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.

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A276530 a(n) = (a(n-1) * a(n-5) + a(n-3)^3) / a(n-6), a(0) = a(1) = ... = a(5) = 1.

Original entry on oeis.org

1, 1, 1, 1, 1, 1, 2, 3, 4, 12, 39, 142, 1077, 21209, 779449, 106636837, 245010524697, 3336696488691229, 1125981890791313205482, 693480182652378523758257457499, 47660918720485535883730945247863294175948, 13387114027268508450553229985503810242341235794343085252
Offset: 0

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Author

Seiichi Manyama, Nov 16 2016

Keywords

Crossrefs

Programs

  • Mathematica
    RecurrenceTable[{a[n] == (a[n - 1] a[n - 5] + a[n - 3]^3)/a[n - 6], a[0] == a[1] == a[2] == a[3] == a[4] == a[5] == 1}, a, {n, 0, 21}] (* Michael De Vlieger, Nov 16 2016 *)
    nxt[{a_,b_,c_,d_,e_,f_}]:={b,c,d,e,f,(f b+d^3)/a}; NestList[nxt,{1,1,1,1,1,1},25][[;;,1]] (* Harvey P. Dale, Apr 21 2023 *)
  • Ruby
    def A(k, m, n)
      a = Array.new(2 * k, 1)
      ary = [1]
      while ary.size < n + 1
        i = a[-1] * a[1] + a[k] ** m
        break if i % a[0] > 0
        a = *a[1..-1], i / a[0]
        ary << a[0]
      end
      ary
    end
    def A276530(n)
      A(3, 3, n)
    end

A375621 a(n) = (a(n-3)*a(n-5) + a(n-1)*a(n-7))/a(n-8) with a(0) = ... = a(7) = 1.

Original entry on oeis.org

1, 1, 1, 1, 1, 1, 1, 1, 2, 3, 4, 6, 9, 17, 35, 106, 210, 385, 1028, 2767, 9761, 32795, 129759, 351733, 1076957, 6165427, 27815973, 148629048, 721531991, 3768314574, 17276660082, 109959356649, 1149560654775, 7208229224331, 53412249630318, 392919259603556
Offset: 0

Views

Author

Mohamed Bensaid, Aug 21 2024

Keywords

Comments

Sequence defined by recursion derived from Sato discrete tau function.

Crossrefs

Programs

  • Maple
    a:= proc(n) option remember; `if`(n<8, 1,
         (a(n-3)*a(n-5) + a(n-1)*a(n-7))/a(n-8))
        end:
    seq(a(n), n=0..35);  # Alois P. Heinz, Aug 24 2024
  • Python
    def calculate_terms(n):
        a = [1] * n
        for l in range(n - 8):
            a[l + 8] = (a[l + 3] * a[l + 5] + a[l + 7] * a[l + 1]) // a[l]
        return a
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