Paul Weisenhorn - cp's OEIS frontend

A346295 a(n) = Sum_{k=0..n} (2^k + 1) * (2^k + 2) / 2.

3, 9, 24, 69, 222, 783, 2928, 11313, 44466, 176307, 702132, 2802357, 11197110, 44763831, 179006136, 715926201, 2863508154, 11453639355, 45813770940, 183253510845, 733010897598, 2932037298879, 11728136612544, 46912521284289, 187650034805442, 750600038558403

Offset: 0

Paul Weisenhorn, Jul 13 2021

All terms are multiples of 3.

Roger B. Nelson, Proof without Words: A Triangular Sum, Mathematics Magazine Vol. 78, No. 5 (December 2005), p. 395.
Index entries for linear recurrences with constant coefficients, signature (8,-21,22,-8).

Cf. A028401 (first differences).

Cf. A006095, A076024.

a:= proc(n) option remember:
if n=0 then 3 else (2^n+1)*(2^n+2)/2+procname(n-1) fi:
end proc:
seq(a(n), n=0..30);

Accumulate @ Table[(2^k + 1)*(2^k + 2)/2, {k, 0, 25}] (* Amiram Eldar, Jul 27 2021 *)
LinearRecurrence[{8,-21,22,-8},{3,9,24,69},30] (* Harvey P. Dale, Nov 21 2021 *)

a(n)=sum(k=0, n, (2^k+1)*(2^k+2)/2); \\ Michel Marcus, Jul 16 2021

a(n) = (2^(n+1) + 4) * (2^(n+1) + 5) / 6 - 4 + n.
More generally: let f(n, b) be the triangular sum Sum_{k=0..n} (2^k+b) * (2^k+b+1) / 2.
f(n, b) = (2^(n+1) + 3*b + 1) * (2^(n+1) + 3*b + 2) / 6 - (b + 1)^2 + b*(b + 1)*n / 2.
G.f.: ((b^2+3*b+2)/2 - (3*b^2+8*b+4)*x + (4*b^2+8*b+3)*x^2) / ((4*x-1) * (2*x-1) * (x-1)^2).
E.g.f.: exp(x) * ((6*b+3)*exp(x) + 2*exp(3*x) + 3(b^2+b)*x/2 + (3*b^2-3*b-4) / 2) / 3.
Then b = -1 gives A006095, b = 0 gives A076024, b = 1 gives A346295, b = 2 gives A346375.
G.f.: 3*(5*x^2 - 5*x + 1) / ((4*x - 1) * (2*x - 1) * (x - 1)^2).
a(n) = 8*a(n-1) - 21*a(n-2) + 22*a(n-3) - 8*a(n-4) for n > 3.
This recurrence is valid for all sequences f(n,b).
E.g.f.: exp(x) * (9*exp(x) + 2*exp(3*x) + 3*x - 2) / 3. - Stefano Spezia, Aug 13 2021

A346375 a(n) = Sum_{k=0..n} (2^k + 2) * (2^k + 3) / 2.

Original entry on oeis.org

6, 16, 37, 92, 263, 858, 3069, 11584, 44995, 177350, 704201, 2806476, 11205327, 44780242, 179038933, 715991768, 2863639259, 11453901534, 45814295265, 183254559460, 733012994791, 2932041493226, 11728145001197, 46912538061552, 187650068359923, 750600105667318, 3002400087124729

Offset: 0

Paul Weisenhorn, Jul 14 2021

Roger B. Nelson, Proof without Words: A Triangular Sum, Mathematics Magazine Vol. 78, No. 5 (December 2005), p. 395.
Index entries for linear recurrences with constant coefficients, signature (8,-21,22,-8).

