Search: id:A056220
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%I A056220
%S A056220 1,1,7,17,31,49,71,97,127,161,199,241,287,337,391,449,511,577,647,721,
%T A056220 799,881,967,1057,1151,1249,1351,1457,1567,1681,1799,1921,2047,2177,
%U A056220 2311,2449,2591,2737,2887,3041,3199,3361,3527,3697,3871,4049,4231,4417
%V A056220 -1,1,7,17,31,49,71,97,127,161,199,241,287,337,391,449,511,577,647,721,
               799,881,967,
%W A056220 1057,1151,1249,1351,1457,1567,1681,1799,1921,2047,2177,2311,2449,2591,
               2737,2887,3041,
%X A056220 3199,3361,3527,3697,3871,4049,4231,4417
%N A056220 2*n^2-1.
%C A056220 Image of squares (A000290) under "little Hankel" transform that sends 
               [c_0, c_1, ...] to [d_0, d_1, ...] where d_n = c_n^2 - c_{n+1}*c_{n-1}.
%C A056220 Also surround numbers of an n X n square - Jason Earls (zevi_35711(AT)yahoo.com), 
               Apr 16 2001
%C A056220 Also 8n + 8 is a square. - Cino Hilliard (hillcino368(AT)gmail.com), 
               Dec 18 2003
%C A056220 The sums of the consecutive integer sequences 2n^2 to 2(n+1)^2-1 are 
               cubes, as 2n^2+...+2(n+1)^2-1 = (1/2)(2(n+1)^2-1-2n^2+1)(2(n+1)^2-1+2n^2)=(2n+1)^3. 
               Eg 2+3+4+5+6+7 = 27 =3^3, then 8+9+10+..+17 = 125 = 5^3. - Andras 
               Erszegi (erszegi.andras(AT)chello.hu), Apr 29 2005
%C A056220 Sequence allows us to find X values of the equation: 2*X^3 + 2*X^2 = 
               Y^2. To find Y values: b(n)=2n(2*n^2 - 1). - Mohamed Bouhamida (bhmd95(AT)yahoo.fr), 
               Nov 06 2007
%C A056220 Average of the squares of two consecutive terms is also a square. In 
               fact: (2*n^2 - 1)^2 + (2*(n+1)^2 - 1)^2 = 2*(2*n^2 + 2*n + 1)^2 [From 
               Matias Saucedo (solomatias(AT)yahoo.com.ar), Aug 18 2008]
%C A056220 Equals row sums of triangle A143593 & binomial transform of [1, 6, 4, 
               0, 0, 0,...]. [From Gary W. Adamson (qntmpkt(AT)yahoo.com), Aug 26 
               2008]
%C A056220 Also, except (-1,1), a(n)=2*n^2+8n+7; fourth column of [A154685] [From 
               Vincenzo Librandi (vincenzo.librandi(AT)tin.it), Jan 25 2009]
%C A056220 sqrt(a(n) + a(n+1) + 1) = 2n+1 [From Doug Bell (bell.doug(AT)gmail.com), 
               Mar 09 2009]
%C A056220 Apart the first term which is -1 the number of units of a(n) belongs 
               to a periodic sequence: 1, 7, 7, 1, 9.We conclude that a(n) and a(n+5) 
               have the same number of units. [From Mohamed Bouhamida (bhmd95(AT)yahoo.fr), 
               Sep 05 2009]
%H A056220 <a href="Sindx_Rea.html#recLCC">Index entries for sequences related to 
               linear recurrences with constant coefficients</a>
%F A056220 G.f.: (-1+4x+x^2)/(1-x)^3.
%F A056220 a(n) = A119258(n+1,2) for n>0. - Reinhard Zumkeller (reinhard.zumkeller(AT)gmail.com), 
               May 11 2006
%F A056220 Starting with offset 1, equals binomial transform of [1, 6, 4, 0, 0, 
               0,...] [From Gary W. Adamson (qntmpkt(AT)yahoo.com), Aug 29 2008]
%F A056220 Contribution from Doug Bell (bell.doug(AT)gmail.com), Mar 08 2009: (Start)
%F A056220 a(0) = -1,
%F A056220 a(n) = sqrt(A001844(n)^2 - A069074(n-1)),
%F A056220 a(n+1) = sqrt(A001844(n)^2 + A069074(n-1)) = sqrt(a(n)^2 + A069074(n-1)*2) 
               (End)
%e A056220 a(0) = 0^2-1*1 = -1, a(1) = 1^2-4*0 = 1, a(2) = 2^2-9*1 = 7 etc.
%e A056220 a(4) = 31 = (1, 3, 3, 1) dot (1, 6, 4, 0) = (1 + 18 + 12 + 0). [From 
               Gary W. Adamson (qntmpkt(AT)yahoo.com), Aug 29 2008]
%t A056220 lst={};Do[AppendTo[lst, 2*n^2-1], {n, 0, 4^2}];lst...or/and... lst={};
               Do[AppendTo[lst, ChebyshevT[2, n]], {n, 0, 4^2}];lst [From Vladimir 
               Orlovsky (4vladimir(AT)gmail.com), Sep 11 2008]
%t A056220 1. Table[2*n^2 + 4*n + 1, {n, -1, 46}] (.) 2. lst = {}; Do[a = 2*n^2 
               + 4*n + 1; AppendTo[lst, a], {n, -1, 46}] lst [From Zerinvary Lajos 
               (zerinvarylajos(AT)yahoo.com), Jul 10 2009]
%o A056220 (PARI) a(n)=if(n<0,0,2*n^2-1)
%Y A056220 Cf. A047875, A000105, A077585.
%Y A056220 Cf. A005563, A046092, A001082, A002378, A036666, A062717, A028347, A087475, 
               A000217.
%Y A056220 Cf. A143593 [From Gary W. Adamson (qntmpkt(AT)yahoo.com), Aug 26 2008]
%Y A056220 Cf. A154685 [From Vincenzo Librandi (vincenzo.librandi(AT)tin.it), Jan 
               25 2009]
%Y A056220 Sequence in context: A046118 A120092 A130284 this_sequence A024840 A024835 
               A144861
%Y A056220 Adjacent sequences: A056217 A056218 A056219 this_sequence A056221 A056222 
               A056223
%K A056220 sign
%O A056220 0,3
%A A056220 N. J. A. Sloane (njas(AT)research.att.com), Aug 06 2000
%E A056220 Formula and additional comments from Henry Bottomley (se16(AT)btinternet.com), 
               Dec 12 2000

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