A001108 -id:A001108 - cp's OEIS frontend

A361670 Squarefree part of the n-th triangular number.

1, 3, 6, 10, 15, 21, 7, 1, 5, 55, 66, 78, 91, 105, 30, 34, 17, 19, 190, 210, 231, 253, 69, 3, 13, 39, 42, 406, 435, 465, 31, 33, 561, 595, 70, 74, 703, 741, 195, 205, 861, 903, 946, 110, 115, 1081, 282, 6, 1, 51, 1326, 1378, 159, 165, 385, 399, 1653, 1711, 1770, 1830, 1891, 217, 14, 130, 2145, 2211, 2278

Offset: 1

R. J. Mathar, Mar 20 2023

a(n) / A083481(n) is either 2 or 1/2 depending on A136480(n) being even or odd, which is indicated by A039963(n).

a(n) = 1 for n>0 in A001108. - Michel Marcus, Mar 22 2023

Cf. A000217, A007913, A083481 (of oblong), A361671 (of tetrahedral).

Cf. A001108, A039963.

a:= n-> mul(i[1]^irem(i[2], 2), i=ifactors(n*(n+1)/2)[2]):
seq(a(n), n=1..70);  # Alois P. Heinz, Mar 20 2023

a(n) = core(n*(n+1)/2); \\ Michel Marcus, Mar 22 2023

from sympy.ntheory.factor_ import core
def A361670(n): return core(n*(n+1)>>1) # Chai Wah Wu, Mar 20 2023

a(n) = A007913(A000217(n)).

A082649 Triangle of coefficients in expansion of sinh^2(n*x) in powers of sinh(x).

Original entry on oeis.org

1, 4, 4, 16, 24, 9, 64, 128, 80, 16, 256, 640, 560, 200, 25, 1024, 3072, 3456, 1792, 420, 36, 4096, 14336, 19712, 13440, 4704, 784, 49, 16384, 65536, 106496, 90112, 42240, 10752, 1344, 64, 65536, 294912, 552960, 559104, 329472, 114048, 22176, 2160, 81, 262144, 1310720, 2785280, 3276800, 2329600

Offset: 1

Gary W. Adamson, May 16 2003, suggested by Herb Conn

Using arcsin(x) = Pi/2 - arccos(x), valid for -1 < = x <= 1, we find sin^2(k*arcsin(x)) = sin^2(k*arccos(x)) for k odd, while sin^2(k*arcsin(x)) = 1 - sin^2(k*arccos(x)) for k even. Thus the expansion of sin^2(n*x) in powers of cos(x) will produce a similar table of coefficients. See the example section below. - Peter Bala, Feb 02 2017

			sinh^2 x = sinh^2 x
sinh^2 2x = 4 sinh^4 x + 4 sinh^2 x
sinh^2 3x = 16 sinh^6 x + 24 sinh^4 x + 9 sinh^2 x
sinh^2 4x = 64 sinh^8 x + 128 sinh^6 x + 80 sinh^4 x + 16 sinh^2 x
sinh^2 5x = 256 sinh^10 x + 640 sinh^8 x + 560 sinh^6 x + 200 sinh^4 x + 25 sinh^2 x
From _Peter Bala_, Feb 02 2016: (Start)
sin^2(x) = 1 - cos^2(x);
sin^2(2*x) = -4*cos^4(x) + 4*cos^2(x);
sin^2(3*x) = 1 - (16*cos^6(x) - 24*cos^4(x) + 9*cos^2(x));
sin^2(4*x) = -64*cos^8(x) + 128*cos^6(x) - 80*cos^4(x) + 16*cos^2(x);
sin^2(5*x) = 1 - (256*cos^10(x) - 640*cos^8(x) + 560*cos^6(x) - 200*cos^4(x) + 25*cos^2(x)). (End)

Robert Israel, Table of n, a(n) for n = 1..10011 (rows 0 to 140, flattened)

A001108 gives row sums.

