A100252 - OEIS

%I #12 Oct 17 2016 02:59:59

%S 36,4,9801,1225,81,225,9,0,196,64,36,441,3025,16,17689,100,484,0,2601,

%T 729,68121,225,25,7225,25921,81,1225,203401,441,1089,4761,196,15376,

%U 36,1936,511225,784,576,55071241,47089,1156,256,529046001,2916,1134225

%N Least square n-gonal number greater than 1, or 0 if none exists.

%C Also, let j be the smallest integer for which 1+(1+1*n)+(1+2*n)+... +(1+j*n)=k^2=s. Then a(n)=s; if no such j exists, then a(n)=0. Basis for sequence is shortest arithmetic series with initial term 1 and difference n that sums to a perfect square.

%C See A100251 and A188898 for the corresponding indices of these terms. Note that a(n) is zero for n = 10, 20, 52 (numbers in A188896). Although the Mathematica program searches only the first 25000 square numbers for n-gonal numbers, the Reduce function can show that there are no square n-gonal numbers (other than 0 and 1) for these n. - _T. D. Noe_, Apr 19 2011

%H Vincenzo Librandi, <a href="/A100252/b100252.txt">Table of n, a(n) for n = 3..100</a>

%H Eric W. Weisstein, <a href="http://mathworld.wolfram.com/PolygonalNumber.html">MathWorld: Polygonal Number</a>

%H Eric W. Weisstein, <a href="http://mathworld.wolfram.com/SquareNumber.html">MathWorld: Square Number</a>

%F 1+(1+1*n)+(1+2*n)+...+(1+A100254(n)*n) = 1+(1+1*n)+(1+2*n)+...+A100253(n) = A100251(n)^2 = a(n).

%e a(3)=9801 since 1 + 4 + 7 +...+ (1+80*3)= 99^2 = 9801 and no other arithmetic series with initial term 1, difference 3 and fewer terms sums to a perfect square.

%t NgonIndex[n_, v_] := (-4 + n + Sqrt[16 - 8*n + n^2 - 16*v + 8*n*v])/(n - 2)/2; Table[k = 2; While[sqr = k^2; i = NgonIndex[n, sqr]; k < 25000 && ! IntegerQ[i], k++]; If[k == 25000, k = sqr = i = 0]; sqr, {n, 3, 64}] (* _T. D. Noe_, Apr 19 2011 *)

%Y Cf. A000290 (squares), A188891 (similar sequence for triangular numbers).

%K nonn

%O 3,1

%A _Charlie Marion_, Nov 21 2004