, 36608, 37050, 37674, 44200, 44850, 49400, 49504, 53248, 53900, 55328, 59800, 63750, 66976, 71250, 72930, 85000, 95000, 95744, 97240, 100100, 107008, 107406, 112112, 117306, 120042, 131274, 142002, 145314, 192500, 208544, 223074, 242250, 252448, 272272, 293250, 311168, 318500, 323000, 369750, 385434, 391000, 395250, 423500, 431250, 450846, 452608, 485982, 493000, 505856, 519498, 527000, 568458, 575000, 612352, 617526, 654500, 660114, 685216, 731500, 735150, 747954

Offset: 1

Matthijs Coster, May 19 2009

nonn

A collection of congruences with distinct moduli, each greater than 1, such that each integer satisfies at least one of the congruences, is said to be a covering system. Let N be the LCM of these moduli. We consider minimal N's, i.e., N is the LCM of some moduli, but none of the divisors has this property.

Hough and Nielsen (2019) proved that each term must be divisible by 2 or 3. - Max Alekseyev, Nov 19 2022

			80 is in the set since 1 mod 2; 2 mod 4; 4 mod 8; 8 mod 16; 4 mod 5; 8 mod 10; 16 mod 20, 32 mod 40; 0 mod 80 is a covering system with LCM 80. None of the divisors has that property.
36 is not minimal since 12 is a divisor and 12 is the LCM of a covering system.

Jai Setty, Table of n, a(n) for n = 1..87
Max Alekseyev, Covering systems corresponding to the terms a(1)-a(41).
Donald Jason Gibson, A covering system with least modulus 25, Math. Comp. 78 (2009), 1127-1146.
Robert D. Hough and Pace P. Nielsen, Covering systems with restricted divisibility, Duke Math. J. 168:17 (2019), 3261-3295. arXiv:1703.02133 [math.NT].
Pace P. Nielsen, A covering system whose smallest modulus is 40, Journal of Number Theory 129 (2009), 640-666.
Pace P. Nielsen, A movie explaining covering systems.
Jai Setty, Covering systems corresponding to the terms a(1)-a(87).

Cf. A160560.

Corrected by Eric Rowland, Oct 24 2018
a(17)-a(23) from Max Alekseyev, Nov 19 2022
a(24)-a(41) from Max Alekseyev, Mar 21 2023
Missing terms a(8) and a(15) inserted and their multiples removed by Jai Setty, May 29 2024

A158918 n, ps(n), ps^2(n), ..., ps^9(n) forms an increasing ps-sequence of length 10.

Original entry on oeis.org

12900, 737100, 772176, 832050, 844032, 844992, 864976, 872208, 879984, 887088, 926400, 939900, 954828, 960372, 962724, 964800, 967500, 969444, 972804, 973296, 975828, 976144, 980000, 982044, 984064

Offset: 1

Matthijs Coster, Mar 30 2009

nonn

ps (read phi-sigma) is the consecution of the sigma (sum of divisors) and Euler totient function.
The problem raises from the aliquot sequences. In the case of aliquot sequences we calculate the next element by calculating the sum of the divisors, and afterwards subtract the original number n.
Instead of subtract n we apply the totient function in order to get ps-sequences. Fast decreases can appear if the sum of divisors consist of many different small primes.
In fact ps-sequences end very fast in cycles, often cycles of small length.
I didn't find an increasing ps-sequence of length > 12.

			12900 = 2^2*3*5^2*43, s(12900) = 7*4*31*44, ps(12900) = 6*30*8*10 = 2^6*3^2*5^2 = 14400;
s(14400) = 127*13*31, ps(14400) = 126*12*30 = 2^4*3^4*5*7 = 45360; ...

#(not fast!)
def phi(L):
    m = 1
    for l in L:
        m = m * (l[0]-1)
        for i in (1..l[1]-1):
            m = m * l[0]
    return m
def sigma(L):
    m = 1
    for l in L:
        q = 1
        for i in (0..l[1]):
            q = q * l[0]
            m = m * (q-1) / (l[0]-1)
    return m
cc = 8
START = 2
END = 10000000
for f0 in (START..END):
    c = 0
    f = f0
    Lf = list(factor(f))
    s = sigma(Lf)
    Ls = list(factor(s))
    f1 = phi(Ls)
    while f < f1:
        c = c + 1
        f = f1
        Lf = list(factor(f))
        s = sigma(Lf)
        Ls = list(factor(s))
        f1 = phi(Ls)
    if c > cc:
        print(c, ":", f0)

A160560 Almost covering numbers.