Closely related to A123583 and A123588.

g:= (1+x*y)/((1-x*y)*(1-(4+2*y)*x+x^2*y^2)):
S:= series(g,x,15):
seq(seq(coeff(coeff(S,x,n),y,k),k=0..n),n=0..14); # Robert Israel, Dec 20 2017

Table[Reverse[CoefficientList[1/x TrigExpand[Sinh[n ArcSinh[Sqrt[x]]]^2], x]], {n, 7}] // Flatten (* Eric W. Weisstein, Apr 05 2017 *)
Abs[Table[CoefficientList[x^n Piecewise[{{1 - ChebyshevT[n, 1/Sqrt[x]]^2, Mod[n, 2] == 0}, {ChebyshevT[n, 1/Sqrt[x]]^2, Mod[n, 2] == 1}}], x], {n, 10}]] // Flatten (* Eric W. Weisstein, Apr 05 2017 *)

Coefficients are: 4^(n-1), (2n)4^(n-2), (2n)(2n-3)4^(n-3)/2!, (2n)(2n-4)(2n-5)4^(n-4)/3!, (2n)(2n-5)(2n-6)(2n-7)4^(n-5)/4!, (2n)(2n-6)(2n-7)(2n-8)(2n-9)4^(n-6)/5!...

G.f. as triangle: (1+x*y)/((1-x*y)*(1-(4+2*y)*x+x^2*y^2)). - Robert Israel, Dec 20 2017

More terms from Robert Israel, Dec 20 2017

A135203 For any integer n >= 1 the sequence gives the minimum power x for which n^x+(n-1)^x+(n-2)^x+...+1^x produces a perfect square.

Original entry on oeis.org

1, 3, 3, 3, 3, 3, 3, 1, 3, 3, 3, 3, 3, 3, 3, 3, 3, 3, 3, 3, 3, 3, 3, 2, 3, 3, 3, 3, 3, 3, 3, 3, 3, 3, 3, 3, 3, 3, 3, 3, 3, 3, 3, 3, 3, 3, 3, 3, 1, 3, 3, 3, 3, 3, 3, 3, 3, 3, 3, 3, 3, 3, 3, 3, 3, 3, 3, 3, 3, 3, 3, 3, 3, 3, 3, 3, 3, 3, 3, 3, 3, 3, 3, 3, 3, 3, 3, 3, 3, 3, 3, 3, 3, 3, 3, 3, 3, 3, 3, 3

Offset: 1

Paolo P. Lava and Giorgio Balzarotti, Nov 28 2007

All 3's apart from 1's in positions given by A001108 and 2 for n=24.

			n=4 -> 4^3+3^3+2^3+1^3 = 64+27+8+1 = 100
n=5 -> 5^3+4^3+3^3+2^3+1^3 = 125+64+27+8+1 = 225
n=6 -> 6^3+5^3+4^3+3^3+2^3+1^3 = 216+125+64+27+8+1 = 441

Antti Karttunen, Table of n, a(n) for n = 1..65537

Cf. A001108.

P:=proc(n) local a,i,k,j,ok,x; for i from 1 by 1 to n do x:=1; ok:=1; while ok=1 do a:=0; k:=i; while k>0 do a:=a+k^x; k:=k-1; od; if (trunc(sqrt(a)))^2=a then print(x); ok:=0; else x:=x+1; fi; od; od; end: P(100);

A135203(n) = for(x=1,oo,if(issquare(sum(k=1,n,k^x)), return(x))); \\ Antti Karttunen, Sep 27 2018

Offset and a typo in the definition corrected by Antti Karttunen, Sep 27 2018

A144844 a(n) = ((2 + sqrt(2))^n - (2 - sqrt(2))^n)^2/8.

Original entry on oeis.org

0, 1, 16, 196, 2304, 26896, 313600, 3655744, 42614784, 496754944, 5790601216, 67500196864, 786839961600, 9172078759936, 106917585289216, 1246322708463616, 14528202160472064, 169353135091941376, 1974124812461670400, 23012085209172803584

Offset: 0

Al Hakanson (hawkuu(AT)gmail.com), Sep 22 2008

Vincenzo Librandi, Table of n, a(n) for n = 0..950
Index entries for linear recurrences with constant coefficients, signature (14,-28,8).

