cp's OEIS Frontend

This is a front-end for the Online Encyclopedia of Integer Sequences, made by Christian Perfect. The idea is to provide OEIS entries in non-ancient HTML, and then to think about how they're presented visually. The source code is on GitHub.

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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.

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

Views

Author

Artur Jasinski, Jan 23 2007

Keywords

Comments

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

Examples

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

Crossrefs

Programs

  • Mathematica
    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
  • PARI
    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

Extensions

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

Views

Author

Artur Jasinski, Jan 23 2007, Jan 29 2007

Keywords

Examples

			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.
		

Crossrefs

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

Programs

  • Mathematica
    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 *)

Formula

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
Showing 1-2 of 2 results.