A157096 -id:A157096 - cp's OEIS frontend

A212336 Expansion of 1/(1 - 23x + 23x^2 - x^3).

1, 23, 506, 11110, 243915, 5355021, 117566548, 2581109036, 56666832245, 1244089200355, 27313295575566, 599648413462098, 13164951800590591, 289029291199530905, 6345479454589089320, 139311518709760434136, 3058507932160140461673

Offset: 0

Bruno Berselli, Jun 08 2012

Partial sums of A077421.

Bruno Berselli, Table of n, a(n) for n = 0..200
Robert K. Moniot, A Family of Integer Sequences, 2021.
Index entries for linear recurrences with constant coefficients, signature (23,-23,1).

Sequences with g.f. of the type 1/(1-k*x+k*x^2-x^3): A334673 (k=24), A212336 (k=23), A212335 (k=22), A097833 (k=21), A097832 (k=20), A049664 (k=19), A097831-A097829 (k=18,17,16), A076139 (k=15), A097828-A097826 (k=14,13,12), A097784 (k=11), A092420 (k=10), A076765 (k=9), A092521 (k=8), A053142 (k=7), A089817(k=6), A061278 (k=5), A027941 (k=4), A000217 (k=3), A021823 (k=2), A133872 (k=1), A079978 (k=0).

m:=17; R:=PowerSeriesRing(Integers(), m); Coefficients(R!(1/(1-23*x+23*x^2-x^3)));

I:=[1,23,506]; [n le 3 select I[n] else 23*Self(n-1)-23*Self(n-2)+Self(n-3): n in [1..20]]; // Vincenzo Librandi, Aug 18 2013

a:= n-> (<<0|1|0>, <0|0|1>, <1|-23|23>>^n. <<1, 23, 506>>)[1, 1]:
seq(a(n), n=0..20);  # Alois P. Heinz, Jun 15 2012

CoefficientList[Series[1/(1 - 23 x + 23 x^2 - x^3), {x, 0, 16}], x]
LinearRecurrence[{23, -23, 1}, {1, 23, 506}, 20] (* Vincenzo Librandi, Aug 18 2013 *)

makelist(coeff(taylor(1/(1-23*x+23*x^2-x^3), x, 0, n), x, n), n, 0, 16);

PARI
```
Vec(1/(1-23*x+23*x^2-x^3)+O(x^17))
```

[(1/20)*(-1 +21*chebyshev_U(n, 11) -chebyshev_U(n-1, 11)) for n in (0..30)] # G. C. Greubel, Feb 07 2022

G.f.: 1/((1-x)*(1 - 22*x + x^2)).
a(n) = (((6+sqrt(30))^(2*n+3) + (6-sqrt(30))^(2*n+3))/6^(n+1) - 12)/240.
a(n) = a(-n-3) = 23*a(n-1) - 23*a(n-2) + a(n-3).
a(n)*a(n+2) = a(n+1)*(a(n+1)-1).
a(n+1) - 11*a(n) = A133285(n+2).
11*a(n+1) - a(n) = (1/5)*A157096(n+2).
a(n) = (1/20)*(-1 + 21*ChebyshevU(n, 11) - ChebyshevU(n-1, 11)). - G. C. Greubel, Feb 07 2022

A157088 Consider all consecutive integer Pythagorean septuples (X, X+1, X+2, X+3, Z-2, Z-1, Z) ordered by increasing Z; sequence gives X values.

Original entry on oeis.org

0, 21, 312, 4365, 60816, 847077, 11798280, 164328861, 2288805792, 31878952245, 444016525656, 6184352406957, 86136917171760, 1199732487997701, 16710117914796072, 232741918319147325, 3241676738553266496, 45150732421426583637, 628868577161418904440, 8759009347838438078541