Z:=PolynomialRing(Integers()); N:=NumberField(x^2-2); [ Integers()!a: a in [ ((2+r2)^n-(2-r2)^n)^2/8: n in [0..19] ] ]; // Klaus Brockhaus, Oct 20 2008

I:=[0,1,16]; [n le 3 select I[n] else 14*Self(n-1)-28*Self(n-2)+8*Self(n-3): n in [1..20]]; // Vincenzo Librandi, Feb 05 2018

Table[ Simplify[ ((2 + Sqrt@2)^n - (2 - Sqrt@2)^n)^2/8], {n, 0, 19}] (* Robert G. Wilson v, Sep 24 2008 *)
CoefficientList[Series[x (1 + 2 x) / ((1 - 2 x) (1 - 12 x + 4 x^2)), {x, 0, 33}], x] (* Vincenzo Librandi, Feb 06 2018 *)
LinearRecurrence[{14,-28,8},{0,1,16},20] (* Harvey P. Dale, Apr 12 2020 *)

my(x='x+O('x^30)); concat([0], Vec(x*(1+2*x)/((1-2*x)*(1-12*x+4*x^2)) )) \\ G. C. Greubel, Sep 27 2018

From R. J. Mathar, Sep 24 2008: (Start)
G.f.: x*(1+2*x)/((1-2*x)*(1-12*x+4*x^2)).
a(n) = 2^(n-2)*(A001109(n+1) - 3*A001109(n) - 1) = 2^(n-1)*A001108(n). (End)
a(n) = 14*a(n-1) - 28*a(n-2) + 8*a(n-3) for n > 2; a(0) = 0, a(1) = 1; a(2) = 16. - Klaus Brockhaus, Jul 15 2009
a(n) = A007070(n-1)^2 = (((2+sqrt(2))^n - (2-sqrt(2))^n) / sqrt(8))^2. - Franklin T. Adams-Watters, Aug 06 2009
a(n) = 2^(n-3)*(Q(2*n) - 2), where Q(m) are the Pell-Lucas numbers (A002203). - G. C. Greubel, Sep 27 2018

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))

A205185 Period 6: repeat [1, 8, 9, 8, 1, 0].

Original entry on oeis.org

1, 8, 9, 8, 1, 0, 1, 8, 9, 8, 1, 0, 1, 8, 9, 8, 1, 0, 1, 8, 9, 8, 1, 0, 1, 8, 9, 8, 1, 0, 1, 8, 9, 8, 1, 0, 1, 8, 9, 8, 1, 0, 1, 8, 9, 8, 1, 0, 1, 8, 9, 8, 1, 0, 1, 8, 9, 8, 1, 0, 1, 8, 9, 8, 1, 0, 1, 8, 9, 8, 1, 0, 1, 8, 9, 8, 1, 0, 1, 8, 9, 8, 1, 0, 1, 8

Offset: 1

Ant King, Jan 24 2012

The terms of this sequence are the units' digits of the indices of those nonzero triangular numbers that are also perfect squares (A001108).

			As the fourth nonzero triangular number that is also a perfect square is A000217(288), and 288 has units' digit A010879(288)=8, then a(4)=8.

Vincenzo Librandi, Table of n, a(n) for n = 1..1000
Index entries for linear recurrences with constant coefficients, signature (0, 0, 0, 0, 0, 1).

Cf. A000217, A000290, A001108, A010879.

&cat[[1, 8, 9, 8, 1, 0]: n in [0..20]]; // Wesley Ivan Hurt, Jun 18 2016

A205185:=n->[1, 8, 9, 8, 1, 0][(n mod 6)+1]: seq(A205185(n), n=0..100); # Wesley Ivan Hurt, Jun 18 2016

LinearRecurrence[{0, 0, 0, 0, 0, 1}, {1, 8, 9, 8, 1, 0}, 86]
PadRight[{}, 100, {1, 8, 9, 8, 1, 0}] (* Vincenzo Librandi, Jun 19 2016 *)

a(n)=[0, 1, 8, 9, 8, 1][n%6+1] \\ Charles R Greathouse IV, Jul 17 2016

G.f.: x*(1+7x+x^2) / ((1-x)*(1+x)*(1-x+x^2)).
a(n) = a(n-6) for n>6.
a(n) = 9-a(n-3) for n>3.
a(n) = a(n-1) - a(n-3) + a(n-4) for n>4.
a(n) = 1/6*(27+(-1)^n*(5-32*cos(2*n*Pi/3))).

A212797 Number of spanning trees in C_2 X C_n.