Cf. A006095, A076024.

a:= proc(n) option remember:
     if n=0 then 6 else procname(n-1)+(2^n+3)*(2^n+2)/2 fi:
    end proc:
seq(a(n), n=0..26);

a[n_]:=Sum[(2^k+2)*(2^k+3)/2,{k,0,n}];Array[a,30,0] (* Giorgos Kalogeropoulos, Jul 27 2021 *)

a(n) = sum(k=0, n, (2^k+2)*(2^k+3)/2); \\ Michel Marcus, Jul 28 2021

a(n) = Sum_{k=0..n} (2^k + 2) * (2^k + 3) / 2.
a(n) = (2^(n+1) + 7) * (2^(n+1) + 8)/6 - 9 + 3*n.
More generally: let f(n, b) = Sum_{k=0..n} (2^k + b) * (2^k + b + 1)/2 then f(n, b) = (2^(n+1) + 3*b + 1) * (2^(n+1) + 3*b + 2) / 6 - (b + 1)^2 + b*(b + 1)*n/2.
G.f.: ((b^2+3*b+2)/2 - (3*b^2+8*b+4)*x + (4*b^2+8*b+3)*x^2) / ((4*x-1) * (2*x-1) * (x-1)^2).
E.g.f.: exp(x)*((6*b+3)*exp(x) + 2*exp(3*x) + 3*(b^2+b)*x/2 +(3*b^2-3*b-4) / 2) / 3.
Then b = -1 gives A006095, b = 0 gives A076024, b = 1 gives A346295, b = 2 gives A346375.
a(n) = 8*a(n-1) - 21*a(n-2) + 22*a(n-3) - 8*a(n-4) with n > 3.
This recurrence is valid for all sequences f(n, b).
G.f.: (35*x^2 - 32*x + 6) / ((4*x - 1) * (2*x - 1) * (x - 1)^2).
E.g.f.: exp(x) * (1 + 15*exp(x) + 2*exp(3*x) + 9*x)/3. - Stefano Spezia, Aug 15 2021

A320773 Numbers (excluding squares) whose square root has a continued fraction with a period < 3.

Original entry on oeis.org

2, 3, 5, 6, 8, 10, 11, 12, 15, 17, 18, 20, 24, 26, 27, 30, 35, 37, 38, 39, 40, 42, 48, 50, 51, 56, 63, 65, 66, 68, 72, 80, 82, 83, 84, 87, 90, 99, 101, 102, 104, 105, 110, 120, 122, 123, 132, 143, 145, 146, 147, 148, 150, 152, 156, 168, 170, 171, 182, 195, 197, 198, 200

Offset: 1

Paul Weisenhorn, Oct 21 2018

nonn

The Heron sequence of every number a(n) has the following relationship: numerator(h(k))^2 - a(n)*denominator(h(k))^2 = 1 for k > 1.
The Heron sequence of every number a(n) has the following relationship with the continued fraction f(s) convergent to sqrt(a(n)): h(k) = f(2^k-1).
From Gerhard Kirchner, Jan 17 2020: (Start)
Numbers k = m^2 + r with m > 0 and 0 < r <= 2m such that r is a divisor of 2m.
Continued fraction: k = [m; 2m/r, 2m, 2m/r, 2m, ...].
The number of terms that are between m^2 and (m+1)^2  is equal to the number of divisors of 2m, which is A099777(m).
Proof see link. The Maxima code below demonstrates the divisor property. Note that there is no divisor of 2m between m and 2m.
(End)

			The continued fraction of sqrt(6) = 2 + 1/(2 + 1/(4 + 1/(2 + 1/(4 + 1/(2 + 1/(4 + ...)))))) = [2; 2, 4, 2, 4, 2, 4, ...] has repeating portion (2, 4) with period 2, so 6 is a term.

