A018930 -id:A018930 - cp's OEIS frontend

A018928 Define {b(n)} by b(1)=3, b(n) (n >= 2) is the smallest number such that b(1)^2 + ... + b(n)^2 = m^2 for some m and all b(i) are distinct. Sequence gives values of m.

3, 5, 13, 85, 157, 12325, 12461, 106285, 276341, 339709, 10363909, 17238541, 1936511509, 51335823965, 133473142309, 872709007405, 1574530008629, 667511933218429, 698925273030725, 707670964169285, 1839944506840141

Offset: 1

Charles Reed (charles.reed(AT)bbs.ewgateway.org)

nonn

Also: Begin with the least length of a Pythagorean Triangle (PT), a(1)=3. Then a(n) is the least hypotenuse of a PT which has a(n-1) as one of its legs. - Robert G. Wilson v, Mar 17 2014

Robert G. Wilson v, Table of n, a(n) for n = 1..165 (first 100 terms from Lei Zhou)

Cf. A018929, A018930.

NextA018928[n_] := Block[{a = n^2, b, l, i, c, d, f}, b = Divisors[a]; l = Length[b]; i = l; While[i--; c = b[[i]]; d = a/c - (c - 1); (d <= 1) || EvenQ[d]]; f = (a/c + (c - 1) + 1)/2]; Table[If[i == 1, a = 3, a = NextA018928[a]]; a, {i, 1, 21}](* Lei Zhou, Feb 20 2014 *)
f[s_List] := Block[{x = s[[-1]]}, Append[s, Transpose[ Solve[ x^2 + y^2 == z^2 && y > 0 && z > 0, {y, z}, Integers]][[-1, 1, 2]]]]; lst = Nest[f, lst, 15] (* Robert G. Wilson v, Mar 17 2014 *)

More terms from David W. Wilson

A018929 Define {b(n)} by b(1) = 3, b(n) (n >= 2) is smallest number such that b(1)^2 + ... + b(n)^2 = m^2 for some m and all b(i) are distinct. Sequence gives values of m^2.

Original entry on oeis.org

9, 25, 169, 7225, 24649, 151905625, 155276521, 11296501225, 76364348281, 115402204681, 107410609760281, 297167295808681, 3750076824489457081, 2635366822165468321225, 17815079717838565851481, 761621011605820344834025

Offset: 1

Charles Reed (charles.reed(AT)bbs.ewgateway.org)

nonn

Cf. A018928, A018930.

a(n) = A018928(n+1)^2 - A018930(n+1)^2. - César Aguilera, Nov 02 2018

More terms from David W. Wilson

A127689 a(1)=3; for n>1, a(n) is least number such that a(n) > a(n-1) and a(1)^2+...+a(n)^2 is a square.

Original entry on oeis.org

3, 4, 12, 84, 132, 12324, 15960, 26280, 27300, 66660, 115188, 9777193284, 23465263884, 48701491080, 40900397690640, 680008604512020, 127049882801497788, 247290967245178188, 335580091290976716, 1045885075937364972, 1607091702050097396, 3419793695168900508, 5138020847719969956, 10059508412964112740

Offset: 1

Artur Jasinski, Jan 23 2007

nonn

Without the a(n) > a(n-1) constraint, the sequence would be A018930.

			a(2)=4 because 3^2+4^2=5^2, a(3)=12 because 3^2+4^2+12^2=13^2 etc.

Cf. A018930, A127690, A127691.

a = {3}; For[k = 1 + a[[Length[a]]], Length[a] < 11, While[ ! IntegerQ[Sqrt[(k)^2 + Sum[(a[[t]])^2, {t, 1, Length[a]}]]], k++ ]; AppendTo[a, k]]; a

q=3; s=9; for(n=1,30, b=0; fordiv(s,d, if(d*d>=s,break); t=(s\d-d)/2; if(t>q,b=t); ); q=b; print1(q,", "); s+=q^2); \\ Max Alekseyev, Nov 23 2012

More terms from Max Alekseyev, Nov 23 2012

A127690 a(1)=3; for n>1, a(n) is such that a(1)^2+...+a(n)^2 = (1+a(n))^2.

Original entry on oeis.org

3, 4, 12, 84, 3612, 6526884, 21300113901612, 226847426110843688722000884, 25729877366557343481074291996721923093306518970391612, 331013294649039928396936390888878360035026305412754995683702777533071737279144813617823976263475290370884

Offset: 1

Artur Jasinski, Jan 23 2007, Jan 29 2007

nonn

			a(2)=4 because (3^2+4^2=5^2) and (4+1=5), a(3)=12 because (3^2+4^2+12^2=13^2) and (12+1=13) a(5)= 3612 because (3^2+4^2+12^2+84^2+3612^2=3613^2) and (3612+1=3613) etc.

Sierpinski W., Elementary theory of numbers, Monografie Matematyczne 42 (1964), Chapter II, p. 63.

Cf. A018930, A127689, A127691.

