A046080 -id:A046080 - cp's OEIS frontend

A101929 Number of Pythagorean triples with hypotenuse < 10^n.

1, 50, 878, 12467, 161431, 1980636, 23471468, 271360645, 3080075423, 34465432849, 381301109908, 4179478903380, 45459467009955, 491241450001314, 5278882299478781, 56453500988940599, 601181789833245614, 6378285697775544212

Offset: 1

Eric W. Weisstein, Dec 21 2004

nonn

There seems to be a relation between A101930 and this sequence: A101930(n) = n + a(n). - Jan Feitsma and Bart Dopheide (dopheide(AT)fmf.nl), Mar 10 2005

A101930(n) - a(n) = A046080(10^n). The formula listed for A046080 supports the relation of Jan Feitsma and Bart Dopheide: A046080(10^n) = n. - Frank Marcoline (fvmarcoline(AT)gmail.com), Dec 10 2008

Eric Weisstein's World of Mathematics, Pythagorean Triple

Cf. A101930.

a(n)=my(t, lim=10^n-1); for(m=2, sqrtint(lim-1), forstep(n=1+m%2, min(sqrtint(lim-m^2), m-1), 2, if(gcd(m, n)==1, t+=lim\(m^2+n^2)))); t \\ Charles R Greathouse IV, Sep 15 2012

a(n) = A101930(n) - n. - Robert G. Wilson v, Mar 20 2014

More terms from Jan Feitsma and Bart Dopheide (dopheide(AT)fmf.nl), Mar 10 2005
a(10)-a(11) from Charles R Greathouse IV, Sep 15 2012
a(12)-a(17) from Hiroaki Yamanouchi, Jul 14 2014
a(18) from Matan M. Atzmoni, Feb 04 2023

A267113 Bitwise-OR of the exponents of all 4k+1 primes in the prime factorization of n.

Original entry on oeis.org

0, 0, 0, 0, 1, 0, 0, 0, 0, 1, 0, 0, 1, 0, 1, 0, 1, 0, 0, 1, 0, 0, 0, 0, 2, 1, 0, 0, 1, 1, 0, 0, 0, 1, 1, 0, 1, 0, 1, 1, 1, 0, 0, 0, 1, 0, 0, 0, 0, 2, 1, 1, 1, 0, 1, 0, 0, 1, 0, 1, 1, 0, 0, 0, 1, 0, 0, 1, 0, 1, 0, 0, 1, 1, 2, 0, 0, 1, 0, 1, 0, 1, 0, 0, 1, 0, 1, 0, 1, 1, 1, 0, 0, 0, 1, 0, 1, 0, 0, 2, 1, 1, 0, 1, 1, 1, 0, 0, 1, 1, 1, 0, 1, 0, 1, 1, 1, 0, 1, 1

Offset: 1

Antti Karttunen, Feb 03 2016

nonn

			For n = 65 = 5 * 13 = (4+1)^1 * ((3*4)+1)^1, bitwise-or of 1 and 1 is 1, thus a(65) = 1.

Antti Karttunen, Table of n, a(n) for n = 1..10000

Cf. A170818, A267116, A260728.
Cf. A004144 (indices of zeros), A009003 (of nonzeros).
Differs from both A046080 and A083025 for the first time at n=65, which here a(65) = 1.

a(n) = A267116(A170818(n)).
Other identities. For all n >= 0:
a(n) = a(A170818(n)). [The result depends only on the prime factors of the form 4k+1.]
a(n) <= A083025(n).

A379830 a(n) is the number of Pythagorean triples (u, v, w) for which w - u = n where u < v < w.

Original entry on oeis.org

0, 0, 1, 0, 1, 0, 1, 0, 2, 2, 1, 0, 1, 0, 1, 0, 2, 0, 4, 0, 1, 0, 1, 0, 2, 3, 1, 2, 1, 0, 1, 0, 5, 0, 1, 0, 4, 0, 1, 0, 2, 0, 1, 0, 1, 2, 1, 0, 2, 4, 7, 0, 1, 0, 4, 0, 2, 0, 1, 0, 1, 0, 1, 2, 5, 0, 1, 0, 1, 0, 1, 0, 8, 0, 1, 3, 1, 0, 1, 0, 2, 6, 1, 0, 1, 0, 1, 0

Offset: 0

Felix Huber, Jan 07 2025

nonn

The difference between the hypotenuse and the short leg of a primitive Pythagorean triple (p^2 - q^2, 2*p*q, p^2 + q^2) (where p > q are coprimes and not both odd) is d = max(2*q^2, (p - q)^2). For every of these primitive Pythagorean triples whose d divides n, there is a Pythagorean triple with w - u = n. Therefore d <= n and it follows that 1 <= q <= sqrt(n/2) and q + 1 <= p <= q + sqrt(n), which means that there is a finite number of Pythagorean triples with w - u = n.