Gerhard Kirchner, Divisor property of a(n)

Digits:=40: nr:=0:
for z from 2 to 200 do
  test:=true: c:=sqrt(z):
  if (c=floor(c)) the test:=false: end if:
  while (test=true) do
    b[0]:=floor(c):
    r[0]:=c:
    for k from 1 to 2 do
      r[k]:=evalf(1/(r[k-1]-b[k-1])):
      b[k]:=floor(r[k]):
    end do:
    if (b[1]=2*b[0]) or (b[2]=2*b[0]) then nr:=nr+1: a[nr]:=z: printf("%4d",z): end if:
    test:=false:
  end do:
end do:

Select[Range[200], !IntegerQ[Sqrt[#]] && Length@ContinuedFraction[Sqrt[#]][[-1]]<3 &] (* Amiram Eldar, Nov 01 2018 *)

block([n: 2, m: 0, r: 0, k: 0, kmax: 10,v: ""],
  while kGerhard Kirchner, Jan 17 2020 */

Edited by Jon E. Schoenfield, Oct 19 2019

A319750 a(n) is the denominator of the Heron sequence with h(0) = 3.

Original entry on oeis.org

1, 3, 33, 3927, 55602393, 11147016454528647, 448011292165037607943004375755833, 723685043824607606355691108666081531638582859833105061571146291527

Offset: 0

Paul Weisenhorn, Sep 27 2018

The numerators of the Heron sequence are in A319749.
There is the following relationship between the denominator of the Heron sequence and the denominator of the continued fraction A041018(n)/ A041019(n) convergent to sqrt(13).
n even: a(n) = A041019((5*2^n-5)/3).
n  odd: a(n) = A041019((5*2^n-1)/3).
General: all numbers c(n) = A078370(n) = (2*n+1)^2 + 4 have the same relationship between the denominator of the Heron sequence and the denominator of the continued fraction convergent to 2*n+1.
sqrt(c(n)) has the continued fraction [2*n+1; n, 1, 1, n, 4*n+2].
hn(n)^2 - c(n)*hd(n)^2 = 4 for n > 1.

			A078370(2) = 29.
hd(0) = A041047(0) = 1, hd(1) = A041047(3) = 5,
hd(2) = A041047(5) = 135, hd(3) = A041047(13) = 38145.

Cf. A041018, A041019, A078370, A010122, A041047, A319749.

hn[0]:=3: hd[0]:=1:
for n from 1 to 6 do
  hn[n]:=(hn[n-1]^2+13*hd[n-1]^2)/2:
  hd[n]:=hn[n-1]*hd[n-1]:
  printf("%5d%40d%40d\n", n, hn[n], hd[n]):
end do:

def aupton(nn):
    hn, hd, alst = 3, 1, [1]
    for n in range(nn):
        hn, hd = (hn**2 + 13*hd**2)//2, hn*hd
        alst.append(hd)
    return alst
print(aupton(7)) # Michael S. Branicky, Mar 15 2022

h(n) = hn(n)/hd(n), hn(0) = 3, hd(0) = 1.
hn(n+1) = (hn(n)^2 + 13*hd(n)^2)/2.
hd(n+1) = hn(n)*hd(n).
A041018(n) = A010122(n)*A041018(n-1) + A041018(n-2).
A041019(n) = A010122(n)*A041019(n-1) + A041019(n-2).
a(0) = 1, a(1) = 3 and a(n) = 2*T(2^(n-2), 11/2)*a(n-1) for n >= 2, where T(n,x) denotes the n-th Chebyshev polynomial of the first kind. - Peter Bala, Mar 16 2022

a(5) corrected and terms a(6) and a(7) added by Peter Bala, Mar 15 2022

A319749 a(n) is the numerator of the Heron sequence with h(0)=3.

