A119032 -id:A119032 - cp's OEIS frontend

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.

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

A107075 Centered square numbers that are also centered pentagonal numbers.

Original entry on oeis.org

1, 181, 58141, 18721081, 6028129801, 1941039074701, 625008553923781, 201250813324382641, 64802136881897286481, 20866086825157601864101, 6718815155563865902953901, 2163437614004739663149291881

Offset: 1

Richard Choulet, Aug 30 2007, Sep 20 2007

The centered square numbers are n^2 + (n+1)^2 while the centered pentagonal numbers are (5*r^2 + 5*r + 2)/2. A number has both properties iff 5*(2*r+1)^2 = (4*n+2)^2 + 1. We solve the equation 5*Y^2 - 1 = X^2 whose solutions in positive integers are given by A075796 and A007805 respectively. The r values are 0,8,..., i.e., A053606. The n values define A119032.

Harvey P. Dale, Table of n, a(n) for n = 1..399
Index entries for linear recurrences with constant coefficients, signature (323,-323,1).

Cf. A007805, A053606, A075796, A119032.

a:= n-> (Matrix([181,1,1]). Matrix([[323,1,0], [ -323,0,1], [1,0,0]])^n)[1,3]: seq(a(n), n=1..20); # Alois P. Heinz, Aug 14 2008

LinearRecurrence[{323,-323,1},{1,181,58141},20] (* Harvey P. Dale, Nov 15 2018 *)

G.f.: (z*(1-142*z+z^2))/((1-z)*(1-322*z+z^2)).
a(n+2) = 322*a(n+1)-a(n)-140 with a(1)=1 and a(2)=181.
a(n+1) = 161*a(n)-70+18*(80*a(n)^2-70*a(n)+15)^0.5.
a(n) = (14+(9-4*sqrt(5))^(2*n-1)+(9+4*sqrt(5))^(2*n-1))/32. - Gerry Martens, Jun 06 2015

More terms from Alois P. Heinz, Aug 14 2008

A276472 Modified Pascal's triangle read by rows: T(n,k) = T(n-1,k) + T(n-1,k-1), 12. T(n,n) = T(n,n-1) + T(n-1,n-1), n>1. T(1,1) = 1, T(2,1) = 1. n>=1.

Original entry on oeis.org

1, 1, 2, 4, 3, 5, 11, 7, 8, 13, 29, 18, 15, 21, 34, 76, 47, 33, 36, 55, 89, 199, 123, 80, 69, 91, 144, 233, 521, 322, 203, 149, 160, 235, 377, 610, 1364, 843, 525, 352, 309, 395, 612, 987, 1597, 3571, 2207, 1368, 877, 661, 704, 1007, 1599, 2584, 4181

Offset: 1

Yuriy Sibirmovsky, Sep 12 2016

The recurrence relations for the border terms are the only way in which this differs from Pascal's triangle.
Column T(2n,n+1) appears to be divisible by 4 for n>=2; T(2n-1,n) divisible by 3 for n>=2; T(2n,n-2) divisible by 2 for n>=3.
The symmetry of T(n,k) can be observed in a hexagonal arrangement (see the links).
Consider T(n,k) mod 3 = q. Terms with q = 0 show reflection symmetry with respect to the central column T(2n-1,n), while q = 1 and q = 2 are mirror images of each other (see the link).

			Triangle T(n,k) begins:
n\k 1    2    3    4   5    6    7    8    9
1   1
2   1    2
3   4    3    5
4   11   7    8    13
5   29   18   15   21   34
6   76   47   33   36   55   89
7   199  123  80   69   91   144 233
8   521  322  203  149  160  235 377  610
9   1364 843  525  352  309  395 612  987  1597
...
In another format:
__________________1__________________
_______________1_____2_______________
____________4_____3_____5____________
________11_____7_____8_____13________
____29_____18_____15____21_____34____
_76_____47____33_____36____55_____89_

Yuriy Sibirmovsky, T(n,k), read by rows as a linear sequence a(j) for j = 1..5050
Yuriy Sibirmovsky, Symmetrical hexagonal arrangement for initial terms of T(n,k)
Yuriy Sibirmovsky, T(n,k) compared with Pascal's triangle
Yuriy Sibirmovsky, Illustration for T(n,k) mod 3

Cf. A000045, A000204, A001519, A001906, A002878, A005248, A054441, A088305, A122367, A258109, A026671, A026726, A026732, A004187, A049685, A033891, A206351, A092521, A081018, A049684, A058038, A081016, A003482, A049629, A133273, A049664, A049683, A119032, A023039, A007805, A033889.

