A072221 -id:A072221 - cp's OEIS frontend

A233474 Numbers m such that 5*T(m)-1 is a square, where T = A000217.

1, 4, 52, 169, 1993, 6436, 75700, 244417, 2874625, 9281428, 109160068, 352449865, 4145207977, 13383813460, 157408743076, 508232461633, 5977387028929, 19299449728612, 226983298356244, 732870857225641, 8619387950508361, 27829793124845764

Offset: 1

Bruno Berselli, Dec 11 2013

Sum_{i=1..infinity} 1/a(i) = 1.2758228304947598524736181699610...

			169 is in the sequence because 5*169*170/2-1 = 268^2.

Bruno Berselli, Table of n, a(n) for n = 1..200
Index entries for linear recurrences with constant coefficients, signature (1,38,-38,-1,1)

Cf. A000217, A129556 (numbers m such that 5*A000217(m)+1 is a square), A132593 (numbers m such that 5*A000217(m) is a square).

Cf. numbers m such that k*A000217(m)-1 is a square: A072221 for k=1; m=1 for k=2; this sequence for k=5.

LinearRecurrence[{1, 38, -38, -1, 1}, {1, 4, 52, 169, 1993}, 30]

G.f.: x*(1 + 3*x + 10*x^2 + 3*x^3 + x^4) / ((1 - x)*(1 - 38*x^2 + x^4)).
a(n) = a(n-1) + 38*a(n-2) - 38*a(n-3) - a(n-4) + a(n-5) for n>5, a(1)=1, a(2)=4, a(3)=52, a(4)=169, a(5)=1993.
a(n) = -1/2 + ( (15 + 4*sqrt(10)*(-1)^n)*(19 - 6*sqrt(10))^floor(n/2) + (15 - 4*sqrt(10)*(-1)^n)*(19 + 6*sqrt(10))^floor(n/2) )/20.

A174541 Baron Münchhausen's Sequence.

Original entry on oeis.org

0, 1, 1, 1, 2, 1, 1, 1, 2, 1, 1, 2, 2, 2, 1, 1, 2, 2, 2, 2, 1, 1, 2, 2, 1, 2, 2, 1, 1, 2, 2, 2, 2, 2, 2, 1, 1, 2, 2, 2, 2, 2, 2, 2, 1, 1, 2, 2, 1, 2, 2, 2, 2, 2, 1, 1, 2, 2, 2, 2, 2, 2, 2, 2, 2, 1, 1, 2, 2, 2, 2, 2, 2, 2, 2, 2, 2, 1, 1, 2, 2, 2, 2, 2, 2, 2, 2, 2, 2, 2, 1, 1, 2, 2, 2, 2, 2, 2, 2, 2, 2, 2, 2, 2, 1

Offset: 1

Tanya Khovanova, Konstantin Knop, and Alexey Radul, Mar 21 2010

Let n coins weighing 1, 2, ..., n grams be given. Suppose Baron Münchhausen knows which coin weighs how much, but his audience does not. Then a(n) is the minimum number of weighings the Baron must conduct on a balance scale, so as to unequivocally demonstrate the weight of at least one of the coins.
After a(1) = 0, a(n) is either 1 or 2 for all n.
a(n) = 1 for n triangular, n triangular-plus-one, T_n a square, and T_n a square-plus-one, where T_n is the n-th triangular number; a(n) = 2 for all other n > 1.

			a(7) = 1 because the weighing 1 + 2 + 3 < 7 conclusively demonstrates the weight of the seven-gram coin.

M. Brand, Tightening the bounds on the Baron's Omni-sequence, Discrete Math., 312 (2012), 1326-1335.
Tanya Khovanova, Konstantin Knop and A. Radul, Baron Münchhausen's Sequence, arXiv:1003.3406 [math.CO], 2010.
Tanya Khovanova, Konstantin Knop, and A. Radul, Baron Münchhausen's Sequence, J. Int. Seq. 13 (2010) # 10.8.7.
Tanya Khovanova and A. Radul, Another Coins Sequence

Cf. A000217, A000124, A001108, A072221, A186313.

triangularQ[n_] := IntegerQ[ Sqrt[8n+1]]; a[1] = 0; a[n_ /; triangularQ[n] || triangularQ[n-1] || IntegerQ[ Sqrt[n*(n+1)/2]] || IntegerQ[ Sqrt[n*(n+1)/2 - 1]]] = 1; a[] = 2; Table[a[n], {n, 1, 105}] (* _Jean-François Alcover, Jul 30 2012, after comments *)

;;; The following Scheme program generates terms of Baron
;;; Münchhausen's Sequence.
(define (acceptable? n)
  (or (triangle? n)
      (= n 2)
      (triangle? (- n 1))
      (square? (triangle n))
      (square? (- (triangle n) 1))))
(stream-map
 (lambda (n)
   (if (= n 1)
       0
       (if (acceptable? n)
           1
           2)))
 (the-integers))

A175492 Numbers m >= 3 such that binomial(m,3) + 1 is a square.

