A106521 Numbers m such that Sum_{k=0..10} (m+k)^2 is a square.

18, 38, 456, 854, 9192, 17132, 183474, 341876, 3660378, 6820478, 73024176, 136067774, 1456823232, 2714535092, 29063440554, 54154634156, 579811987938, 1080378148118, 11567176318296, 21553408328294

Offset: 1

Author

Ralf Stephan, May 30 2005

Keywords

• nonn
• easy

Comments

Equivalently, 11*a(n)^2 + 110*a(n) + 385 is a square.

11*((m+5)^2+10) is a square iff the second factor is divisible by 11 and the quotient is a square, i.e., iff m = 11*k - 4 or m = 11*k - 6 and 11*k^2 +- 2 k + 1 is a square. Thus a(n) == (7,5,5,7,7,5,5,7,...) (mod 11), repeating with period 4 and the values are obtained by solving these Pell-type equations (cf. link to Dario Alpern's quadratic solver). The corresponding recurrence equations (see PARI code) should make it possible to prove the conjectured g.f. - M. F. Hasler, Jan 27 2008

All sequences of this type (i.e., sequences with fixed offset k, and a discernible pattern: k=0...10 for this sequence, k=0..1 for A001652) can be continued using a formula such as x(n) = a*x(n-p) - x(n-2p) + b, where a and b are various constants, and p is the period of the series. Alternatively 'p' can be considered the number of concurrent series. - Daniel Mondot, Aug 05 2016

Examples

			Since 18^2 + 19^2 + ... + 28^2 = 5929 = 77^2, 18 is in the sequence. - _Michael B. Porter_, Aug 07 2016
		

Links

• Vincenzo Librandi, Table of n, a(n) for n = 1..500
• Dario Alpern, Quadratic two integer variable equation solver
• Index entries for linear recurrences with constant coefficients, signature (1,20,-20,-1,1).

Crossrefs

Cf. A001032, A094196.

Programs

Mathematica

LinearRecurrence[{1,20,-20,-1,1},{18,38,456,854,9192},30] (* Harvey P. Dale, May 07 2011 *)

PARI

A106521(n)={local(xy=[ -4-2*(n%2);11],PQRS=[10,3;33,10],KL=[45;165]);until(0>=n-=2,xy=PQRS*xy+KL);xy[1]} \\ M. F. Hasler, Jan 27 2008

Formula

G.f.: 2*x*(9+10*x+29*x^2-x^3-2*x^4)/(1-x)/(1-20*x^2+x^4). - Vladeta Jovovic, May 31 2005; adapted to the offset by Bruno Berselli, May 16 2011

a(1)=18, a(2)=38, a(3)=456, a(4)=854, a(5)=9192; thereafter a(n)=a(n-1)+20*a(n-2)- 20*a(n-3)-a(n-4)+a(n-5). - Harvey P. Dale, May 07 2011

a(n) = A198949(n+1)-5. - Bruno Berselli, Feb 12 2012

a(1)=18, a(2)=38, a(3)=456, a(4)=854; thereafter a(n) = 20*a(n-2) - a(n-4) + 90. - Daniel Mondot, Aug 05 2016

Extensions

Edited and extended by M. F. Hasler, Jan 27 2008