Original entry on oeis.org

3, 11, 119, 14159, 200477279, 40191139395243839, 1615327685887921300502934267457919, 2609283532796026943395592527806764363779539144932833602430435810559

Offset: 0

Paul Weisenhorn, Sep 27 2018

The denominator of the Heron sequence is in A319750.
The following relationship holds between the numerator of the Heron sequence and the numerator of the continued fraction A041018(n)/A041019(n) convergent to sqrt(13).
n even: a(n)=A041018((5*2^n-5)/3).
n  odd: a(n)=A041018((5*2^n-1)/3).
More generally, all numbers c(n)=A078370(n)=(2n+1)^2+4 have the same relationship between the numerator of the Heron sequence and the numerator of the continued fraction convergent to 2n+1.
sqrt(c(n)) has the continued fraction 2n+1; n,1,1,n,4n+2.
hn(n)^2-c(n)*hd(n)^2=4 for n>1.
From Peter Bala, Mar 29 2022: (Start)
Applying Heron's method (sometimes called the Babylonian method) to approximate the square root of the function x^2 + 4, starting with a guess equal to x, produces the sequence of rational functions [x, 2*T(1,(x^2+2)/2)/x, 2*T(2,(x^2+2)/2)/( 2*x*T(1,(x^2+2)/2) ), 2*T(4,(x^2+2)/2)/( 4*x*T(1,(x^2+2)/2)*T(2,(x^2+2)/2) ), 2*T(8,(x^2+2)/2)/( 8*x*T(1,(x^2+2)/2)*T(2,(x^2+2)/2)*T(4,(x^2+2)/2) ), ...], where T(n,x) denotes the n-th Chebyshev polynomial of the first kind. The present sequence is the case x = 3. Cf. A001566 and A058635 (case x = 1), A081459 and A081460 (essentially the case x = 4). (End)

			A078370(2)=29.
hn(0)=A041046(0)=5; hn(1)=A041046(3)=27; hn(2)=A041046(5)=727;
hn(3)=A041046(13)=528527.

P. Liardet and P. Stambul, Séries d'Engel et fractions continuées, Journal de Théorie des Nombres de Bordeaux 12 (2000), 37-68.
Wikipedia, Engel Expansion
Wikipedia, Methods of computing square roots
Index entries for sequences of form a(n+1)=a(n)^2 + ...

2*T(2^n,x/2) modulo differences of offset: A001566 (x = 3 and x = 7), A003010 (x = 4), A003487 (x = 5), A003423 (x = 6), A346625 (x = 8), A135927 (x = 10), A228933 (x = 18).

Cf. A041018, A041019, A057076, A078370, A081459, A010122, A319750, A041046.

hn[0]:=3:  hd[0]:=1:
for n from 1 to 6 do
hn[n]:=(hn[n-1]^2+13*hd[n-1]^2)/2:
hd[n]:=hn[n-1]*hd[n-1]:
   printf("%5d%40d%40d\n", n, hn[n], hd[n]):
end do:
#alternative program
a := n -> if n = 0 then 3 else simplify( 2*ChebyshevT(2^(n-1), 11/2) ) end if:
seq(a(n), n = 0..7); # Peter Bala, Mar 16 2022

def aupton(nn):
    hn, hd, alst = 3, 1, [3]
    for n in range(nn):
        hn, hd = (hn**2 + 13*hd**2)//2, hn*hd
        alst.append(hn)
    return alst
print(aupton(7)) # Michael S. Branicky, Mar 16 2022

h(n) = hn(n)/hd(n); hn(0)=3; hd(0)=1.
hn(n+1) = (hn(n)^2+13*hd(n)^2)/2.
hd(n+1) = hn(n)*hd(n).
A041018(n) = A010122(n)*A041018(n-1) + A041018(n-2).
A041019(n) = A010122(n)*A041019(n-1) + A041019(n-2).
From Peter Bala, Mar 16 2022: (Start)
a(n) = 2*T(2^(n-1),11/2) for n >= 1, where T(n,x) denotes the n-th Chebyshev polynomial of the first kind.
a(n) = 2*T(2^n, 3*sqrt(-1)/2) for n >= 2.
a(n) = ((11 + 3*sqrt(13))/2)^(2^(n-1)) + ((11 - 3*sqrt(13))/2)^(2^(n-1)) for n >= 1.
a(n+1) = a(n)^2 - 2 for n >= 1.
a(n) = A057076(2^(n-1)) for n >= 1.
Engel expansion of (1/6)*(13 - 3*sqrt(13)); that is, (1/6)*(13 - 3*sqrt(13)) = 1/3 + 1/(3*11) + 1/(3*11*119) + .... (Define L(n) = (1/2)*(n - sqrt(n^2 - 4)) for n >= 2 and show L(n) = 1/n + L(n^2-2)/n. Iterate this relation with n = 11. See also Liardet and Stambul, Section 4.)
sqrt(13) = 6*Product_{n >= 0} (1 - 1/a(n)).
sqrt(13) = (9/5)*Product_{n >= 0} (1 + 2/a(n)). See A001566. (End)