Original entry on oeis.org

7, 10, 24, 26, 65, 13777

Offset: 1

Ctibor O. Zizka, May 29 2010

Related sequences:
Numbers m such that binomial(m,2) is a square: A055997;
Numbers m such that binomial(m,2) + 1 is a square: A006451 + 1;
Numbers m such that binomial(m,2) - 1 is a square: A072221 + 1;
Numbers m >= 3 such that binomial(m,3) is a square: {3, 4, 50} (Proved by A. J. Meyl in 1878);
Numbers m >= 4 such that binomial(m,4) + 1 is a square: {6, 7, 45, 55, ...};
Numbers m >= 7 such that binomial(m,7) + 1 is a square: {8, 10, 21, 143, ...}.
No additional terms up to 10 million. - Harvey P. Dale, Apr 04 2017
No additional terms up to 10 billion. - Jon E. Schoenfield, Mar 18 2022
No additional terms up to 1 trillion. The sequence is finite by Siegel's theorem on integral points. - David Radcliffe, Jan 01 2024

Wikipedia, Tetrahedral number

Cf. A006451, A007318, A055997, A072221.

Cf. A216268 (values of binomial(m, 3)) and A216269 (square roots of binomial(m, 3) + 1).

lst = {}; k = 3; While[k < 10^6, If[ IntegerQ@ Sqrt[ Binomial[k, 3] + 1], AppendTo[lst, k]]; k++ ]; lst (* Robert G. Wilson v, Jun 11 2010 *)
Select[Range[3,14000],IntegerQ[Sqrt[Binomial[#,3]+1]]&] (* Harvey P. Dale, Apr 04 2017 *)

isok(m) = (m>=3) && issquare(binomial(m,3)+1); \\ Michel Marcus, Mar 15 2022

from sympy import binomial
from sympy.ntheory.primetest import is_square
for m in range(3, 10**6):
    if is_square(binomial(m,3)+1):
        print(m) # Mohammed Yaseen, Mar 18 2022

A229081 Numbers n such that there exists a square m^2 with 3n^2 - n <= m^2 <= 3n^2 + n.

Original entry on oeis.org

1, 3, 4, 7, 8, 11, 12, 14, 15, 16, 18, 19, 22, 23, 26, 27, 29, 30, 33, 34, 37, 38, 40, 41, 42, 44, 45, 48, 49, 52, 53, 55, 56, 57, 59, 60, 63, 64, 67, 68, 70, 71, 74, 75, 78, 79, 82, 83, 85, 86, 89, 90, 93, 94, 96, 97, 98, 100, 101, 104, 105, 108, 109, 111, 112, 113, 115, 116, 119, 120, 123, 124, 126

Offset: 1

Ralf Stephan, Sep 13 2013

nonn

			There is a square between 3*4^2-4 and 3*4^2+4 (44<=49<=52) but not between 3*5^2-5=70 and 3*5^2+5=80, so 4 is in sequence but not 5.

Robert Israel, Table of n, a(n) for n = 1..10000

Cf. A006451, A072221.

[n: n in [1..130] | exists{3*n^2+i: i in [-n..n] | IsSquare(3*n^2+i)}]; // Bruno Berselli, Sep 13 2013

filter:= n -> ceil(sqrt(3*n^2-n))<=floor(sqrt(3*n^2+n)):
select(filter, [$1..200]); # Robert Israel, Jan 05 2020

for(n=1,200,for(i=-n,n,f=0;if(issquare(3*n*n+i),f=1;break));if(f,print1(n,",")))

A233231 a(n) = 10*a(n-3) - a(n-6) + 4 for n>5, a(0)=2, a(1)=3, a(2)=5, a(3)=12, a(4)=29, a(5)=51.

