A271115 Numbers whose square is a chiliagonal (or 1000-gonal) number.
1, 50049, 410924801, 491565199, 4035971329551, 33137139500710799, 39640013290309201, 325462334331581751249, 2672192117918839703333201, 3196586448455823020136799, 26245412174507812354285027551, 215486635921438132237851754543199
Offset: 1
Examples
50049 is in the sequence because 50049^2 = 2504902401, which is the 2241st 1000-gonal number. - _Colin Barker_, Apr 01 2016
Links
- Colin Barker, Table of n, a(n) for n = 1..350
- M. A. Asiru, All square chiliagonal numbers, Int J Math Edu Sci Technol, 47:7(2016), 1123-1134.
- Index entries for linear recurrences with constant coefficients, signature (0,0,80640398,0,0,-1).
Programs
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GAP
g:=1000; Q0:=(g-4)^2; D1:=2*g-4; S:=[2*[ 500, 1 ], 4*[ 1022201, 22880 ], 498*[ 8980, 201 ], 996*[ 1, 0 ],-2*[- 500, 1 ], -4*[- 1022201, 22880 ]];; S1:=Filtered(S,i->IsInt((i[1]+g-4)/(2*g-4)));; S2:=Filtered([1..Length(S)],i->IsInt((S[i][1]+g-4)/(2*g-4)));; S3:=List(S2,i->S[i]);; u:=40320199;; v:=902490;; G:=[[u,2*(g-2)*v],[v,u]];; A:=List([1..Length(S3)],s->List(List([0..6],i->G^i*TransposedMat([S3[s]])),Concatenation));; Length(A); D1:=Union(List([1..Length(A)],k->A[k]));; Length(D1); D2:=List(D1,i-> [(i[1]+(g-4))/(2*(g-2)),i[2]/2] );; Length(D2); D3:=Filtered(D2,i->IsInt(i[1])); D4:=Filtered(D3,i->i[2]>0); D5:=List(D4,i->i[2]); # indices of square numbers for square 1000 gonal numbers (or square chiliagonal numbers)
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Mathematica
Rest@ CoefficientList[Series[x (1 + x) (1 + 50048 x + 410874753 x^2 + 50048 x^3 + x^4)/(1 - 80640398 x^3 + x^6), {x, 0, 12}], x] (* Michael De Vlieger, Apr 01 2016 *) LinearRecurrence[{0,0,80640398,0,0,-1},{1,50049,410924801,491565199,4035971329551,33137139500710799},20] (* Harvey P. Dale, Sep 01 2025 *)
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PARI
Vec(x*(1+x)*(1+50048*x+410874753*x^2+50048*x^3+x^4)/(1-80640398*x^3+x^6) + O(x^15)) /* Colin Barker, Apr 01 2016 */
Formula
a(n)^2 = A271105(n).
a(n) = 80640398*a(n-3)-a(n-6) for n>6. - Colin Barker, Apr 01 2016
G.f.: x*(1+x)*(1+50048*x+410874753*x^2+50048*x^3+x^4) / (1-80640398*x^3+x^6). - Colin Barker, Apr 01 2016
a(n) = 40320199*a(n-3) + 900685020*A271470(n-3) - 449440020 for n>3. - Muniru A Asiru, Apr 09 2016
A010888(a(3*n)) = A131598(n-1) where A131598 has period 3: repeat [2, 5, 8] and A010888 is digital root. - Michel Marcus, Dec 04 2014
Extensions
Merged with identical sequence submitted by Colin Barker, Apr 01 2016. - N. J. A. Sloane, Apr 06 2016
Comments