Original entry on oeis.org

2, 32, 294, 2304, 16810, 117600, 799694, 5326848, 34928082, 226195360, 1450199542, 9220780800, 58221203066, 365440965344, 2282085037470, 14187697422336, 87860208024994, 542209573735200, 3335797263902918, 20465738163774720, 125247216613782858

Offset: 1

N. J. A. Sloane, May 27 2012

nonn

From Harry Richman and Alois P. Heinz, Jan 31 2023: (Start)
Row 2 of array in A212796.
a(n) is a divisibility sequence, i.e., if m divides n then a(m) divides a(n), since A001108 is one. (End)

Germain Kreweras, Complexité et circuits Eulériens dans les sommes tensorielles de graphes, J. Combin. Theory, B 24 (1978), 202-212. See p. 210, Parag. 4.
Index entries for linear recurrences with constant coefficients, signature (14,-63,100,-63,14,-1).

Cf. A001108, A212796.

default(realprecision, 120);
{a(n) = round(n*2^(2*n-1)*prod(k=1, n-1, 1+sin(k*Pi/n)^2))} \\ Seiichi Manyama, Jan 13 2021

my(N=66, x='x+O('x^N)); Vec(2*x*deriv(x*(1+x)/((1-x)*(1-6*x+x^2)))) \\ Seiichi Manyama, Jan 13 2021

G.f.: 2*x*(1+2*x-14*x^2+2*x^3+x^4)/((x-1)^2*(1-6*x+x^2)^2) . - R. J. Mathar, Apr 16 2018
Conjecture: a(n) = 16*A001109(n+1) +3*A001109(n) -16*(A144133(n)-6*A144133(n-1)) -n = 3*A144133(n-1) -2*A144133(n-2) +3*A144133(n-3) -n. - R. J. Mathar, Apr 16 2018
From Seiichi Manyama, Jan 13 2021: (Start)
a(n) = 2 * n * A001108(n).
a(n) = 14*a(n-1) - 63*a(n-2) + 100*a(n-3) - 63*a(n-4) + 14*a(n-5) - a(n-6) for n > 6. (End)

More terms from Seiichi Manyama, Jan 13 2021

A214728 Least k such that n + (n+1) + ... + (n+k-1) is a square.

Original entry on oeis.org

1, 1, 3, 5, 1, 9, 11, 13, 15, 1, 19, 3, 2, 25, 27, 29, 1, 33, 5, 37, 39, 8, 43, 45, 2, 1, 3, 53, 55, 57, 59, 61, 9, 65, 67, 6, 1, 8, 75, 11, 2, 81, 83, 5, 87, 9, 13, 3, 95, 1, 99, 101, 18, 15, 107, 109, 111, 8, 10, 117, 2, 121, 24, 125, 1, 129, 131, 19, 135, 25, 139, 6

Offset: 0

Alex Ratushnyak, Jul 27 2012

a(n) is the number of consecutive integers starting from n needed to sum up to a perfect square.
Indices of 1's: A000290.
Indices of 2's: A046092(k), k!=A001108(m).
If a(n) is bigger than previous terms then a(n)=n*2-1, for example a(5)=9 is bigger than previous maximum, and a(5)=5*2-1.
Terms of A108269 never appear in a(n).

			a(2): 2+3+4 = 9, three summands, so a(2)=3.
a(3): 3+4+5+6+7 = 25, five summands, so a(3)=5.
a(12): 12+13 = 25, so a(12)=2.

Harvey P. Dale, Table of n, a(n) for n = 0..1000

Cf. A000290, A108269.

int main() {  // OK with GCC
  unsigned long long i, n, sum, sr;
  for (n=0; n<333; ++n) {
    for (sum=0, i=n; i==n || sr*sr!=sum; ++i)  sr=sqrt(sum+=i);
    printf("%llu, ", i-n);
  }
}

lks[n_]:=Module[{k=1},While[!IntegerQ[Sqrt[Total[Range[n,n+k-1]]]], k++];k]; lks/@Range[0,80] (* Harvey P. Dale, Mar 14 2016 *)

A308085 a(n) is the least positive number k such that n(n-1)/2 + k(k-1)/2 is a square.