Original entry on oeis.org

2, 3, 5, 12, 29, 51, 122, 291, 509, 1212, 2885, 5043, 12002, 28563, 49925, 118812, 282749, 494211, 1176122, 2798931, 4892189, 11642412, 27706565, 48427683, 115248002, 274266723, 479384645, 1140837612, 2714960669, 4745418771, 11293128122, 26875339971

Offset: 0

Frank M Jackson, Dec 06 2013

Apart from a(0), a(n) such that the triple (5,a(n),a(n)+1) forms a Heronian triangle. Equivalently, a(n) such that 6*(a(n)+3)*(a(n)-2) is a square. Note that this sequence generates all Heronian triples with a fixed side of 5 except (5,5,8) which is the only solution to Heronian triples of the form (5,x,x+3).

			a(5)=29 as the triangle with sides (5,29,30) has integer area 72.

Index entries for linear recurrences with constant coefficients, signature (1,0,10,-10,0,-1,1 ).

Cf. A072221, A230666.

seq[n_] := seq[n]=Which[n==0, 2, n==1, 3, n==2, 5, n==3, 12, n==4, 29, n==5, 51, True, 10seq[n-3]-seq[n-6]+4]; Table[seq[m], {m, 0, 100}]
LinearRecurrence[{1, 0, 10, -10, 0, -1, 1}, {2, 3, 5, 12, 29, 51, 122}, 30] (* T. D. Noe, Dec 09 2013 *)

G.f.: (2 + x + 2*x^2 - 13*x^3 + 7*x^4 + 2*x^5 + 3*x^6)/((1 - x)*(1 - 10*x^3 + x^6)). [Bruno Berselli, Dec 09 2013]

a(n) = a(n-1) + 10*a(n-3) - 10*a(n-4) - a(n-6) + a(n-7) for n>6. [Bruno Berselli, Dec 09 2013]

A233232 Primes p such that a Heronian triangle with a fixed side length of 5 contains p as another side length.

Original entry on oeis.org

3, 5, 13, 29, 509, 1213, 4892189, 111790443613, 8585304626467575518951161931678989213, 196181078582773936644856635510156388051229, 18087581947968179558090719299773079036323829315869

Offset: 1

Frank M Jackson, Dec 06 2013

nonn

The triangle inequality requires that any integer triangle with a fixed side length of 5 can have remaining side lengths of (x, x+1),...,(x, x+4). The constraint that primitive Heronian triangles have only one even side will just permit (x, x+1) and (x, x+3). Also there is only one solution for the triple (5, x, x+3) namely (5, 5, 8). So a(n) = x or x+1 whenever the Heronian triangle has x or x+1 as a prime and n is the ordered occurrence of this prime. The ordered sequence of x is given by A233231 where x(n) = 10x(n-3)-x(n-6)+4, x(0)=2, x(1)=3, x(2)=5, x(3)=12, x(4)=29, x(5)=51 starting at x(1).

			a(4)=29 because the triangle with sides (5, 29, 30) is Heronian, 29 is prime and is the 4th occurrence of such a prime.

Cf. A072221, A230666, A233231.

seq[n_] := seq[n]=Which[n==0, 2, n==1, 3, n==2, 5, n==3, 12, n==4, 29, n==5, 51, True, 10seq[n-3]-seq[n-6]+4]; lst={}; Do[Which[PrimeQ[seq[m]], AppendTo[lst, seq[m]], PrimeQ[seq[m]+1], AppendTo[lst, seq[m]+1], True, Null], {m, 1, 400}]; lst
t = LinearRecurrence[{1, 0, 10, -10, 0, -1, 1}, {2, 3, 5, 12, 29, 51, 122}, 400]; Select[Union[t, t + 1], PrimeQ[#] &] (* T. D. Noe, Dec 09 2013 *)

Primes of the form x(m) or x(m)+1 where x(m) is given by x(m) = 10x(m-3)-x(m-6)+4, x(0)=2, x(1)=3, x(2)=5, x(3)=12, x(4)=29, x(5)=51 starting at x(1).

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A233474 Numbers m such that 5*T(m)-1 is a square, where T = A000217.

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A174541 Baron Münchhausen's Sequence.

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A175492 Numbers m >= 3 such that binomial(m,3) + 1 is a square.

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A229081 Numbers n such that there exists a square m^2 with 3n^2 - n <= m^2 <= 3n^2 + n.

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A233231 a(n) = 10*a(n-3) - a(n-6) + 4 for n>5, a(0)=2, a(1)=3, a(2)=5, a(3)=12, a(4)=29, a(5)=51.

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A233232 Primes p such that a Heronian triangle with a fixed side length of 5 contains p as another side length.

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