a(6) and a(7) added by Peter Bala, Mar 16 2022

A201632 If the sum of the squares of 4 consecutive numbers is a triangular number t(u), then a(n) is its index u.

Original entry on oeis.org

35, 83, 1203, 2835, 40883, 96323, 1388835, 3272163, 47179523, 111157235, 1602714963, 3776073843, 54445129235, 128275353443, 1849531679043, 4357585943235, 62829631958243, 148029646716563, 2134357954901235, 5028650402419923

Offset: 1

Paul Weisenhorn, Jan 09 2013

Sum_{(e(n)+j)^2,j=0..3} = a(n)*(a(n)+1)/2=t(a(n)) give the Pell equation c(n)^2 - 32*d(n)^2 = 41 with 2*a(n) + 1 = c(n) and e(n) + 1.5 = d(n). e(n) = A201633(n).

In general, for the sum of the squares of k consecutive numbers, one get an analog sequence with k in {4, 5, 6, 7, 11, 15, 17, 19, 23,...}. It gives the Pell equation c(n)^2 - 8k*d(n)^2 = 4*binomial((k+1),3) + 1 with 2*a(n) + 1 = c(n) and e(n) + (k-1)/2 = d(n).

			For n=2: a(2)=83; t(83)=83*84/2=3486.
A201633(2)=e(2)=28; 28^2+29^2+30^2+31^2=3486.

Harvey P. Dale, Table of n, a(n) for n = 1..1000
Index entries for linear recurrences with constant coefficients, signature (1,34,-34,-1,1).

Cf. A201633, A218395.

LinearRecurrence[{1,34,-34,-1,1},{35,83,1203,2835,40883},30] (* Harvey P. Dale, Dec 10 2024 *)

G.f.: (35*x+48*x^2-70*x^3+3*x^5)/((1-x)*(1-34*x^2+x^4)).
a(n+4) = 34*a(n+2) - a(n) + 16.
a(n+5) = a(n+4) + 34*a(n+3) - 34*a(n+2) - a(n+1) + a(n).
eigenvalues ej: {1,(3+2r),-(3+2r),(3-2r),-(3-2r)}.
a(n+1) = (k1*e1 + k2*e2^n + k3*e3^n + k4*e4^n + k5*e5^n)/4 for k1=-2; k2=50+35r; k3=21+15r; k4=50-35r; k5=21-15r, where r = sqrt(2).

Corrected by R. J. Mathar, Jun 14 2016

A187789 a(n) is the start position for a Sankt-Petrus-game with n white and n black stones and the least step A187788(n).

Original entry on oeis.org

2, 3, 2, 8, 5, 5, 2, 7, 1, 17, 15, 4, 8, 1, 2, 30, 26, 11, 35, 7, 26, 27, 23, 44, 24, 30, 6, 39, 53, 18, 2, 15, 61, 40, 30, 68, 44, 32, 78, 29, 81, 15, 19, 76, 51, 67, 40, 19, 53, 42, 53, 3, 74, 103, 73, 35, 105, 78, 110, 105, 76, 61, 2, 5, 48, 128, 82, 36, 37, 63, 88, 87, 31, 123, 93, 126, 2, 1, 156, 89, 33, 160, 90, 135, 124, 136, 145, 79, 42, 26, 104, 94, 67, 44, 186, 30, 133, 137, 40, 118

Offset: 1

Paul Weisenhorn, Jan 06 2013

nonn

Beginning at the position a(n) with the least step A187788(n) (n-1) white stones were eliminated; then starting at the position 1 of the last white stone, n black stones were eliminated.

			n=8; WWBBWBBWWBBWWBWB; step=A187788(8)=3; start=a(8)=7; elimination: white stones: {9,12,15,2,5,8,13}, black stones: {4,10,16,6,14,7,3,11}.