			The a(18) = 4 Pythagorean triples are (27, 36, 45), (16, 30, 34), (40, 42, 58), (7, 24, 25) because 45 - 27 = 34 - 16 = 58 - 40 = 25 - 7 = 18.
See also linked Maple program "Pythagorean triples for which w - u = n".

Felix Huber, Table of n, a(n) for n = 0..10000
Felix Huber, Pythagorean triples for which w - u = n
Eric Weisstein's World of Mathematics, Pythagorean Triple

Cf. A024361, A024362, A024363, A046079, A046080, A046081, A096033, A224921.

A379830:=proc(n)
    local a,p,q;
    a:=0;
    for q to isqrt(floor(n/2)) do
        for p from q+1 to q+isqrt(n) do
            if igcd(p,q)=1 and (is(p,even) or is(q,even)) and n mod max((p-q)^2,2*q^2)=0 then
                a:=a+1
            fi
        od
    od;
    return a
end proc;
seq(A379830(n),n=0..87);

A088959 Lowest numbers which are d-Pythagorean decomposable, i.e., square is expressible as sum of two positive squares in more ways than for any smaller number.

Original entry on oeis.org

1, 5, 25, 65, 325, 1105, 5525, 27625, 32045, 160225, 801125, 1185665, 5928325, 29641625, 48612265, 243061325, 1215306625, 2576450045, 12882250225, 64411251125, 157163452745, 785817263725, 3929086318625, 10215624428425, 11472932050385, 51078122142125

Offset: 1

Lekraj Beedassy, Dec 01 2003

nonn

These are also the integer radii of circles around the origin that contain record numbers of lattice points. See A071383 for radii that are not necessarily integer. - Günter Rote, Sep 14 2023

			From _Petros Hadjicostas_, Jul 21 2019: (Start)
Squares 1^2, 2^2, 3^2, and 4^2 have 0 representations as the sum of two positive squares. (Thus, A088111(1) = 0 for the number of representations of 1^2.) Thus a(1) = 1.
Square 5^2 can be written as 3^2 + 4^2 only (here A088111(2) = 1). Thus, a(2) = 5.
Looking at sequence A046080, we see that for 5 <= n <= 24, only n^2 = 5^2, 10^2, 13^2, 15^2, 17^2, 20^2 can be written as a sum of two positive squares (in a single way) because 5^2 = 3^2 + 4^2, 10^2 = 6^2 + 8^2, 13^2 = 5^2 + 12^2, 17^2 = 8^2 + 15^2, and 20^2 = 12^2 + 16^2.
Since A046080(25) = 2 and A088111(3) = 2, we have that 25^2 can be written as a sum of two positive squares in two ways. Indeed, 25^2 = 7^2 + 24^2 = 15^2 + 20^2. Thus, a(3) = 25.
For 26 <= n <= 64, we see from sequence A046080 that n^2 cannot be written in more than 2 ways as a sum of two positive squares.
Since A046080(65) = 4, we see that 65^2 can be written as the sum of two positive squares in 4 ways. Indeed, 65^2 = 16^2 + 63^2 = 25^2 + 60^2 = 33^2 + 56^2 = 39^2 + 52^2. Thus, a(4) = 65.
(End)

R. M. Sternheimer, Additional Remarks Concerning The Pythagorean Triplets, Journal of Recreational Mathematics, Vol. 30, No. 1, pp. 45-48, 1999-2000, Baywood NY.

Ray Chandler, Table of n, a(n) for n = 1..307

Cf. A052199. Subsequence of A054994. Number of ways: see A088111. Where records occur in A046080.

from math import prod
from sympy import isprime
primes_congruent_1_mod_4 = [5]
def prime_4k_plus_1(i): # the i-th prime that is congruent to 1 mod 4
    while i>=len(primes_congruent_1_mod_4): # generate primes on demand
        n = primes_congruent_1_mod_4[-1]+4
        while not isprime(n): n += 4
        primes_congruent_1_mod_4.append(n)
    return primes_congruent_1_mod_4[i]
def generate_A054994():
    TO_DO = {(1,())}
    while True:
        radius, exponents = min(TO_DO)
        yield radius, exponents
        TO_DO.remove((radius, exponents))
        TO_DO.update(successors(radius,exponents))
def successors(r,exponents):
    for i,e in enumerate(exponents):
        if i==0 or exponents[i-1]>e:
            yield (r*prime_4k_plus_1(i), exponents[:i]+(e+1,)+exponents[i+1:])
    if exponents==() or exponents[-1]>0:
        yield (r*prime_4k_plus_1(len(exponents)), exponents+(1,))
n,record=0,-1
for radius,expo in generate_A054994():
    num_pyt = (prod((2*e+1) for e in expo)-1)//2
    if num_pyt>record:
        record = num_pyt
        n += 1
        print(radius, end="") # or record, for A088111
        if n==26: break # stop after 26 entries
        print(end=", ")
print() # Günter Rote, Sep 13 2023

