A003153 -id:A003153 - cp's OEIS frontend

A082845 Partial sums of A082844.

2, 5, 7, 10, 12, 14, 17, 19, 22, 24, 26, 29, 31, 34, 36, 39, 41, 43, 46, 48, 51, 53, 55, 58, 60, 63, 65, 68, 70, 72, 75, 77, 80, 82, 84, 87, 89, 92, 94, 96, 99, 101, 104, 106, 109, 111, 113, 116, 118, 121, 123, 125, 128, 130, 133, 135, 138, 140, 142, 145, 147, 150, 152

Offset: 1

Benoit Cloitre, Apr 14 2003

nonn

Michael De Vlieger, Table of n, a(n) for n = 1..10000
Robbert Fokkink, The Pell Tower and Ostronometry, arXiv:2309.01644 [math.CO], 2023.

Not the same as A003153.

With[{s = Sqrt[2]}, Array[1 + Floor[# (1 + s) - (2 - s)] &, 120]] (* Michael De Vlieger, Sep 09 2023 *)

a(n) = 1+floor((1+sqrt(2))*n-(2-sqrt(2))).

A175155 Numbers m satisfying m^2 + 1 = x^2 * y^3 for positive integers x and y.

Original entry on oeis.org

0, 682, 1268860318, 1459639851109444, 2360712083917682, 86149711981264908618, 4392100110703410665318, 8171493471761113423918890682, 15203047261220215902863544865414318, 5484296027914919579181500526692857773246, 28285239023397517753374058381589688919682, 12439333951782387734360136352377558500557329868

Offset: 1

Michel Lagneau, Feb 27 2010

nonn

This sequence is infinite. The fundamental solution of m^2 + 1 = x^2 y^3 is (m,x,y) = (682,61,5), which means the Pellian equation m^2 - 125x^2 = -1 has the solution (m,x) = (682,61) = (m(1),x(1)). This Pellian equation admits an infinity of solutions (m(2k+1),x(2k+1)), k=1,2,..., given by the following recursive relation, starting with m(1)=682, x(1)= 61: m(2k+1) + x(2k+1)*sqrt(125) = (m(1) + x(1)*sqrt(125))^(2k+1).
Squares of these terms are in A060355, since both a(n)^2 and a(n)^2 + 1 are powerful (A001694). - Charles R Greathouse IV, Nov 16 2012
It appears that y = A077426. - Robert G. Wilson v, Nov 16 2012
Also m^2 + 1 is powerful. Other solutions arise from solutions x to x^2 - k^3*y^2 = -1. - Georgi Guninski, Nov 17 2012
Although it is believed that the b-file is complete for all terms m < 10^100, the search only looked for y < 100000. - Robert G. Wilson v, Nov 17 2012

			For m=682, m^2 + 1 = 465125 = 61^2 * 5^3.

Albert H. Beiler, "The Pellian" (Chap. 22), Recreations in the Theory of Numbers, 2nd ed. NY: Dover, 1966.
A. Cayley, Report of a committee appointed for the purpose of carrying on the tables connected with the Pellian equation ..., Collected Mathematical Papers. Vols. 1-13, Cambridge Univ. Press, London, 1889-1897, Vol. 13, pp. 430-443.
J. M. De Koninck, Ces nombres qui nous fascinent, Ellipses, 2008, p. 108.

Robert G. Wilson v, Table of n, a(n) for n = 1..292. [This list has not been proved to be complete! - _N. J. A. Sloane_, Nov 17 2012]
E. E. Whitford, The Pell equation, New York, 1912.
Wikipedia, Pell's equation

Pellian equation: A006704, A006702, A006705, A006703, A003153, A001571, A001570.

C:=array(0..20,0..20):C[1,1]=1: C[2,1]=1: n1:=682:x1:=61:for nn from 1 by 2 to 15 do:s:=0:for i from 2 to 15 do:for j from 1 to i do:C[i,j]:= C[i-1,j] + C[i-1,j-1]: od:od:for n from 1 by 2 to nn+1 do:s:=s + C[nn+1,n] * n1^(nn-n+1)*x1^(n-1)*125^((n-1)/2):od:print (s):od: # Michel Lagneau
# 2nd program R. J. Mathar, Mar 16 2016:
# print (nonsorted!) all solutions of A175155 up to search limit
with(numtheory):
# upper limit for solutions n
nsearchlim := 10^40 :
A175155y := proc(y::integer)
    local disc;
    disc := y^3 ;
    cfrac(sqrt(disc),periodic,quotients) ;
end proc:
for y from 2 do
    if issqrfree(y) then
        # find continued fraction for x^2-(y^3=disc)*y^2=-1, sqrt(disc)
        cf := A175155y(y) ;
        nlen :=  nops(op(2,cf)) ;
        if type(nlen,odd) then
            # fundamental solution
            fuso := numtheory[nthconver](cf,nlen-1) ;
            fusolx := numer(fuso) ;
            fusoly := denom(fuso) ;
            solx := fusolx ;
            soly := fusoly ;
            while solx <= nsearchlim do
                rhhs := solx^2-y^3*soly^2 ;
                if rhhs = -1 then
                    # print("n=",solx,"x=",soly,"y=",y^3) ;
                    print(solx) ;
                end if;
                # solutions from fundamental solution
                tempx := fusolx*solx+y^3*fusoly*soly ;
                tempy := fusolx*soly+fusoly*solx ;
                solx := tempx ;
                soly := tempy ;
            end do;
        end if;
    fi;
end do:

nmax = 10^50; ymax = 100; instances = 10; fi[y_] := n /. FindInstance[0 <= n <= nmax && x > 0 && n^2 + 1 == x^2*y^3, {n, x}, Integers, instances]; yy = Select[Range[1, ymax, 2], !IntegerQ[Sqrt[#]] && OddQ[ Length[ ContinuedFraction[Sqrt[#]][[2]]]]&]; Join[{0}, fi /@ yy // Flatten // Union // Most] (* Jean-François Alcover, Jul 12 2017 *)

is(n)=ispowerful(n^2+1) \\ Charles R Greathouse IV, Nov 16 2012

m(1)=682, x(1) = 61 and m(2k+1) + x(2k+1)*sqrt(125) = (m(1) + x(1)*sqrt(125))^(2k+1) m(2k+1) = C(2k+1,0) * m(1)^(2k+1) + C(2k+1,2)*m(1)^(2k-1)*x(1)^2 + ...

Added condition that x and y must be positive. Added missing initial term 0. Added warning that b-file has not been proved to be correct - there could be missing entries. - N. J. A. Sloane, Nov 17 2012

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A082845 Partial sums of A082844.

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A175155 Numbers m satisfying m^2 + 1 = x^2 * y^3 for positive integers x and y.

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