Offset: 0

Charlie Marion, Mar 12 2009

In general, the first terms of consecutive integer Pythagorean 2k+1-tuples may be found as follows: let first(0)=0, first(1) = k*(2k+1) and, for n > 1, first(n) = (4k+2)*first(n-1) - first(n-2) + 2*k^2; e.g., if k=4, then first(2) = 680 = 18*36 - 0 + 32.
In general, the first and last terms of consecutive integer Pythagorean 2k+1-tuples may be found as follows: let first(0)=0 and last(0)=k; for n > 0, let first(n) = (2k+1)*first(n-1) + 2k*last(n-1) + k and last(n) = (2k+2)*first(n-1) + (2k+1)*last(n-1) + 2k; e.g., if k=4 and n=2, first(2) = 680 = 9*36 + 8*44 + 4 and last(2) = 764 = 10*36 + 9*44 + 8.
In general, the first terms of consecutive integer Pythagorean 2k+1-tuples may be found as follows: first(n) = (k^(n+1)((1+sqrt((k+1)/k))^(2n+1) + (1-sqrt((k+1)/k))^(2n+1)) - 2*k)/4; e.g., if k=4 and n=2, then first(2) = 680 = (4^3((1+sqrt(5/4)^5 + (1-sqrt(5/4))^5)-2*4)/4.
In general, if u(n) is the numerator and e(n) is the denominator of the n-th continued fraction convergent to sqrt((k+1)/k), then the first terms of consecutive integer Pythagorean 2k+1-tuples may be found as follows: first(2n+1) = k*u(2n)*u(2n+1) and, for n > 0, first(2n) = (k+1)*e(2n-1)*e(2n); e.g., a(3) = 4365 = 3*15*97 and a(4) = 60816 = 4*84*181.
In general, if first(n) is the first term of the n-th consecutive integer Pythagorean 2k+1-tuple, then lim_{n->inf} first(n+1)/first(n) = k*(1+sqrt((k+1)/k))^2 = 2k + 1 + 2*sqrt(k^2+k).

			a(2)=312 since 312^2 + 313^2 + 314^2 + 315^2 = 361^2 + 361^2 + 363^2.

A. H. Beiler, Recreations in the Theory of Numbers. New York: Dover, 1964, pp. 122-125.
L. E. Dickson, History of the Theory of Numbers, Vol. II, Diophantine Analysis. Dover Publications, Inc., Mineola, NY, 2005, pp. 181-183.
W. Sierpinski, Pythagorean Triangles. Dover Publications, Mineola NY, 2003, pp. 16-22.

G. C. Greubel, Table of n, a(n) for n = 0..870
Tanya Khovanova, Recursive Sequences
Ron Knott, Pythagorean Triples and Online Calculators
Index entries for linear recurrences with constant coefficients, signature (15,-15,1).

Cf. A001652, A157084, A157092, A157096.

[Round((3^(n+1)*((1+Sqrt(4/3))^(2*n+1)+(1-Sqrt(4/3))^(2*n+1))-2*3)/4): n in [0..50]]; // G. C. Greubel, Nov 04 2017

CoefficientList[Series[3*x*(-7 + x)/((x - 1)*(x^2 - 14*x + 1)), {x, 0, 50}], x] (* G. C. Greubel, Nov 04 2017 *)

my(x='x+O('x^50)); concat([0], Vec(3*x*(-7+x)/((x-1)*(x^2-14*x+1)))) \\ G. C. Greubel, Nov 04 2017

For n > 1, a(n) = 14*a(n-1) - a(n-2) + 18.
For n > 0, a(n) = 7*a(n-1) + 6*A157089(n-1) + 3.
Limit_{n->oo} a(n+1)/a(n) = 3*(1+sqrt(4/3))^2 = 7 + 2*sqrt(12).
a(n) = (3^(n+1)*((1+sqrt(4/3))^(2*n+1) + (1-sqrt(4/3))^(2*n+1)) - 2*3)/4.
From R. J. Mathar, Mar 19 2009: (Start)
G.f.: 3*x*(-7+x)/((x-1)*(x^2-14*x+1)).
a(n) = 15*a(n-1) - 15*a(n-2) + a(n-3) = 3*A001921(n). (End)

A157092 Consider all consecutive integer Pythagorean 9-tuples (X, X+1, X+2, X+3, X+4, Z-3, Z-2, Z-1, Z) ordered by increasing Z; sequence gives X values.