W. Ahrens, Das Josephusspiel, Archiv für Kulturgeschichte, Jg 11(1913), 129-151.

R. Baumann, Das Josephus-Problem, LOG IN, Heft Nr. 165, pp. 68-71, 2010 (in German).
Index entries for sequences related to the Josephus Problem

Cf. A187788, A206602.

stone:=proc(n1)
local n,j,k,h,z,zp: global a,m,s:
n:=2*n1: m:=m+1:
for j from 1 to n-1 do z[j]:=z[j]+1: end do:
z[n]:=1: zp:=1:
for j from 1 to n1 do
for k from 1 to (s-2) do  zp:=z[zp]: end do:
  h:=z[zp]: z[zp]:=z[z[zp]]: zp:=z[zp]:
end do:
if (h=1) then a[n1]:=1: else a[n1]:=n+2-h: end if:
end proc:
m:=0: s:=1:
while (m < 100) do
s1:=s: s:=s+1: c:=1:
for p from 2 to 100 by 2 do  p1:=p-1: p2:=p+1:
  b:=(c+s1) mod p +1:
  if (b=1) and (a[p1]=0) then stone(p1): end if:
  c:=(b+s1) mod p2 +1:
  if (c=1) and (a[p]=0) then stone(p): end if:
end do:
end do:

A187788 a(n) is the least step for the Sankt-Petrus-game with n white and n black stones.

Original entry on oeis.org

2, 5, 2, 4, 3, 11, 2, 3, 8, 16, 4, 21, 6, 5, 2, 11, 20, 34, 8, 15, 10, 7, 13, 11, 13, 45, 18, 23, 8, 3, 2, 25, 75, 42, 13, 5, 23, 13, 50, 16, 18, 89, 38, 8, 39, 30, 29, 38, 7, 45, 23, 137, 46, 63, 17, 48, 5, 46, 34, 140, 33, 39, 2, 28, 29, 79, 33, 48, 3, 10, 46, 120, 6, 37, 17, 8, 44, 15, 160, 20, 35, 144, 104, 179, 153, 24, 8, 265, 19, 9, 62, 7, 139, 19, 44, 93, 182, 27, 158, 185

Offset: 1

Paul Weisenhorn, Jan 06 2013

nonn

Beginning at the position A187789(n) with step a(n), (n-1) white stones were eliminated; then from the position 1 of the last white stone, n black stones were eliminated.

			n=4; WBWWBBWB; startposition=A187789(4)=8; least step=a(4)=4; elimination: white stones: {3,7,4}; black stones: {6,5,8,2}.

W. Ahrens, Das Josephusspiel, Archiv für Kulturgeschichte, Jg 11(1913), 129-151.

R. Baumann, Das Josephus-Problem, LOG IN, Heft Nr. 165, pp. 68-71, 2010 (in German).
Index entries for sequences related to the Josephus Problem

Cf. A187789, A206602.

First column in A321781.

s:=1: M:={}:
for n from 1 to 100 do M:=M union {n}:
while (M <> {}) do
  s1:=s: s:=s+1: f[1]:=1:
  for n from 2 to 101 do  n1:=n-1:
    f[n]:=(f[n1]+s1) mod n +1:
    if (f[n]=1) and (n1 in M) then
      a[n1]:=s: M:=M minus {n1}:
    end if:
  end do:
end do:

A201633 Numbers k such that Sum_{j=0..3} (k + j)^2 is a triangular number.

