A192629
Numerators of the Fermat-Euler rational Diophantine m-tuple.
Original entry on oeis.org
0, 1, 3, 8, 120, 777480
Offset: 0
0/1, 1/1, 3/1, 8/1, 120/1, 777480/8288641.
1 + 1*(777480/8288641) = (3011/2879)^2.
- A. Dujella, Rational Diophantine m-tuples
- E. Herrmann, A. Pethoe and H. G. Zimmer, On Fermat's quadruple equations, Abh. Math. Sem. Univ. Hamburg 69 (1999), 283-291.
- Michael Stoll, Diagonal genus 5 curves, elliptic curves over Q(t), and rational diophantine quintuples, Acta Arith. 190 (2019), 239-261.
A192630
Denominators of the Fermat-Euler rational Diophantine m-tuple.
Original entry on oeis.org
1, 1, 1, 1, 1, 8288641
Offset: 0
0/1, 1/1, 3/1, 8/1, 120/1, 777480/8288641.
1 + 1*(777480/8288641) = (3011/2879)^2.
A192631
Numerators of the Diophantus-Dujella rational Diophantine quintuple: 1 + the product of any two distinct terms is a square.
Original entry on oeis.org
1, 33, 17, 105, 549120
Offset: 1
1/16, 33/16, 17/4, 105/16, 549120/10201.
1 + (1/16)*(33/16) = (17/16)^2.
1 + (33/16)*(549120/10201) = (1069/101)^2.
- E. Herrmann, A. Pethoe and H. G. Zimmer, On Fermat's quadruple equations, Abh. Math. Sem. Univ. Hamburg 69 (1999), 283-291.
A192632
Denominators of the Diophantus-Dujella rational Diophantine quintuple: 1 + the product of any two distinct terms is a square.
Original entry on oeis.org
16, 16, 4, 16, 10201
Offset: 1
1/16, 33/16, 17/4, 105/16, 549120/10201.
1 + (1/16)*(33/16) = (17/16)^2.
1 + (33/16)*(549120/10201) = (1069/101)^2.
A180928
1 + product of any two terms is a triangular number.
Original entry on oeis.org
(0*1)+1 = 1 is triangular.
(0*5)+1 = 1 is triangular.
(1*5)+1 = 6 is triangular.
(0*27)+1 = 1 is triangular.
(1*27)+1 = 28 is triangular.
(5*27)+1 = 136 is triangular.
(0*70)+1 = 1 is triangular.
(1*70)+1 = 71 is NOT triangular, so 70 is not the next value.
(5*70)+1 = 351 is triangular.
(27*70)+1 = 1891 is triangular.
A219953
a(1) = 1; for n > 1, a(n) = smallest integer > a(n-1) such that a(n)*a(i)+1 is semiprime for all 1 <= i <= n-1.
Original entry on oeis.org
1, 3, 8, 38, 86, 318, 504, 3600, 8132, 83160, 116850, 202272, 399126, 6190086, 8756916, 25253676, 309709400, 1112878446, 1478724036, 11062089360, 97331025386
Offset: 1
a(1) = 1 by definition.
a(2) = 3: 3 > 1, and 1*3 + 1 = 4 = 2^2 is semiprime.
a(3) = 8: 8 > 3, and 1*8 + 1 = 9 = 3^2 is semiprime, and 3*8 + 1 = 25 = 5^2 is semiprime.
a(4) = 38: 38 > 8, and 1*38 + 1 = 39 = 3*13 is semiprime, and 3*38 + 1 = 115 = 5*23 is semiprime, and 8*38 + 1 = 305 = 5*61 is semiprime.
From _Michel Marcus_, Jul 26 2015: (Start)
The resulting semiprimes are:
4;
9, 25;
39, 115, 305;
87, 259, 689, 3269;
319, 955, 2545, 12085, 27349;
...