Original entry on oeis.org

0, 36, 680, 12236, 219600, 3940596, 70711160, 1268860316, 22768774560, 408569081796, 7331474697800, 131557975478636, 2360712083917680, 42361259535039636, 760141959546795800, 13640194012307284796, 244763350261984330560, 4392100110703410665316, 78813038642399407645160

Offset: 0

Charlie Marion, Mar 12 2009

nonn

In general, the first terms of consecutive integer Pythagorean 2k+1-tuples may be found as follows: let first(0)=0, first(1) = k*(2k+1) and, for n > 1, first(n) = (4k+2)*first(n-1) - first(n-2) + 2*k^2; e.g., if k=5, then first(2) = 1260 = 22*55 - 0 + 50.
In general, the first and last terms of consecutive integer Pythagorean 2k+1-tuples may be found as follows: let first(0)=0 and last(0)=k; for n > 0, let first(n) = (2k+1)*first(n-1) + 2k*last(n-1) + k and last(n) = (2k+2)*first(n-1) + (2k+1)*last(n-1) + 2k; e.g., if k=5 and n=2, then first(2) = 1260 = 11*55 + 10*65 + 5 and last(2) = 1385 = 12*55 + 11*65 + 10.
In general, the first terms of consecutive integer Pythagorean 2k+1-tuples may be found as follows: first(n) = (k^(n+1)((1+sqrt((k+1)/k))^(2n+1) + (1-sqrt((k+1)/k))^(2n+1)) - 2*k)/4; e.g., if k=5 and n=2, then first(2) = 1260 = (5^3((1+sqrt((6/5))^5 + (1-sqrt(6/5))^5) - 2*5)/4.
In general, if u(n) is the numerator and e(n) is the denominator of the n-th continued fraction convergent to sqrt((k+1)/k), then the first terms of consecutive integer Pythagorean 2k+1-tuples may be found as follows: first(2n+1) = k*u(2n)*u(2n+1) and, for n > 0, first(2n) = (k+1)*e(2n-1)*e(2n); e.g., a(1) = 36 = 4*1*9 and a(2) = 680 = 5*8*17.
In general, if first(n) is the first term of the n-th consecutive integer Pythagorean 2k+1-tuple, then lim_{n->inf} first(n+1)/first(n) = k*(1+sqrt((k+1)/k))^2 = 2k + 1 + 2*sqrt(k^2+k).

			a(2)=680 since 680^2 + 681^2 + 682^2 + 683^2 + 684^2 = 761^2 + 762^2 + 763^2 + 764^2.

A. H. Beiler, Recreations in the Theory of Numbers. New York: Dover, 1964, pp. 122-125.
L. E. Dickson, History of the Theory of Numbers, Vol. II, Diophantine Analysis. Dover Publications, Inc., Mineola, NY, 2005, pp. 181-183.
W. Sierpinski, Pythagorean Triangles. Dover Publications, Mineola NY, 2003, pp. 16-22.

G. C. Greubel, Table of n, a(n) for n = 0..790
Tanya Khovanova, Recursive Sequences
Ron Knott, Pythagorean Triples and Online Calculators
Index entries for linear recurrences with constant coefficients, signature (19,-19,1).

Cf. A001652, A157084, A157088, A157096.

[Round((4^(n+1)*((1+Sqrt(5/4))^(2*n+1) + (1-Sqrt(5/4))^(2*n+1)) - 2*4)/4): n in [0..50]]; // G. C. Greubel, Nov 04 2017

RecurrenceTable[{a[0]==0,a[1]==36,a[n]==18a[n-1]-a[n-2]+32},a,{n,20}] (* or *) LinearRecurrence[{19,-19,1},{0,36,680},20] (* Harvey P. Dale, Oct 09 2012 *)

x='x+O('x^50); concat([0], Vec(4*x*(-9+x)/((x-1)*(x^2-18*x+1)))) \\ G. C. Greubel, Nov 04 2017

