Hein van Winkel - cp's OEIS frontend

User: Hein van Winkel

Hein van Winkel has authored 4 sequences.

A338940 a(n) is the number of solutions to the Diophantine equation p * x * (x + n) = y^2 with p = a*b a perfect square and a+b = n.

0, 0, 0, 0, 1, 1, 0, 1, 0, 3, 0, 1, 1, 1, 4, 2, 1, 2, 0, 3, 0, 1, 0, 4, 4, 3, 0, 1, 1, 12, 0, 3, 0, 3, 4, 2, 1, 1, 4, 12, 1, 4, 0, 1, 7, 1, 0, 7, 0, 10, 4, 3, 1, 3, 4, 4, 0, 3, 0, 12, 1, 1, 0, 4, 16, 4, 0, 3, 0, 12, 0, 7, 1, 3, 14, 1, 0, 12, 0, 21, 0, 3, 0, 4, 16, 1, 4, 4, 1, 21, 4, 1, 0, 1, 4, 10, 1, 2, 0, 10

Offset: 1

Hein van Winkel, Nov 16 2020

nonn

Related to Heron triangles with a partition point on a side of length n where the incircle is tangent. Some partitions correspond to a finite number of Heron triangles. The numbers a(n) in this sequence are the numbers of Heron triangles that match these 'finite' partitions.

			n = 25 = 5 + 20 = 9 + 16 gives 100 * x * (x + 25) = y^2 or 144 * x * (x + 25) = y^2 or 144 * x * (x + 25) = y^2. And the solutions are (x,y) = (144,1560) or (20,300) or (144,1872) or (20,360).

Chai Wah Wu, Table of n, a(n) for n = 1..10000
Hein van Winkel, Heron triangles, See the sections SQO and SQE.

Cf. A115878, A338939.

Let n = 2^t * p_1^a_1 * p_2^a_2 *...* p_r^a_r * q_1^b_1 * q_2^b_2 *...* q_s^b_s with t>=0, a_i>=0 for i=1..r, where p_i == 1 (mod 4) for i=1..r and q_j = -1 (mod 4) for j=1..s.
Further let A = (2a_1 + 1) * (2a_2 + 1) *...* (2a_r + 1) and B = A * (2b_1 + 1) * (2b_2 + 1) *...* (2b_s + 1).
Then a(n) = (A-1) * (B-1) / 4 for t = 0 and a(n) = A * (B-1) / 2 for t = 1 AND t = 2 and a(n) = (2*t - 3) * A * (B+1) / 2 for t > 2.
a(n) = A338939(n) * A115878(n).

A338939 a(n) is the number of partitions n = a + b such that a*b is a perfect square.

Original entry on oeis.org

0, 1, 0, 1, 1, 1, 0, 1, 0, 3, 0, 1, 1, 1, 1, 1, 1, 1, 0, 3, 0, 1, 0, 1, 2, 3, 0, 1, 1, 3, 0, 1, 0, 3, 1, 1, 1, 1, 1, 3, 1, 1, 0, 1, 1, 1, 0, 1, 0, 5, 1, 3, 1, 1, 1, 1, 0, 3, 0, 3, 1, 1, 0, 1, 4, 1, 0, 3, 0, 3, 0, 1, 1, 3, 2, 1, 0, 3, 0, 3, 0, 3, 0, 1, 4, 1, 1, 1, 1, 3, 1, 1, 0, 1, 1, 1, 1, 1, 0, 5

Offset: 1

Hein van Winkel, Nov 16 2020

nonn

Number of ways to regroup the unit squares of a rectangle with semiperimeter n into a square.

a(n) > 0 for n in A337140.

			n = 10 = 1 + 9 = 2 + 8 = 5 + 5 with 1*9 = 3^2 and 2*8 = 4^2 and 5*5 = 5^2. Then a(10) = 3. Also 10 = 2^1 * 5^1. So t=1, a_1=1 and a(n) = A = 2*1+1 = 3.

David A. Corneth, Table of n, a(n) for n = 1..10000

Cf. A115878, A337140, A338940.