Original entry on oeis.org

2, 4, 6, 8, 16, 18, 30, 32, 40, 54, 64, 126, 128, 150, 162, 200, 224, 256, 486, 512, 750, 882, 1000, 1024, 1458, 1568, 1638, 1782, 1950, 2048, 2600, 2912, 3750, 4096, 4374, 5000, 5632, 6174, 6318, 8192, 10976

Offset: 1

Matthijs Coster, May 19 2009

A collection of congruences with distinct moduli, each greater than 1, such that each integer satisfies at least one of the congruences, is said to be a covering system. Let N be the lcm of these moduli. If modulo N we cannot cover all residues, but can cover all but one residue, then N is an almost covering number.
We denote by T(N) the number of divisors of N and by R(N) the smallest number of uncovered numbers modulo N. Suppose N = p^k * M, where gcd(p,M)=1, p is prime, R(M) = 1, and T(M) = p-1, then R(N) = 1 as well.
Some further terms that are obtained this way include: 12750, 13122, 16384, 16758, 17000, 18750, 19602, 21294, 25000, 25350, 26624, 29792, 32768, 33800, 37856, 39366, 43218, 61952, 65536, 74358, 76832, 82134, 93750. - Max Alekseyev, Feb 12 2025

			30 is an almost covering number since 1 mod 2; 2 mod 3; 4 mod 5; 4 mod 6; 8 mod 10; 12 mod 15 and 6 mod 30 covers all numbers modulo 30 except 30-folds.

Donald Jason Gibson, A covering system with least modulus 25, Math. Comp. 78, (2009), 1127-1146.
Nathan McNew and Jai Setty, On the densities of covering numbers and abundant numbers, arXiv:2507.23041 [math.NT], 2025. See p. 7.
Pace P. Nielsen, A covering system whose smallest modulus is 40, Journal of Number Theory 129, (2009), 640-666.
Pace P. Nielsen, A movie explaining covering systems.

Cf. A160559.

Corrected by Eric Rowland, Oct 24 2018

Edited, missing term a(27)=1638 inserted, and a(38)-a(41) added by Max Alekseyev, Feb 08 2025

A141621 Numbers that begin a run of 5 consecutive integers of the form p^2*q where p and q are distinct primes.

Original entry on oeis.org

10093613546512321, 14414905793929921, 266667848769941521, 562672865058083521, 1579571757660876721, 1841337567664174321, 2737837351207392721, 4456162869973433521, 4683238426747860721, 4993613853242910721, 5037980611623036721, 5174116847290255921

Offset: 1

Matthijs Coster, Aug 23 2008

Old name was "The first number of a series of 5 consecutive numbers with the same signature, i.e., all numbers have the format p^2*q, where p and q are primes. Therefore the number of divisors is the same (6)." [That name could have been confusing in that not every sequence of 5 consecutive integers having the same prime signature has the prime signature p^2*q; e.g., 204323 is the first of 5 consecutive numbers of the form p^2*q*r. - Jon E. Schoenfield, Jun 05 2018]
Each of the five numbers in each such sequence has 6 divisors.
It is easy to prove that any number in this sequence must be congruent to 1 modulo 240. The program below calculates only an element of the sequence. Since the reference A119479 it is the smallest one. If we assume that the first element has the format 7^2*n49, the second number has the format 2*p^2, the third element has the format 3^2*n9 and the fifth element has the format 5^2*n25, then p must be modulo 22050 one out of 1181, 3719, 4219, 9119, 12931, 17831, 18331 or 20869.
It is unclear if these numbers are the smallest ones. - Matthijs Coster, Aug 28 2008 [The terms listed in the Data section are, in fact, the smallest numbers matching the definition. - Jon E. Schoenfield, Jun 05 2018]
The first quintuple not of the aforementioned form starts with 5344962129269790721 = 23^2*prime. - Ivan Neretin, Feb 08 2016
Among the first 200 terms, the frequency with which the squared prime factor p is {7, 17, 23, 31, 41, 47, 73, 127, 193, 1039, 1399} is {171, 10, 6, 4, 3, 1, 1, 1, 1, 1, 1}, respectively. - Jon E. Schoenfield, Jun 09 2018

			a(1) = 10093613546512321, because
10093613546512321 = 7^2 * 205992113194129,
10093613546512322 =   2 * 71040881^2,
10093613546512323 = 3^2 * 1121512616279147,
10093613546512324 = 2^2 * 2523403386628081, and
10093613546512325 = 5^2 * 403744541860493,
so each of the five consecutive integers is of the form p^2*q, and no smaller run of five consecutive integers has this property. [corrected by _Jon E. Schoenfield_, Jun 05 2018]