For n > 1, a(n) = 18*a(n-1) - a(n-2) + 32.
For n > 0, a(n) = 9*a(n-1) + 8*A157093(n-1) + 4.
a(n) = (4^(n+1)((1+sqrt(5/4))^(2n+1) + (1-sqrt(5/4))^(2n+1)) - 2*4)/4.
Lim_{n->inf} a(n+1)/a(n) = 4*(1+sqrt(5/4))^2 = 9 + 2*sqrt(20).
From R. J. Mathar, Mar 19 2009: (Start)
G.f.: 4*x*(-9+x)/((x-1)*(x^2-18*x+1)).
a(n) = 19*a(n-1) - 19*a(n-2) + a(n-3).
a(n) = 4*A119032(n+1). (End)
For n > 0, 1/a(n) = Sum_{k>=1} F(3*k)/phi^(6*k*n + 3*k), where F(n) = A000045(n) and phi = A001622 = (sqrt(5)+1)/2. - Diego Rattaggi, Dec 28 2019
E.g.f.: (1/2)*((2 + sqrt(5))*exp((9+4*sqrt(5))*x) + (2 - sqrt(5))*exp((9-4*sqrt(5))*x) - 4*exp(x)). - Stefano Spezia, Dec 29 2019

Terms a(15) onward added by G. C. Greubel, Nov 06 2017

A157084 Consider all consecutive integer Pythagorean quintuples (X, X+1, X+2, Z-1, Z) ordered by increasing Z; sequence gives X values.

Original entry on oeis.org

0, 10, 108, 1078, 10680, 105730, 1046628, 10360558, 102558960, 1015229050, 10049731548, 99482086438, 984771132840, 9748229241970, 96497521286868, 955226983626718, 9455772314980320

Offset: 0

Charlie Marion, Mar 12 2009

In general, the first terms of consecutive integer Pythagorean 2k+1-tuples may be found as follows: let first(0)=0, first(1) = k*(2k+1) and, for n > 1, first(n) = (4k+2)*first(n-1) - first(n-2) + 2*k^2; e.g., if k=3, then first(2) = 312 = 14*21 - 0 + 18.
In general, the first and last terms of consecutive integer Pythagorean 2k+1-tuples may be found as follows: let first(0)=0 and last(0)=k; for n > 0, let first(n) = (2k+1)*first(n-1) + 2k*last(n-1) + k and last(n) = (2k+2)*first(n-1) + (2k+1)*last(n-1) + 2k; e.g., if k=3 and n=2, first(2) = 312 = 7*21 + 6*27 + 3 and last(2) = 363 = 8*21 + 7*27 + 6. a(n) = (2^(n+1)((1+sqrt(3/2))^(2n+1) + (1-sqrt(3/2))^(2n+1)) - 2*2)/4; e.g., 108 = (2^3((1+sqrt(3/2))^5 + (1-sqrt(3/2))^5) - 2*2)/4.
In general, the first terms of consecutive integer Pythagorean 2k+1-tuples may be found as follows: first(n) = (k^(n+1)((1+sqrt((k+1)/k))^(2n+1) + (1-sqrt((k+1)/k))^(2n+1)) - 2*k)/4; e.g., if k=3 and n=2, then first(2) = 312 = (3^3((1+sqrt((4/3))^(5) + (1-sqrt(4/3)^5) - 2*3)/4.
In general, if u(n) is the numerator and e(n) is the denominator of the n-th continued fraction convergent to sqrt((k+1)/k), then the first terms of Consecutive Integer Pythagorean 2k+1-tuples may be found as follows: first(2n+1) = k*u(2n)*u(2n+1) and, for n > 0, first(2n) = (k+1)*e(2n-1)*e(2n); e.g., a(3) = 1078 = 2*11*49 and a(4) = 10680 = 3*40*89.
In general, if first(n) is the first term of the n-th Consecutive Integer Pythagorean 2k+1-tuple, then lim n->inf first(n+1)/first(n) = k*(1 + sqrt((k+1)/k))^2 = 2k + 1 + 2*sqrt(k^2+k).

			a(2) = 108 since 108^2 + 109^2 + 110^2 = 133^2 + 134^2.

A. H. Beiler, Recreations in the Theory of Numbers. New York: Dover, 1964, pp. 122-125.
L. E. Dickson, History of the Theory of Numbers, Vol. II, Diophantine Analysis. Dover Publications, Inc., Mineola, NY, 2005, pp. 181-183.
W. Sierpinski, Pythagorean Triangles. Dover Publications, Mineola NY, 2003, pp. 16-22.

