0,2

Numbers simultaneously square and centered square. E.g. a(1)=25 because 25 is the fourth centered square number and the fifth square number. - Steven Schlicker (schlicks(AT)gvsu.edu), Apr 24 2007

Solutions to A007913(x)=A007913(2x-1) - Benoit Cloitre (benoit7848c(AT)orange.fr), Apr 07 2002

A. H. Beiler, Recreations in the Theory of Numbers, Dover, NY, 1964, p. 256.

Fielder, Daniel C.; Special integer sequences controlled by three parameters. Fibonacci Quart 6 1968 64-70.

Problem 47, Amer. Math. Monthly, 4 (1897), 25-28.

S. Schlicker, Numbers Simultaneously Polygonal and Centered Polygonal, submitted.

Eric Weisstein's World of Mathematics, Link to a section of The World of Mathematics.

a(n)=A078522(n)+1. a(n)=ceiling(A*B^n) where A=(3+2*sqrt(2))/8 and B=17+12*sqrt(2). - Benoit Cloitre (benoit7848c(AT)orange.fr), Jan 19 2003

G.f.: (1-10x+x^2)/((1-x)(1-34x+x^2)).

a(n) = ceiling(A046176(n)/sqrt(2)) - Helge Robitzsch (hrobi(AT)math.uni-goettingen.de), Jul 28 2000

a(n+1)=17*a(n)-4+6*sqrt(8*a(n)^2-4*a(n)). - Richard Choulet (richardchoulet(AT)yahoo.fr), Sep 14 2007

Define x(n) + y(n)*sqrt(8) = (4+sqrt(8))*(3+sqrt(8))^n, s(n) = (y(n)+1)/2; then a(n) = (1/2)*(2+4*(s(n)^2-s(n))) - Steven Schlicker (schlicks(AT)gvsu.edu), Apr 24 2007

CP := n -> 1+1/2*4*(n^2-n): N:=10: u:=3: v:=1: x:=4: y:=1: k_pcp:=[1]: for i from 1 to N do tempx:=x; tempy:=y; x:=tempx*u+8*tempy*v: y:=tempx*v+tempy*u: s:=(y+1)/2: k_pcp:=[op(k_pcp), CP(s)]: end do: k_pcp; - Steven Schlicker (schlicks(AT)gvsu.edu), Apr 24 2007

(PARI) a(n)=if(n<0, 0, sqr(subst(poltchebi(n+1)+poltchebi(n), x, 3)/4))

Cf. A000290, A001844, A007913.

Sequence in context: A012835 A132540 A122142 this_sequence A139163 A012692 A066852

Adjacent sequences: A008841 A008842 A008843 this_sequence A008845 A008846 A008847

nonn

njas

Entry edited by njas, Sep 14 2007