cp's OEIS Frontend

This is a front-end for the Online Encyclopedia of Integer Sequences, made by Christian Perfect. The idea is to provide OEIS entries in non-ancient HTML, and then to think about how they're presented visually. The source code is on GitHub.

A060984 a(1) = 1; a(n+1) = a(n) + (largest square <= a(n)).

Original entry on oeis.org

1, 2, 3, 4, 8, 12, 21, 37, 73, 137, 258, 514, 998, 1959, 3895, 7739, 15308, 30437, 60713, 121229, 242333, 484397, 967422, 1933711, 3865811, 7730967, 15459367, 30912128, 61814609, 123625653, 247235577, 494448306, 988888002, 1977738918, 3955408759, 7910812423
Offset: 1

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Author

R. K. Guy, May 11 2001

Keywords

Comments

Arises in analyzing "put-or-take" games (see Winning Ways, 484-486, 501-503), the prototype being Epstein's Put-or-Take-a-Square game.

References

  • E. R. Berlekamp, J. H. Conway and R. K. Guy, Winning Ways, Academic Press, NY, 2 vols., 1982.
  • R. K. Guy, Unsolved Problems in Number Theory, Springer, 1st edition, 1981. See section E26.

Crossrefs

Programs

  • Haskell
    a060984 n = a060984_list !! (n-1)
    a060984_list = iterate (\x -> x + a048760 x) 1
    -- Reinhard Zumkeller, Dec 24 2013
    
  • Mathematica
    a[1] = 1; a[n_] := a[n] = a[n - 1] + Floor[ Sqrt[ a[n - 1] ] ]^2; Table[ a[n], {n, 1, 40} ]
    RecurrenceTable[{a[1]==1,a[n]==a[n-1]+Floor[Sqrt[a[n-1]]]^2},a,{n,40}] (* Harvey P. Dale, Nov 19 2011 *)
    NestList[#+Floor[Sqrt[#]]^2&,1,40] (* Harvey P. Dale, Jan 22 2013 *)
  • PARI
    { default(realprecision, 100); for (n=1, 500, if (n==1, a=1, a+=floor(sqrt(a))^2); write("b060984.txt", n, " ", a); ) } \\ Harry J. Smith, Jul 15 2009
    
  • Python
    from sympy import integer_nthroot
    A060984_list = [1]
    for i in range(10**3): A060984_list.append(A060984_list[-1]+integer_nthroot(A060984_list[-1],2)[0]**2) # Chai Wah Wu, Apr 02 2021
    
  • Python
    from math import isqrt
    from itertools import accumulate
    def f(an, _): return an + isqrt(an)**2
    print(list(accumulate([1]*36, f))) # Michael S. Branicky, Apr 02 2021

Formula

a(n+1) = a(n)+[sqrt(a(n))]^2 = a(n)+A061886(n) = a(n)+A048760(a(n)) = A061887(a(n)). - Henry Bottomley, May 12 2001
a(n) ~ c * 2^n, where c = 0.11511532187216693... (see A237888). - Vaclav Kotesovec, Feb 15 2014

Extensions

More terms from David W. Wilson, Henry Bottomley and Robert G. Wilson v, May 12 2001