G. C. Greubel, Table of n, a(n) for n = 0..1000
Tanya Khovanova, Recursive Sequences
Ron Knott, Pythagorean Triples and Online Calculators
Index entries for linear recurrences with constant coefficients, signature (11, -11, 1).

Cf. A001652, A157088, A157092, A157096.

[(1/2 + Sqrt(6)/4)*(5 + 2*Sqrt(6))^n - (Sqrt(6)/4 - 1/2)*(5 - 2*Sqrt(6))^n - 1: n in [0..50]]; // G. C. Greubel, Nov 04 2017

CoefficientList[Series[2*x*(-5 + x)/((x - 1)*(x^2 - 10*x + 1)), {x, 0, 50}], x] (* G. C. Greubel, Nov 04 2017 *)

x='x+O('x^50); concat([0], Vec(2*(x - 5)/((x-1)*(x^2 - 10*x + 1)))) \\ G. C. Greubel, Nov 04 2017

For n > 1, a(n) = 10*a(n-1) - a(n-2) + 8.
For n > 0, a(n) = 5*a(n-1) + 4*A157085(n-1) + 2.
Lim_{n->inf} a(n+1)/a(n) = 2(1 + sqrt(3/2))^2 = 5 + 2*sqrt(6).
From R. J. Mathar, Mar 19 2009: (Start)
G.f.: 2*x*(x - 5)/((x-1)*(x^2 - 10*x + 1)).
a(n) = 11*a(n-1) - 11*a(n-2) + a(n-3).
a(n) = 2*A087125(n) = A054320(n) - 1. (End)
From Sergei N. Gladkovskii, Jan 12 2012: (Start)
G.f.: 1/(x-1) + (x+1)/(x^2-10*x+1).
a(n) = (1/2 + sqrt(6)/4)*(5 + 2*sqrt(6))^n - (sqrt(6)/4 - 1/2)*(5 - 2*sqrt(6))^n - 1. (End)

A157097 Consider all Consecutive Integer Pythagorean 11-tuples (X, X+1, X+2, X+3, X+4, X+5, Z-4, Z-3, Z-2, Z-1, Z) ordered by increasing Z; sequence gives Z values.

Original entry on oeis.org

5, 65, 1385, 30365, 666605, 14634905, 321301265, 7053992885, 154866542165, 3400009934705, 74645352021305, 1638797734533965, 35978904807725885, 789897108035435465, 17341757471971854305, 380728767275345359205, 8358691122585626048165, 183510475929608427700385, 4028871779328799783360265