Original entry on oeis.org

11, 28, 424, 1001, 14453, 34054, 491026, 1156883, 16680479, 39300016, 566645308, 1335043709, 19249260041, 45352186138, 653908196134, 1540639285031, 22213629408563, 52336383504964, 754609491695056, 1777896399883793, 25634509088223389, 60396141212544046

Offset: 1

Paul Weisenhorn, Jan 09 2013

Sum_{j=0..3} (a(n)+j)^2 = u(n)*(u(n)+1)/2 = t(u(n)) with A201632(n) = u(n) give the Pell equation c(n)^2 - 32*d(n)^2 = 41. 2*u(n)+1 = c(n) and a(n) + 1.5 = d(n).

Also integers k such that k^2 + (k+1)^2 + (k+2)^2 + (k+3)^2 is equal to a hexagonal number. - Colin Barker, Dec 21 2014

			For n=3: a(3)=424; 424^2+425^2+426^2+427^2=724206.
u(3)=A201632(3)=1203; t(1203)=1203*1204/2=724206.

Colin Barker, Table of n, a(n) for n = 1..1000
Index entries for linear recurrences with constant coefficients, signature (1,34,-34,-1,1).

Cf. A201632, A218395, A252747.

LinearRecurrence[{1,34,-34,-1,1},{11,28,424,1001,14453},30] (* Harvey P. Dale, Apr 16 2013 *)

Vec(x*(x^4+x^3-22*x^2-17*x-11)/((x-1)*(x^2-6*x+1)*(x^2+6*x+1)) + O(x^30)) \\ Colin Barker, Dec 21 2014

from functools import cache
@cache
def a(n):
    if n < 6: return [11, 28, 424, 1001, 14453][n-1]
    return a(n-1) + 34*a(n-2) - 34*a(n-3) - a(n-4) + a(n-5)
print([a(n) for n in range(1, 23)]) # Michael S. Branicky, Nov 28 2021

G.f.: (11*x+17*x^2+22*x^3-x^4-x^5)/((1-x)*(1-34*x^2+x^4)). [corrected by Georg Fischer, May 11 2019]
a(n+4) = 34*a(n-2) - a(n-4) + 48; r=sqrt(2).
a(n+5) = a(n+4) + 34*a(n+3) - 34*a(n+2) - a(n+1) + a(n).
Eigenvalues ej: {1,(3+2r),-(3+2r),(3-2*r),-(3-2*r)}.
a(n+1) = (k1*e1+k2*e2^n+k3*e3^n+k4*e4^n+k5*e5^n)/16 for k1=-24, k2=70+50r, k3=30+21r, k4=70-50r, k5=30-21r.

More terms from Colin Barker, Dec 21 2014

A218395 Numbers whose square is the sum of the squares of 11 consecutive integers.

Original entry on oeis.org

11, 77, 143, 1529, 2849, 30503, 56837, 608531, 1133891, 12140117, 22620983, 242193809, 451285769, 4831736063, 9003094397, 96392527451, 179610602171, 1923018812957, 3583208949023, 38363983731689, 71484568378289, 765356655820823, 1426108158616757

Offset: 0

Paul Weisenhorn, Oct 28 2012

a(n)^2 = Sum_{j=0..10} (x(n)+j)^2 = 11*(x(n)+5)^2 + 110 and b(n) = x(n)+5 give the Pell equation a(n)^2 - 11*b(n)^2 = 110 with the 2 fundamental solutions (11; 1) and (77; 23) and the solution (10; 3) for the unit form. A198949(n+1) = b(n); A106521(n) = x(n) and x(0) = -4.

