

A106521


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


10



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

1,1


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 Pelltype 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(np)  x(n2p) + 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


LINKS



FORMULA

a(1)=18, a(2)=38, a(3)=456, a(4)=854, a(5)=9192; thereafter a(n)=a(n1)+20*a(n2) 20*a(n3)a(n4)+a(n5).  Harvey P. Dale, May 07 2011
a(1)=18, a(2)=38, a(3)=456, a(4)=854; thereafter a(n) = 20*a(n2)  a(n4) + 90.  Daniel Mondot, Aug 05 2016


EXAMPLE

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


MATHEMATICA

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


PROG

(PARI) A106521(n)={local(xy=[ 42*(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


CROSSREFS



KEYWORD

nonn,easy


AUTHOR



EXTENSIONS



STATUS

approved