Offset: 0

Charlie Marion, Mar 12 2009

In general, the last terms of Consecutive Integer Pythagorean 2k+1-tuples may be found as follows: let last(0)=0, last(1)=k*(2k+3) and, for n > 1, last(n) = (4k+2)*last(n-1) - last(n-2)-2*k*(k-1); e.g., if k=6, then last(2) = 2274 = 26*90 - 6 - 60.
In general, the first and last terms of Consecutive Integer Pythagorean 2k+1-tuples may be found as follows: let first(0)=0 and last(0)=k; for n > 0, let first(n) = (2k+1)*first(n-1) + 2k*last(n-1) + k and last(n) = (2k+2)*first(n-1) + (2k+1)*last(n-1) + 2k; e.g., if k=6 and n=2, then first(2) = 2100 = 13*78 + 12*90 + 6 and last(2) = 2274 = 14*78 + 13*90 + 12.
In general, the last terms of Consecutive Integer Pythagorean 2k+1-tuples may be found as follows: if q=(k+1)/k, then last(n) = k^n*(k+1)*((1+sqrt(q))^(2*n+1) - (1-sqrt(q))^(2*n+1))/(4*sqrt(q)) + (k-1)/2; e.g., if k=6 and n=2, then last(2) = 2274 = 6^2*7*((1+sqrt(7/6))^5 - (1-sqrt(7/6))^5)/(4*sqrt(7/6)) + 5/2.
In general, if u(n) is the numerator and e(n) is the denominator of the n-th continued fraction convergent to sqrt((k+1)/k), then the last terms of Consecutive Integer Pythagorean 2k+1-tuples may be found as follows: last(2n+1) = (e(2n+1)^2 + k^2*e(2n)^2 + k*(k-1)*e(2n+1)*e(n))/k and, for n > 0, last(2n) = (k*(u(2n)^2 + u(2n-1)^2 + (k-1)*u(2n)*u(2n-1)))/(k+1); e.g., a(3) = 30365 = (220^2 + 5^2*21^2 + 5*4*220*21)/5 and a(4) = 666605 = (5*(505^2 + 241^2 + 4*505*241))/6.
In general, if b(0)=1, b(1)=4k+2 and, for n > 1, b(n) = (4k+2)*b(n-1) - b(n-2), and last(n) is the last term of the n-th Consecutive Integer Pythagorean 2k+1-tuple as defined above, then Sum_{i=0..n} (k*last(i) - k(k-1)/2) = k(k+1)/2*b(n); e.g., if n=3, then 1+2+3+4+5+61+62+63+64+65+1381+1382+1383+1384+1385 = 7245 = 15*483.
In general, if last(n) is the last term of the n-th Consecutive Integer Pythagorean 2k+1-tuple, then lim_{n->infinity} last(n+1)/last(n) = k*(1+sqrt((k+1)/k))^2 = 2k + 1 + 2*sqrt(k^2+k).

			a(2)=65 since 55^2 + 56^2 + 57^2 + 58^2 + 59^2 + 60^2 = 61^2 + 62^2 + 63^2 + 64^2 + 65^2.

A. H. Beiler, Recreations in the Theory of Numbers. New York: Dover, 1964, pp. 122-125.
L. E. Dickson, History of the Theory of Numbers, Vol. II, Diophantine Analysis. Dover Publications, Inc., Mineola, NY, 2005, pp. 181-183.
W. Sierpinski, Pythagorean Triangles. Dover Publications, Mineola NY, 2003, pp. 16-22.

Paolo Xausa, Table of n, a(n) for n = 0..500
Tanya Khovanova, Recursive Sequences
Ron Knott, Pythagorean Triples and Online Calculators
Index entries for linear recurrences with constant coefficients, signature (23,-23,1).

Cf. A001653, A157085, A157089, A157093.

LinearRecurrence[{23, -23, 1}, {5, 65, 1385}, 25] (* Paolo Xausa, May 29 2025 *)

For n > 1, a(n) = 22*a(n-1) - a(n-2) - 40.
For n > 0, a(n) = 12*A157096(n-1) + 11*a(n-1) + 10.
a(n) = 5^n*6*((1+sqrt(6/5))^(2*n+1) - (1-sqrt(6/5))^(2*n+1))/(4*sqrt(6/5)) + 4/2; e.g., 1385 = 5^2*6*((1+sqrt(6/5))^5 - (1-sqrt(6/5))^5)/(4*sqrt(6/5)) + 4/2.
Limit_{n->oo} a(n+1)/a(n) = 5*(1+sqrt(6/5))^2 = 11 + 2*sqrt(30).
G.f.: 5*(1-10*x+x^2)/((1-x)*(1-22*x+x^2)). - Colin Barker, Mar 27 2012
a(n) = 23*a(n-1) - 23*a(n-2) + a(n-3). - Wesley Ivan Hurt, Oct 26 2020

a(13), a(15) corrected by Georg Fischer, Oct 26 2020

A262017 The first of five consecutive positive integers the sum of the squares of which is equal to the sum of the squares of six consecutive positive integers.

Original entry on oeis.org

61, 1381, 30361, 666601, 14634901, 321301261, 7053992881, 154866542161, 3400009934701, 74645352021301, 1638797734533961, 35978904807725881, 789897108035435461, 17341757471971854301, 380728767275345359201, 8358691122585626048161, 183510475929608427700381

Offset: 1

Colin Barker, Sep 08 2015

For the first of the corresponding six consecutive positive integers, see A157096.

			61 is in the sequence because 61^2 + ... + 65^2 = 19855 = 55^2 + ... + 60^2.