(End)
-
A219953 := proc(n)
option remember;
if n= 1 then
1;
else
for a from procname(n-1)+1 do
issp := true ;
for i from 1 to n-1 do
if numtheory[bigomega]( a*procname(n-i)+1) = 2 then
;
else
issp := false;
break ;
end if;
end do:
if issp then
return a;
end if;
end do:
end if;
end proc: # R. J. Mathar, Dec 15 2012
-
a = {1}; Do[k = a[[n - 1]] + 1; While[! AllTrue[(k a[[n - #]] + 1) & /@ Range@ (n - 1), Total[Last /@ FactorInteger@ #] == 2 &], k++]; AppendTo[a, k], {n, 2, 13}]; a (* Michael De Vlieger, Jul 26 2015, Version 10 *)
-
ok(v, n, k) = {v[n] = k; for (j=1, n-1, if (bigomega(1+v[n]*v[j]) != 2, return (0));); return (1);}
lista(nn) = {print1(k=1, ", "); v = [k]; for (n=2, nn, k = v[n-1]+1; v = concat(v, k); while (! ok(v, n, k), k++); v[n] = k; print1(k, ", "););} \\ Michel Marcus, Jul 26 2015
A274694
Variation on Fermat's Diophantine m-tuple: 1 + the product of any two distinct terms is a prime power.
Original entry on oeis.org
1, 2, 3, 4, 6, 12, 211050, 3848880, 20333040, 125038830, 2978699430
Offset: 1
After a(1)=1, a(2)=2, a(3)=3, we want m, the smallest number > 3 such that m+1, 2m+1 and 3m+1 are all prime powers: this is m = 4 = a(4).
-
seq = [1]
prev_element = 1
max_n = 8
for n in range(2, max_n+1):
next_element = prev_element + 1
while True:
all_match = True
for element in seq:
x = element * next_element + 1
if not x.is_prime_power():
all_match = False
break
if all_match:
seq.append( next_element )
break
next_element += 1
prev_element = next_element
print(seq)
A274696
Variation on Fermat's Diophantine m-tuple: 1 + the LCM of any two distinct terms is a square.
Original entry on oeis.org
0, 1, 3, 8, 15, 24, 120, 168, 840, 1680, 5040, 201600, 256032000
Offset: 1
After a(1)=0, a(2)=1, a(3)=3, we want m, the smallest number > 3 such that lcm(0,m)+1, lcm(2,m)+1 and lcm(3,m)+1 are squares: this is m = 8 = a(4).
-
a = {0}; Do[AppendTo[a, SelectFirst[Range[Max@ a + 1, 3*10^5], Function[k, Times @@ Boole@ Map[IntegerQ@ Sqrt[LCM[a[[#]], k] + 1] &, Range[n - 1]] == 1]]], {n, 2, 12}]; a (* Michael De Vlieger, Jul 05 2016, Version 10 *)
-
seq = [0]
prev_element = 0
max_n = 13
for n in range(2, max_n+1):
next_element = prev_element + 1
while True:
all_match = True
for element in seq:
x = lcm( element, next_element ) + 1
if not is_square(x):
all_match = False
break
if all_match:
seq.append( next_element )
print(seq)
break
next_element += 1
prev_element = next_element
print(seq)
A274697
Variation on Fermat's Diophantine m-tuple: 1 + the GCD of any two distinct terms is a square.
Original entry on oeis.org
0, 3, 15, 24, 48, 63, 120, 195, 255, 528, 960, 3024, 3363, 3480, 3720, 3843, 4095, 4623, 5475, 12099, 16383, 19599, 24963, 37635, 38415, 44943, 56643, 62499, 65535, 69168, 71823, 85263, 94863, 114243, 168099
Offset: 1
After a(1)=0, a(2)=3, a(3)=15, we want m, the smallest number > 15 such that GCD(0,m)+1, GCD(3,m)+1 and GCD(15,m)+1 are squares: this is m = 24 = a(4).
-
seq = []
prev_element = 0
seq.append( prev_element )
max_n = 35
for n in range(2, max_n+1):
next_element = prev_element + 1
while True:
all_match = True
for element in seq:
x = gcd( element, next_element ) + 1
if not ( is_square(x) ):
all_match = False
break
if all_match:
seq.append( next_element )
print(seq)
break
next_element = next_element + 1
prev_element = next_element
print(seq)
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