A214963 -id:A214963 - cp's OEIS frontend

A214961 a(0)=a(1)=1, a(n) = least k > a(n-1) such that k*a(n-2) is a triangular number.

1, 1, 3, 6, 7, 11, 13, 21, 25, 30, 49, 59, 97, 117, 193, 233, 385, 465, 492, 596, 983, 1191, 1965, 2381, 2516, 4761, 5031, 5761, 6290, 8466, 9795, 15470, 15867, 17403, 20559, 24170, 26945, 27192, 27755, 30130, 35235, 43537, 45100, 56805, 58717, 58739, 91000, 117477

Offset: 0

Alex Ratushnyak, Aug 03 2012

nonn

Robert Israel, Table of n, a(n) for n = 0..2100

Cf. A214914, A214915, A214916, A214963, A213005.

f:= proc(a,b)
  local s;
  s:= map(t -> rhs(op(t)), [msolve(x^2=1, 8*a)]);
  min(select(`>`, map(t -> (t^2-1)/(8*a), s), b))
end proc:
A[0]:= 1: A[1]:= 1:
for nn from 2 to 100 do
  A[nn]:= f(A[nn-2],A[nn-1])
od:
seq(A[i],i=0..100); # Robert Israel, Jun 17 2020

a[0]=a[1]=1;a[n_]:=a[n]=(k=a[n-1]+1;While[!IntegerQ@Sqrt[1+8*a[n-2]k],k++];k);Array[a,50,0] (* Giorgos Kalogeropoulos, May 21 2021 *)
lktn[{a_,b_}]:=Module[{k=b+1},While[!OddQ[Sqrt[8a k+1]],k++];{b,k}]; NestList[lktn,{1,1},50][[;;,1]] (* Harvey P. Dale, Sep 09 2023 *)

prpr = prev = 1
for n in range(1, 55):
    print(prpr, end=', ')
    b = k = 0
    while k<=prev:
        d = b*(b+1)//2
        k = 0
        if d%prpr==0:
            k = d // prpr
        b += 1
    prpr = prev
    prev = k

A214916 a(0) = a(1) = 1, a(n) = n! / a(n-2).

Original entry on oeis.org

1, 1, 2, 6, 12, 20, 60, 252, 672, 1440, 5400, 27720, 88704, 224640, 982800, 5821200, 21288960, 61102080, 300736800, 1990850400, 8089804800, 25662873600, 138940401600, 1007370302400, 4465572249600, 15397724160000, 90311261040000, 707173952284800, 3375972620697600

Offset: 0

Alex Ratushnyak, Aug 03 2012

nonn

a(n) is least k > a(n-1) such that k*a(n-2) is a factorial.

Two periodic subsets of these numbers appear in the coefficients of a series involved in a solution of a Riccati-type differential equation addressed by the Bernoulli brothers: z = 1 - x^4/12 + x^8/672 - x^12/88704 + ... = 1 - 2 * x^4/4! + 60 * x^8/8! - 5400 * x^12/12! + ... . See the MathOverflow question. - Tom Copeland, Jan 24 2017

Joerg Arndt, Table of n, a(n) for n = 0..100
MathOverflow, Link of a power series by the Bernoulli brothers for a Riccati equation to zonotopes?, a MathOverflow question by Tom Copeland, 2017.

Cf. A214914, A214915, A214961, A214963, A007662.

[1] cat  [n le 2 select n  else  Factorial(n)  div  Self(n-2): n in [1..30]]; // Vincenzo Librandi, Jan 25 2017

import math
prpr = prev = 1
for n in range(2, 33):
    print(prpr, end=', ')
    cur = math.factorial(n) // prpr
    prpr = prev
    prev = cur

a(0) = a(1) = 1, for n>=2, a(n) = n! / a(n-2).

a(n) = n!!!! * (n-1)!!!! = A007662(n) * A007662(n-1), for n >= 1. - David Radcliffe, Jun 05 2025

A214915 a(0)=a(1)=1, a(n) = least k > a(n-1) such that k*a(n-2) is a Lucas number.

Original entry on oeis.org

1, 1, 2, 3, 9, 41, 642, 45378, 300467529, 141919565128041, 23889121268674415030529282, 1882004232433803432495089121844337244861378, 15480785260333522531462377514739365882017797041068613651703152050407493402249

Offset: 0

Alex Ratushnyak, Aug 03 2012

nonn

Cf. A214914, A214916, A214961, A214963.

cp's OEIS Frontend

A214961 a(0)=a(1)=1, a(n) = least k > a(n-1) such that k*a(n-2) is a triangular number.

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A214916 a(0) = a(1) = 1, a(n) = n! / a(n-2).

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A214915 a(0)=a(1)=1, a(n) = least k > a(n-1) such that k*a(n-2) is a Lucas number.

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