Corrected and extended by Ray Chandler, Jan 12 2012

Name edited by Petros Hadjicostas, Jul 21 2019

A088111 Number of ways associated with A088959.

Original entry on oeis.org

0, 1, 2, 4, 7, 13, 22, 31, 40, 67, 94, 121, 202, 283, 364, 607, 850, 1093, 1822, 2551, 3280, 5467, 7654, 9112, 9841, 12757, 16402, 22963, 27337, 29524, 38272, 49207, 68890, 82012, 88573, 114817, 147622, 206671, 246037, 265720, 344452, 442867, 620014, 738112

Offset: 1

Lekraj Beedassy, Apr 24 2004

nonn

Ray Chandler, Table of n, a(n) for n = 1..307

Cf. A088959. Records in A046080.

Corrected and extended by Ray Chandler, Jan 12 2012

A316359 a(n) is the number of solutions to the Diophantine equation i^3 + j^3 + k^3 = n^3, where 0 < i <= j <= k.

Original entry on oeis.org

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

Offset: 1

Arlu Genesis A. Padilla, Jun 30 2018

nonn

The first number to have a nonzero number of solutions is 6, which is 3^3 + 4^3 + 5^3 = 6^3. Its cube 216 has been called Plato's number in reference to this.

First occurrence of k=0,1,2...: 0, 6, 18, 54, 87, 108, 216, 174, 348, 396, 324, 696, 864, 492, etc. - Robert G. Wilson v, Jul 02 2018

			a(18)=2, because 18^3 = 9^3 + 12^3 + 15^3 = 2^3 + 12^3 + 16^3.

Arlu Genesis A. Padilla, Table of n, a(n) for n = 1..10000

Cf. A046080.

Array[Count[PowersRepresentations[#^3, 3, 3], ?(FreeQ[Differences@ #, 0] &)] &, 105] (* _Michael De Vlieger, Jun 30 2018 *)

a(n) = sum(i=1, n, sum(j=1, i, sum(k=1, j, i^3 + j^3 + k^3 == n^3))); \\ Michel Marcus, Jul 02 2018

a(n)={sum(i=1, n, sum(j=1, i, my(k); ispower(n^3-j^3-i^3, 3, &k) && k>=1 && k<=j ))} \\ Andrew Howroyd, Jul 07 2018

from sympy.solvers.diophantine.diophantine import power_representation
def A316359(n): return len(list(power_representation(n**3,3,3))) # Chai Wah Wu, Nov 19 2024

A056137 Number of ways in which n can be the longer leg (middle side) of an integer-sided right triangle.

Original entry on oeis.org

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

Offset: 1

Henry Bottomley, Jun 15 2000

nonn

Cf. A009023, A055523.

a(n) = A046079(n) - A056138(n) = A046081(n) - A046080(n) - A056138(n).

A211340 Number of integer pairs (x,y) such that 1

Original entry on oeis.org

0, 1, 3, 5, 9, 13, 17, 23, 30, 38, 45, 53, 64, 74, 86, 97, 110, 123, 138, 154, 168, 186, 203, 220, 241, 261, 282, 302, 324, 348, 370, 396, 421, 448, 476, 501, 531, 558, 591, 622, 651, 684, 717, 753, 788, 821, 858, 894, 933, 973, 1014, 1054, 1093, 1135

Offset: 1

Clark Kimberling, Apr 08 2012

nonn

For a guide to related sequences, see A211266.

Robert Israel, Table of n, a(n) for n = 1..2000

Cf. A046080, A211266.