Apart from the initial term, the sequence is the same as A053631.

a = {3}; For[k = 1 + a[[Length[a]]], Length[a] < 5, While[ ! ((IntegerQ[Sqrt[(k)^2 + Sum[(a[[t]])^2, {t, 1, Length[a]}]]]) && (Sqrt[(k)^2 + Sum[(a[[t]])^2, {t, 1, Length[a]}]] == k + 1)), k++ ]; AppendTo[a, k]]; a
a = {3}; For[k = 1 + a[[Length[a]]], Length[a] < 12, s2 = Plus @@ (a^2); t = Reduce[{y^2 + s2 == (y + 1)^2}, y, Integers]; t = t /. {Equal -> Rule}; k = y /. t; AppendTo[a, k]]; a (* Daniel Huber *)

For n>2, a(n) = (a(1)^2 + a(2)^2 + ... + a(n-1)^2 - 1)/2 = ((a(n-1) + 1)^2 - 1)/2. - Max Alekseyev, Nov 23 2012

a(n) = A053630(n-1)-1 for n>=2. - R. J. Mathar, Apr 23 2007

A072470 a(0) = 0, a(1) = 9; for n > 1 a(n) = smallest positive square (possibly required to be greater than a(n-1)?) such that a(0) + a(1) + ... + a(n) is a square.

Original entry on oeis.org

0, 9, 16, 144, 7056, 17424, 151880976, 3370896, 11141224704, 65067847056, 39037856400, 107295207555600, 189756686048400, 3749779657193648400, 2631616745340978864144, 15179712895673097530256

Offset: 0

Amarnath Murthy, Jun 19 2002

nonn

Sequence is infinite as every partial sum (n>0) is odd, say 2k + 1 and then k^2 is a candidate for the next term.

			a(3) = 16 as a(1) + a(2) + a(3) = 25 is also a square.
a(4) = 144 as 0 + 9 + 16 + 144 = 169 is also a square.

a[0] = 0; a[1] = 9; a[n_] := a[n] = (k = Sqrt[a[n - 1]] + 1; s = Sum[a[i], {i, 0, n - 1}]; While[ !IntegerQ[ Sqrt[s + k^2]], k++ ]; k^2);

a(n) = A018930(n)^2. - Benoit Cloitre, Jun 21 2002

a(n) = A018929(n+1) - A018929(n) for n > 1. - César Aguilera, Nov 10 2018

Edited by N. J. A. Sloane and Robert G. Wilson v, Jun 21 2002

More terms from Benoit Cloitre, Jun 21 2002

A307077 Let a(1)=3; for n > 1, let a(n) be the least positive integer k such that k > a(n-1), a(1)^2 + ... + a(n-1)^2 + k^2 is a square and the Pythagorean triple sqrt(a(1)^2 + ... + a(n-1)^2), a(n), sqrt(a(1)^2 + ... + a(n)^2) is primitive.

Original entry on oeis.org

3, 4, 12, 84, 132, 12324, 89892, 2447844, 28350372, 295742791596, 171480834409712412, 633511848768467916, 1616599508725767821225590810932, 4158520496012961741299012805876, 115366949386695884000892071516523067413910188

Offset: 1

Rohan Hemasingha, May 30 2019

nonn

For n > 1, a(n) is the even value of a primitive Pythagorean triple where the larger odd value of the triple equals the smaller odd value of a primitive Pythagorean triple with even value a(n+1) (see A239381). - Torlach Rush, Jan 27 2023

Cf. A018930, A127689, A239381.

lista(NN) = s=9;k=3;print1(k);for(n=1,NN-1,v=divisors(s);j=#v;while(v[j]*(v[j]+2*k)>s,j--);while(gcd((s-v[j]^2)/(2*v[j]), s)!=1, j--);print1(", ", k=(s-v[j]^2)/(2*v[j]));s+=k^2); \\ Jinyuan Wang, May 31 2019

The numbers are generated by using the well-known characterization of primitive Pythagorean triples, namely (a,b,c) is a PPT iff there are positive integers j,k of opposite parity with j > k, and gcd(j,k)=1 such that a = j^2 - k^2, b = 2jk, c = j^2 + k^2.

a(14)-a(15) from Jinyuan Wang, Jun 01 2019

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A018928 Define {b(n)} by b(1)=3, b(n) (n >= 2) is the smallest number such that b(1)^2 + ... + b(n)^2 = m^2 for some m and all b(i) are distinct. Sequence gives values of m.

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A018929 Define {b(n)} by b(1) = 3, b(n) (n >= 2) is smallest number such that b(1)^2 + ... + b(n)^2 = m^2 for some m and all b(i) are distinct. Sequence gives values of m^2.

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A127689 a(1)=3; for n>1, a(n) is least number such that a(n) > a(n-1) and a(1)^2+...+a(n)^2 is a square.

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A127690 a(1)=3; for n>1, a(n) is such that a(1)^2+...+a(n)^2 = (1+a(n))^2.

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A072470 a(0) = 0, a(1) = 9; for n > 1 a(n) = smallest positive square (possibly required to be greater than a(n-1)?) such that a(0) + a(1) + ... + a(n) is a square.

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A307077 Let a(1)=3; for n > 1, let a(n) be the least positive integer k such that k > a(n-1), a(1)^2 + ... + a(n-1)^2 + k^2 is a square and the Pythagorean triple sqrt(a(1)^2 + ... + a(n-1)^2), a(n), sqrt(a(1)^2 + ... + a(n)^2) is primitive.

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Extensions