A338939:=n->map[fold=(`+`,0)](i->if(issqr(i*(n-i)),1,0),[$1..1/2*n]); seq(A338939(n),n=1..100); # Felix Huber, Oct 02 2024

a[n_] := Count[IntegerPartitions[n, {2}], ?(IntegerQ @ Sqrt[Times @@ #] &)]; Array[a, 100] (* _Amiram Eldar, Nov 22 2020 *)

a(n)=my(c=0);for(i=1,n-1,if((2*i<=n)&&issquare(i*(n-i)),c++));c
for(n=1,100,print1(a(n),", ")) \\ Derek Orr, Nov 18 2020

a(n) = sum(i=1, n-1, (2*i<=n) && issquare(i*(n-i))); \\ Michel Marcus, Dec 21 2020

first(n) = {my(res = vector(n)); for(i = 1, n, d = divisors(i^2); for(i = (#d + 1)\2, #d, c = d[i] + d[#d + 1 - i]; if(c <= n, res[c]++ , next(2) ) ) ); res } \\ David A. Corneth, Dec 21 2020

from sympy.abc import x,y
from sympy.solvers.diophantine.diophantine import diop_quadratic
def A338939(n): return len(diop_quadratic(x*(n-x)-y**2))>>2 # Chai Wah Wu, Aug 21 2024

Let n = 2^t * p_1^a_1 * p_2^a_2 * ... * p_r^a_r * q_1^b_1 * q_2^b_2 *...* q_s^b_s with t >= 0, a_i >= 0 for i = 1..r, where p_i == 1 (mod 4) and q_j == -1 (mod 4) for j = 1..s. Further, let A = (2a_1 + 1)*(2a_2 + 1)*...*(2a_r + 1). Then a(n) = (A-1)/2 for odd n and a(n) = A for even n.

a(n) = A338940(n) / A115878(n).

A337140 Numbers m = a + b with a and b positive integers whose product a*b = k^2 is a square.

Original entry on oeis.org

2, 4, 5, 6, 8, 10, 12, 13, 14, 15, 16, 17, 18, 20, 22, 24, 25, 26, 28, 29, 30, 32, 34, 35, 36, 37, 38, 39, 40, 41, 42, 44, 45, 46, 48, 50, 51, 52, 53, 54, 55, 56, 58, 60, 61, 62, 64, 65, 66, 68, 70, 72, 73, 74, 75, 76, 78, 80, 82, 84, 85, 86, 87, 88, 89, 90, 91

Offset: 1

Hein van Winkel, Aug 18 2020

Related to Heron triangles with a partition point on one of the sides. Calculations become quite different when the partition a + b = m gives the perfect square k^2 = a*b.
These numbers coincide with the numbers > 1 not in A004614.
Let m = 2^t * p_1^a_1 * p_2^a_2 * ... * p_r^a_r * q_1^b_1 * q_2^b_2 * ... * q_s^b_s with t >= 0, a_i >= 0 for i=1..r, where p_i == 1 (mod 4) for i=1..r and q_j == -1 (mod 4) for j=1..s.
Even numbers (A005843) belong to this sequence: m = 2*k and p = k^2.
Numbers divisible by a prime q congruent to 1 (mod 4) (cf. A004613) belong to this sequence: m = q * m_1 = (u^2 + v^2) * m_1 and p = (u*v*q)^2.
The other numbers are divisible only by primes congruent to 3 (mod 4) (cf. A004614).
If a term m is not in the union of A005843 and A004613, then m = q_1^b_1 * q_2^b_2 * ... * q_s^b_s is a term of A018825 (numbers not the sum of two nonzero squares) = q_i * m_1 = q_i *(u^2 + v) and p = q_i^2 * u^2 * v for all u^2 < m_1 and v nonsquare. And so m is not a term: A contradiction.

			Even numbers m = 2*k give a = b = k. For example, 94 = 47+47 and k^2 = 47^2.
Numbers which are divisible by a prime q congruent to 1 (mod 4) give m = q*m' = (u^2 + v^2)*m' and p = (u*v*m')^2. For example, 87 = 3*29 = 3*(25 + 4) = (5*4*3)^2 = 60^2.

Cf. A004614.

Cf. A004613, A005843, A018825.

Select[Range[100], Length @ Select[Times @@@ IntegerPartitions[#, {2}], IntegerQ @ Sqrt[#1] &] > 0 &] (* Amiram Eldar, Aug 26 2020 *)

upto(n) = { my(res = List(vector(n\2, i, 2*i))); forstep(i = 1, n, 2, c = core(i); for(k = 1, sqrtint((n-i)\c), listput(res, i + c*k^2); ) ); listsort(res, 1); res } \\ David A. Corneth, Aug 26 2020

is(n) = for(i = 1, n\2 + 1, if(issquare(i * (n-i)), return(n>1))); 0 \\ David A. Corneth, Aug 26 2020

from itertools import count, islice
from sympy import primefactors
def A337140_gen(startvalue=2): # generator of terms >= startvalue
    return filter(lambda n: n&1^1 or not all(p&2 for p in primefactors(n>>(~n & n-1).bit_length())), count(max(startvalue,2)))
A337140_list = list(islice(A337140_gen(),30)) # Chai Wah Wu, Aug 21 2024

A180622 Number of distinct sums i+j, absolute differences |i-j|, products ij and quotients i/j and j/i with 1 <= i, j <= n.