N:= 100: # for a(1)..a(N)
V:= Vector(N):
for y from 1 to N-1 do
  for x from 1 to y do
    r:= x^2 + y^2;
    if r > N^2 then break fi;
    t:= ceil(sqrt(r));
    V[t]:= V[t]+1
od od:
ListTools:-PartialSums(convert(V,list)); # Robert Israel, Jun 04 2019

a = 1; b = n; z1 = 120;
t[n_] := t[n] = Flatten[Table[x^2 + y^2, {x, a, b - 1}, {y, x, b}]] (* 1<=x<=y<=n *)
c[n_, k_] := c[n, k] = Count[t[n], k]
TableForm[Table[c[n, k], {n, 1, 7}, {k, 1, n^2}]]
Table[c[n, n^2], {n, 1, z1}]    (* A046080 *)
c1[n_, m_] := c1[n, m] = Sum[c[n, k], {k, a, m}]
Table[c1[n, n^2], {n, 1, z1/2}] (* A211340 *)

A247589 Number of integer-sided obtuse triangles with largest side n.

Original entry on oeis.org

0, 0, 1, 1, 2, 4, 5, 7, 10, 12, 15, 17, 21, 25, 29, 33, 37, 42, 48, 53, 58, 65, 71, 76, 83, 91, 100, 106, 113, 122, 130, 140, 149, 158, 169, 177, 188, 197, 210, 221, 230, 243, 255, 269, 281, 292, 306, 318, 333, 346

Offset: 1

Vladimir Letsko, Sep 20 2014

nonn

			a(5) = 2 because there are 2 integer-sided acute triangles with largest side 5: (2,4,5); (3,3,5).

Vladimir Letsko, Mathematical Marathon, problem 192 (in Russian).

Cf. A046080, A002623, A224921, A247586, A247587, A247588.

tr_o:=proc(n) local a,b,t,d;t:=0:
for a to n do
for b from max(a,n+1-a) to n do
d:=a^2+b^2-n^2:
if d<0 then t:=t+1 fi
od od;
t; end;

a(n) = k*(k + (1+(-1)^n)/2) + Sum_{j=1..floor(n*(1-sqrt(2)/2))} floor(sqrt(2*j*n - j^2 - 1) - j), where k = floor((2*n*(sqrt(2) - 1) + 1 - (-1)^n)/4) (it appears that k(n) is A070098(n)). - Anton Nikonov, Sep 29 2014

A309332 Number of ways the n-th triangular number T(n) = A000217(n) can be written as the sum of two positive triangular numbers.

Original entry on oeis.org

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

Offset: 1

Alois P. Heinz, Aug 01 2019

nonn

The order doesn't matter. 21 = 6+15 = 15+6 are not counted as distinct solutions. - N. J. A. Sloane, Feb 22 2020

			a(3) = 1: 2*3/2 + 2*3/2 = 3*4/2.
a(21) = 2: 6*7/2 + 20*21/2 = 12*13/2 + 17*18/2 = 21*22/2.
a(23) = 3: 9*10/2 + 21*22/2 = 11*12/2 + 20*21/2 = 14*15/2 + 18*19/2 = 23*24/2.

Alois P. Heinz, Table of n, a(n) for n = 1..20000

Cf. A000217, A001652, A012132, A027861, A046080 (the same for squares), A053141, A062301 (the same for primes), A108769, A309507.

a:= proc(n) local h, j, r, w; h, r:= n*(n+1), 0;
      for j from n-1 by -1 do w:= j*(j+1);
        if 2*w

a[n_] := Module[{h = n(n+1), j, r = 0, w}, For[j = n-1, True, j--, w = j(j+1); If[2w < h, Break[]]; If[ IntegerQ[Sqrt[4(h-w)+1]], r++]]; r];
Table[a[n], {n, 1, 120}] (* Jean-François Alcover, Nov 16 2022, after Alois P. Heinz *)

a(n) > 0 <=> n in { A012132 }.
a(n) = 0 <=> n in { A027861 }.
a(n) = 1 <=> n in { A108769 }.

cp's OEIS Frontend

A101929 Number of Pythagorean triples with hypotenuse < 10^n.

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A267113 Bitwise-OR of the exponents of all 4k+1 primes in the prime factorization of n.

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A379830 a(n) is the number of Pythagorean triples (u, v, w) for which w - u = n where u < v < w.

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Maple

A088959 Lowest numbers which are d-Pythagorean decomposable, i.e., square is expressible as sum of two positive squares in more ways than for any smaller number.

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A088111 Number of ways associated with A088959.

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Extensions

A316359 a(n) is the number of solutions to the Diophantine equation i^3 + j^3 + k^3 = n^3, where 0 < i <= j <= k.

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Mathematica

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PARI

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A056137 Number of ways in which n can be the longer leg (middle side) of an integer-sided right triangle.

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A211340 Number of integer pairs (x,y) such that 1

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Maple

Mathematica

A247589 Number of integer-sided obtuse triangles with largest side n.

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