A054994 Numbers of the form q1^b1 * q2^b2 * q3^b3 * q4^b4 * q5^b5 * ... where q1=5, q2=13, q3=17, q4=29, q5=37, ... (A002144) and b1 >= b2 >= b3 >= b4 >= b5 >= ....

1, 5, 25, 65, 125, 325, 625, 1105, 1625, 3125, 4225, 5525, 8125, 15625, 21125, 27625, 32045, 40625, 71825, 78125, 105625, 138125, 160225, 203125, 274625, 359125, 390625, 528125, 690625, 801125, 1015625, 1185665, 1221025, 1373125, 1795625

Offset: 1

Author

Bernard Altschuler (Altschuler_B(AT)bls.gov), May 30 2000

Keywords

• easy
• nonn

Comments

This sequence is related to Pythagorean triples regarding the number of hypotenuses which are in a particular number of total Pythagorean triples and a particular number of primitive Pythagorean triples.

Least integer "mod 4 prime signature" values that are the hypotenuse of at least one primitive Pythagorean triple. - Ray Chandler, Aug 26 2004

See A097751 for definition of "mod 4 prime signature"; terms of A097752 with all prime factors of form 4*k+1.

Sequence A006339 (Least hypotenuse of n distinct Pythagorean triangles) is a subset of this sequence. - Ruediger Jehn, Jan 13 2022

Examples

			1=5^0, 5=5^1, 25=5^2, 65=5*13, 125=5^3, 325=5^2*13, 625=5^4, etc.
		

Links

• Ray Chandler, Table of n, a(n) for n = 1..10000
• Eric Weisstein's World of Mathematics, Pythagorean Triple

Crossrefs

Subsequence of A097752.

Cf. A002144, A006278, A006339, A071383, A097751, A097752, A097753, A097754, A097755, A097756.

Programs

Mathematica

maxTerm = 10^15;(* this limit gives ~ 500 terms *) maxNumberOfExponents = 9;(* this limit has to be increased until the number of reaped terms no longer changes *) bmax = Ceiling[ Log[ maxTerm]/Log[q]]; q = Reap[For[k = 0 ; cnt = 0, cnt <= maxNumberOfExponents, k++, If[PrimeQ[4*k + 1], Sow[4*k + 1]; cnt++]]][[2, 1]]; Clear[b]; b[maxNumberOfExponents + 1] = 0; iter = Sequence @@ Table[{b[k], b[k + 1], bmax[[k]]}, {k, maxNumberOfExponents, 1, -1}]; Reap[ Do[an = Product[q[[k]]^b[k], {k, 1, maxNumberOfExponents}]; If[an <= maxTerm, Print[an]; Sow[an]], Evaluate[iter]]][[2, 1]] // Flatten // Union (* Jean-François Alcover, Jan 18 2013 *)

PARI

list(lim)=
{
  my(u=[1], v=List(), w=v, pr, t=1);
  forprime(p=5,,
    if(p%4>1, next);
    t*=p;
    if(t>lim, break);
    listput(w,t)
  );
  for(i=1,#w,
    pr=1;
    for(e=1,logint(lim\=1,w[i]),
      pr*=w[i];
      for(j=1,#u,
        t=pr*u[j];
        if(t>lim, break);
        listput(v,t)
      )
    );
    if(w[i]^2Charles R Greathouse IV, Dec 11 2016

Python

def generate_A054994():
    """generate arbitrarily many elements of the sequence.
    TO_DO is a list of pairs (radius, exponents) where
    "exponents" is a weakly decreasing sequence, and
    radius == prod(prime_4k_plus_1(i)**j for i,j in enumerate(exponents))
    An example entry is (5525, (2, 1, 1)) because 5525 = 5**2 * 13 * 17.
    """
    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(radius,exponents):
    # try to increase each exponent by 1 if possible
    for i,e in enumerate(exponents):
        if i==0 or exponents[i-1]>e:
            # can add 1 in position i without violating monotonicity
            yield (radius*prime_4k_plus_1(i), exponents[:i]+(e+1,)+exponents[i+1:])
    if exponents==() or exponents[-1]>0: # add new exponent 1 at the end:
        yield (radius*prime_4k_plus_1(len(exponents)), exponents+(1,))
from sympy import isprime
primes_congruent_1_mod_4 = [5] # will be filled with 5,13,17,29,37,...
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]
for n,radius in enumerate(generate_A054994()):
    if n==34:
        print(radius)
        break # print the first 35 elements
    print(radius, end=", ")
# Günter Rote, Sep 12 2023

Formula

Sum_{n>=1} 1/a(n) = Product_{n>=1} 1/(1 - 1/A006278(n)) = 1.2707219403... - Amiram Eldar, Oct 20 2020

Extensions

More terms from Henry Bottomley, Mar 14 2001