General: If the sum of the squares of c neighboring numbers is a square with c = 3*k^2-1 and 1 <= k, then a(n)^2 = Sum_{j=0..c-1} (x(n)+j)^2 and b(n) = 2*x(n)+c-1 give the Pell equation a(n)^2 - c*(b(n)/2)^2 = binomial(c+1,3)/2. a(n) = 2*e1*a(n-k) - a(n-2*k); b(n) = 2*e1*b(n-k) - b(n-2*k); a(n) = e1*a(n-k) + c*e2*b(n-k); b(n) = e2*a(n-k) + e1*b(n-k) with the solution (e1; e2) for the unit form.

			For n=6, Sum_{z=17132..17142} z^2 = 3230444569;
a(6) = sqrt(3230444569) = 56837;
b(6) = sqrt((a(6)^2-110)/11) = 17137; x(6) = b(6)-5 = 17132.

V. Pletser, Finding all squared integers expressible as the sum of consecutive squared integers using generalized Pell equation solutions with Chebyshev polynomials, arXiv preprint arXiv:1409.7972 [math.NT], 2014. See Table 1 p. 7.
Index entries for linear recurrences with constant coefficients, signature (0,20,0,-1).

c=2: A001653(n+1) = a(n); A002315(n) = b(n); A001652(n) = x(n).

Cf. A001032 (11 is a term of that sequence), A198947.

s:=0: n:=-1:
for j from -5 to 5 do s:=s+j^2: end do:
for z from -4 to 100000 do
  s:=s-(z-1)^2+(z+10)^2: r:=sqrt(s):
  if (r=floor(r)) then
    n:=n+1: a(n):=r: x(n):=z:
    b(n):=sqrt((s-110)/11):
    print(n,a(n),b(n),x(n)):
  end if:
end do:

LinearRecurrence[{0,20,0,-1},{11,77,143,1529},30] (* Harvey P. Dale, Aug 15 2022 *)

a(n) = 20*a(n-2) - a(n-4); b(n) = 20*b(n-2) - b(n-4);
a(n) = 10*a(n-2) + 33*b(n-2); b(n) = 3*a(n-2) + 10*b(n-2).
a(n) = a(n-1) + 20*a(n-2) - 20*a(n-3) - a(n-4) + a(n-5).
G.f.: 11 * (1-x)*(1+8*x+x^2) / (1 - 20*x^2 + x^4).
With r=sqrt(11); s=10+3*r; t=10-3*r:
a(2*n) = ((11+r)*s^n + (11-r)*t^n)/2.
a(2*n+1) = ((77+23*r) * s^n + (77-23*r)*t^n)/2.
a(n) = 11 * A198947(n+1). - Bill McEachen, Dec 01 2022

cp's OEIS Frontend

User: Paul Weisenhorn

A346295 a(n) = Sum_{k=0..n} (2^k + 1) * (2^k + 2) / 2.

Views

Author

Keywords

Comments

Links

Crossrefs

Programs

Maple

Mathematica

PARI

Formula

A346375 a(n) = Sum_{k=0..n} (2^k + 2) * (2^k + 3) / 2.

Views

Author

Keywords

Links

Crossrefs

Programs

Maple

Mathematica

PARI

Formula

A320773 Numbers (excluding squares) whose square root has a continued fraction with a period < 3.

Views

Author

Keywords

Comments

Examples

Links

Programs

Maple

Mathematica

Maxima

Extensions

A319750 a(n) is the denominator of the Heron sequence with h(0) = 3.

Views

Author

Keywords

Comments

Examples

Crossrefs

Programs

Maple

Python

Formula

Extensions

A319749 a(n) is the numerator of the Heron sequence with h(0)=3.

Views

Author

Keywords

Comments

Examples

Links

Crossrefs

Programs

Maple

Python

Formula

Extensions

A201632 If the sum of the squares of 4 consecutive numbers is a triangular number t(u), then a(n) is its index u.

Views

Author

Keywords

Comments

Examples

Links

Crossrefs

Programs

Mathematica

Formula

Extensions

A187789 a(n) is the start position for a Sankt-Petrus-game with n white and n black stones and the least step A187788(n).

Views

Author

Keywords

Comments

Examples