Original entry on oeis.org

1, 1, 2, 3, 4, 2, 6, 7, 1, 9, 10, 6, 3, 13, 14, 2, 16, 17, 18, 4, 6, 21, 3, 23, 24, 9, 5, 27, 13, 4, 30, 31, 2, 6, 34, 35, 5, 37, 38, 16, 7, 30, 23, 6, 44, 20, 46, 8, 16, 1, 7, 51, 12, 53, 9, 42, 23, 8, 58, 59, 60, 10, 27, 63, 9, 65, 20, 67, 11, 69, 6, 10, 72, 3, 44, 12, 21, 77, 11, 34, 80, 46

Offset: 1

J. M. Bergot and Robert Israel, Jun 05 2019

a(n) <= n-1 if n > 1, because (n-1)*(n-2)/2 + n*(n-1)/2 = (n-1)^2.
a(7*k+2) <= k and a(7*k+6) <= k+2, because (7*k+2)*(7*k+1)/2 + k*(k-1)/2 = (5*k+1)^2 and (7*k+6)*(7*k+5)/2 + (k+2)*(k+1)/2 = (5*k+4)^2.
From Bernard Schott, Jun 27 2019: (Start)
a(m) = 1 iff the triangular number t(m-1) = (m-1)*m/2 is a square, so iff m-1 is in A001108, or m in A055997.
a(m) = 2 iff the triangular number t(m-1) + 1 is a square, so iff m-1 is in A006451. (End)

			a(5) = 4 because 4*3/2 + 5*4/2 = 4^2 and none of 1*0/2 + 5*4/2, 2*1/2 + 5*4/2, 3*2/2 + 5*4/2 are squares.

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

Cf. A000217, A055997 (a(n)=1).

f:= proc(n) local k;
  for k from 1 do
    if issqr((k*(k-1)+n*(n-1))/2) then return k fi
  od
end proc:
map(f, [$1..100]);

a(n) = {my(k=1); while (!issquare(n*(n-1)/2 + k*(k-1)/2), k++); k;} \\ Michel Marcus, Jun 27 2019

A334009 Triangle read by rows: T(n, k) = binomial(n + k - 1, 2k - 1) 4^(k - 1) * n/k, 1 <= k <= n.

Original entry on oeis.org

1, 4, 4, 9, 24, 16, 16, 80, 128, 64, 25, 200, 560, 640, 256, 36, 420, 1792, 3456, 3072, 1024, 49, 784, 4704, 13440, 19712, 14336, 4096, 64, 1344, 10752, 42240, 90112, 106496, 65536, 16384, 81, 2160, 22176, 114048, 329472, 559104, 552960, 294912, 65536, 100

Offset: 1

Michael Somos, Apr 12 2020

Let P(n, x) := Sum_{k=1..n} T(n, k)*x^k. Then P(n, P(m, x)) = P(n*m, x) for all n and m in Z.

The r=4 case of the Logistic Map is 4*x*(1 - x) = -P(1, -x). The r=2 case leads to A193862.

			First four rows:
.1
.4...4
.9..24..16
16..80.128..64

Cf. A001108, A039598, A123588, A193862.

T[ n_, k_] := If[k == 0, 0, Binomial[n + k - 1, 2 k - 1] 4^(k - 1) n / k];

{T(n, k) = if(k == 0, 0, binomial(n + k - 1, 2*k - 1) * 4^(k - 1) * n/k)};

P(n, x) = sinh(n * arcsinh(sqrt(x)))^2 = (hypergeom([-n, n], [1/2], -x) - 1)/2 are the row polynomials.
G.f.: Sum_{n, m} T(n, k) * x^k * y^n = x * y * (1 + y) / ((1 - y) * (1 - (2 + 4*x)*y + y^2)).
Row sums are A001108.
T(n, k) = (-1)^n * (-4)^(k-1) * A039598(-k - 1, n - 1) for all n in Z if k<0.
T(n, k) = -(-1)^(n+k) * A123588(n,k) if 1 <= k <= n.

cp's OEIS Frontend

A361670 Squarefree part of the n-th triangular number.

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A082649 Triangle of coefficients in expansion of sinh^2(n*x) in powers of sinh(x).

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A135203 For any integer n >= 1 the sequence gives the minimum power x for which n^x+(n-1)^x+(n-2)^x+...+1^x produces a perfect square.

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A144844 a(n) = ((2 + sqrt(2))^n - (2 - sqrt(2))^n)^2/8.

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A174541 Baron Münchhausen's Sequence.

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Scheme

A205185 Period 6: repeat [1, 8, 9, 8, 1, 0].

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A212797 Number of spanning trees in C_2 X C_n.

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A308085 a(n) is the least positive number k such that n(n-1)/2 + k(k-1)/2 is a square.

A334009 Triangle read by rows: T(n, k) = binomial(n + k - 1, 2k - 1) 4^(k - 1) * n/k, 1 <= k <= n.