Colin Barker, Table of n, a(n) for n = 1..744
Index entries for linear recurrences with constant coefficients, signature (23,-23,1).

Cf. A157096, A262018, A262019.

Cf. A027578, A027865.

Vec(-x*(x^2-22*x+61)/((x-1)*(x^2-22*x+1)) + O(x^40))

a(n) = 23*a(n-1)-23*a(n-2)+a(n-3) for n>3.

G.f.: -x*(x^2-22*x+61) / ((x-1)*(x^2-22*x+1)).

A262018 The first of five consecutive positive integers the sum of the squares of which is equal to the sum of the squares of eleven consecutive positive integers.

Original entry on oeis.org

28, 5308, 945148, 168231388, 29944242268, 5329906892668, 948693482652988, 168862110005339548, 30056506887467786908, 5349889363859260730428, 952250250260060942229628, 169495194656926988456143708, 30169192398682743884251350748, 5369946751770871484408284289788

Offset: 1

Colin Barker, Sep 08 2015

For the first of the corresponding eleven consecutive positive integers, see A262019.

			28 is in the sequence because 28^2 + ... + 32^2 = 4510 = 15^2 + ... + 25^2.

Colin Barker, Table of n, a(n) for n = 1..444
Index entries for linear recurrences with constant coefficients, signature (179,-179,1).

Cf. A157096, A262017, A262019.

LinearRecurrence[{179,-179,1},{28,5308,945148},30] (* Harvey P. Dale, May 16 2019 *)

Vec(-4*x*(7*x^2+74*x+7)/((x-1)*(x^2-178*x+1)) + O(x^20))

a(n) = 179*a(n-1)-179*a(n-2)+a(n-3) for n>3.

G.f.: -4*x*(7*x^2+74*x+7) / ((x-1)*(x^2-178*x+1)).

A262019 The first of eleven consecutive positive integers the sum of the squares of which is equal to the sum of the squares of five consecutive positive integers.

Original entry on oeis.org

15, 3575, 637215, 113421575, 20188404015, 3593422493975, 639609015524415, 113846811340852775, 20264092809656270415, 3606894673307475281975, 642006987755920943922015, 114273636925880620542837575, 20340065365818994535681167215, 3620417361478855146730704927575

Offset: 1

Colin Barker, Sep 08 2015

For the first of the corresponding five consecutive positive integers, see A262018.

			15 is in the sequence because 15^2 + ... + 25^2 = 4510 = 28^2 + ... + 32^2.

Colin Barker, Table of n, a(n) for n = 1..444
Index entries for linear recurrences with constant coefficients, signature (179,-179,1).

Cf. A157096, A262017, A262018.

Vec(5*x*(5*x^2-178*x-3)/((x-1)*(x^2-178*x+1)) + O(x^20))

a(n) = 179*a(n-1)-179*a(n-2)+a(n-3) for n>3.

G.f.: 5*x*(5*x^2-178*x-3) / ((x-1)*(x^2-178*x+1)).

cp's OEIS Frontend

A212336 Expansion of 1/(1 - 23*x + 23*x^2 - x^3).

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A157088 Consider all consecutive integer Pythagorean septuples (X, X+1, X+2, X+3, Z-2, Z-1, Z) ordered by increasing Z; sequence gives X values.

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A157092 Consider all consecutive integer Pythagorean 9-tuples (X, X+1, X+2, X+3, X+4, Z-3, Z-2, Z-1, Z) ordered by increasing Z; sequence gives X values.

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A157084 Consider all consecutive integer Pythagorean quintuples (X, X+1, X+2, Z-1, Z) ordered by increasing Z; sequence gives X values.

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A157097 Consider all Consecutive Integer Pythagorean 11-tuples (X, X+1, X+2, X+3, X+4, X+5, Z-4, Z-3, Z-2, Z-1, Z) ordered by increasing Z; sequence gives Z values.

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A262017 The first of five consecutive positive integers the sum of the squares of which is equal to the sum of the squares of six consecutive positive integers.

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A212336 Expansion of 1/(1 - 23x + 23x^2 - x^3).