Original entry on oeis.org

3, 6, 12, 19, 30, 38, 56, 68, 86, 100, 129, 143, 178, 197, 222, 246, 293, 313, 367, 392, 428, 460, 525, 551, 606, 643, 694, 730, 813, 841, 931, 977, 1034, 1084, 1151, 1188, 1296, 1351, 1419, 1467, 1586, 1627, 1752, 1811, 1880, 1947, 2084, 2132, 2247, 2308

Offset: 1

Hein van Winkel (hein65(AT)duizendknoop.com), Sep 12 2010

nonn

I was inspired by the 24-game. How many results can you get from two numbers by addition, subtraction, multiplication and division?

			For n = 3 sums 2, 3, 4, 5, 6 differences 0, 1, 2 multiplications 1, 2, 3, 4, 6, 9 divisions 1/2, 1/3, 2/3, 2, 3, 3/2 different results are 2, 3, 4, 5, 6, 0, 1, 9, 1/2, 1/3, 2/3, 3/2 or ordered 0, 1/3, 1/2, 2/3, 1, 3/2, 2, 3, 4, 5, 6, 9 so f(3) = 12.

Chai Wah Wu, Table of n, a(n) for n = 1..10000 (terms 1..2000 from Owen Whitby)

Cf. A002088, A018805, A027424, A263995.

A180622 := proc(n) s := {} ; for i from 1 to n do for j from 1 to n do s := s union {i+j} ; s := s union {abs(i-j)} ; s := s union {i*j} ; s := s union {i/j} ; s := s union {j/i} ; end do: end do: nops(s) ; end proc: seq(A180622(n),n=1..83) ; # R. J. Mathar, Sep 19 2010

a180622[maxn_] := Module[{seq = {}, vals = {}, vnew, an, n1}, Do[vnew = {}; n1 = n - 1; Do[vnew = vnew~Join~{i + n, n - i, i*n, i/n, n/i}, {i, n1}]; vnew = vnew~Join~{n + n, 0, n*n, 1}; vals = Union[vals, vnew]; an = Length[vals]; AppendTo[seq, an], {n, maxn}]; seq] (* Owen Whitby, Nov 03 2010 *)

from fractions import Fraction
def A180622(n): return len(set(range((n<<1)+1))|set().union(*({i*j,Fraction(i,j),Fraction(j,i)} for i in range(1,n+1) for j in range(1,i+1)))) # Chai Wah Wu, Aug 21 2024

# faster program
from functools import lru_cache
from sympy import primepi
def A180622(n):
    @lru_cache(maxsize=None)
    def f(n): # based on second formula in A018805
        if n == 0:
            return 0
        c, j = 0, 2
        k1 = n//j
        while k1 > 1:
            j2 = n//k1 + 1
            c += (j2-j)*(f(k1)-1)
            j, k1 = j2, n//j2
        return (n*(n-1)-c+j)
    return len({i*j for i in range(1,n+1) for j in range(1,i+1)})+f(n)+primepi(n<<1)-primepi(n)-n # Chai Wah Wu, Aug 21 2024

a(n) = A263995(n) + 2*A002088(n) - n = A263995(n) + A018805(n) - n + 1. To see this, terms of the form |i-j| are covered by terms of the form i+j or i*j with the exception of 0. Thus i+j, |i-j| and ij result in A263995(n)+1 distinct terms. Terms i/j where gcd(i,j) > 1 can be reduced to a term i'/j' where gcd(i',j')=1. The only terms left are i/j with gcd(i,j) = 1 for which there are A018805(n) such terms. We need to subtract n terms for the cases where j=1 since i/j=i/1 are integers. This results in the formula. - Chai Wah Wu, Aug 21 2024

More terms from R. J. Mathar, Sep 19 2010

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User: Hein van Winkel

A338940 a(n) is the number of solutions to the Diophantine equation p * x * (x + n) = y^2 with p = a*b a perfect square and a+b = n.

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A338939 a(n) is the number of partitions n = a + b such that a*b is a perfect square.

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A337140 Numbers m = a + b with a and b positive integers whose product a*b = k^2 is a square.

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A180622 Number of distinct sums i+j, absolute differences |i-j|, products ij and quotients i/j and j/i with 1 <= i, j <= n.

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