Nm=12;
T=Table[0,{n,1,Nm},{k,1,n}];
T[[1,1]]=1;
T[[2,1]]=1;
T[[2,2]]=2;
Do[T[[n,1]]=T[[n-1,1]]+T[[n,2]];
T[[n,n]]=T[[n-1,n-1]]+T[[n,n-1]];
If[k!=1&&k!=n,T[[n,k]]=T[[n-1,k]]+T[[n-1,k-1]]],{n,3,Nm},{k,1,n}];
{Row[#,"\t"]}&/@T//Grid

T(n,k) = if (k==1, if (n==1, 1, if (n==2, 1, T(n-1,1) + T(n,2))), if (kMichel Marcus, Sep 14 2016

Conjectures:
Relations with other sequences:
T(n+1,1) = A002878(n-1), n>=1.
T(n,n) = A001519(n) = A122367(n-1), n>=1.
T(n+1,2) = A005248(n-1), n>=1.
T(n+1,n) = A001906(n) = A088305(n), n>=1.
T(2n-1,n) = 3*A054441(n-1), n>=2. [the central column].
Sum_{k=1..n} T(n,k) = 3*A105693(n-1), n>=2. [row sums].
Sum_{k=1..n} T(n,k)-T(n,1)-T(n,n) = 3*A258109(n), n>=2.
T(2n,n+1) - T(2n,n) = A026671(n), n>=1.
T(2n,n-1) - T(2n,n) = 2*A026726(n-1), n>=2.
T(n,ceiling(n/2)) - T(n-1,floor(n/2)) = 2*A026732(n-3), n>=3.
T(2n+1,2n) = 3*A004187(n), n>=1.
T(2n+1,2) = 3*A049685(n-1), n>=1.
T(2n+1,2n) + T(2n+1,2) = 3*A033891(n-1), n>=1.
T(2n+1,3) = 5*A206351(n), n>=1.
T(2n+1,2n)/3 - T(2n+1,3)/5 = 4*A092521(n-1), n>=2.
T(2n,1) = 1 + 5*A081018(n-1), n>=1.
T(2n,2) = 2 + 5*A049684(n-1), n>=1.
T(2n+1,2) = 3 + 5*A058038(n-1), n>=1.
T(2n,3) = 3 + 5*A081016(n-2), n>=2.
T(2n+1,1) = 4 + 5*A003482(n-1), n>=1.
T(3n,1) = 4*A049629(n-1), n>=1.
T(3n,1) = 4 + 8*A119032(n), n>=1.
T(3n+1,3) = 8*A133273(n), n>=1.
T(3n+2,3n+2) = 2 + 32*A049664(n), n>=1.
T(3n,3n-2) = 4 + 32*A049664(n-1), n>=1.
T(3n+2,2) = 2 + 16*A049683(n), n>=1.
T(3n+2,2) = 2*A023039(n), n>=1.
T(2n-1,2n-1) = A033889(n-1), n>=1.
T(3n-1,3n-1) = 2*A007805(n-1), n>=1.
T(5n-1,1) = 11*A097842(n-1), n>=1.
T(4n+5,3) - T(4n+1,3) = 15*A000045(8n+1), n>=1.
T(5n+4,3) - T(5n-1,3) = 11*A000204(10n-2), n>=1.
Relations between left and right sides:
T(n,1) = T(n,n) - T(n-2,n-2), n>=3.
T(n,2) = T(n,n-1) - T(n-2,n-3), n>=4.
T(n,1) + T(n,n) = 3*T(n,n-1), n>=2.

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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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A107075 Centered square numbers that are also centered pentagonal numbers.

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A276472 Modified Pascal's triangle read by rows: T(n,k) = T(n-1,k) + T(n-1,k-1), 12. T(n,n) = T(n,n-1) + T(n-1,n-1), n>1. T(1,1) = 1, T(2,1) = 1. n>=1.

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