Ray Chandler, Table of n, a(n) for n = 1..2000 (first 25 terms from Ivan Neretin, through 200 terms from Jon E. Schoenfield)
Richard I. Hess, Problem 1231, Crux Mathematicorum, Vol. 13, No. 4, p. 118, 1987. (takes a long time to download)
Richard I. Hess, Puzzles from Around the World, p. 63, H17.
Carlos Rivera, Problem 20.- Divisors (II) K consecutive numbers with the same number of divisors, The Prime Puzzles & Problems Connection.
StackExchange, Sequence of numbers with prime factorization pq^2
wu :: forums, Same Number of Divisors, Oct 05 2007.
Index to sequences related to prime signature

Cf. A119479, A006558, A005237, A005238, A006601.

Cf. A054753, A074172, A178032, A308683.

## Warning: this program appears to be incorrect [Joerg Arndt, Feb 29 2016]
for m in range(5000):
    p = 22050*m+17831
    if is_prime(p):
        n = 2*p^2-2
        n4 = n/4+1
        if is_prime(n4):
            n49 = floor((n+1)/49)
            if (49*n49 == n+1) and is_prime(n49):
                n9 = floor((n+3)/9)
                if (9*n9 == n+3) and is_prime(n9):
                    n25 = floor((n+5)/25)
                    if (25*n25 == n+5) and is_prime(n25):
                        print(n+1, n49, p, n9, n4, n25)

Two more terms Matthijs Coster, Aug 28 2008
Missing terms added and extended by Ivan Neretin, Feb 08 2016
New name from Jon E. Schoenfield, Jun 05 2018

A106007 Coster numbers: similar to Friedman numbers (A036057), but here the operations + - * / are allowed. All digits of a number have to be used exactly twice.

Original entry on oeis.org

0, 1, 13, 14, 15, 18, 19, 24, 25, 26, 28, 29, 35, 36, 37, 39, 45, 46, 48, 49, 52, 55, 59, 64, 65, 66, 69, 73, 75, 78, 79, 81, 82, 85, 89, 91, 95, 99, 125, 126, 127, 128, 129, 134, 135, 136, 137, 138, 143, 144, 145, 146, 147, 148, 149, 152, 153, 154, 155, 156, 157, 158

Offset: 0

Matthijs Coster, Dec 31 2006, Feb 02 2007

These numbers were introduced in 2006 in the yearly number puzzle in the Dutch Journal "Pythagoras" (see http://www.pythagoras.nu). It is known that there are infinitely many Coster numbers (result of David Kloet, Albert Hendriks and Arjen Stolk, unpublished).

			36 is in the sequence since 3*6+3*6 = 36

Matthijs Coster, Feb 02 2007, Table of n, a(n) for n = 0..89
Matthijs Coster, www.wiskundemeisjes.nl/20061030/coster-getallen/ (in Dutch)
Matthijs Coster, www.matcos.nl/sequences
Pythagoras, www.pythagoras.nu (in Dutch)

Cf. A036057.

cp's OEIS Frontend

User: Matthijs Coster

A252496 Numbers k such that arctan(1/k) = arctan(1/x) - arctan(1/y) for some integers 0

Views

Author

Keywords

Comments

Examples

Links

Programs

Maple

SageMath

A248979 Numbers n such that 11 is not a divisor of A002805(11*n).

Views

Author

Keywords

Comments

Examples

Crossrefs

Programs

PARI

Sage

A208893 First number in smaller of two distinct sets of n consecutive numbers with identical corresponding prime signatures.

Views

Author

Keywords

Comments

Crossrefs

Extensions

A208894 First number in larger of two distinct sets of n consecutive numbers with identical corresponding prime signatures.

Views

Author

Keywords

Comments

Crossrefs

Extensions

A208892 Two distinct sets of n consecutive numbers with identical corresponding prime signatures.

Views

Author

Keywords

Comments

Examples

Links

Crossrefs

A160559 Minimal covering numbers.

Views

Author

Keywords

Comments

Examples

Links

Crossrefs

Extensions

A158918 n, ps(n), ps^2(n), ..., ps^9(n) forms an increasing ps-sequence of length 10.

Views

Author

Keywords

Comments

Examples

Programs

Sage

A160560 Almost covering numbers.

Views

Author

Keywords

Comments

Examples

Links

Crossrefs

Extensions

A141621 Numbers that begin a run of 5 consecutive integers of the form p^2*q where p and q are distinct primes.

Views

Author

Keywords

Comments

Examples

Links

Crossrefs

Programs

Sage